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An acute-angled scalene triangle $A B C$ is given, with $A C>B C$. Let $O$ be its circumcentre, $H$ its orthocentre, and $F$ the foot of the altitude from $C$. Let $P$ be the point (other than $A$ ) on the line $A B$ such that $A F=P F$, and $M$ be the midpoint of $A C$. We denote the intersection of $P H$ and $B C$ by $X$, the intersection of $O M$ and $F X$ by $Y$, and the intersection of $O F$ and $A C$ by $Z$. Prove that the points $F, M, Y$ and $Z$ are concyclic.
|
It is enough to show that $\mathrm{O} F \perp F X$.
Let $\mathrm{OE} \perp \mathrm{AB}$, then it is trivial that :
$$
C \mathrm{H}=2 \mathrm{OE} .
$$
Since from the hypothesis we have $\mathrm{P} F=\mathrm{A} F$ then we take $\mathrm{PB}=\mathrm{P} F-\mathrm{B} F$ or
$$
\mathrm{PB}=\mathrm{A} F-\mathrm{B} F
$$
Also, $\angle X P B=\angle H A P$ and $\angle H A P=\angle H C X$ since AFGC in inscribable (where G is the foot of the altidude from A),
so $\angle X P B=\angle H C X$ and since $\angle B X P=\angle H X C$, the triangles XHC and XBP are similar.
If XL and XD are respectively the heights of the triangles XHC and XBP we have:
$$
\frac{X D}{X L}=\frac{P B}{C H},
$$
and from (1) and (2) we get:
$$
\frac{X D}{X L}=\frac{A F-B F}{2 O E}=\frac{F E}{O E} \Rightarrow \frac{X D}{F D}=\frac{F E}{O E}
$$
Therefore the triangles XFD, OEF are similar and we get:
$\angle O F X=\angle O F C+\angle L F X=\angle F O E+\angle F X D=\angle X F D+\angle F X D=90^{\circ}$, so $\mathrm{O} F \perp F X$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
An acute-angled scalene triangle $A B C$ is given, with $A C>B C$. Let $O$ be its circumcentre, $H$ its orthocentre, and $F$ the foot of the altitude from $C$. Let $P$ be the point (other than $A$ ) on the line $A B$ such that $A F=P F$, and $M$ be the midpoint of $A C$. We denote the intersection of $P H$ and $B C$ by $X$, the intersection of $O M$ and $F X$ by $Y$, and the intersection of $O F$ and $A C$ by $Z$. Prove that the points $F, M, Y$ and $Z$ are concyclic.
|
It is enough to show that $\mathrm{O} F \perp F X$.
Let $\mathrm{OE} \perp \mathrm{AB}$, then it is trivial that :
$$
C \mathrm{H}=2 \mathrm{OE} .
$$
Since from the hypothesis we have $\mathrm{P} F=\mathrm{A} F$ then we take $\mathrm{PB}=\mathrm{P} F-\mathrm{B} F$ or
$$
\mathrm{PB}=\mathrm{A} F-\mathrm{B} F
$$
Also, $\angle X P B=\angle H A P$ and $\angle H A P=\angle H C X$ since AFGC in inscribable (where G is the foot of the altidude from A),
so $\angle X P B=\angle H C X$ and since $\angle B X P=\angle H X C$, the triangles XHC and XBP are similar.
If XL and XD are respectively the heights of the triangles XHC and XBP we have:
$$
\frac{X D}{X L}=\frac{P B}{C H},
$$
and from (1) and (2) we get:
$$
\frac{X D}{X L}=\frac{A F-B F}{2 O E}=\frac{F E}{O E} \Rightarrow \frac{X D}{F D}=\frac{F E}{O E}
$$
Therefore the triangles XFD, OEF are similar and we get:
$\angle O F X=\angle O F C+\angle L F X=\angle F O E+\angle F X D=\angle X F D+\angle F X D=90^{\circ}$, so $\mathrm{O} F \perp F X$.
|
{
"resource_path": "Balkan_MO/segmented/en-2008-BMO-type1.jsonl",
"problem_match": "# Problem 1",
"solution_match": "# Solution:"
}
|
35af2f2a-6f2d-5324-93f3-2afadcad0676
| 606,353
|
Does there exist a sequence $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ of positive real numbers satisfying both of the following conditions:
(i) $\sum_{i=1}^{n} a_{i} \leq n^{2}$, for every positive integer $n$;
(ii) $\sum_{i=1}^{n} \frac{1}{a_{i}} \leq 2008$, for every positive integer $n$ ?
|
The answer is no.
It is enough to show that
if $\sum_{i=1}^{n} a_{i} \leq n^{2}$ for any $n$, then $\sum_{i=2}^{2^{n}} \frac{1}{a_{i}}>\frac{n}{4}$. (or any other precise estimate)
For this, we use that $\sum_{i=2^{k}+1}^{2^{k+1}} a_{i} \sum_{i=2^{k}+1}^{2^{k+1}} \frac{1}{a_{i}} \geq 2^{2 k}$ for any $k \geq 0$ by the arithmetic-harmonic mean inequality.
Since $\sum_{i=2^{k}+1}^{2^{k+1}} a_{i}<\sum_{i=1}^{2^{k+1}} a_{i} \leq 2^{2 k+2}$, it follows that $\sum_{i=2^{k}+1}^{2^{k+1}} \frac{1}{a_{i}}>\frac{1}{4}$ and hence $\sum_{i=2}^{2^{n}} \frac{1}{a^{i}}>\sum_{k=0}^{n-1} \sum_{i=2^{k}+1}^{2^{k+1}} \frac{1}{a_{i}}>\frac{n}{4}$. (it can be stated in words)
## Remark: no points for using some inequality, that doesn't lead to solution
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Does there exist a sequence $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ of positive real numbers satisfying both of the following conditions:
(i) $\sum_{i=1}^{n} a_{i} \leq n^{2}$, for every positive integer $n$;
(ii) $\sum_{i=1}^{n} \frac{1}{a_{i}} \leq 2008$, for every positive integer $n$ ?
|
The answer is no.
It is enough to show that
if $\sum_{i=1}^{n} a_{i} \leq n^{2}$ for any $n$, then $\sum_{i=2}^{2^{n}} \frac{1}{a_{i}}>\frac{n}{4}$. (or any other precise estimate)
For this, we use that $\sum_{i=2^{k}+1}^{2^{k+1}} a_{i} \sum_{i=2^{k}+1}^{2^{k+1}} \frac{1}{a_{i}} \geq 2^{2 k}$ for any $k \geq 0$ by the arithmetic-harmonic mean inequality.
Since $\sum_{i=2^{k}+1}^{2^{k+1}} a_{i}<\sum_{i=1}^{2^{k+1}} a_{i} \leq 2^{2 k+2}$, it follows that $\sum_{i=2^{k}+1}^{2^{k+1}} \frac{1}{a_{i}}>\frac{1}{4}$ and hence $\sum_{i=2}^{2^{n}} \frac{1}{a^{i}}>\sum_{k=0}^{n-1} \sum_{i=2^{k}+1}^{2^{k+1}} \frac{1}{a_{i}}>\frac{n}{4}$. (it can be stated in words)
## Remark: no points for using some inequality, that doesn't lead to solution
|
{
"resource_path": "Balkan_MO/segmented/en-2008-BMO-type1.jsonl",
"problem_match": "# Problem 2",
"solution_match": "# Solution."
}
|
00c532a9-9fbd-577e-9a17-29c6128a4942
| 606,356
|
Let $n$ be a positive integer. The rectangle $A B C D$ with side lengths $A B=90 n+1$ and $B C=90 n+5$ is partitioned into unit squares with sides parallel to the sides of $A B C D$. Let $S$ be the set of all points which are vertices of these unit squares. Prove that the number of lines which pass through at least two points from $S$ is divisible by 4.
|
Denote $90 n+1=m$. We investigate the number of the lines modulo 4 consecutively reducing different types of lines.
The vertical and horizontal lines are
$(m+5)+(m+1)=2(m+3)$ which is divisible to 4.
Moreover, every line which makes an acute angle to the axe $O x$ (i.e. that line has a positive angular coefficient) corresponds to unique line with an obtuse angle (consider the symmetry with respect to the line through the midpoints of $A B$ and $C D$ ). Therefore it is enough to prove that the lines with acute angles are an even number.
Every line which does not pass through the center $O$ of the rectangle corresponds to another line with the same angular coefficent(consider the symmetry with respect to $O$ ). Therefore it is enough to consider the lines through $O$.
Every line through $O$ has an angular coefficient $\frac{p}{q}$, where $(p, q)=1, p$ and $q$ are odd positive integers. (To see this, consider the two nearest, from the two sides, to $O$ points of the line).
If $p \neq 1, q \neq 1, \quad p \leq m$ and $q \leq m$, the line with angular coefficient $\frac{p}{q}$, uniquely corresponds to the line with angular coefficient $\frac{q}{p}$. It remains to prove that the number of the remaining lines is even.
The last number is
$$
1+\frac{\varphi(m+2)}{2}+\frac{\varphi(m+4)}{2}-1=\frac{\varphi(m+2)+\varphi(m+4)}{2}
$$
because we have:
1) one line with $p=q=1$;
2) $\frac{\varphi(m+2)}{2}$ lines with angular coefficient $\frac{p}{m+2}, p \leq m$ is odd and $(p, m+2)=1$;
3) $\frac{\varphi(m+4)}{2}-1$ lines with angular coefficient $\frac{p}{m+4}, p \leq m$ is odd and $(p, m+4)=1$.
Now the assertion follows from the fact that the number $\varphi(m+2)+\varphi(m+4)=\varphi(90 n+3)+\varphi(90 n+5)$ is divisible to 4.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Let $n$ be a positive integer. The rectangle $A B C D$ with side lengths $A B=90 n+1$ and $B C=90 n+5$ is partitioned into unit squares with sides parallel to the sides of $A B C D$. Let $S$ be the set of all points which are vertices of these unit squares. Prove that the number of lines which pass through at least two points from $S$ is divisible by 4.
|
Denote $90 n+1=m$. We investigate the number of the lines modulo 4 consecutively reducing different types of lines.
The vertical and horizontal lines are
$(m+5)+(m+1)=2(m+3)$ which is divisible to 4.
Moreover, every line which makes an acute angle to the axe $O x$ (i.e. that line has a positive angular coefficient) corresponds to unique line with an obtuse angle (consider the symmetry with respect to the line through the midpoints of $A B$ and $C D$ ). Therefore it is enough to prove that the lines with acute angles are an even number.
Every line which does not pass through the center $O$ of the rectangle corresponds to another line with the same angular coefficent(consider the symmetry with respect to $O$ ). Therefore it is enough to consider the lines through $O$.
Every line through $O$ has an angular coefficient $\frac{p}{q}$, where $(p, q)=1, p$ and $q$ are odd positive integers. (To see this, consider the two nearest, from the two sides, to $O$ points of the line).
If $p \neq 1, q \neq 1, \quad p \leq m$ and $q \leq m$, the line with angular coefficient $\frac{p}{q}$, uniquely corresponds to the line with angular coefficient $\frac{q}{p}$. It remains to prove that the number of the remaining lines is even.
The last number is
$$
1+\frac{\varphi(m+2)}{2}+\frac{\varphi(m+4)}{2}-1=\frac{\varphi(m+2)+\varphi(m+4)}{2}
$$
because we have:
1) one line with $p=q=1$;
2) $\frac{\varphi(m+2)}{2}$ lines with angular coefficient $\frac{p}{m+2}, p \leq m$ is odd and $(p, m+2)=1$;
3) $\frac{\varphi(m+4)}{2}-1$ lines with angular coefficient $\frac{p}{m+4}, p \leq m$ is odd and $(p, m+4)=1$.
Now the assertion follows from the fact that the number $\varphi(m+2)+\varphi(m+4)=\varphi(90 n+3)+\varphi(90 n+5)$ is divisible to 4.
|
{
"resource_path": "Balkan_MO/segmented/en-2008-BMO-type1.jsonl",
"problem_match": "# Problem 3",
"solution_match": "# Solution."
}
|
186884e0-6654-547b-9767-01bd3ec6c6ef
| 606,358
|
Let $c$ be a positive integer. The sequence $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ is defined by $a_{1}=c$, and $a_{n+1}=a_{n}^{2}+a_{n}+c^{3}$, for every positive integer $n$. Find all values of $c$ for which there exist some integers $k \geq 1$ and $m \geq 2$, such that $a_{k}^{2}+c^{3}$ is the $m^{\text {th }}$ power of some positive integer.
|
First, notice:
$$
a_{n+1}^{2}+c^{3}=\left(a_{n}^{2}+a_{n}+c^{3}\right)^{2}+c^{3}=\left(a_{n}^{2}+c^{3}\right)\left(a_{n}^{2}+2 a_{n}+1+c^{3}\right)
$$
We first prove that $a_{n}^{2}+c^{3}$ and $a_{n}^{2}+2 a_{n}+1+c^{3}$ are coprime.
We prove by induction that $4 c^{3}+1$ is coprime with $2 a_{n}+1$, for every $n \geq 1$.
Let $n=1$ and $p$ be a prime divisor of $4 c^{3}+1$ and $2 a_{1}+1=2 c+1$. Then $p$ divides $2\left(4 c^{3}+1\right)=(2 c+1)\left(4 c^{2}-2 c+1\right)+1$, hence $p$ divides 1 , a contradiction. Assume now that $\left(4 c^{3}+1,2 a_{n}+1\right)=1$ for some $n \geq 1$ and the prime $p$ divides $4 c^{3}+1$ and $2 a_{n+1}+1$. Then $p$ divides $4 a_{n+1}+2=\left(2 a_{n}+1\right)^{2}+4 c^{3}+1$, which gives a contradiction.
Assume that for some $n \geq 1$ the number
$$
a_{n+1}^{2}+c^{3}=\left(a_{n}^{2}+a_{n}+c^{3}\right)^{2}+c^{3}=\left(a_{n}^{2}+c^{3}\right)\left(a_{n}^{2}+2 a_{n}+1+c^{3}\right)
$$
is a power. Since $a_{n}^{2}+c^{3}$ and $a_{n}^{2}+2 a_{n}+1+c^{3}$ are coprime, than $a_{n}^{2}+c^{3}$ is a power as well.
The same argument can be further applied giving that $a_{1}^{2}+c^{3}=c^{2}+c^{3}=c^{2}(c+1)$ is a power.
If $a^{2}(a+1)=t^{m}$ with odd $m \geq 3$, then $a=t_{1}^{m}$ and $a+1=t_{2}^{m}$, which is impossible. If $a^{2}(a+1)=t^{2 m_{1}}$ with $m_{1} \geq 2$, then $a=t_{1}^{m_{1}}$ and $a+1=t_{2}^{m_{1}}$, which is impossible.
Therefore $a^{2}(a+1)=t^{2}$ whence we obtain the solutions $a=s^{2}-1, s \geq 2, s \in \mathbb{N}$.
|
c=s^{2}-1, s \geq 2, s \in \mathbb{N}
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $c$ be a positive integer. The sequence $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ is defined by $a_{1}=c$, and $a_{n+1}=a_{n}^{2}+a_{n}+c^{3}$, for every positive integer $n$. Find all values of $c$ for which there exist some integers $k \geq 1$ and $m \geq 2$, such that $a_{k}^{2}+c^{3}$ is the $m^{\text {th }}$ power of some positive integer.
|
First, notice:
$$
a_{n+1}^{2}+c^{3}=\left(a_{n}^{2}+a_{n}+c^{3}\right)^{2}+c^{3}=\left(a_{n}^{2}+c^{3}\right)\left(a_{n}^{2}+2 a_{n}+1+c^{3}\right)
$$
We first prove that $a_{n}^{2}+c^{3}$ and $a_{n}^{2}+2 a_{n}+1+c^{3}$ are coprime.
We prove by induction that $4 c^{3}+1$ is coprime with $2 a_{n}+1$, for every $n \geq 1$.
Let $n=1$ and $p$ be a prime divisor of $4 c^{3}+1$ and $2 a_{1}+1=2 c+1$. Then $p$ divides $2\left(4 c^{3}+1\right)=(2 c+1)\left(4 c^{2}-2 c+1\right)+1$, hence $p$ divides 1 , a contradiction. Assume now that $\left(4 c^{3}+1,2 a_{n}+1\right)=1$ for some $n \geq 1$ and the prime $p$ divides $4 c^{3}+1$ and $2 a_{n+1}+1$. Then $p$ divides $4 a_{n+1}+2=\left(2 a_{n}+1\right)^{2}+4 c^{3}+1$, which gives a contradiction.
Assume that for some $n \geq 1$ the number
$$
a_{n+1}^{2}+c^{3}=\left(a_{n}^{2}+a_{n}+c^{3}\right)^{2}+c^{3}=\left(a_{n}^{2}+c^{3}\right)\left(a_{n}^{2}+2 a_{n}+1+c^{3}\right)
$$
is a power. Since $a_{n}^{2}+c^{3}$ and $a_{n}^{2}+2 a_{n}+1+c^{3}$ are coprime, than $a_{n}^{2}+c^{3}$ is a power as well.
The same argument can be further applied giving that $a_{1}^{2}+c^{3}=c^{2}+c^{3}=c^{2}(c+1)$ is a power.
If $a^{2}(a+1)=t^{m}$ with odd $m \geq 3$, then $a=t_{1}^{m}$ and $a+1=t_{2}^{m}$, which is impossible. If $a^{2}(a+1)=t^{2 m_{1}}$ with $m_{1} \geq 2$, then $a=t_{1}^{m_{1}}$ and $a+1=t_{2}^{m_{1}}$, which is impossible.
Therefore $a^{2}(a+1)=t^{2}$ whence we obtain the solutions $a=s^{2}-1, s \geq 2, s \in \mathbb{N}$.
|
{
"resource_path": "Balkan_MO/segmented/en-2008-BMO-type1.jsonl",
"problem_match": "# Problem 4",
"solution_match": "# Solution."
}
|
d2369d3d-5ee6-5dc0-88f6-2022f869ca09
| 606,360
|
We start by observing that $z$ must be even, so $z^{2}=3^{x}-5^{y} \equiv(-1)^{x}-1(\bmod 4)$ is divisible by 4 , which implies that $x$ is even, say $x=2 t$. Then our equation can be rewritten as $\left(3^{t}-z\right)\left(3^{t}+z\right)=5^{y}$, which means that both $3^{t}-z=5^{k}$ and $3^{t}+z=5^{y-k}$ for some nonnegative integer $k$. Since $5^{k}+5^{y-k}=2 \cdot 3^{t}$ is not divisible by 5 , it follows that $k=0$ and
$$
2 \cdot 3^{t}=5^{y}+1
$$
Suppose that $t \geq 2$. Then $5^{y}+1$ is divisible by 9 , which is only possible if $y \equiv 3$ $(\bmod 6)$. However, in this case $5^{y}+1 \equiv 5^{3}+1 \equiv 0(\bmod 7)$, so $5^{y}+1$ is also divisible by 7 , which is impossible.
Therefore we must have $t \leq 1$, which yields a (unique) solution $(x, y, z)=(2,1,2)$.
|
We start by observing that $f$ is injective. From the known identity
$$
\left(a^{2}+2 b^{2}\right)\left(c^{2}+2 d^{2}\right)=(a c \pm 2 b d)^{2}+2(a d \mp b c)^{2}
$$
we obtain $f(a c+2 b d)^{2}+2 f(a d-b c)^{2}=f(a c-2 b d)^{2}+2 f(a d+b c)^{2}$, assuming that the arguments are positive integers. Specially, for $b=c=d=1$ and $a \geq 3$ we have $f(a+2)^{2}+2 f(a-1)^{2}=f(a-2)^{2}+2 f(a+1)^{2}$. Denoting $g(n)=f(n)^{2}$ we get a recurrent relation $g(a+2)-2 g(a+1)+2 g(a-1)-g(a-2)=0$ whose characteristic polynomial is $(x+1)(x-1)^{3}$, which leads to
$$
g(n)=A(-1)^{n}+B+C n+D n^{2} .
$$
Substituting $m=n$ in the original equation yields $g(3 g(n))=9 n^{4}$, which together with $(\dagger)$ gives us
$$
\begin{aligned}
L=9 n^{4} & =A(-1)^{3\left(A(-1)^{n}+B+C n+D n^{2}\right)}+B+\underbrace{3 C\left[A(-1)^{n}+B+C n+D n^{2}\right]} \\
& +9 D\left[A(-1)^{n}+B+C n+D n^{2}\right]^{2}=R .
\end{aligned}
$$
Since $0=\lim _{n \rightarrow \infty} \frac{R-L}{n^{4}}=9 D^{3}-9$, we have $D^{3}=1$ (so $D \neq 0$ ); similarly, $0=\lim _{n \rightarrow \infty} \frac{R-L}{n^{3}}=18 D^{2} C$, so $C=0$. Moreover, for $n=2 k$ and $n=2 k+1$
respectively we obtain $0=\lim _{k \rightarrow \infty} \frac{R-L}{(2 k)^{2}}=18 D^{2}(A+B)$ and $0=\lim _{k \rightarrow \infty} \frac{R-L}{(2 k+1)^{2}}=$ $18 D^{2}(-A+B)$, implying $A+B=-A+B=0$; hence $A=B=0$.
Finally, $g(n)=D n^{2}, D^{3}=1$ and $g: \mathbb{N} \rightarrow \mathbb{N}$ gives us $D=1$, i.e. $f(n)=n$. It is directly verified that this function satisfies the conditions of the problem.
Remark. Using limits can be avoided. Since the rigth-hand side in ( $\dagger$ ) takes only integer values, it follows that $A, B, C, D$ are rational, so taking suitable multiples of integers for $n$ eliminates the powers of -1 and leaves a polynomial equality.
|
(x, y, z)=(2,1,2)
|
Yes
|
Incomplete
|
proof
|
Number Theory
|
We start by observing that $z$ must be even, so $z^{2}=3^{x}-5^{y} \equiv(-1)^{x}-1(\bmod 4)$ is divisible by 4 , which implies that $x$ is even, say $x=2 t$. Then our equation can be rewritten as $\left(3^{t}-z\right)\left(3^{t}+z\right)=5^{y}$, which means that both $3^{t}-z=5^{k}$ and $3^{t}+z=5^{y-k}$ for some nonnegative integer $k$. Since $5^{k}+5^{y-k}=2 \cdot 3^{t}$ is not divisible by 5 , it follows that $k=0$ and
$$
2 \cdot 3^{t}=5^{y}+1
$$
Suppose that $t \geq 2$. Then $5^{y}+1$ is divisible by 9 , which is only possible if $y \equiv 3$ $(\bmod 6)$. However, in this case $5^{y}+1 \equiv 5^{3}+1 \equiv 0(\bmod 7)$, so $5^{y}+1$ is also divisible by 7 , which is impossible.
Therefore we must have $t \leq 1$, which yields a (unique) solution $(x, y, z)=(2,1,2)$.
|
We start by observing that $f$ is injective. From the known identity
$$
\left(a^{2}+2 b^{2}\right)\left(c^{2}+2 d^{2}\right)=(a c \pm 2 b d)^{2}+2(a d \mp b c)^{2}
$$
we obtain $f(a c+2 b d)^{2}+2 f(a d-b c)^{2}=f(a c-2 b d)^{2}+2 f(a d+b c)^{2}$, assuming that the arguments are positive integers. Specially, for $b=c=d=1$ and $a \geq 3$ we have $f(a+2)^{2}+2 f(a-1)^{2}=f(a-2)^{2}+2 f(a+1)^{2}$. Denoting $g(n)=f(n)^{2}$ we get a recurrent relation $g(a+2)-2 g(a+1)+2 g(a-1)-g(a-2)=0$ whose characteristic polynomial is $(x+1)(x-1)^{3}$, which leads to
$$
g(n)=A(-1)^{n}+B+C n+D n^{2} .
$$
Substituting $m=n$ in the original equation yields $g(3 g(n))=9 n^{4}$, which together with $(\dagger)$ gives us
$$
\begin{aligned}
L=9 n^{4} & =A(-1)^{3\left(A(-1)^{n}+B+C n+D n^{2}\right)}+B+\underbrace{3 C\left[A(-1)^{n}+B+C n+D n^{2}\right]} \\
& +9 D\left[A(-1)^{n}+B+C n+D n^{2}\right]^{2}=R .
\end{aligned}
$$
Since $0=\lim _{n \rightarrow \infty} \frac{R-L}{n^{4}}=9 D^{3}-9$, we have $D^{3}=1$ (so $D \neq 0$ ); similarly, $0=\lim _{n \rightarrow \infty} \frac{R-L}{n^{3}}=18 D^{2} C$, so $C=0$. Moreover, for $n=2 k$ and $n=2 k+1$
respectively we obtain $0=\lim _{k \rightarrow \infty} \frac{R-L}{(2 k)^{2}}=18 D^{2}(A+B)$ and $0=\lim _{k \rightarrow \infty} \frac{R-L}{(2 k+1)^{2}}=$ $18 D^{2}(-A+B)$, implying $A+B=-A+B=0$; hence $A=B=0$.
Finally, $g(n)=D n^{2}, D^{3}=1$ and $g: \mathbb{N} \rightarrow \mathbb{N}$ gives us $D=1$, i.e. $f(n)=n$. It is directly verified that this function satisfies the conditions of the problem.
Remark. Using limits can be avoided. Since the rigth-hand side in ( $\dagger$ ) takes only integer values, it follows that $A, B, C, D$ are rational, so taking suitable multiples of integers for $n$ eliminates the powers of -1 and leaves a polynomial equality.
|
{
"resource_path": "Balkan_MO/segmented/en-2009-BMO-type2.jsonl",
"problem_match": "\n1.",
"solution_match": "\n4."
}
|
69400796-5fd3-5ea6-922f-93a331343b00
| 606,362
|
In a triangle $A B C$, points $M$ and $N$ on the sides $A B$ and $A C$ respectively are such that $M N \| B C$. Let $B N$ and $C M$ intersect at point $P$. The circumcircles of triangles $B M P$ and $C N P$ intersect at two distinct points $P$ and $Q$. Prove that $\angle B A Q=\angle C A P$.
(Moldova)
|
Since the quadrilaterals $B M P Q$ and $C N P Q$ are cyclic, we have $\angle B Q N=\angle B Q P+$ $\angle P Q N=\angle A M C+\angle M C A=180^{\circ}-$ $\angle C A B$, so $A B Q N$ is cyclic as well. Hence $\frac{\sin \angle B A Q}{\sin \angle N A Q}=\frac{B Q}{N Q}$. Moreover, triangles $M B Q$ and $C N Q$ are similar, so
$$
\frac{\sin \angle B A Q}{\sin \angle C A Q}=\frac{B Q}{N Q}=\frac{B M}{C N}=\frac{A B}{A C}
$$
On the other hand, if $A P$ meets $B C$ at $A_{1}$, then by the Cheva theorem $\frac{B A_{1}}{A_{1} C}=$
$\frac{B M}{M A} \cdot \frac{A N}{N C}=1$, so $A_{1}$ is the midpoint of $B C$ and
$$
\frac{\sin \angle C A P}{\sin \angle B A P}=\frac{A B}{A C} \cdot \frac{A C \cdot A A_{1} \sin \angle C A P}{A B \cdot A A_{1} \sin \angle B A P}=\frac{A B}{A C} \cdot \frac{S_{\triangle C A A_{1}}}{S_{\triangle B A A_{1}}}=\frac{A B}{A C}
$$
Therefore, if we denote $\angle C A P=\varphi, \angle B A Q=\psi$ and $\angle B A C=\alpha$, we have $\frac{\sin \psi}{\sin (\alpha-\psi)}=\frac{\sin \varphi}{\sin (\alpha-\varphi)}$, which is equivalent to $\sin \psi \sin (\alpha-\varphi)=\sin \varphi \sin (\alpha-\psi)$. The addition formulas reduce the last equality to $0=\sin \alpha(\sin \varphi \cos \psi-\sin \psi \cos \varphi)=$ $\sin \alpha \sin (\varphi-\psi)$, from which we conclude that $\psi=\varphi$, as desired.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
In a triangle $A B C$, points $M$ and $N$ on the sides $A B$ and $A C$ respectively are such that $M N \| B C$. Let $B N$ and $C M$ intersect at point $P$. The circumcircles of triangles $B M P$ and $C N P$ intersect at two distinct points $P$ and $Q$. Prove that $\angle B A Q=\angle C A P$.
(Moldova)
|
Since the quadrilaterals $B M P Q$ and $C N P Q$ are cyclic, we have $\angle B Q N=\angle B Q P+$ $\angle P Q N=\angle A M C+\angle M C A=180^{\circ}-$ $\angle C A B$, so $A B Q N$ is cyclic as well. Hence $\frac{\sin \angle B A Q}{\sin \angle N A Q}=\frac{B Q}{N Q}$. Moreover, triangles $M B Q$ and $C N Q$ are similar, so
$$
\frac{\sin \angle B A Q}{\sin \angle C A Q}=\frac{B Q}{N Q}=\frac{B M}{C N}=\frac{A B}{A C}
$$
On the other hand, if $A P$ meets $B C$ at $A_{1}$, then by the Cheva theorem $\frac{B A_{1}}{A_{1} C}=$
$\frac{B M}{M A} \cdot \frac{A N}{N C}=1$, so $A_{1}$ is the midpoint of $B C$ and
$$
\frac{\sin \angle C A P}{\sin \angle B A P}=\frac{A B}{A C} \cdot \frac{A C \cdot A A_{1} \sin \angle C A P}{A B \cdot A A_{1} \sin \angle B A P}=\frac{A B}{A C} \cdot \frac{S_{\triangle C A A_{1}}}{S_{\triangle B A A_{1}}}=\frac{A B}{A C}
$$
Therefore, if we denote $\angle C A P=\varphi, \angle B A Q=\psi$ and $\angle B A C=\alpha$, we have $\frac{\sin \psi}{\sin (\alpha-\psi)}=\frac{\sin \varphi}{\sin (\alpha-\varphi)}$, which is equivalent to $\sin \psi \sin (\alpha-\varphi)=\sin \varphi \sin (\alpha-\psi)$. The addition formulas reduce the last equality to $0=\sin \alpha(\sin \varphi \cos \psi-\sin \psi \cos \varphi)=$ $\sin \alpha \sin (\varphi-\psi)$, from which we conclude that $\psi=\varphi$, as desired.
|
{
"resource_path": "Balkan_MO/segmented/en-2009-BMO-type2.jsonl",
"problem_match": "\n2.",
"solution_match": "\n2."
}
|
8fabb8e6-48c6-5fce-8b91-cdbd07e436bc
| 606,363
|
A $9 \times 12$ rectangle is divided into unit squares. The centers of all the unit squares, except the four corner squares and the eight squares adjacent (by side) to them, are colored red. Is it possible to numerate the red centers by $C_{1}, C_{2}, \ldots, C_{96}$ so that the following two conditions are fulfilled:
$1^{\circ}$ All segments $C_{1} C_{2}, C_{2} C_{3}, \ldots C_{95} C_{96}, C_{96} C_{1}$ have the length $\sqrt{13}$;
$2^{\circ}$ The poligonal line $C_{1} C_{2} \ldots C_{96} C_{1}$ is centrally symmetric?
(Bulgaria)
|
Place the given rectangle into the coordinate plane so that the center of the square at the intersection of $i$-th column and $j$-th row has the coordinates $(i, j)$. Suppose that a desired numeration of the red points exists; it corresponds to a path, i.e. a closed poligonal line consisting of 96 segments of length $\sqrt{13}$, passing through each red point exactly once. Note that points $(i, j)$ and $(k, l)$ are adjacent in the path if and only if $\{|i-k|,|j-l|\}=\{2,3\}$.
The center of symmetry must be at point $C\left(5 \frac{1}{2}, 5\right)$. Consider the points $A(2,2)$, $B(11,8)$. These two points are symmetric with respect to $C$ and divide the path into two parts $\gamma_{1}$ and $\gamma_{2}$. Note that, if the rectangular board is colored alternately white and black (like a chessboard), $A$ and $B$ are of different colors, and each segment connects two squares of different colors. It follows that each of $\gamma_{1}, \gamma_{2}$ consists of an odd number of segments. Thus these two parts are of different lengths and cannot be symmetric to each other. Therefore each
of $\gamma_{1}, \gamma_{2}$ is centrally symmetric itself.
Being of an odd length, each of the parts $\gamma_{1}, \gamma_{2}$ must contain a segment which is centrally symmetric with respect to $C$. There are only two such segments one connecting $(5,4)$ and $(8,6)$, and one connecting $(5,6)$ and $(8,4)$, so these two segments must be parts of our path. Moreover, point $(2,2)$ is connected with only two points, namely $(4,5)$ and $(5,4)$, so these three points are directly connected. Analogous conclusions can be made about points $(2,8),(11,2)$ and $(11,8)$, so the closed path $(5,4)-(2,2)-(4,5)-(2,8)-(5,6)-(8,4)-(11,2)-(9,5)-(11,8)-$ $(8,6)-(5,4)$ is entirely contained in our path, which is clearly a contradiction.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
A $9 \times 12$ rectangle is divided into unit squares. The centers of all the unit squares, except the four corner squares and the eight squares adjacent (by side) to them, are colored red. Is it possible to numerate the red centers by $C_{1}, C_{2}, \ldots, C_{96}$ so that the following two conditions are fulfilled:
$1^{\circ}$ All segments $C_{1} C_{2}, C_{2} C_{3}, \ldots C_{95} C_{96}, C_{96} C_{1}$ have the length $\sqrt{13}$;
$2^{\circ}$ The poligonal line $C_{1} C_{2} \ldots C_{96} C_{1}$ is centrally symmetric?
(Bulgaria)
|
Place the given rectangle into the coordinate plane so that the center of the square at the intersection of $i$-th column and $j$-th row has the coordinates $(i, j)$. Suppose that a desired numeration of the red points exists; it corresponds to a path, i.e. a closed poligonal line consisting of 96 segments of length $\sqrt{13}$, passing through each red point exactly once. Note that points $(i, j)$ and $(k, l)$ are adjacent in the path if and only if $\{|i-k|,|j-l|\}=\{2,3\}$.
The center of symmetry must be at point $C\left(5 \frac{1}{2}, 5\right)$. Consider the points $A(2,2)$, $B(11,8)$. These two points are symmetric with respect to $C$ and divide the path into two parts $\gamma_{1}$ and $\gamma_{2}$. Note that, if the rectangular board is colored alternately white and black (like a chessboard), $A$ and $B$ are of different colors, and each segment connects two squares of different colors. It follows that each of $\gamma_{1}, \gamma_{2}$ consists of an odd number of segments. Thus these two parts are of different lengths and cannot be symmetric to each other. Therefore each
of $\gamma_{1}, \gamma_{2}$ is centrally symmetric itself.
Being of an odd length, each of the parts $\gamma_{1}, \gamma_{2}$ must contain a segment which is centrally symmetric with respect to $C$. There are only two such segments one connecting $(5,4)$ and $(8,6)$, and one connecting $(5,6)$ and $(8,4)$, so these two segments must be parts of our path. Moreover, point $(2,2)$ is connected with only two points, namely $(4,5)$ and $(5,4)$, so these three points are directly connected. Analogous conclusions can be made about points $(2,8),(11,2)$ and $(11,8)$, so the closed path $(5,4)-(2,2)-(4,5)-(2,8)-(5,6)-(8,4)-(11,2)-(9,5)-(11,8)-$ $(8,6)-(5,4)$ is entirely contained in our path, which is clearly a contradiction.
|
{
"resource_path": "Balkan_MO/segmented/en-2009-BMO-type2.jsonl",
"problem_match": "\n3.",
"solution_match": "\n3."
}
|
bf7ce674-0b20-5b87-bdfd-859a0056894f
| 606,367
|
Determine all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ satisfying
$$
f\left(f(m)^{2}+2 f(n)^{2}\right)=m^{2}+2 n^{2} \quad \text { for all } m, n \in \mathbb{N} . \quad \text { (Bulgaria) }
$$
Time allowed: 270 minutes.
Each problem is worth 10 points.
## SOLUTIONS
|
We start by observing that $f$ is injective. From the known identity
$$
\left(a^{2}+2 b^{2}\right)\left(c^{2}+2 d^{2}\right)=(a c \pm 2 b d)^{2}+2(a d \mp b c)^{2}
$$
we obtain $f(a c+2 b d)^{2}+2 f(a d-b c)^{2}=f(a c-2 b d)^{2}+2 f(a d+b c)^{2}$, assuming that the arguments are positive integers. Specially, for $b=c=d=1$ and $a \geq 3$ we have $f(a+2)^{2}+2 f(a-1)^{2}=f(a-2)^{2}+2 f(a+1)^{2}$. Denoting $g(n)=f(n)^{2}$ we get a recurrent relation $g(a+2)-2 g(a+1)+2 g(a-1)-g(a-2)=0$ whose characteristic polynomial is $(x+1)(x-1)^{3}$, which leads to
$$
g(n)=A(-1)^{n}+B+C n+D n^{2} .
$$
Substituting $m=n$ in the original equation yields $g(3 g(n))=9 n^{4}$, which together with $(\dagger)$ gives us
$$
\begin{aligned}
L=9 n^{4} & =A(-1)^{3\left(A(-1)^{n}+B+C n+D n^{2}\right)}+B+\underbrace{3 C\left[A(-1)^{n}+B+C n+D n^{2}\right]} \\
& +9 D\left[A(-1)^{n}+B+C n+D n^{2}\right]^{2}=R .
\end{aligned}
$$
Since $0=\lim _{n \rightarrow \infty} \frac{R-L}{n^{4}}=9 D^{3}-9$, we have $D^{3}=1$ (so $D \neq 0$ ); similarly, $0=\lim _{n \rightarrow \infty} \frac{R-L}{n^{3}}=18 D^{2} C$, so $C=0$. Moreover, for $n=2 k$ and $n=2 k+1$
respectively we obtain $0=\lim _{k \rightarrow \infty} \frac{R-L}{(2 k)^{2}}=18 D^{2}(A+B)$ and $0=\lim _{k \rightarrow \infty} \frac{R-L}{(2 k+1)^{2}}=$ $18 D^{2}(-A+B)$, implying $A+B=-A+B=0$; hence $A=B=0$.
Finally, $g(n)=D n^{2}, D^{3}=1$ and $g: \mathbb{N} \rightarrow \mathbb{N}$ gives us $D=1$, i.e. $f(n)=n$. It is directly verified that this function satisfies the conditions of the problem.
Remark. Using limits can be avoided. Since the rigth-hand side in ( $\dagger$ ) takes only integer values, it follows that $A, B, C, D$ are rational, so taking suitable multiples of integers for $n$ eliminates the powers of -1 and leaves a polynomial equality.
|
f(n)=n
|
Yes
|
Yes
|
proof
|
Algebra
|
Determine all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ satisfying
$$
f\left(f(m)^{2}+2 f(n)^{2}\right)=m^{2}+2 n^{2} \quad \text { for all } m, n \in \mathbb{N} . \quad \text { (Bulgaria) }
$$
Time allowed: 270 minutes.
Each problem is worth 10 points.
## SOLUTIONS
|
We start by observing that $f$ is injective. From the known identity
$$
\left(a^{2}+2 b^{2}\right)\left(c^{2}+2 d^{2}\right)=(a c \pm 2 b d)^{2}+2(a d \mp b c)^{2}
$$
we obtain $f(a c+2 b d)^{2}+2 f(a d-b c)^{2}=f(a c-2 b d)^{2}+2 f(a d+b c)^{2}$, assuming that the arguments are positive integers. Specially, for $b=c=d=1$ and $a \geq 3$ we have $f(a+2)^{2}+2 f(a-1)^{2}=f(a-2)^{2}+2 f(a+1)^{2}$. Denoting $g(n)=f(n)^{2}$ we get a recurrent relation $g(a+2)-2 g(a+1)+2 g(a-1)-g(a-2)=0$ whose characteristic polynomial is $(x+1)(x-1)^{3}$, which leads to
$$
g(n)=A(-1)^{n}+B+C n+D n^{2} .
$$
Substituting $m=n$ in the original equation yields $g(3 g(n))=9 n^{4}$, which together with $(\dagger)$ gives us
$$
\begin{aligned}
L=9 n^{4} & =A(-1)^{3\left(A(-1)^{n}+B+C n+D n^{2}\right)}+B+\underbrace{3 C\left[A(-1)^{n}+B+C n+D n^{2}\right]} \\
& +9 D\left[A(-1)^{n}+B+C n+D n^{2}\right]^{2}=R .
\end{aligned}
$$
Since $0=\lim _{n \rightarrow \infty} \frac{R-L}{n^{4}}=9 D^{3}-9$, we have $D^{3}=1$ (so $D \neq 0$ ); similarly, $0=\lim _{n \rightarrow \infty} \frac{R-L}{n^{3}}=18 D^{2} C$, so $C=0$. Moreover, for $n=2 k$ and $n=2 k+1$
respectively we obtain $0=\lim _{k \rightarrow \infty} \frac{R-L}{(2 k)^{2}}=18 D^{2}(A+B)$ and $0=\lim _{k \rightarrow \infty} \frac{R-L}{(2 k+1)^{2}}=$ $18 D^{2}(-A+B)$, implying $A+B=-A+B=0$; hence $A=B=0$.
Finally, $g(n)=D n^{2}, D^{3}=1$ and $g: \mathbb{N} \rightarrow \mathbb{N}$ gives us $D=1$, i.e. $f(n)=n$. It is directly verified that this function satisfies the conditions of the problem.
Remark. Using limits can be avoided. Since the rigth-hand side in ( $\dagger$ ) takes only integer values, it follows that $A, B, C, D$ are rational, so taking suitable multiples of integers for $n$ eliminates the powers of -1 and leaves a polynomial equality.
|
{
"resource_path": "Balkan_MO/segmented/en-2009-BMO-type2.jsonl",
"problem_match": "\n4.",
"solution_match": "\n4."
}
|
b40aab4f-ceaa-50ac-acdd-1658201a4bcf
| 606,370
|
The left-hand side is equal to
$$
\frac{a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}-a^{3} b^{2} c-b^{3} c^{2} a-c^{3} a^{2} b}{(a+b)(b+c)(c+a)}
$$
so it is enough to show that $a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3} \geq a^{3} b^{2} c+b^{3} c^{2} a+c^{3} a^{2} b$. The AM-GM inequality gives us $a^{3} b^{3}+a^{3} b^{3}+a^{3} c^{3} \geq 3 \sqrt[3]{a^{3} b^{3} \cdot a^{3} b^{3} \cdot a^{3} c^{3}}=3 a^{3} b^{2} c$; summing this inequality and its cyclic analogs yields the desired inequality. Equality holds if and only if $a=b=c$.
|
There are $n+1-\varphi(n)$ nonnegative integers not coprime with $n$, and whenever $r$ is among them, so is $n-r$. This gives us the formula $f(n)=\frac{1}{2} n(n+1-\varphi(n))$. Suppose that $f(n)=f(n+p)$. We observe first that $n$ and $n+p$ divide $2 f(n)<n(n+p)$, so $n$ and $n+p$ are not coprime, which implies that $n=k p$ for some $k \in \mathbb{N}$. Then the equality $f(n)=f(n+p)$ is equivalent to $k(k p+1-\varphi(k p))=(k+1)(k p+p+$ $1-\varphi(k p+p)$ ), so
$$
k p+1-\varphi(k p)=(k+1) x \quad \text { and } \quad k p+p+1-\varphi(k p+p)=k x
$$
for some $x \in \mathbb{N}, x<p$. Subtraction gives us $x=\varphi(k p+p)-\varphi(k p)-p$. Since $\varphi(k p)$ and $\varphi(k p+p)$ are both divisible by $p-1$ (by the formula for $\varphi(n)$ ), we obtain $x \equiv-1(\bmod p-1)$.
If $p=2$ then $x=1$ and $\varphi(2 k+2)=k+3$, which is impossible because $\varphi(2 k+2) \leq$ $k+1$. If $p=3$ then $x=1$ and $\varphi(3 k+3)=2 k+4$, again impossible because $\varphi(3 k+3) \leq 2 k+2$. Therefore $p \geq 5$, so $x \equiv-1(\bmod p-1)$ implies $x=p-2$. Plugging this value in (1) leads to
$$
\varphi(k p)=2 k+3-p \quad \text { and } \quad \varphi(k p+p)=2 k+1+p
$$
If $k$ is divisible by $p$, then $\varphi(k p)$ is also divisible by $p$, so $p \mid 2 k+3$ and hence $p \mid 3$, a contradiction. Similarly, $p \nmid k+1$. It follows that $\varphi(k p)=(p-1) \varphi(k)$ and $\varphi(k p+p)=(p-1) \varphi(k+1)$ which together with (1) yields
$$
\varphi(k)=\frac{2 k+2}{p-1}-1 \quad \text { and } \quad \varphi(k+1)=\frac{2 k+2}{p-1}+1
$$
From here we see that $\varphi(t)$ is not divisible by 4 either for $t=k$ or for $t=k+1$, which is only possible if $t=q^{i}$ or $t=2 q^{i}$ for some odd prime $q$ and $i \in \mathbb{N}$, or $t \in\{1,2,4\}$. The cases $t=1,2,4$ are easily ruled out, so either $k$ or $k+1$ is of the form $q^{i}$ or $2 q^{i}$. For $t=k=q^{i}, \varphi\left(q^{i}\right)+1=q^{i-1}(q-1)+1$ divides $2 q^{i}+2$ which is impossible because $q^{i}+1>q^{i-1}(q-1)+1>\frac{2}{3}\left(2 q^{i}+2\right)$. The other three cases are similarly shown to be impossible.
|
not found
|
Yes
|
Incomplete
|
proof
|
Inequalities
|
The left-hand side is equal to
$$
\frac{a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}-a^{3} b^{2} c-b^{3} c^{2} a-c^{3} a^{2} b}{(a+b)(b+c)(c+a)}
$$
so it is enough to show that $a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3} \geq a^{3} b^{2} c+b^{3} c^{2} a+c^{3} a^{2} b$. The AM-GM inequality gives us $a^{3} b^{3}+a^{3} b^{3}+a^{3} c^{3} \geq 3 \sqrt[3]{a^{3} b^{3} \cdot a^{3} b^{3} \cdot a^{3} c^{3}}=3 a^{3} b^{2} c$; summing this inequality and its cyclic analogs yields the desired inequality. Equality holds if and only if $a=b=c$.
|
There are $n+1-\varphi(n)$ nonnegative integers not coprime with $n$, and whenever $r$ is among them, so is $n-r$. This gives us the formula $f(n)=\frac{1}{2} n(n+1-\varphi(n))$. Suppose that $f(n)=f(n+p)$. We observe first that $n$ and $n+p$ divide $2 f(n)<n(n+p)$, so $n$ and $n+p$ are not coprime, which implies that $n=k p$ for some $k \in \mathbb{N}$. Then the equality $f(n)=f(n+p)$ is equivalent to $k(k p+1-\varphi(k p))=(k+1)(k p+p+$ $1-\varphi(k p+p)$ ), so
$$
k p+1-\varphi(k p)=(k+1) x \quad \text { and } \quad k p+p+1-\varphi(k p+p)=k x
$$
for some $x \in \mathbb{N}, x<p$. Subtraction gives us $x=\varphi(k p+p)-\varphi(k p)-p$. Since $\varphi(k p)$ and $\varphi(k p+p)$ are both divisible by $p-1$ (by the formula for $\varphi(n)$ ), we obtain $x \equiv-1(\bmod p-1)$.
If $p=2$ then $x=1$ and $\varphi(2 k+2)=k+3$, which is impossible because $\varphi(2 k+2) \leq$ $k+1$. If $p=3$ then $x=1$ and $\varphi(3 k+3)=2 k+4$, again impossible because $\varphi(3 k+3) \leq 2 k+2$. Therefore $p \geq 5$, so $x \equiv-1(\bmod p-1)$ implies $x=p-2$. Plugging this value in (1) leads to
$$
\varphi(k p)=2 k+3-p \quad \text { and } \quad \varphi(k p+p)=2 k+1+p
$$
If $k$ is divisible by $p$, then $\varphi(k p)$ is also divisible by $p$, so $p \mid 2 k+3$ and hence $p \mid 3$, a contradiction. Similarly, $p \nmid k+1$. It follows that $\varphi(k p)=(p-1) \varphi(k)$ and $\varphi(k p+p)=(p-1) \varphi(k+1)$ which together with (1) yields
$$
\varphi(k)=\frac{2 k+2}{p-1}-1 \quad \text { and } \quad \varphi(k+1)=\frac{2 k+2}{p-1}+1
$$
From here we see that $\varphi(t)$ is not divisible by 4 either for $t=k$ or for $t=k+1$, which is only possible if $t=q^{i}$ or $t=2 q^{i}$ for some odd prime $q$ and $i \in \mathbb{N}$, or $t \in\{1,2,4\}$. The cases $t=1,2,4$ are easily ruled out, so either $k$ or $k+1$ is of the form $q^{i}$ or $2 q^{i}$. For $t=k=q^{i}, \varphi\left(q^{i}\right)+1=q^{i-1}(q-1)+1$ divides $2 q^{i}+2$ which is impossible because $q^{i}+1>q^{i-1}(q-1)+1>\frac{2}{3}\left(2 q^{i}+2\right)$. The other three cases are similarly shown to be impossible.
|
{
"resource_path": "Balkan_MO/segmented/en-2010-BMO-type2.jsonl",
"problem_match": "\n1.",
"solution_match": "\n4."
}
|
6a76abdf-9cc8-5d98-8782-bbf27b5965b9
| 606,374
|
Let $A B C$ be an acute-angled triangle with orthocenter $H$ and let $M$ be the midpoint of $A C$. The foot of the altitude from $C$ is $C_{1}$. Point $H_{1}$ is symmetric to $H$ in $A B$. The projections of $C_{1}$ on lines $A H_{1}, A C$ and $B C$ are $P, Q$ and $R$ respectively. If $M_{1}$ is the circumcenter of triangle $P Q R$, prove that the point symmetric to $M$ with respect to $M_{1}$ lies on line $B H_{1}$.
(Serbia)
|
We shall use the following simple statement.
Lemma. Let $A_{1} A_{2} A_{3} A_{4}$ be a convex cyclic quadrilateral whose diagonals are orthogonal and meet at $X$. If $B_{i}$ is the midpoint of side $A_{i} A_{i+1}$ and $X_{i}$ the projection of $X$ on this side $\left(A_{5}=A_{1}\right)$, then the eight points $B_{i}, X_{i}$ $(i=1,2,3,4)$ lie on a circle.
Proof. Quadrilateral $B_{1} B_{2} B_{3} B_{4}$ is a rectangle because $B_{1} B_{2}\left\|B_{3} B_{4}\right\| A_{1} A_{3}$ and $B_{2} B_{3}\left\|B_{4} B_{1}\right\| A_{2} A_{4}$. Denote by $k$ the circumcircle of $B_{1} B_{2} B_{3} B_{4}$. Since $\angle B_{3} X A_{3}=\angle A_{4} A_{3} A_{1}=\angle A_{4} A_{2} A_{1}=\angle A_{1} X X_{1}$, points $B_{3}, X, X_{1}$ are collinear, so $X_{1}$ lies on the circle with diameter $B_{1} B_{3}$, i.e. $k$. Similarly, $X_{2}, X_{3}, X_{4}$ lie on $k$.
It is known that $H_{1}$ lies on the circumcircle of $A B C$. By the lemma, points $P, Q, R$ all lie on the circle with diameter $M N$, where $N$ is the midpoint of $B H_{1}$. Therefore $N$ is symmetric to $M$ with respect to $M_{1}$ and lies on $B H_{1}$ as desired.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute-angled triangle with orthocenter $H$ and let $M$ be the midpoint of $A C$. The foot of the altitude from $C$ is $C_{1}$. Point $H_{1}$ is symmetric to $H$ in $A B$. The projections of $C_{1}$ on lines $A H_{1}, A C$ and $B C$ are $P, Q$ and $R$ respectively. If $M_{1}$ is the circumcenter of triangle $P Q R$, prove that the point symmetric to $M$ with respect to $M_{1}$ lies on line $B H_{1}$.
(Serbia)
|
We shall use the following simple statement.
Lemma. Let $A_{1} A_{2} A_{3} A_{4}$ be a convex cyclic quadrilateral whose diagonals are orthogonal and meet at $X$. If $B_{i}$ is the midpoint of side $A_{i} A_{i+1}$ and $X_{i}$ the projection of $X$ on this side $\left(A_{5}=A_{1}\right)$, then the eight points $B_{i}, X_{i}$ $(i=1,2,3,4)$ lie on a circle.
Proof. Quadrilateral $B_{1} B_{2} B_{3} B_{4}$ is a rectangle because $B_{1} B_{2}\left\|B_{3} B_{4}\right\| A_{1} A_{3}$ and $B_{2} B_{3}\left\|B_{4} B_{1}\right\| A_{2} A_{4}$. Denote by $k$ the circumcircle of $B_{1} B_{2} B_{3} B_{4}$. Since $\angle B_{3} X A_{3}=\angle A_{4} A_{3} A_{1}=\angle A_{4} A_{2} A_{1}=\angle A_{1} X X_{1}$, points $B_{3}, X, X_{1}$ are collinear, so $X_{1}$ lies on the circle with diameter $B_{1} B_{3}$, i.e. $k$. Similarly, $X_{2}, X_{3}, X_{4}$ lie on $k$.
It is known that $H_{1}$ lies on the circumcircle of $A B C$. By the lemma, points $P, Q, R$ all lie on the circle with diameter $M N$, where $N$ is the midpoint of $B H_{1}$. Therefore $N$ is symmetric to $M$ with respect to $M_{1}$ and lies on $B H_{1}$ as desired.
|
{
"resource_path": "Balkan_MO/segmented/en-2010-BMO-type2.jsonl",
"problem_match": "\n2.",
"solution_match": "\n2."
}
|
358e2bc6-d7e2-5225-8247-19b941730f1b
| 606,377
|
We define a $w$-strip as the set of all points in the plane that are between or on two parallel lines on a mutual distance $w$. Let $S$ be a set of $n$ points in the plane such that any three points from $S$ can be covered by a 1 -strip. Show that the entire set $S$ can be covered by a 2 -strip.
(Romania)
|
Of all triangles with the vertices in $S$, consider one with a maximum area, say $\triangle A B C$. Let $A^{\prime}, B^{\prime}, C^{\prime}$ be the points symmetric to $A, B, C$ with respect to the midpoints of $B C, C A, A B$, respectively. We claim that all points from $S$ must lie inside or on the boundary of $\triangle A^{\prime} B^{\prime} C^{\prime}$. Indeed, if $X \in S$ is outside $\triangle A^{\prime} B^{\prime} C^{\prime}$, we can assume without loss of generality that $X$ and $B C$ are on different sides of $B^{\prime} C^{\prime}$, and then $\triangle B C X$ has an area greater than $\triangle A B C$, a contradiction.
The triangle $A B C$ can be covered by a 1-strip, so the triangle $A^{\prime} B^{\prime} C^{\prime}$, being similar to $A B C$ with ratio 2 , can be covered by a 2 -strip, also covering all of $S$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
We define a $w$-strip as the set of all points in the plane that are between or on two parallel lines on a mutual distance $w$. Let $S$ be a set of $n$ points in the plane such that any three points from $S$ can be covered by a 1 -strip. Show that the entire set $S$ can be covered by a 2 -strip.
(Romania)
|
Of all triangles with the vertices in $S$, consider one with a maximum area, say $\triangle A B C$. Let $A^{\prime}, B^{\prime}, C^{\prime}$ be the points symmetric to $A, B, C$ with respect to the midpoints of $B C, C A, A B$, respectively. We claim that all points from $S$ must lie inside or on the boundary of $\triangle A^{\prime} B^{\prime} C^{\prime}$. Indeed, if $X \in S$ is outside $\triangle A^{\prime} B^{\prime} C^{\prime}$, we can assume without loss of generality that $X$ and $B C$ are on different sides of $B^{\prime} C^{\prime}$, and then $\triangle B C X$ has an area greater than $\triangle A B C$, a contradiction.
The triangle $A B C$ can be covered by a 1-strip, so the triangle $A^{\prime} B^{\prime} C^{\prime}$, being similar to $A B C$ with ratio 2 , can be covered by a 2 -strip, also covering all of $S$.
|
{
"resource_path": "Balkan_MO/segmented/en-2010-BMO-type2.jsonl",
"problem_match": "\n3.",
"solution_match": "\n3."
}
|
f382c2f9-bdfe-5942-a009-da357933120f
| 606,380
|
For every integer $n \geq 2$, denote by $f(n)$ the sum of positive integers not exceeding $n$ that are not coprime to $n$. Prove that $f(n+p) \neq f(n)$ for any such $n$ and any prime number $p$.
(Turkey)
Time allowed: 270 minutes.
Each problem is worth 10 points.
## SOLUTIONS
|
There are $n+1-\varphi(n)$ nonnegative integers not coprime with $n$, and whenever $r$ is among them, so is $n-r$. This gives us the formula $f(n)=\frac{1}{2} n(n+1-\varphi(n))$. Suppose that $f(n)=f(n+p)$. We observe first that $n$ and $n+p$ divide $2 f(n)<n(n+p)$, so $n$ and $n+p$ are not coprime, which implies that $n=k p$ for some $k \in \mathbb{N}$. Then the equality $f(n)=f(n+p)$ is equivalent to $k(k p+1-\varphi(k p))=(k+1)(k p+p+$ $1-\varphi(k p+p)$ ), so
$$
k p+1-\varphi(k p)=(k+1) x \quad \text { and } \quad k p+p+1-\varphi(k p+p)=k x
$$
for some $x \in \mathbb{N}, x<p$. Subtraction gives us $x=\varphi(k p+p)-\varphi(k p)-p$. Since $\varphi(k p)$ and $\varphi(k p+p)$ are both divisible by $p-1$ (by the formula for $\varphi(n)$ ), we obtain $x \equiv-1(\bmod p-1)$.
If $p=2$ then $x=1$ and $\varphi(2 k+2)=k+3$, which is impossible because $\varphi(2 k+2) \leq$ $k+1$. If $p=3$ then $x=1$ and $\varphi(3 k+3)=2 k+4$, again impossible because $\varphi(3 k+3) \leq 2 k+2$. Therefore $p \geq 5$, so $x \equiv-1(\bmod p-1)$ implies $x=p-2$. Plugging this value in (1) leads to
$$
\varphi(k p)=2 k+3-p \quad \text { and } \quad \varphi(k p+p)=2 k+1+p
$$
If $k$ is divisible by $p$, then $\varphi(k p)$ is also divisible by $p$, so $p \mid 2 k+3$ and hence $p \mid 3$, a contradiction. Similarly, $p \nmid k+1$. It follows that $\varphi(k p)=(p-1) \varphi(k)$ and $\varphi(k p+p)=(p-1) \varphi(k+1)$ which together with (1) yields
$$
\varphi(k)=\frac{2 k+2}{p-1}-1 \quad \text { and } \quad \varphi(k+1)=\frac{2 k+2}{p-1}+1
$$
From here we see that $\varphi(t)$ is not divisible by 4 either for $t=k$ or for $t=k+1$, which is only possible if $t=q^{i}$ or $t=2 q^{i}$ for some odd prime $q$ and $i \in \mathbb{N}$, or $t \in\{1,2,4\}$. The cases $t=1,2,4$ are easily ruled out, so either $k$ or $k+1$ is of the form $q^{i}$ or $2 q^{i}$. For $t=k=q^{i}, \varphi\left(q^{i}\right)+1=q^{i-1}(q-1)+1$ divides $2 q^{i}+2$ which is impossible because $q^{i}+1>q^{i-1}(q-1)+1>\frac{2}{3}\left(2 q^{i}+2\right)$. The other three cases are similarly shown to be impossible.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
For every integer $n \geq 2$, denote by $f(n)$ the sum of positive integers not exceeding $n$ that are not coprime to $n$. Prove that $f(n+p) \neq f(n)$ for any such $n$ and any prime number $p$.
(Turkey)
Time allowed: 270 minutes.
Each problem is worth 10 points.
## SOLUTIONS
|
There are $n+1-\varphi(n)$ nonnegative integers not coprime with $n$, and whenever $r$ is among them, so is $n-r$. This gives us the formula $f(n)=\frac{1}{2} n(n+1-\varphi(n))$. Suppose that $f(n)=f(n+p)$. We observe first that $n$ and $n+p$ divide $2 f(n)<n(n+p)$, so $n$ and $n+p$ are not coprime, which implies that $n=k p$ for some $k \in \mathbb{N}$. Then the equality $f(n)=f(n+p)$ is equivalent to $k(k p+1-\varphi(k p))=(k+1)(k p+p+$ $1-\varphi(k p+p)$ ), so
$$
k p+1-\varphi(k p)=(k+1) x \quad \text { and } \quad k p+p+1-\varphi(k p+p)=k x
$$
for some $x \in \mathbb{N}, x<p$. Subtraction gives us $x=\varphi(k p+p)-\varphi(k p)-p$. Since $\varphi(k p)$ and $\varphi(k p+p)$ are both divisible by $p-1$ (by the formula for $\varphi(n)$ ), we obtain $x \equiv-1(\bmod p-1)$.
If $p=2$ then $x=1$ and $\varphi(2 k+2)=k+3$, which is impossible because $\varphi(2 k+2) \leq$ $k+1$. If $p=3$ then $x=1$ and $\varphi(3 k+3)=2 k+4$, again impossible because $\varphi(3 k+3) \leq 2 k+2$. Therefore $p \geq 5$, so $x \equiv-1(\bmod p-1)$ implies $x=p-2$. Plugging this value in (1) leads to
$$
\varphi(k p)=2 k+3-p \quad \text { and } \quad \varphi(k p+p)=2 k+1+p
$$
If $k$ is divisible by $p$, then $\varphi(k p)$ is also divisible by $p$, so $p \mid 2 k+3$ and hence $p \mid 3$, a contradiction. Similarly, $p \nmid k+1$. It follows that $\varphi(k p)=(p-1) \varphi(k)$ and $\varphi(k p+p)=(p-1) \varphi(k+1)$ which together with (1) yields
$$
\varphi(k)=\frac{2 k+2}{p-1}-1 \quad \text { and } \quad \varphi(k+1)=\frac{2 k+2}{p-1}+1
$$
From here we see that $\varphi(t)$ is not divisible by 4 either for $t=k$ or for $t=k+1$, which is only possible if $t=q^{i}$ or $t=2 q^{i}$ for some odd prime $q$ and $i \in \mathbb{N}$, or $t \in\{1,2,4\}$. The cases $t=1,2,4$ are easily ruled out, so either $k$ or $k+1$ is of the form $q^{i}$ or $2 q^{i}$. For $t=k=q^{i}, \varphi\left(q^{i}\right)+1=q^{i-1}(q-1)+1$ divides $2 q^{i}+2$ which is impossible because $q^{i}+1>q^{i-1}(q-1)+1>\frac{2}{3}\left(2 q^{i}+2\right)$. The other three cases are similarly shown to be impossible.
|
{
"resource_path": "Balkan_MO/segmented/en-2010-BMO-type2.jsonl",
"problem_match": "\n4.",
"solution_match": "\n4."
}
|
7f330dd6-2174-5725-85a5-49570da9e6c3
| 606,383
|
Let $A B C D$ be a cyclic quadrilateral which is not a trapezoid and whose diagonals meet at $E$. The midpoints of $A B$ and $C D$ are $F$ and $G$ respectively, and $\ell$ is the line through $G$ parallel to $A B$. The feet of the perpendiculars from $E$ onto the lines $\ell$ and $C D$ are $H$ and $K$, respectively. Prove that the lines $E F$ and $H K$ are perpendicular.
|
The points $E, K, H, G$ are on the circle of diameter $G E$, so
$$
\angle E H K=\angle E G K
$$
Also, from $\angle D C A=\angle D B A$ and $\frac{C E}{C D}=\frac{B E}{B A}$ it follows
$$
\frac{C E}{C G}=\frac{2 C E}{C D}=\frac{2 B E}{B A}=\frac{B E}{B F},
$$
therefore $\triangle C G E \sim \triangle B F E$. In particular, $\angle E G C=\angle B F E$, so by $(\dagger)$
$$
\angle E H K=\angle B F E .
$$
But $H E \perp F B$ and so, since $F E$ and $H K$ are obtained by rotations of these lines by the same (directed) angle, $F E \perp H K$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C D$ be a cyclic quadrilateral which is not a trapezoid and whose diagonals meet at $E$. The midpoints of $A B$ and $C D$ are $F$ and $G$ respectively, and $\ell$ is the line through $G$ parallel to $A B$. The feet of the perpendiculars from $E$ onto the lines $\ell$ and $C D$ are $H$ and $K$, respectively. Prove that the lines $E F$ and $H K$ are perpendicular.
|
The points $E, K, H, G$ are on the circle of diameter $G E$, so
$$
\angle E H K=\angle E G K
$$
Also, from $\angle D C A=\angle D B A$ and $\frac{C E}{C D}=\frac{B E}{B A}$ it follows
$$
\frac{C E}{C G}=\frac{2 C E}{C D}=\frac{2 B E}{B A}=\frac{B E}{B F},
$$
therefore $\triangle C G E \sim \triangle B F E$. In particular, $\angle E G C=\angle B F E$, so by $(\dagger)$
$$
\angle E H K=\angle B F E .
$$
But $H E \perp F B$ and so, since $F E$ and $H K$ are obtained by rotations of these lines by the same (directed) angle, $F E \perp H K$.
|
{
"resource_path": "Balkan_MO/segmented/en-2011-BMO-type1.jsonl",
"problem_match": "# PROBLEM 1",
"solution_match": "\nSolution."
}
|
f89c0e6d-4465-5955-b8ae-be529bb4869b
| 604,161
|
Given real numbers $x, y, z$ such that $x+y+z=0$, show that
$$
\frac{x(x+2)}{2 x^{2}+1}+\frac{y(y+2)}{2 y^{2}+1}+\frac{z(z+2)}{2 z^{2}+1} \geq 0 .
$$
When does equality hold?
|
The inequality is clear if $x y z=0$, in which case equality holds if and only if $x=y=z=0$.
Henceforth assume $x y z \neq 0$ and rewrite the inequality as
$$
\frac{(2 x+1)^{2}}{2 x^{2}+1}+\frac{(2 y+1)^{2}}{2 y^{2}+1}+\frac{(2 z+1)^{2}}{2 z^{2}+1} \geq 3 .
$$
Notice that (exactly) one of the products $x y, y z, z x$ is positive, say $y z>0$, to get
$$
\begin{array}{rlr}
\frac{(2 y+1)^{2}}{2 y^{2}+1}+\frac{(2 z+1)^{2}}{2 z^{2}+1} & \geq \frac{2(y+z+1)^{2}}{y^{2}+z^{2}+1} & \text { (by Jensen) } \\
& =\frac{2(x-1)^{2}}{x^{2}-2 y z+1} & (\text { for } x+y+z=0) \\
& \geq \frac{2(x-1)^{2}}{x^{2}+1} . & (\text { for } y z>0)
\end{array}
$$
Here equality holds if and only if $x=1$ and $y=z=-1 / 2$. Finally, since
$$
\frac{(2 x+1)^{2}}{2 x^{2}+1}+\frac{2(x-1)^{2}}{x^{2}+1}-3=\frac{2 x^{2}(x-1)^{2}}{\left(2 x^{2}+1\right)\left(x^{2}+1\right)} \geq 0, \quad x \in \mathbb{R},
$$
the conclusion follows. Clearly, equality holds if and only if $x=1$, so $y=z=-1 / 2$. Therefore, if $x y z \neq 0$, equality holds if and only if one of the numbers is 1 , and the other two are $-1 / 2$.
Marking Scheme. Proving the inequality and identifying the equality case when one of the variables vanishes
Applying Jensen or Cauchy-Schwarz inequality to the fractions involving the pair of numbers of the same sign 3p
Producing the corresponding lower bound in the third variable ................... 3p
Remark. Any partial or equivalent approach should be marked accordingly.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Given real numbers $x, y, z$ such that $x+y+z=0$, show that
$$
\frac{x(x+2)}{2 x^{2}+1}+\frac{y(y+2)}{2 y^{2}+1}+\frac{z(z+2)}{2 z^{2}+1} \geq 0 .
$$
When does equality hold?
|
The inequality is clear if $x y z=0$, in which case equality holds if and only if $x=y=z=0$.
Henceforth assume $x y z \neq 0$ and rewrite the inequality as
$$
\frac{(2 x+1)^{2}}{2 x^{2}+1}+\frac{(2 y+1)^{2}}{2 y^{2}+1}+\frac{(2 z+1)^{2}}{2 z^{2}+1} \geq 3 .
$$
Notice that (exactly) one of the products $x y, y z, z x$ is positive, say $y z>0$, to get
$$
\begin{array}{rlr}
\frac{(2 y+1)^{2}}{2 y^{2}+1}+\frac{(2 z+1)^{2}}{2 z^{2}+1} & \geq \frac{2(y+z+1)^{2}}{y^{2}+z^{2}+1} & \text { (by Jensen) } \\
& =\frac{2(x-1)^{2}}{x^{2}-2 y z+1} & (\text { for } x+y+z=0) \\
& \geq \frac{2(x-1)^{2}}{x^{2}+1} . & (\text { for } y z>0)
\end{array}
$$
Here equality holds if and only if $x=1$ and $y=z=-1 / 2$. Finally, since
$$
\frac{(2 x+1)^{2}}{2 x^{2}+1}+\frac{2(x-1)^{2}}{x^{2}+1}-3=\frac{2 x^{2}(x-1)^{2}}{\left(2 x^{2}+1\right)\left(x^{2}+1\right)} \geq 0, \quad x \in \mathbb{R},
$$
the conclusion follows. Clearly, equality holds if and only if $x=1$, so $y=z=-1 / 2$. Therefore, if $x y z \neq 0$, equality holds if and only if one of the numbers is 1 , and the other two are $-1 / 2$.
Marking Scheme. Proving the inequality and identifying the equality case when one of the variables vanishes
Applying Jensen or Cauchy-Schwarz inequality to the fractions involving the pair of numbers of the same sign 3p
Producing the corresponding lower bound in the third variable ................... 3p
Remark. Any partial or equivalent approach should be marked accordingly.
|
{
"resource_path": "Balkan_MO/segmented/en-2011-BMO-type1.jsonl",
"problem_match": "# PROBLEM 2",
"solution_match": "\nSolution."
}
|
4cd97ab9-6970-5d06-9ffb-c65f1d1a2670
| 604,175
|
Let $S$ be a finite set of positive integers which has the following property: if $x$ is a member of $S$, then so are all positive divisors of $x$. A non-empty subset $T$ of $S$ is good if whenever $x, y \in T$ and $x<y$, the ratio $y / x$ is a power of a prime number. A non-empty subset $T$ of $S$ is bad if whenever $x, y \in T$ and $x<y$, the ratio $y / x$ is not a power of a prime number. We agree that a singleton subset of $S$ is both good and bad. Let $k$ be the largest possible size of a good subset of $S$. Prove that $k$ is also the smallest number of pairwise-disjoint bad subsets whose union is $S$.
|
Notice first that a bad subset of $S$ contains at most one element from a good one, to deduce that a partition of $S$ into bad subsets has at least as many members as a maximal good subset.
Notice further that the elements of a good subset of $S$ must be among the terms of a geometric sequence whose ratio is a prime: if $x<y<z$ are elements of a good subset of $S$, then $y=x p^{\alpha}$ and $z=y q^{\beta}=x p^{\alpha} q^{\beta}$ for some primes $p$ and $q$ and some positive integers $\alpha$ and $\beta$, so $p=q$ for $z / x$ to be a power of a prime.
Next, let $P=\{2,3,5,7,11, \cdots\}$ denote the set of all primes, let
$$
m=\max \left\{\exp _{p} x: x \in S \text { and } p \in P\right\}
$$
where $\exp _{p} x$ is the exponent of the prime $p$ in the canonical decomposition of $x$, and notice that a maximal good subset of $S$ must be of the form $\left\{a, a p, \cdots, a p^{m}\right\}$ for some prime $p$ and some positive integer $a$ which is not divisible by $p$. Consequently, a maximal good subset of $S$ has $m+1$ elements, so a partition of $S$ into bad subsets has at least $m+1$ members.
Finally, notice by maximality of $m$ that the sets
$$
S_{k}=\left\{x: x \in S \text { and } \sum_{p \in P} \exp _{p} x \equiv k(\bmod m+1)\right\}, \quad k=0,1, \cdots, m
$$
form a partition of $S$ into $m+1$ bad subsets. The conclusion follows.
 Considering the maximal exponent $m$ of a prime and deriving $k=m+1 \ldots \ldots \mathbf{1 p}$ Noticing that the intersection of a bad set and a good set contains at most one element and infering that a partition of $S$ into bad sets has at least $k$ members.....2p
Producing a partition of $S$ into $k$ bad subsets . . . . . . . . . . . . . . . . . . . . . . .
Remark. Any partial or equivalent approach should be marked accordingly.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Let $S$ be a finite set of positive integers which has the following property: if $x$ is a member of $S$, then so are all positive divisors of $x$. A non-empty subset $T$ of $S$ is good if whenever $x, y \in T$ and $x<y$, the ratio $y / x$ is a power of a prime number. A non-empty subset $T$ of $S$ is bad if whenever $x, y \in T$ and $x<y$, the ratio $y / x$ is not a power of a prime number. We agree that a singleton subset of $S$ is both good and bad. Let $k$ be the largest possible size of a good subset of $S$. Prove that $k$ is also the smallest number of pairwise-disjoint bad subsets whose union is $S$.
|
Notice first that a bad subset of $S$ contains at most one element from a good one, to deduce that a partition of $S$ into bad subsets has at least as many members as a maximal good subset.
Notice further that the elements of a good subset of $S$ must be among the terms of a geometric sequence whose ratio is a prime: if $x<y<z$ are elements of a good subset of $S$, then $y=x p^{\alpha}$ and $z=y q^{\beta}=x p^{\alpha} q^{\beta}$ for some primes $p$ and $q$ and some positive integers $\alpha$ and $\beta$, so $p=q$ for $z / x$ to be a power of a prime.
Next, let $P=\{2,3,5,7,11, \cdots\}$ denote the set of all primes, let
$$
m=\max \left\{\exp _{p} x: x \in S \text { and } p \in P\right\}
$$
where $\exp _{p} x$ is the exponent of the prime $p$ in the canonical decomposition of $x$, and notice that a maximal good subset of $S$ must be of the form $\left\{a, a p, \cdots, a p^{m}\right\}$ for some prime $p$ and some positive integer $a$ which is not divisible by $p$. Consequently, a maximal good subset of $S$ has $m+1$ elements, so a partition of $S$ into bad subsets has at least $m+1$ members.
Finally, notice by maximality of $m$ that the sets
$$
S_{k}=\left\{x: x \in S \text { and } \sum_{p \in P} \exp _{p} x \equiv k(\bmod m+1)\right\}, \quad k=0,1, \cdots, m
$$
form a partition of $S$ into $m+1$ bad subsets. The conclusion follows.
 Considering the maximal exponent $m$ of a prime and deriving $k=m+1 \ldots \ldots \mathbf{1 p}$ Noticing that the intersection of a bad set and a good set contains at most one element and infering that a partition of $S$ into bad sets has at least $k$ members.....2p
Producing a partition of $S$ into $k$ bad subsets . . . . . . . . . . . . . . . . . . . . . . .
Remark. Any partial or equivalent approach should be marked accordingly.
|
{
"resource_path": "Balkan_MO/segmented/en-2011-BMO-type1.jsonl",
"problem_match": "# PROBLEM 3",
"solution_match": "\nSolution."
}
|
7e22dae2-c48e-5c68-a7b6-a13101d8f996
| 604,187
|
Let $A B C D E F$ be a convex hexagon of area 1 , whose opposite sides are parallel. The lines $A B, C D$ and $E F$ meet in pairs to determine the vertices of a triangle. Similarly, the lines $B C, D E$ and $F A$ meet in pairs to determine the vertices of another triangle. Show that the area of at least one of these two triangles is at least $3 / 2$.
|
Unless otherwise stated, throughout the proof indices take on values from 0 to 5 and are reduced modulo 6 . Label the vertices of the hexagon in circular order, $A_{0}, A_{1}, \cdots, A_{5}$, and let the lines of support of the alternate sides $A_{i} A_{i+1}$ and $A_{i+2} A_{i+3}$ meet at $B_{i}$. To show that the area of at least one of the triangles $B_{0} B_{2} B_{4}, B_{1} B_{3} B_{5}$ is greater than or equal to $3 / 2$, it is sufficient to prove that the total area of the six triangles $A_{i+1} B_{i} A_{i+2}$ is at least 1:
$$
\sum_{i=0}^{5} \operatorname{area} A_{i+1} B_{i} A_{i+2} \geq 1
$$
To begin with, reflect each $B_{i}$ through the midpoint of the segment $A_{i+1} A_{i+2}$ to get the points $B_{i}^{\prime}$. We shall prove that the six triangles $A_{i+1} B_{i}^{\prime} A_{i+2}$ cover the hexagon. To this end, reflect $A_{2 i+1}$ through the midpoint of the segment $A_{2 i} A_{2 i+2}$ to get the points $A_{2 i+1}^{\prime}$, $i=0,1,2$. The hexagon splits into three parallelograms, $A_{2 i} A_{2 i+1} A_{2 i+2} A_{2 i+1}^{\prime}, i=0,1,2$, and a (possibly degenerate) triangle, $A_{1}^{\prime} A_{3}^{\prime} A_{5}^{\prime}$. Notice first that each parallelogram $A_{2 i} A_{2 i+1} A_{2 i+2} A_{2 i+1}^{\prime}$ is covered by the pair of triangles $\left(A_{2 i} B_{2 i+5}^{\prime} A_{2 i+1}, A_{2 i+1} B_{2 i}^{\prime} A_{2 i+2}\right)$, $i=0,1,2$. The proof is completed by showing that at least one of these pairs contains a triangle that covers the triangle $A_{1}^{\prime} A_{3}^{\prime} A_{5}^{\prime}$. To this end, it is sufficient to prove that $A_{2 i} B_{2 i+5}^{\prime} \geq A_{2 i} A_{2 i+5}^{\prime}$ and $A_{2 j+2} B_{2 j}^{\prime} \geq A_{2 j+2} A_{2 j+3}^{\prime}$ for some indices $i, j \in\{0,1,2\}$. To establish the first inequality, notice that
$$
\begin{gathered}
A_{2 i} B_{2 i+5}^{\prime}=A_{2 i+1} B_{2 i+5}, \quad A_{2 i} A_{2 i+5}^{\prime}=A_{2 i+4} A_{2 i+5}, \quad i=0,1,2, \\
\\
\frac{A_{1} B_{5}}{A_{4} A_{5}}=\frac{A_{0} B_{5}}{A_{5} B_{3}} \quad \text { and } \quad \frac{A_{3} B_{1}}{A_{0} A_{1}}=\frac{A_{2} A_{3}}{A_{0} B_{5}},
\end{gathered}
$$
to get
$$
\prod_{i=0}^{2} \frac{A_{2 i} B_{2 i+5}^{\prime}}{A_{2 i} A_{2 i+5}^{\prime}}=1
$$
Similarly,
$$
\prod_{j=0}^{2} \frac{A_{2 j+2} B_{2 j}^{\prime}}{A_{2 j+2} A_{2 j+3}^{\prime}}=1
$$
whence the conclusion.
Marking Scheme. Stating that the total area of the small triangles $\geq 1 \ldots \ldots . . \mathbf{1 p}$
Decomposition of the hexagon into three adequate parallelograms and a triangle $\mathbf{1 p}$
Proving that each pair of triangles adjacent to a parallelogram covers that parallelogram
Proving the central triangle also covered 5p
Remark. Any partial or equivalent approach should be marked accordingly.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C D E F$ be a convex hexagon of area 1 , whose opposite sides are parallel. The lines $A B, C D$ and $E F$ meet in pairs to determine the vertices of a triangle. Similarly, the lines $B C, D E$ and $F A$ meet in pairs to determine the vertices of another triangle. Show that the area of at least one of these two triangles is at least $3 / 2$.
|
Unless otherwise stated, throughout the proof indices take on values from 0 to 5 and are reduced modulo 6 . Label the vertices of the hexagon in circular order, $A_{0}, A_{1}, \cdots, A_{5}$, and let the lines of support of the alternate sides $A_{i} A_{i+1}$ and $A_{i+2} A_{i+3}$ meet at $B_{i}$. To show that the area of at least one of the triangles $B_{0} B_{2} B_{4}, B_{1} B_{3} B_{5}$ is greater than or equal to $3 / 2$, it is sufficient to prove that the total area of the six triangles $A_{i+1} B_{i} A_{i+2}$ is at least 1:
$$
\sum_{i=0}^{5} \operatorname{area} A_{i+1} B_{i} A_{i+2} \geq 1
$$
To begin with, reflect each $B_{i}$ through the midpoint of the segment $A_{i+1} A_{i+2}$ to get the points $B_{i}^{\prime}$. We shall prove that the six triangles $A_{i+1} B_{i}^{\prime} A_{i+2}$ cover the hexagon. To this end, reflect $A_{2 i+1}$ through the midpoint of the segment $A_{2 i} A_{2 i+2}$ to get the points $A_{2 i+1}^{\prime}$, $i=0,1,2$. The hexagon splits into three parallelograms, $A_{2 i} A_{2 i+1} A_{2 i+2} A_{2 i+1}^{\prime}, i=0,1,2$, and a (possibly degenerate) triangle, $A_{1}^{\prime} A_{3}^{\prime} A_{5}^{\prime}$. Notice first that each parallelogram $A_{2 i} A_{2 i+1} A_{2 i+2} A_{2 i+1}^{\prime}$ is covered by the pair of triangles $\left(A_{2 i} B_{2 i+5}^{\prime} A_{2 i+1}, A_{2 i+1} B_{2 i}^{\prime} A_{2 i+2}\right)$, $i=0,1,2$. The proof is completed by showing that at least one of these pairs contains a triangle that covers the triangle $A_{1}^{\prime} A_{3}^{\prime} A_{5}^{\prime}$. To this end, it is sufficient to prove that $A_{2 i} B_{2 i+5}^{\prime} \geq A_{2 i} A_{2 i+5}^{\prime}$ and $A_{2 j+2} B_{2 j}^{\prime} \geq A_{2 j+2} A_{2 j+3}^{\prime}$ for some indices $i, j \in\{0,1,2\}$. To establish the first inequality, notice that
$$
\begin{gathered}
A_{2 i} B_{2 i+5}^{\prime}=A_{2 i+1} B_{2 i+5}, \quad A_{2 i} A_{2 i+5}^{\prime}=A_{2 i+4} A_{2 i+5}, \quad i=0,1,2, \\
\\
\frac{A_{1} B_{5}}{A_{4} A_{5}}=\frac{A_{0} B_{5}}{A_{5} B_{3}} \quad \text { and } \quad \frac{A_{3} B_{1}}{A_{0} A_{1}}=\frac{A_{2} A_{3}}{A_{0} B_{5}},
\end{gathered}
$$
to get
$$
\prod_{i=0}^{2} \frac{A_{2 i} B_{2 i+5}^{\prime}}{A_{2 i} A_{2 i+5}^{\prime}}=1
$$
Similarly,
$$
\prod_{j=0}^{2} \frac{A_{2 j+2} B_{2 j}^{\prime}}{A_{2 j+2} A_{2 j+3}^{\prime}}=1
$$
whence the conclusion.
Marking Scheme. Stating that the total area of the small triangles $\geq 1 \ldots \ldots . . \mathbf{1 p}$
Decomposition of the hexagon into three adequate parallelograms and a triangle $\mathbf{1 p}$
Proving that each pair of triangles adjacent to a parallelogram covers that parallelogram
Proving the central triangle also covered 5p
Remark. Any partial or equivalent approach should be marked accordingly.
|
{
"resource_path": "Balkan_MO/segmented/en-2011-BMO-type1.jsonl",
"problem_match": "# PROBLEM 4",
"solution_match": "\nSolution."
}
|
a9756494-36d3-5f6c-86db-46a9b48f012d
| 604,201
|
\quad$ Let $A, B$ and $C$ be points lying on a circle $\Gamma$ with centre $O$. Assume that $\angle A B C>90$. Let $D$ be the point of intersection of the line $A B$ with the line perpendicular to $A C$ at $C$. Let $l$ be the line through $D$ which is perpendicular to $A O$. Let $E$ be the point of intersection of $l$ with the line $A C$, and let $F$ be the point of intersection of $\Gamma$ with $l$ that lies between $D$ and $E$.
Prove that the circumcircles of triangles $B F E$ and $C F D$ are tangent at $F$.
|
Let $\ell \cap A O=\{K\}$ and $G$ be the other end point of the diameter of $\Gamma$ through $A$. Then $D, C, G$ are collinear. Moreover, $E$ is the orthocenter of triangle $A D G$. Therefore $G E \perp A D$ and $G, E, B$ are collinear.
As $\angle C D F=\angle G D K=\angle G A C=\angle G F C, F G$ is tangent to the circumcircle of triangle $C F D$ at $F$. As $\angle F B E=\angle F B G=\angle F A G=\angle G F K=\angle G F E, F G$ is also tangent to the circumcircle of $B F E$ at $F$. Hence the circumcircles of the triangles $C F D$ and $B F E$ are tangent at $F$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
\quad$ Let $A, B$ and $C$ be points lying on a circle $\Gamma$ with centre $O$. Assume that $\angle A B C>90$. Let $D$ be the point of intersection of the line $A B$ with the line perpendicular to $A C$ at $C$. Let $l$ be the line through $D$ which is perpendicular to $A O$. Let $E$ be the point of intersection of $l$ with the line $A C$, and let $F$ be the point of intersection of $\Gamma$ with $l$ that lies between $D$ and $E$.
Prove that the circumcircles of triangles $B F E$ and $C F D$ are tangent at $F$.
|
Let $\ell \cap A O=\{K\}$ and $G$ be the other end point of the diameter of $\Gamma$ through $A$. Then $D, C, G$ are collinear. Moreover, $E$ is the orthocenter of triangle $A D G$. Therefore $G E \perp A D$ and $G, E, B$ are collinear.
As $\angle C D F=\angle G D K=\angle G A C=\angle G F C, F G$ is tangent to the circumcircle of triangle $C F D$ at $F$. As $\angle F B E=\angle F B G=\angle F A G=\angle G F K=\angle G F E, F G$ is also tangent to the circumcircle of $B F E$ at $F$. Hence the circumcircles of the triangles $C F D$ and $B F E$ are tangent at $F$.
|
{
"resource_path": "Balkan_MO/segmented/en-2012-BMO-type3.jsonl",
"problem_match": "\n1. ",
"solution_match": "\nSolution."
}
|
8ad74077-a892-5b78-8b16-c68a5f491b10
| 604,215
|
Prove that
$$
\sum_{c y c}(x+y) \sqrt{(z+x)(z+y)} \geq 4(x y+y z+z x),
$$
for all positive real numbers $x, y$ and $z$.
|
We will obtain the inequality by adding the inequalities
$$
(x+y) \sqrt{(z+x)(z+y)} \geq 2 x y+y z+z x
$$
for cyclic permutation of $x, y, z$.
Squaring both sides of this inequality we obtain
$$
(x+y)^{2}(z+x)(z+y) \geq 4 x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+4 x y^{2} z+4 x^{2} y z+2 x y z^{2}
$$
which is equivalent to
$$
x^{3} y+x y^{3}+z\left(x^{3}+y^{3}\right) \geq 2 x^{2} y^{2}+x y z(x+y)
$$
which can be rearranged to
$$
(x y+y z+z x)(x-y)^{2} \geq 0
$$
which is clearly true.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Prove that
$$
\sum_{c y c}(x+y) \sqrt{(z+x)(z+y)} \geq 4(x y+y z+z x),
$$
for all positive real numbers $x, y$ and $z$.
|
We will obtain the inequality by adding the inequalities
$$
(x+y) \sqrt{(z+x)(z+y)} \geq 2 x y+y z+z x
$$
for cyclic permutation of $x, y, z$.
Squaring both sides of this inequality we obtain
$$
(x+y)^{2}(z+x)(z+y) \geq 4 x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+4 x y^{2} z+4 x^{2} y z+2 x y z^{2}
$$
which is equivalent to
$$
x^{3} y+x y^{3}+z\left(x^{3}+y^{3}\right) \geq 2 x^{2} y^{2}+x y z(x+y)
$$
which can be rearranged to
$$
(x y+y z+z x)(x-y)^{2} \geq 0
$$
which is clearly true.
|
{
"resource_path": "Balkan_MO/segmented/en-2012-BMO-type3.jsonl",
"problem_match": "\n2. ",
"solution_match": "\nSolution 1."
}
|
36d8b6b0-4757-56f8-81f8-55a19f290550
| 604,232
|
Prove that
$$
\sum_{c y c}(x+y) \sqrt{(z+x)(z+y)} \geq 4(x y+y z+z x),
$$
for all positive real numbers $x, y$ and $z$.
|
For positive real numbers $x, y, z$ there exists a triangle with the side lengths $\sqrt{x+y}, \sqrt{y+z}, \sqrt{z+x}$ and the area $K=\sqrt{x y+y z+z x} / 2$.
The existence of the triangle is clear by simple checking of the triangle inequality. To prove the area formula, we have
$$
K=\frac{1}{2} \sqrt{x+y} \sqrt{z+x} \sin \alpha
$$
where $\alpha$ is the angle between the sides of length $\sqrt{x+y}$ and $\sqrt{z+x}$. On the other hand, from the law of cosines we have
$$
\cos \alpha=\frac{x+y+z+x-y-z}{2 \sqrt{(x+y)(z+x)}}=\frac{x}{\sqrt{(x+y)(z+x)}}
$$
and
$$
\sin \alpha=\sqrt{1-\cos ^{2} \alpha}=\frac{\sqrt{x y+y z+z x}}{\sqrt{(x+y)(z+x)}}
$$
Now the inequality is equivalent to
$$
\sqrt{x+y} \sqrt{y+z} \sqrt{z+x} \sum_{c y c} \sqrt{x+y} \geq 16 K^{2}
$$
This can be rewritten as
$$
\frac{\sqrt{x+y} \sqrt{y+z} \sqrt{z+x}}{4 K} \geq 2 \frac{K}{\sum_{c y c} \sqrt{x+y} / 2}
$$
to become the Euler inequality $R \geq 2 r$.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Prove that
$$
\sum_{c y c}(x+y) \sqrt{(z+x)(z+y)} \geq 4(x y+y z+z x),
$$
for all positive real numbers $x, y$ and $z$.
|
For positive real numbers $x, y, z$ there exists a triangle with the side lengths $\sqrt{x+y}, \sqrt{y+z}, \sqrt{z+x}$ and the area $K=\sqrt{x y+y z+z x} / 2$.
The existence of the triangle is clear by simple checking of the triangle inequality. To prove the area formula, we have
$$
K=\frac{1}{2} \sqrt{x+y} \sqrt{z+x} \sin \alpha
$$
where $\alpha$ is the angle between the sides of length $\sqrt{x+y}$ and $\sqrt{z+x}$. On the other hand, from the law of cosines we have
$$
\cos \alpha=\frac{x+y+z+x-y-z}{2 \sqrt{(x+y)(z+x)}}=\frac{x}{\sqrt{(x+y)(z+x)}}
$$
and
$$
\sin \alpha=\sqrt{1-\cos ^{2} \alpha}=\frac{\sqrt{x y+y z+z x}}{\sqrt{(x+y)(z+x)}}
$$
Now the inequality is equivalent to
$$
\sqrt{x+y} \sqrt{y+z} \sqrt{z+x} \sum_{c y c} \sqrt{x+y} \geq 16 K^{2}
$$
This can be rewritten as
$$
\frac{\sqrt{x+y} \sqrt{y+z} \sqrt{z+x}}{4 K} \geq 2 \frac{K}{\sum_{c y c} \sqrt{x+y} / 2}
$$
to become the Euler inequality $R \geq 2 r$.
|
{
"resource_path": "Balkan_MO/segmented/en-2012-BMO-type3.jsonl",
"problem_match": "\n2. ",
"solution_match": "\nSolution 2."
}
|
36d8b6b0-4757-56f8-81f8-55a19f290550
| 604,232
|
Let $n$ be a positive integer. Let $P_{n}=\left\{2^{n}, 2^{n-1} \cdot 3,2^{n-2} \cdot 3^{2}, \ldots, 3^{n}\right\}$. For each subset $X$ of $P_{n}$, we write $S_{X}$ for the sum of all elements of $X$, with the convention that $S_{\emptyset}=0$ where $\emptyset$ is the empty set. Suppose that $y$ is a real number with $0 \leq y \leq 3^{n+1}-2^{n+1}$. Prove that there is a subset $Y$ of $P_{n}$ such that $0 \leq y-S_{Y}<2^{n}$
|
Let $\alpha=3 / 2$ so $1+\alpha>\alpha^{2}$.
Given $y$, we construct $Y$ algorithmically. Let $Y=\varnothing$ and of course $S_{\varnothing}=0$. For $i=0$ to $m$, perform the following operation:
$$
\text { If } S_{Y}+2^{i} 3^{m-i} \leq y \text {, then replace } Y \text { by } Y \cup\left\{2^{i} 3^{m-i}\right\}
$$
When this process is finished, we have a subset $Y$ of $P_{m}$ such that $S_{Y} \leq y$.
Notice that the elements of $P_{m}$ are in ascending order of size as given, and may alternatively be described as $2^{m}, 2^{m} \alpha, 2^{m} \alpha^{2}, \ldots, 2^{m} \alpha^{m}$. If any member of this list is not in $Y$, then no two consecutive members of the list to the left of the omitted member can both be in $Y$. This is because $1+\alpha>\alpha^{2}$, and the greedy nature of the process used to construct $Y$.
Therefore either $Y=P_{m}$, in which case $y=3^{m+1}-2^{m+1}$ and all is well, or at least one of the two leftmost elements of the list is omitted from $Y$.
If $2^{m}$ is not omitted from $Y$, then the algorithmic process ensures that $\left(S_{Y}-2^{m}\right)+2^{m-1} 3>y$, and so $y-S_{Y}<2^{m}$. On the other hand, if $2^{m}$ is omitted from $Y$, then $y-S_{Y}<2^{m}$ ).
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $n$ be a positive integer. Let $P_{n}=\left\{2^{n}, 2^{n-1} \cdot 3,2^{n-2} \cdot 3^{2}, \ldots, 3^{n}\right\}$. For each subset $X$ of $P_{n}$, we write $S_{X}$ for the sum of all elements of $X$, with the convention that $S_{\emptyset}=0$ where $\emptyset$ is the empty set. Suppose that $y$ is a real number with $0 \leq y \leq 3^{n+1}-2^{n+1}$. Prove that there is a subset $Y$ of $P_{n}$ such that $0 \leq y-S_{Y}<2^{n}$
|
Let $\alpha=3 / 2$ so $1+\alpha>\alpha^{2}$.
Given $y$, we construct $Y$ algorithmically. Let $Y=\varnothing$ and of course $S_{\varnothing}=0$. For $i=0$ to $m$, perform the following operation:
$$
\text { If } S_{Y}+2^{i} 3^{m-i} \leq y \text {, then replace } Y \text { by } Y \cup\left\{2^{i} 3^{m-i}\right\}
$$
When this process is finished, we have a subset $Y$ of $P_{m}$ such that $S_{Y} \leq y$.
Notice that the elements of $P_{m}$ are in ascending order of size as given, and may alternatively be described as $2^{m}, 2^{m} \alpha, 2^{m} \alpha^{2}, \ldots, 2^{m} \alpha^{m}$. If any member of this list is not in $Y$, then no two consecutive members of the list to the left of the omitted member can both be in $Y$. This is because $1+\alpha>\alpha^{2}$, and the greedy nature of the process used to construct $Y$.
Therefore either $Y=P_{m}$, in which case $y=3^{m+1}-2^{m+1}$ and all is well, or at least one of the two leftmost elements of the list is omitted from $Y$.
If $2^{m}$ is not omitted from $Y$, then the algorithmic process ensures that $\left(S_{Y}-2^{m}\right)+2^{m-1} 3>y$, and so $y-S_{Y}<2^{m}$. On the other hand, if $2^{m}$ is omitted from $Y$, then $y-S_{Y}<2^{m}$ ).
|
{
"resource_path": "Balkan_MO/segmented/en-2012-BMO-type3.jsonl",
"problem_match": "\n3. ",
"solution_match": "\nSolution 1."
}
|
c43dfea6-df01-5761-9e08-5cb6650d2b24
| 604,248
|
Let $n$ be a positive integer. Let $P_{n}=\left\{2^{n}, 2^{n-1} \cdot 3,2^{n-2} \cdot 3^{2}, \ldots, 3^{n}\right\}$. For each subset $X$ of $P_{n}$, we write $S_{X}$ for the sum of all elements of $X$, with the convention that $S_{\emptyset}=0$ where $\emptyset$ is the empty set. Suppose that $y$ is a real number with $0 \leq y \leq 3^{n+1}-2^{n+1}$. Prove that there is a subset $Y$ of $P_{n}$ such that $0 \leq y-S_{Y}<2^{n}$
|
Note that $3^{m+1}-2^{m+1}=(3-2)\left(3^{m}+3^{m-1} \cdot 2+\cdots+3 \cdot 2^{m-1}+2^{m}\right)=S_{P_{m}}$. Dividing every element of $P_{m}$ by $2^{m}$ gives us the following equivalent problem:
Let $m$ be a positive integer, $a=3 / 2$, and $Q_{m}=\left\{1, a, a^{2}, \ldots, a^{m}\right\}$. Show that for any real number $x$ satisfying $0 \leq x \leq 1+a+a^{2}+\cdots+a^{m}$, there exists a subset $X$ of $Q_{m}$ such that $0 \leq x-S_{X}<1$.
We will prove this problem by induction on $m$. When $m=1, S_{\varnothing}=0, S_{\{1\}}=1, S_{\{a\}}=3 / 2$, $S_{\{1, a\}}=5 / 2$. Since the difference between any two consecutive of them is at most 1, the claim is true.
Suppose that the statement is true for positive integer $m$. Let $x$ be a real number with $0 \leq$ $x \leq 1+a+a^{2}+\cdots+a^{m+1}$. If $0 \leq x \leq 1+a+a^{2}+\cdots+a^{m}$, then by the induction hypothesis there exists a subset $X$ of $Q_{m} \subset Q_{m+1}$ such that $0 \leq x-S_{X}<1$.
If $\frac{a^{m+1}-1}{a-1}=1+a+a^{2}+\cdots+a^{m}<x$, then $x>a^{m+1}$ as
$$
\frac{a^{m+1}-1}{a-1}=2\left(a^{m+1}-1\right)=a^{m+1}+\left(a^{m+1}-2\right) \geq a^{m+1}+a^{2}-2=a^{m+1}+\frac{1}{4} .
$$
Therefore $0<\left(x-a^{m+1}\right) \leq 1+a+a^{2}+\cdots+a^{m}$. Again by the induction hypothesis, there exists a subset $X$ of $Q_{m}$ satisfying $0 \leq\left(x-a^{m+1}\right)-S_{X}<1$. Hence $0 \leq x-S_{X^{\prime}}<1$ where $X^{\prime}=X \cup\left\{a^{m+1}\right\} \subset Q_{m+1}$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $n$ be a positive integer. Let $P_{n}=\left\{2^{n}, 2^{n-1} \cdot 3,2^{n-2} \cdot 3^{2}, \ldots, 3^{n}\right\}$. For each subset $X$ of $P_{n}$, we write $S_{X}$ for the sum of all elements of $X$, with the convention that $S_{\emptyset}=0$ where $\emptyset$ is the empty set. Suppose that $y$ is a real number with $0 \leq y \leq 3^{n+1}-2^{n+1}$. Prove that there is a subset $Y$ of $P_{n}$ such that $0 \leq y-S_{Y}<2^{n}$
|
Note that $3^{m+1}-2^{m+1}=(3-2)\left(3^{m}+3^{m-1} \cdot 2+\cdots+3 \cdot 2^{m-1}+2^{m}\right)=S_{P_{m}}$. Dividing every element of $P_{m}$ by $2^{m}$ gives us the following equivalent problem:
Let $m$ be a positive integer, $a=3 / 2$, and $Q_{m}=\left\{1, a, a^{2}, \ldots, a^{m}\right\}$. Show that for any real number $x$ satisfying $0 \leq x \leq 1+a+a^{2}+\cdots+a^{m}$, there exists a subset $X$ of $Q_{m}$ such that $0 \leq x-S_{X}<1$.
We will prove this problem by induction on $m$. When $m=1, S_{\varnothing}=0, S_{\{1\}}=1, S_{\{a\}}=3 / 2$, $S_{\{1, a\}}=5 / 2$. Since the difference between any two consecutive of them is at most 1, the claim is true.
Suppose that the statement is true for positive integer $m$. Let $x$ be a real number with $0 \leq$ $x \leq 1+a+a^{2}+\cdots+a^{m+1}$. If $0 \leq x \leq 1+a+a^{2}+\cdots+a^{m}$, then by the induction hypothesis there exists a subset $X$ of $Q_{m} \subset Q_{m+1}$ such that $0 \leq x-S_{X}<1$.
If $\frac{a^{m+1}-1}{a-1}=1+a+a^{2}+\cdots+a^{m}<x$, then $x>a^{m+1}$ as
$$
\frac{a^{m+1}-1}{a-1}=2\left(a^{m+1}-1\right)=a^{m+1}+\left(a^{m+1}-2\right) \geq a^{m+1}+a^{2}-2=a^{m+1}+\frac{1}{4} .
$$
Therefore $0<\left(x-a^{m+1}\right) \leq 1+a+a^{2}+\cdots+a^{m}$. Again by the induction hypothesis, there exists a subset $X$ of $Q_{m}$ satisfying $0 \leq\left(x-a^{m+1}\right)-S_{X}<1$. Hence $0 \leq x-S_{X^{\prime}}<1$ where $X^{\prime}=X \cup\left\{a^{m+1}\right\} \subset Q_{m+1}$.
|
{
"resource_path": "Balkan_MO/segmented/en-2012-BMO-type3.jsonl",
"problem_match": "\n3. ",
"solution_match": "\nSolution 2."
}
|
c43dfea6-df01-5761-9e08-5cb6650d2b24
| 604,248
|
\quad$ Let $\mathbb{Z}^{+}$be the set of positive integers. Find all functions $f: \mathbb{Z}^{+} \rightarrow \mathbb{Z}^{+}$such that the following conditions both hold:
(i) $f(n!)=f(n)$ ! for every positive integer $n$,
(ii) $m-n$ divides $f(m)-f(n)$ whenever $m$ and $n$ are different positive integers.
|
There are three such functions: the constant functions 1, 2 and the identity function $\mathrm{id}_{\mathbf{Z}^{+}}$. These functions clearly satisfy the conditions in the hypothesis. Let us prove that there are only ones.
Consider such a function $f$ and suppose that it has a fixed point $a \geq 3$, that is $f(a)=a$. Then $a!,(a!)!, \cdots$ are all fixed points of $f$, hence the function $f$ has a strictly increasing sequence $a_{1}<a_{2}<\cdots<a_{k}<\cdots$ of fixed points. For a positive integer $n$, $a_{k}-n$ divides $a_{k}-f(n)=$ $f\left(a_{k}\right)-f(n)$ for every $k \in \mathbf{Z}^{+}$. Also $a_{k}-n$ divides $a_{k}-n$, so it divides $a_{k}-f(n)-\left(a_{k}-n\right)=$ $n-f(n)$. This is possible only if $f(n)=n$, hence in this case we get $f=\mathrm{id}_{\mathbf{Z}^{+}}$.
Now suppose that $f$ has no fixed points greater than 2 . Let $p \geq 5$ be a prime and notice that by Wilson's Theorem we have $(p-2)!\equiv 1(\bmod p)$. Therefore $p$ divides $(p-2)!-1$. But $(p-2)!-1$ divides $f((p-2)!)-f(1)$, hence $p$ divides $f((p-2)!)-f(1)=(f(p-2))!-f(1)$. Clearly we have $f(1)=1$ or $f(1)=2$. As $p \geq 5$, the fact that $p$ divides $(f(p-2))!-f(1)$ implies that $f(p-2)<p$. It is easy to check, again by Wilson's Theorem, that $p$ does not divide $(p-1)!-1$ and $(p-1)!-2$, hence we deduce that $f(p-2) \leq p-2$. On the other hand, $p-3=(p-2)-1$ divides $f(p-2)-f(1) \leq(p-2)-1$. Thus either $f(p-2)=f(1)$ or $f(p-2)=p-2$. As $p-2 \geq 3$, the last case is excluded, since the function $f$ has no fixed points greater than 2 . It follows $f(p-2)=f(1)$ and this property holds for all primes $p \geq 5$. Taking $n$ any positive integer, we deduce that $p-2-n$ divides $f(p-2)-f(n)=f(1)-f(n)$ for all primes $p \geq 5$. Thus $f(n)=f(1)$, hence $f$ is the constant function 1 or 2 .
|
proof
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
\quad$ Let $\mathbb{Z}^{+}$be the set of positive integers. Find all functions $f: \mathbb{Z}^{+} \rightarrow \mathbb{Z}^{+}$such that the following conditions both hold:
(i) $f(n!)=f(n)$ ! for every positive integer $n$,
(ii) $m-n$ divides $f(m)-f(n)$ whenever $m$ and $n$ are different positive integers.
|
There are three such functions: the constant functions 1, 2 and the identity function $\mathrm{id}_{\mathbf{Z}^{+}}$. These functions clearly satisfy the conditions in the hypothesis. Let us prove that there are only ones.
Consider such a function $f$ and suppose that it has a fixed point $a \geq 3$, that is $f(a)=a$. Then $a!,(a!)!, \cdots$ are all fixed points of $f$, hence the function $f$ has a strictly increasing sequence $a_{1}<a_{2}<\cdots<a_{k}<\cdots$ of fixed points. For a positive integer $n$, $a_{k}-n$ divides $a_{k}-f(n)=$ $f\left(a_{k}\right)-f(n)$ for every $k \in \mathbf{Z}^{+}$. Also $a_{k}-n$ divides $a_{k}-n$, so it divides $a_{k}-f(n)-\left(a_{k}-n\right)=$ $n-f(n)$. This is possible only if $f(n)=n$, hence in this case we get $f=\mathrm{id}_{\mathbf{Z}^{+}}$.
Now suppose that $f$ has no fixed points greater than 2 . Let $p \geq 5$ be a prime and notice that by Wilson's Theorem we have $(p-2)!\equiv 1(\bmod p)$. Therefore $p$ divides $(p-2)!-1$. But $(p-2)!-1$ divides $f((p-2)!)-f(1)$, hence $p$ divides $f((p-2)!)-f(1)=(f(p-2))!-f(1)$. Clearly we have $f(1)=1$ or $f(1)=2$. As $p \geq 5$, the fact that $p$ divides $(f(p-2))!-f(1)$ implies that $f(p-2)<p$. It is easy to check, again by Wilson's Theorem, that $p$ does not divide $(p-1)!-1$ and $(p-1)!-2$, hence we deduce that $f(p-2) \leq p-2$. On the other hand, $p-3=(p-2)-1$ divides $f(p-2)-f(1) \leq(p-2)-1$. Thus either $f(p-2)=f(1)$ or $f(p-2)=p-2$. As $p-2 \geq 3$, the last case is excluded, since the function $f$ has no fixed points greater than 2 . It follows $f(p-2)=f(1)$ and this property holds for all primes $p \geq 5$. Taking $n$ any positive integer, we deduce that $p-2-n$ divides $f(p-2)-f(n)=f(1)-f(n)$ for all primes $p \geq 5$. Thus $f(n)=f(1)$, hence $f$ is the constant function 1 or 2 .
|
{
"resource_path": "Balkan_MO/segmented/en-2012-BMO-type3.jsonl",
"problem_match": "\n4. ",
"solution_match": "\nSolution 1."
}
|
6880311a-8721-5e50-bb73-074a536e973a
| 604,273
|
\quad$ Let $\mathbb{Z}^{+}$be the set of positive integers. Find all functions $f: \mathbb{Z}^{+} \rightarrow \mathbb{Z}^{+}$such that the following conditions both hold:
(i) $f(n!)=f(n)$ ! for every positive integer $n$,
(ii) $m-n$ divides $f(m)-f(n)$ whenever $m$ and $n$ are different positive integers.
|
Note first that if $f\left(n_{0}\right)=n_{0}$, then $m-n_{0} \mid f(m)-m$ for all $m \in \mathbf{Z}^{+}$. If $f\left(n_{0}\right)=n_{0}$ for infinitely many $n_{0} \in \mathbf{Z}^{+}$, then $f(m)-m$ has infinitely many divisors, hence $f(m)=m$ for all $m \in \mathbf{Z}^{+}$. On the other hand, if $f\left(n_{0}\right)=n_{0}$ for some $n_{0} \geq 3$, then $f$ fixes each term of the sequence $\left(n_{k}\right)_{k=0}^{\infty}$, which is recursively defined by $n_{k}=n_{k-1}!$. Hence if $f(3)=3$, then $f(n)=n$ for all $n \in \mathbf{Z}^{+}$.
We may assume that $f(3) \neq 3$. Since $f(1)=f(1)!$, and $f(2)=f(2)!, f(1), f(2) \in\{1,2\}$. We have $4=3!-2 \mid f(3)!-f(2)$. This together with $f(3) \neq 3$ implies that $f(3) \in\{1,2\}$. Let $n>3$, then $n!-3 \mid f(n)!-f(3)$ and $3 \nmid f(n)!$, i.e. $f(n)!\in\{1,2\}$. Hence we conclude that $f(n) \in\{1,2\}$ for all $n \in \mathbf{Z}^{+}$.
If $f$ is not constant, then there exist positive integers $m, n$ with $\{f(n), f(m)\}=\{1,2\}$. Let $k=2+\max \{m, n\}$. If $f(k) \neq f(m)$, then $k-m \mid f(k)-f(m)$. This is a contradiction as $|f(k)-f(m)|=1$ and $k-m \geq 2$.
Therefore the functions satisfying the conditions are $f \equiv 1, f \equiv 2, f=\mathrm{id}_{\mathbf{z}^{+}}$.
|
f \equiv 1, f \equiv 2, f=\mathrm{id}_{\mathbf{z}^{+}}
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
\quad$ Let $\mathbb{Z}^{+}$be the set of positive integers. Find all functions $f: \mathbb{Z}^{+} \rightarrow \mathbb{Z}^{+}$such that the following conditions both hold:
(i) $f(n!)=f(n)$ ! for every positive integer $n$,
(ii) $m-n$ divides $f(m)-f(n)$ whenever $m$ and $n$ are different positive integers.
|
Note first that if $f\left(n_{0}\right)=n_{0}$, then $m-n_{0} \mid f(m)-m$ for all $m \in \mathbf{Z}^{+}$. If $f\left(n_{0}\right)=n_{0}$ for infinitely many $n_{0} \in \mathbf{Z}^{+}$, then $f(m)-m$ has infinitely many divisors, hence $f(m)=m$ for all $m \in \mathbf{Z}^{+}$. On the other hand, if $f\left(n_{0}\right)=n_{0}$ for some $n_{0} \geq 3$, then $f$ fixes each term of the sequence $\left(n_{k}\right)_{k=0}^{\infty}$, which is recursively defined by $n_{k}=n_{k-1}!$. Hence if $f(3)=3$, then $f(n)=n$ for all $n \in \mathbf{Z}^{+}$.
We may assume that $f(3) \neq 3$. Since $f(1)=f(1)!$, and $f(2)=f(2)!, f(1), f(2) \in\{1,2\}$. We have $4=3!-2 \mid f(3)!-f(2)$. This together with $f(3) \neq 3$ implies that $f(3) \in\{1,2\}$. Let $n>3$, then $n!-3 \mid f(n)!-f(3)$ and $3 \nmid f(n)!$, i.e. $f(n)!\in\{1,2\}$. Hence we conclude that $f(n) \in\{1,2\}$ for all $n \in \mathbf{Z}^{+}$.
If $f$ is not constant, then there exist positive integers $m, n$ with $\{f(n), f(m)\}=\{1,2\}$. Let $k=2+\max \{m, n\}$. If $f(k) \neq f(m)$, then $k-m \mid f(k)-f(m)$. This is a contradiction as $|f(k)-f(m)|=1$ and $k-m \geq 2$.
Therefore the functions satisfying the conditions are $f \equiv 1, f \equiv 2, f=\mathrm{id}_{\mathbf{z}^{+}}$.
|
{
"resource_path": "Balkan_MO/segmented/en-2012-BMO-type3.jsonl",
"problem_match": "\n4. ",
"solution_match": "\nSolution 2."
}
|
6880311a-8721-5e50-bb73-074a536e973a
| 604,273
|
We denote the angles of the triangle by $\alpha, \beta$ and $\gamma$ as usual. Since $\angle K M P=90^{\circ}-\frac{\beta}{2}$, it suffices to prove that $\angle K L P=90^{\circ}+\frac{\beta}{2}$, which is equivalent to $\angle K L C=\frac{\beta}{2}$.
Let $I$ be the incenter of triangle $A B C$ and let $D$ be the tangency point of the incircle with $A B$. Since $C K \| I B$ and $C L \| I A$, it holds that $\angle K C L=\angle A I B$. Moreover, from $C N=A D=\frac{b+c-a}{2}$ and $\angle K C N=\frac{\beta}{2}$ we obtain $C K=C N \cos \frac{\beta}{2}=$ $A D \cos \frac{\beta}{2}=A I \cos \frac{\alpha}{2} \cos \frac{\beta}{2}$ and analogously $C L=B I \cos \frac{\alpha}{2} \cos \frac{\beta}{2}$, which imply $\frac{C K}{C L}=\frac{A I}{B I}$. Hence the triangles $K C L$ and $A I B$ are similar, and thus $\angle K L C=$ $\angle A B I=\frac{\beta}{2}$.
|
Consider the graph $\mathcal{G}$ whose vertices are the contestants, where there is an edge between two contestants if and only if they are not friends.
Lemma. There is a vertex in graph $\mathcal{G}$ with degree at most 2 .
Proof. Suppose that each vertex has a degree at least three. Consider the longest induced path $P=u_{0} u_{1} u_{2} \ldots u_{k}$ in the graph (that is, the path in which no two nonadjacent vertices are connected by an edge). The vertex $u_{0}$ is connected to another two vertices $v$ and $w$, which must be outside the path $P$. Since $P$ is the longest induced path, $v$ and $w$ have neighbors in it. Let $u_{i}$ and $u_{j}$ be the neighbors of $v$ and $w$ respectively with the smallest $i$ and $j$; assume without loss of generality that $i \geq j$. Then $u_{0}, u_{1}, \ldots, u_{i}, v$ form a weakly friendly cycle, but $w$ has two neighbors in it ( $u_{0}$ and $u_{j}$ ), a contradiction.
We now prove the problem statement by induction on the number $n$ of vertices in $\mathcal{G}$. For $n \leq 3$ the statement is trivial; assume that it holds for $n-1$. By the Lemma, there is a vertex $v$ in $\mathcal{G}$ of degree at most two. Graph $\mathcal{G}^{\prime}$, obtained by removing vertex $v$ (and all edges incident to it), clearly satisfies the problem conditions, so its vertices can be partitioned into three rooms in a desired way. Since $v$ has no neighbors in at least one of the rooms, we can place $v$ in that room, finishing the proof.
|
proof
|
Yes
|
Incomplete
|
proof
|
Geometry
|
We denote the angles of the triangle by $\alpha, \beta$ and $\gamma$ as usual. Since $\angle K M P=90^{\circ}-\frac{\beta}{2}$, it suffices to prove that $\angle K L P=90^{\circ}+\frac{\beta}{2}$, which is equivalent to $\angle K L C=\frac{\beta}{2}$.
Let $I$ be the incenter of triangle $A B C$ and let $D$ be the tangency point of the incircle with $A B$. Since $C K \| I B$ and $C L \| I A$, it holds that $\angle K C L=\angle A I B$. Moreover, from $C N=A D=\frac{b+c-a}{2}$ and $\angle K C N=\frac{\beta}{2}$ we obtain $C K=C N \cos \frac{\beta}{2}=$ $A D \cos \frac{\beta}{2}=A I \cos \frac{\alpha}{2} \cos \frac{\beta}{2}$ and analogously $C L=B I \cos \frac{\alpha}{2} \cos \frac{\beta}{2}$, which imply $\frac{C K}{C L}=\frac{A I}{B I}$. Hence the triangles $K C L$ and $A I B$ are similar, and thus $\angle K L C=$ $\angle A B I=\frac{\beta}{2}$.
|
Consider the graph $\mathcal{G}$ whose vertices are the contestants, where there is an edge between two contestants if and only if they are not friends.
Lemma. There is a vertex in graph $\mathcal{G}$ with degree at most 2 .
Proof. Suppose that each vertex has a degree at least three. Consider the longest induced path $P=u_{0} u_{1} u_{2} \ldots u_{k}$ in the graph (that is, the path in which no two nonadjacent vertices are connected by an edge). The vertex $u_{0}$ is connected to another two vertices $v$ and $w$, which must be outside the path $P$. Since $P$ is the longest induced path, $v$ and $w$ have neighbors in it. Let $u_{i}$ and $u_{j}$ be the neighbors of $v$ and $w$ respectively with the smallest $i$ and $j$; assume without loss of generality that $i \geq j$. Then $u_{0}, u_{1}, \ldots, u_{i}, v$ form a weakly friendly cycle, but $w$ has two neighbors in it ( $u_{0}$ and $u_{j}$ ), a contradiction.
We now prove the problem statement by induction on the number $n$ of vertices in $\mathcal{G}$. For $n \leq 3$ the statement is trivial; assume that it holds for $n-1$. By the Lemma, there is a vertex $v$ in $\mathcal{G}$ of degree at most two. Graph $\mathcal{G}^{\prime}$, obtained by removing vertex $v$ (and all edges incident to it), clearly satisfies the problem conditions, so its vertices can be partitioned into three rooms in a desired way. Since $v$ has no neighbors in at least one of the rooms, we can place $v$ in that room, finishing the proof.
|
{
"resource_path": "Balkan_MO/segmented/en-2013-BMO-type2.jsonl",
"problem_match": "\n1.",
"solution_match": "\n4."
}
|
a5afdaf5-b152-56cc-b3b7-aff89fef5728
| 604,293
|
Determine all positive integers $x, y$ and $z$ such that
$$
x^{5}+4^{y}=2013^{z}
$$
|
Reducing modulo 11 yields $x^{5}+4^{y} \equiv 0(\bmod 11)$, where $x^{5} \equiv \pm 1(\bmod 11)$, so we also have $4^{y} \equiv \pm 1(\bmod 11)$. Congruence $4^{y} \equiv-1(\bmod 11)$ does not hold for any $y$, whereas $4^{y} \equiv 1(\bmod 11)$ holds if and only if $5 \mid y$.
Setting $t=4^{y / 5}$, the equation becomes $x^{5}+t^{5}=A \cdot B=2013^{z}$, where $(x, t)=1$ and $A=x+t, B=x^{4}-x^{3} t+x^{2} t^{2}-x t^{3}+t^{4}$. Furthermore, from $B=A\left(x^{3}-\right.$ $\left.2 x^{2} t+3 x t^{2}-4 t^{3}\right)+5 t^{4}$ we deduce $(A, B)=\left(A, 5 t^{4}\right) \mid 5$, but $5 \nmid 2013^{z}$, so we must have $(A, B)=1$. Therefore $A=a^{z}$ and $B=b^{z}$ for some positive integers $a$ and $b$ with $a \cdot b=2013$.
On the other hand, from $\frac{1}{16} A^{4} \leq B \leq A^{4}$ (which is a simple consequence of the mean inequality) we obtain $\frac{1}{16} a^{4} \leq b \leq a^{4}$, i.e. $\frac{1}{16} a^{5} \leq a b=2013 \leq a^{5}$. Therefore $5 \leq a \leq 8$, which is impossible because 2013 has no divisors in the interval [5,8].
|
not found
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Determine all positive integers $x, y$ and $z$ such that
$$
x^{5}+4^{y}=2013^{z}
$$
|
Reducing modulo 11 yields $x^{5}+4^{y} \equiv 0(\bmod 11)$, where $x^{5} \equiv \pm 1(\bmod 11)$, so we also have $4^{y} \equiv \pm 1(\bmod 11)$. Congruence $4^{y} \equiv-1(\bmod 11)$ does not hold for any $y$, whereas $4^{y} \equiv 1(\bmod 11)$ holds if and only if $5 \mid y$.
Setting $t=4^{y / 5}$, the equation becomes $x^{5}+t^{5}=A \cdot B=2013^{z}$, where $(x, t)=1$ and $A=x+t, B=x^{4}-x^{3} t+x^{2} t^{2}-x t^{3}+t^{4}$. Furthermore, from $B=A\left(x^{3}-\right.$ $\left.2 x^{2} t+3 x t^{2}-4 t^{3}\right)+5 t^{4}$ we deduce $(A, B)=\left(A, 5 t^{4}\right) \mid 5$, but $5 \nmid 2013^{z}$, so we must have $(A, B)=1$. Therefore $A=a^{z}$ and $B=b^{z}$ for some positive integers $a$ and $b$ with $a \cdot b=2013$.
On the other hand, from $\frac{1}{16} A^{4} \leq B \leq A^{4}$ (which is a simple consequence of the mean inequality) we obtain $\frac{1}{16} a^{4} \leq b \leq a^{4}$, i.e. $\frac{1}{16} a^{5} \leq a b=2013 \leq a^{5}$. Therefore $5 \leq a \leq 8$, which is impossible because 2013 has no divisors in the interval [5,8].
|
{
"resource_path": "Balkan_MO/segmented/en-2013-BMO-type2.jsonl",
"problem_match": "\n2.",
"solution_match": "\n2."
}
|
91220246-539e-5771-875e-c3b016a80b04
| 604,305
|
Let $S$ be the set of positive real numbers. Find all functions $f: S^{3} \rightarrow S$ such that, for all positive real numbers $x, y, z$ and $k$, the following three conditions are satisfied:
(i) $x f(x, y, z)=z f(z, y, x)$;
(ii) $f\left(x, k y, k^{2} z\right)=k f(x, y, z)$;
(iii) $f(1, k, k+1)=k+1$.
(United Kingdom)
|
It follows from the properties of function $f$ that, for all $x, y, z, a, b>0$, $f\left(a^{2} x, a b y, b^{2} z\right)=b f\left(a^{2} x, a y, z\right)=b \cdot \frac{z}{a^{2} x} f\left(z, a y, a^{2} x\right)=\frac{b z}{a x} f(z, y, x)=\frac{b}{a} f(x, y, z)$.
We shall choose $a$ and $b$ in such a way that the triple $\left(a^{2} x, a b y, b^{2} z\right)$ is of the form $(1, k, k+1)$ for some $k$ : namely, we take $a=\frac{1}{\sqrt{x}}$ and $b$ satisfying $b^{2} z-a b y=1$, which upon solving the quadratic equation yields $b=\frac{y+\sqrt{y^{2}+4 x z}}{2 z \sqrt{x}}$ and $k=\frac{y\left(y+\sqrt{y^{2}+4 x z}\right)}{2 x z}$. Now we easily obtain
$$
f(x, y, z)=\frac{a}{b} f\left(a^{2} x, a b y, b^{2} z\right)=\frac{a}{b} f(1, k, k+1)=\frac{a}{b}(k+1)=\frac{y+\sqrt{y^{2}+4 x z}}{2 x} .
$$
It is directly verified that $f$ satisfies the problem conditions.
|
\frac{y+\sqrt{y^{2}+4 x z}}{2 x}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $S$ be the set of positive real numbers. Find all functions $f: S^{3} \rightarrow S$ such that, for all positive real numbers $x, y, z$ and $k$, the following three conditions are satisfied:
(i) $x f(x, y, z)=z f(z, y, x)$;
(ii) $f\left(x, k y, k^{2} z\right)=k f(x, y, z)$;
(iii) $f(1, k, k+1)=k+1$.
(United Kingdom)
|
It follows from the properties of function $f$ that, for all $x, y, z, a, b>0$, $f\left(a^{2} x, a b y, b^{2} z\right)=b f\left(a^{2} x, a y, z\right)=b \cdot \frac{z}{a^{2} x} f\left(z, a y, a^{2} x\right)=\frac{b z}{a x} f(z, y, x)=\frac{b}{a} f(x, y, z)$.
We shall choose $a$ and $b$ in such a way that the triple $\left(a^{2} x, a b y, b^{2} z\right)$ is of the form $(1, k, k+1)$ for some $k$ : namely, we take $a=\frac{1}{\sqrt{x}}$ and $b$ satisfying $b^{2} z-a b y=1$, which upon solving the quadratic equation yields $b=\frac{y+\sqrt{y^{2}+4 x z}}{2 z \sqrt{x}}$ and $k=\frac{y\left(y+\sqrt{y^{2}+4 x z}\right)}{2 x z}$. Now we easily obtain
$$
f(x, y, z)=\frac{a}{b} f\left(a^{2} x, a b y, b^{2} z\right)=\frac{a}{b} f(1, k, k+1)=\frac{a}{b}(k+1)=\frac{y+\sqrt{y^{2}+4 x z}}{2 x} .
$$
It is directly verified that $f$ satisfies the problem conditions.
|
{
"resource_path": "Balkan_MO/segmented/en-2013-BMO-type2.jsonl",
"problem_match": "\n3.",
"solution_match": "\n3."
}
|
9a3cfccb-47cd-5f74-833c-faf45dc1786a
| 604,314
|
In a mathematical competition some competitors are friends; friendship is always mutual, that is to say that when $A$ is a friend of $B$, then also $B$ is a friend of $A$. We say that $n \geq 3$ different competitors $A_{1}, A_{2}, \ldots, A_{n}$ form a weakly-friendly cycle if $A_{i}$ is not a friend of $A_{i+1}$ for $1 \leq i \leq n\left(A_{n+1}=A_{1}\right)$, and there are no other pairs of non-friends among the components of this cycle.
The following property is satisfied:
for every competitor $C$, and every weakly-friendly cycle $\mathscr{S}$ of competitors not including $C$, the set of competitors $D$ in $\mathscr{S}$ which are not friends of $C$ has at most one element.
Prove that all competitors of this mathematical competition can be arranged into three rooms, such that every two competitors that are in the same room are friends.
(Serbia)
Time allowed: 270 minutes.
Each problem is worth 10 points.
## SOLUTIONS
|
Consider the graph $\mathcal{G}$ whose vertices are the contestants, where there is an edge between two contestants if and only if they are not friends.
Lemma. There is a vertex in graph $\mathcal{G}$ with degree at most 2 .
Proof. Suppose that each vertex has a degree at least three. Consider the longest induced path $P=u_{0} u_{1} u_{2} \ldots u_{k}$ in the graph (that is, the path in which no two nonadjacent vertices are connected by an edge). The vertex $u_{0}$ is connected to another two vertices $v$ and $w$, which must be outside the path $P$. Since $P$ is the longest induced path, $v$ and $w$ have neighbors in it. Let $u_{i}$ and $u_{j}$ be the neighbors of $v$ and $w$ respectively with the smallest $i$ and $j$; assume without loss of generality that $i \geq j$. Then $u_{0}, u_{1}, \ldots, u_{i}, v$ form a weakly friendly cycle, but $w$ has two neighbors in it ( $u_{0}$ and $u_{j}$ ), a contradiction.
We now prove the problem statement by induction on the number $n$ of vertices in $\mathcal{G}$. For $n \leq 3$ the statement is trivial; assume that it holds for $n-1$. By the Lemma, there is a vertex $v$ in $\mathcal{G}$ of degree at most two. Graph $\mathcal{G}^{\prime}$, obtained by removing vertex $v$ (and all edges incident to it), clearly satisfies the problem conditions, so its vertices can be partitioned into three rooms in a desired way. Since $v$ has no neighbors in at least one of the rooms, we can place $v$ in that room, finishing the proof.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
In a mathematical competition some competitors are friends; friendship is always mutual, that is to say that when $A$ is a friend of $B$, then also $B$ is a friend of $A$. We say that $n \geq 3$ different competitors $A_{1}, A_{2}, \ldots, A_{n}$ form a weakly-friendly cycle if $A_{i}$ is not a friend of $A_{i+1}$ for $1 \leq i \leq n\left(A_{n+1}=A_{1}\right)$, and there are no other pairs of non-friends among the components of this cycle.
The following property is satisfied:
for every competitor $C$, and every weakly-friendly cycle $\mathscr{S}$ of competitors not including $C$, the set of competitors $D$ in $\mathscr{S}$ which are not friends of $C$ has at most one element.
Prove that all competitors of this mathematical competition can be arranged into three rooms, such that every two competitors that are in the same room are friends.
(Serbia)
Time allowed: 270 minutes.
Each problem is worth 10 points.
## SOLUTIONS
|
Consider the graph $\mathcal{G}$ whose vertices are the contestants, where there is an edge between two contestants if and only if they are not friends.
Lemma. There is a vertex in graph $\mathcal{G}$ with degree at most 2 .
Proof. Suppose that each vertex has a degree at least three. Consider the longest induced path $P=u_{0} u_{1} u_{2} \ldots u_{k}$ in the graph (that is, the path in which no two nonadjacent vertices are connected by an edge). The vertex $u_{0}$ is connected to another two vertices $v$ and $w$, which must be outside the path $P$. Since $P$ is the longest induced path, $v$ and $w$ have neighbors in it. Let $u_{i}$ and $u_{j}$ be the neighbors of $v$ and $w$ respectively with the smallest $i$ and $j$; assume without loss of generality that $i \geq j$. Then $u_{0}, u_{1}, \ldots, u_{i}, v$ form a weakly friendly cycle, but $w$ has two neighbors in it ( $u_{0}$ and $u_{j}$ ), a contradiction.
We now prove the problem statement by induction on the number $n$ of vertices in $\mathcal{G}$. For $n \leq 3$ the statement is trivial; assume that it holds for $n-1$. By the Lemma, there is a vertex $v$ in $\mathcal{G}$ of degree at most two. Graph $\mathcal{G}^{\prime}$, obtained by removing vertex $v$ (and all edges incident to it), clearly satisfies the problem conditions, so its vertices can be partitioned into three rooms in a desired way. Since $v$ has no neighbors in at least one of the rooms, we can place $v$ in that room, finishing the proof.
|
{
"resource_path": "Balkan_MO/segmented/en-2013-BMO-type2.jsonl",
"problem_match": "\n4.",
"solution_match": "\n4."
}
|
2dd7b8ff-3346-5ccf-8ce1-adb515f40080
| 604,325
|
Let $x, y$ and $z$ be positive real numbers such that $x y+y z+z x=3 x y z$. Prove that
$$
x^{2} y+y^{2} z+z^{2} x \geq 2(x+y+z)-3
$$
and determine when equality holds.
|
The given condition can be rearranged to $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3$. Using this, we obtain:
$$
\begin{aligned}
x^{2} y+y^{2} z+z^{2} x-2(x+y+z)+3 & =x^{2} y-2 x+\frac{1}{y}+y^{2} z-2 y+\frac{1}{z}+z^{2} x-2 x+\frac{1}{x}= \\
& =y\left(x-\frac{1}{y}\right)^{2}+z\left(y-\frac{1}{z}\right)^{2}+x\left(z-\frac{1}{z}\right)^{2} \geq 0
\end{aligned}
$$
Equality holds if and only if we have $x y=y z=z x=1$, or, in other words, $x=y=z=1$.
Alternative solution. It follows from $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3$ and Cauchy-Schwarz inequality that
$$
\begin{aligned}
3\left(x^{2} y+y^{2} z+z^{2} x\right) & =\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(x^{2} y+y^{2} z+z^{2} x\right) \\
& \left.=\left(\left(\frac{1}{\sqrt{y}}\right)^{2}+\left(\frac{1}{\sqrt{z}}\right)^{2}+\left(\frac{1}{\sqrt{x}}\right)^{2}\right)\left((x \sqrt{y})^{2}\right)+(y \sqrt{z})^{2}+(z \sqrt{x})^{2}\right) \\
& \geq(x+y+z)^{2}
\end{aligned}
$$
Therefore, $x^{2} y+y^{2} z+z^{2} x \geq \frac{(x+y+z)^{2}}{3}$ and if $x+y+z=t$ it suffices to show that $\frac{t^{2}}{3} \geq 2 t-3$. The latter is equivalent to $(t-3)^{2} \geq 0$. Equality holds when
$$
x \sqrt{y} \sqrt{y}=y \sqrt{z} \sqrt{z}=z \sqrt{x} \sqrt{x},
$$
i.e. $x y=y z=z x$ and $t=x+y+z=3$. Hence, $x=y=z=1$.
Comment. The inequality is true with the condition $x y+y z+z x \leq 3 x y z$.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $x, y$ and $z$ be positive real numbers such that $x y+y z+z x=3 x y z$. Prove that
$$
x^{2} y+y^{2} z+z^{2} x \geq 2(x+y+z)-3
$$
and determine when equality holds.
|
The given condition can be rearranged to $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3$. Using this, we obtain:
$$
\begin{aligned}
x^{2} y+y^{2} z+z^{2} x-2(x+y+z)+3 & =x^{2} y-2 x+\frac{1}{y}+y^{2} z-2 y+\frac{1}{z}+z^{2} x-2 x+\frac{1}{x}= \\
& =y\left(x-\frac{1}{y}\right)^{2}+z\left(y-\frac{1}{z}\right)^{2}+x\left(z-\frac{1}{z}\right)^{2} \geq 0
\end{aligned}
$$
Equality holds if and only if we have $x y=y z=z x=1$, or, in other words, $x=y=z=1$.
Alternative solution. It follows from $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3$ and Cauchy-Schwarz inequality that
$$
\begin{aligned}
3\left(x^{2} y+y^{2} z+z^{2} x\right) & =\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(x^{2} y+y^{2} z+z^{2} x\right) \\
& \left.=\left(\left(\frac{1}{\sqrt{y}}\right)^{2}+\left(\frac{1}{\sqrt{z}}\right)^{2}+\left(\frac{1}{\sqrt{x}}\right)^{2}\right)\left((x \sqrt{y})^{2}\right)+(y \sqrt{z})^{2}+(z \sqrt{x})^{2}\right) \\
& \geq(x+y+z)^{2}
\end{aligned}
$$
Therefore, $x^{2} y+y^{2} z+z^{2} x \geq \frac{(x+y+z)^{2}}{3}$ and if $x+y+z=t$ it suffices to show that $\frac{t^{2}}{3} \geq 2 t-3$. The latter is equivalent to $(t-3)^{2} \geq 0$. Equality holds when
$$
x \sqrt{y} \sqrt{y}=y \sqrt{z} \sqrt{z}=z \sqrt{x} \sqrt{x},
$$
i.e. $x y=y z=z x$ and $t=x+y+z=3$. Hence, $x=y=z=1$.
Comment. The inequality is true with the condition $x y+y z+z x \leq 3 x y z$.
|
{
"resource_path": "Balkan_MO/segmented/en-2014-BMO-type1.jsonl",
"problem_match": "\nProblem 1.",
"solution_match": "\nSolution."
}
|
eb0844f2-1223-523a-9ddd-c2697944cf95
| 604,335
|
A special number is a positive integer $n$ for which there exist positive integers $a, b, c$ and $d$ with
$$
n=\frac{a^{3}+2 b^{3}}{c^{3}+2 d^{3}}
$$
Prove that:
(a) there are infinitely many special numbers;
(b) 2014 is not a special number.
|
(a) Every perfect cube $k^{3}$ of a positive integer is special because we can write
$$
k^{3}=k^{3} \frac{a^{3}+2 b^{3}}{a^{3}+2 b^{3}}=\frac{(k a)^{3}+2(k b)^{3}}{a^{3}+2 b^{3}}
$$
for some positive integers $a, b$.
(b) Observe that $2014=2.19 .53$. If 2014 is special, then we have,
$$
x^{3}+2 y^{3}=2014\left(u^{3}+2 v^{3}\right)
$$
for some positive integers $x, y, u, v$. We may assume that $x^{3}+2 y^{3}$ is minimal with this property. Now, we will use the fact that if 19 divides $x^{3}+2 y^{3}$, then it divides both $x$ and $y$. Indeed, if 19 does not divide $x$, then it does not divide $y$ too. The relation $x^{3} \equiv-2 y^{3}(\bmod 19)$ implies $\left(x^{3}\right)^{6} \equiv\left(-2 y^{3}\right)^{6}(\bmod 19)$. The latter congruence is equivalent to $x^{18} \equiv 2^{6} y^{18}(\bmod 19)$. Now, according to the Fermat's Little Theorem, we obtain $1 \equiv 2^{6}(\bmod 19)$, that is 19 divides 63 , not possible.
It follows $x=19 x_{1}, y=19 y_{1}$, for some positive integers $x_{1}$ and $y_{1}$. Replacing in (1) we get
$$
19^{2}\left(x_{1}^{3}+2 y_{1}^{3}\right)=2.53\left(u^{3}+2 v^{3}\right)
$$
i.e. $19 \mid u^{3}+2 v^{3}$. It follows $u=19 u_{1}$ and $v=19 v_{1}$, and replacing in (2) we get
$$
x_{1}^{3}+2 y_{1}^{3}=2014\left(u_{1}^{3}+2 v_{1}^{3}\right) .
$$
Clearly, $x_{1}^{3}+2 y_{1}^{3}<x^{3}+2 y^{3}$, contradicting the minimality of $x^{3}+2 y^{3}$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
A special number is a positive integer $n$ for which there exist positive integers $a, b, c$ and $d$ with
$$
n=\frac{a^{3}+2 b^{3}}{c^{3}+2 d^{3}}
$$
Prove that:
(a) there are infinitely many special numbers;
(b) 2014 is not a special number.
|
(a) Every perfect cube $k^{3}$ of a positive integer is special because we can write
$$
k^{3}=k^{3} \frac{a^{3}+2 b^{3}}{a^{3}+2 b^{3}}=\frac{(k a)^{3}+2(k b)^{3}}{a^{3}+2 b^{3}}
$$
for some positive integers $a, b$.
(b) Observe that $2014=2.19 .53$. If 2014 is special, then we have,
$$
x^{3}+2 y^{3}=2014\left(u^{3}+2 v^{3}\right)
$$
for some positive integers $x, y, u, v$. We may assume that $x^{3}+2 y^{3}$ is minimal with this property. Now, we will use the fact that if 19 divides $x^{3}+2 y^{3}$, then it divides both $x$ and $y$. Indeed, if 19 does not divide $x$, then it does not divide $y$ too. The relation $x^{3} \equiv-2 y^{3}(\bmod 19)$ implies $\left(x^{3}\right)^{6} \equiv\left(-2 y^{3}\right)^{6}(\bmod 19)$. The latter congruence is equivalent to $x^{18} \equiv 2^{6} y^{18}(\bmod 19)$. Now, according to the Fermat's Little Theorem, we obtain $1 \equiv 2^{6}(\bmod 19)$, that is 19 divides 63 , not possible.
It follows $x=19 x_{1}, y=19 y_{1}$, for some positive integers $x_{1}$ and $y_{1}$. Replacing in (1) we get
$$
19^{2}\left(x_{1}^{3}+2 y_{1}^{3}\right)=2.53\left(u^{3}+2 v^{3}\right)
$$
i.e. $19 \mid u^{3}+2 v^{3}$. It follows $u=19 u_{1}$ and $v=19 v_{1}$, and replacing in (2) we get
$$
x_{1}^{3}+2 y_{1}^{3}=2014\left(u_{1}^{3}+2 v_{1}^{3}\right) .
$$
Clearly, $x_{1}^{3}+2 y_{1}^{3}<x^{3}+2 y^{3}$, contradicting the minimality of $x^{3}+2 y^{3}$.
|
{
"resource_path": "Balkan_MO/segmented/en-2014-BMO-type1.jsonl",
"problem_match": "\nProblem 2.",
"solution_match": "\nSolution."
}
|
9173244f-9b99-5fbe-af3d-a1f1a809e70c
| 604,346
|
Let $A B C D$ be a trapezium inscribed in a circle $\Gamma$ with diameter $A B$. Let $E$ be the intersection point of the diagonals $A C$ and $B D$. The circle with center $B$ and radius $B E$ meets $\Gamma$ at the points $K$ and $L$, where $K$ is on the same side of $A B$ as $C$. The line perpendicular to $B D$ at $E$ intersects $C D$ at $M$.
Prove that $K M$ is perpendicular to $D L$.
|
Since $A B \| C D$, we have that $A B C D$ is isosceles trapezium. Let $O$ be the center of $k$ and $E M$ meets $A B$ at point $Q$. Then, from the right angled triangle $B E Q$, we have $B E^{2}=B O \cdot B Q$. Since $B E=B K$, we get $B K^{2}=B O \cdot B Q$ (1). Suppose that $K L$ meets $A B$ at $P$. Then, from the right angled triangle $B A K$, we have $B K^{2}=B P . B A$ (2)
From (1) and (2) we get $\frac{B P}{B Q}=\frac{B O}{B A}=\frac{1}{2}$, and therefore $P$ is the midpoint of $B Q$ (3). However, $D M \| A Q$ and $M Q \| A D$ (both are perpendicular to $D B$ ). Hence, $A Q M D$ is parallelogram and thus $M Q=A D=B C$. We conclude that $Q B C M$ is isosceles trapezium. It follows from (3) that $K L$ is the perpendicular bisector of $B Q$ and $C M$, that is, $M$ is symmetric to $C$ with respect to $K L$. Finally, we get that $M$ is the orthocenter
of the triangle $D L K$ by using the well-known result that the reflection of the orthocenter of a triangle to every side belongs to the circumcircle of the triangle and vise versa.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C D$ be a trapezium inscribed in a circle $\Gamma$ with diameter $A B$. Let $E$ be the intersection point of the diagonals $A C$ and $B D$. The circle with center $B$ and radius $B E$ meets $\Gamma$ at the points $K$ and $L$, where $K$ is on the same side of $A B$ as $C$. The line perpendicular to $B D$ at $E$ intersects $C D$ at $M$.
Prove that $K M$ is perpendicular to $D L$.
|
Since $A B \| C D$, we have that $A B C D$ is isosceles trapezium. Let $O$ be the center of $k$ and $E M$ meets $A B$ at point $Q$. Then, from the right angled triangle $B E Q$, we have $B E^{2}=B O \cdot B Q$. Since $B E=B K$, we get $B K^{2}=B O \cdot B Q$ (1). Suppose that $K L$ meets $A B$ at $P$. Then, from the right angled triangle $B A K$, we have $B K^{2}=B P . B A$ (2)
From (1) and (2) we get $\frac{B P}{B Q}=\frac{B O}{B A}=\frac{1}{2}$, and therefore $P$ is the midpoint of $B Q$ (3). However, $D M \| A Q$ and $M Q \| A D$ (both are perpendicular to $D B$ ). Hence, $A Q M D$ is parallelogram and thus $M Q=A D=B C$. We conclude that $Q B C M$ is isosceles trapezium. It follows from (3) that $K L$ is the perpendicular bisector of $B Q$ and $C M$, that is, $M$ is symmetric to $C$ with respect to $K L$. Finally, we get that $M$ is the orthocenter
of the triangle $D L K$ by using the well-known result that the reflection of the orthocenter of a triangle to every side belongs to the circumcircle of the triangle and vise versa.
|
{
"resource_path": "Balkan_MO/segmented/en-2014-BMO-type1.jsonl",
"problem_match": "\nProblem 3.",
"solution_match": "\nSolution."
}
|
714c3ebf-74fa-576e-9d75-b265371566b3
| 604,360
|
Let $n$ be a positive integer. A regular hexagon with side length $n$ is divided into equilateral triangles with side length 1 by lines parallel to its sides.
Find the number of regular hexagons all of whose vertices are among the vertices of the equilateral triangles.
|
By a lattice hexagon we will mean a regular hexagon whose sides run along edges of the lattice. Given any regular hexagon $H$, we construct a lattice hexagon whose edges pass through the vertices of $H$, as shown in the figure, which we will call the enveloping lattice hexagon of $H$. Given a lattice hexagon $G$ of side length $m$, the number of regular hexagons whose enveloping lattice hexagon is $G$ is exactly $m$.
Yet also there are precisely $3(n-m)(n-m+1)+1$ lattice hexagons of side length $m$ in our lattice: they are those with centres lying at most $n-m$ steps from the centre of the lattice. In particular, the total number of regular hexagons equals
$N=\sum_{m=1}^{n}(3(n-m)(n-m+1)+1) m=\left(3 n^{2}+3 n\right) \sum_{m=1}^{n} m-3(2 m+1) \sum_{m=1}^{n} m^{2}+3 \sum_{m=1}^{n} m^{3}$.
Since $\sum_{m=1}^{n} m=\frac{n(n+1)}{2}, \sum_{m=1}^{n} m^{2}=\frac{n(n+1)(2 n+1)}{6}$ and $\sum_{m=1}^{n} m^{3}=\left(\frac{n(n+1)}{2}\right)^{2}$ it is easily checked that $N=\left(\frac{n(n+1)}{2}\right)^{2}$.
|
\left(\frac{n(n+1)}{2}\right)^{2}
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Let $n$ be a positive integer. A regular hexagon with side length $n$ is divided into equilateral triangles with side length 1 by lines parallel to its sides.
Find the number of regular hexagons all of whose vertices are among the vertices of the equilateral triangles.
|
By a lattice hexagon we will mean a regular hexagon whose sides run along edges of the lattice. Given any regular hexagon $H$, we construct a lattice hexagon whose edges pass through the vertices of $H$, as shown in the figure, which we will call the enveloping lattice hexagon of $H$. Given a lattice hexagon $G$ of side length $m$, the number of regular hexagons whose enveloping lattice hexagon is $G$ is exactly $m$.
Yet also there are precisely $3(n-m)(n-m+1)+1$ lattice hexagons of side length $m$ in our lattice: they are those with centres lying at most $n-m$ steps from the centre of the lattice. In particular, the total number of regular hexagons equals
$N=\sum_{m=1}^{n}(3(n-m)(n-m+1)+1) m=\left(3 n^{2}+3 n\right) \sum_{m=1}^{n} m-3(2 m+1) \sum_{m=1}^{n} m^{2}+3 \sum_{m=1}^{n} m^{3}$.
Since $\sum_{m=1}^{n} m=\frac{n(n+1)}{2}, \sum_{m=1}^{n} m^{2}=\frac{n(n+1)(2 n+1)}{6}$ and $\sum_{m=1}^{n} m^{3}=\left(\frac{n(n+1)}{2}\right)^{2}$ it is easily checked that $N=\left(\frac{n(n+1)}{2}\right)^{2}$.
|
{
"resource_path": "Balkan_MO/segmented/en-2014-BMO-type1.jsonl",
"problem_match": "\nProblem 4.",
"solution_match": "\nSolution."
}
|
c2a0bdb9-1b48-5eb4-b675-fc82852509ab
| 604,375
|
Let $a, b$ and $c$ be positive real numbers. Prove that
$$
a^{3} b^{6}+b^{3} c^{6}+c^{3} a^{6}+3 a^{3} b^{3} c^{3} \geq a b c\left(a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}\right)+a^{2} b^{2} c^{2}\left(a^{3}+b^{3}+c^{3}\right)
$$
|
After dividing both sides of the given inequality by $a^{3} b^{3} c^{3}$ it becomes
$$
\left(\frac{b}{c}\right)^{3}+\left(\frac{c}{a}\right)^{3}+\left(\frac{a}{b}\right)^{3}+3 \geq\left(\frac{a}{c} \cdot \frac{b}{c}+\frac{b}{a} \cdot \frac{c}{a}+\frac{c}{b} \cdot \frac{a}{b}\right)+\left(\frac{a}{b} \cdot \frac{a}{c}+\frac{b}{a} \cdot \frac{b}{c}+\frac{c}{a} \cdot \frac{c}{b}\right) .
$$
Set
$$
\frac{b}{a}=\frac{1}{x}, \quad \frac{c}{b}=\frac{1}{y}, \quad \frac{a}{c}=\frac{1}{z} .
$$
Then we have that $x y z=1$ and by substituting (2) into (1), we find that
$$
x^{3}+y^{3}+z^{3}+3 \geq\left(\frac{y}{z}+\frac{z}{x}+\frac{x}{y}\right)+\left(\frac{x}{z}+\frac{y}{x}+\frac{z}{y}\right) .
$$
Multiplying the inequality (3) by $x y z$, and using the fact that $x y z=1$, the inequality is equivalent to
$$
x^{3}+y^{3}+z^{3}+3 x y z-x y^{2}-y z^{2}-z x^{2}-y x^{2}-z y^{2}-x z^{2} \geq 0 .
$$
Finally, notice that by the special case of Schur's inequality
$$
x^{r}(x-y)(x-z)+y^{r}(y-x)(y-z)+z^{r}(z-y)(z-x) \geq 0, \quad x, y, z \geq 0, r>0,
$$
with $r=1$ there holds
$$
x(x-y)(x-z)+y(y-x)(y-z)+z(z-y)(z-x) \geq 0
$$
which after expansion actually coincides with the congruence (4).
Remark 1. The inequality (5) immediately follows by supposing (without loss of generality) that $x \geq y \geq z$, and then writing the left hand side of the inequality (5) in the form
$$
(x-y)(x(x-z)-y(y-z))+z(y-z)(x-z)
$$
which is obviously $\geq 0$.
Remark 2. One can obtain the relation (4) using also the substitution $x=a b^{2}, y=b c^{2}$ and $z=c a^{2}$.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $a, b$ and $c$ be positive real numbers. Prove that
$$
a^{3} b^{6}+b^{3} c^{6}+c^{3} a^{6}+3 a^{3} b^{3} c^{3} \geq a b c\left(a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}\right)+a^{2} b^{2} c^{2}\left(a^{3}+b^{3}+c^{3}\right)
$$
|
After dividing both sides of the given inequality by $a^{3} b^{3} c^{3}$ it becomes
$$
\left(\frac{b}{c}\right)^{3}+\left(\frac{c}{a}\right)^{3}+\left(\frac{a}{b}\right)^{3}+3 \geq\left(\frac{a}{c} \cdot \frac{b}{c}+\frac{b}{a} \cdot \frac{c}{a}+\frac{c}{b} \cdot \frac{a}{b}\right)+\left(\frac{a}{b} \cdot \frac{a}{c}+\frac{b}{a} \cdot \frac{b}{c}+\frac{c}{a} \cdot \frac{c}{b}\right) .
$$
Set
$$
\frac{b}{a}=\frac{1}{x}, \quad \frac{c}{b}=\frac{1}{y}, \quad \frac{a}{c}=\frac{1}{z} .
$$
Then we have that $x y z=1$ and by substituting (2) into (1), we find that
$$
x^{3}+y^{3}+z^{3}+3 \geq\left(\frac{y}{z}+\frac{z}{x}+\frac{x}{y}\right)+\left(\frac{x}{z}+\frac{y}{x}+\frac{z}{y}\right) .
$$
Multiplying the inequality (3) by $x y z$, and using the fact that $x y z=1$, the inequality is equivalent to
$$
x^{3}+y^{3}+z^{3}+3 x y z-x y^{2}-y z^{2}-z x^{2}-y x^{2}-z y^{2}-x z^{2} \geq 0 .
$$
Finally, notice that by the special case of Schur's inequality
$$
x^{r}(x-y)(x-z)+y^{r}(y-x)(y-z)+z^{r}(z-y)(z-x) \geq 0, \quad x, y, z \geq 0, r>0,
$$
with $r=1$ there holds
$$
x(x-y)(x-z)+y(y-x)(y-z)+z(z-y)(z-x) \geq 0
$$
which after expansion actually coincides with the congruence (4).
Remark 1. The inequality (5) immediately follows by supposing (without loss of generality) that $x \geq y \geq z$, and then writing the left hand side of the inequality (5) in the form
$$
(x-y)(x(x-z)-y(y-z))+z(y-z)(x-z)
$$
which is obviously $\geq 0$.
Remark 2. One can obtain the relation (4) using also the substitution $x=a b^{2}, y=b c^{2}$ and $z=c a^{2}$.
|
{
"resource_path": "Balkan_MO/segmented/en-2015-BMO-type1.jsonl",
"problem_match": "\nProblem 1.",
"solution_match": "\nSolution."
}
|
0bedad70-1390-5602-9f96-118630d84815
| 604,388
|
Let $A B C$ be a scalene triangle with incentre $I$ and circumcircle ( $\omega$ ). The lines $A I, B I, C I$ intersect $(\omega)$ for the second time at the points $D, E, F$, respectively. The lines through $I$ parallel to the sides $B C, A C, A B$ intersect the lines $E F, D F, D E$ at the points $K, L, M$, respectively. Prove that the points $K, L, M$ are collinear.
|
First we will prove that $K A$ is tangent to $(\omega)$.
Indeed, it is a well-known fact that $F A=F B=F I$ and $E A=E C=E I$, so $F E$ is the perpendicular bisector of $A I$. It follows that $K A=K I$ and
$$
\angle K A F=\angle K I F=\angle F C B=\angle F E B=\angle F E A,
$$
so $K A$ is tangent to $(\omega)$. Similarly we can prove that $L B, M C$ are tangent to $(\omega)$ as well.
Let $A^{\prime}, B^{\prime}, C^{\prime}$ the intersections of $A I, B I, C I$ with $B C, C A, A B$ respectively. From Pascal's Theorem on the cyclic hexagon $A A C D E B$ we get $K, C^{\prime}, B^{\prime}$ collinear. Similarly $L, C^{\prime}, A^{\prime}$ collinear and $M, B^{\prime}, A^{\prime}$ collinear.
Then from Desargues' Theorem for $\triangle D E F, \triangle A^{\prime} B^{\prime} C^{\prime}$ which are perspective from the point $I$, we get that points $K, L, M$ of the intersection of their corresponding sides are collinear as wanted.
Remark (P.S.C.). After proving that $K A, L B, M C$ are tangent to ( $\omega$ ), we can argue as follows:
It readily follows that $\triangle K A F \sim \triangle K A E$ and so $\frac{K A}{K E}=\frac{K F}{K A}=\frac{A F}{A E}$, thus $\frac{K F}{K E}=\left(\frac{A F}{A E}\right)^{2}$. In a similar way we can find that $\frac{M E}{M D}=\left(\frac{C E}{C D}\right)^{2}$ and $\frac{L D}{L F}=\left(\frac{B D}{B F}\right)^{2}$. Multiplying we obtain $\frac{K F}{K E} \cdot \frac{M E}{M D} \cdot \frac{L D}{L F}=1$, so by the converse of Menelaus theorem applied in the triangle $D E F$ we get that the points $K, L, M$ are collinear.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a scalene triangle with incentre $I$ and circumcircle ( $\omega$ ). The lines $A I, B I, C I$ intersect $(\omega)$ for the second time at the points $D, E, F$, respectively. The lines through $I$ parallel to the sides $B C, A C, A B$ intersect the lines $E F, D F, D E$ at the points $K, L, M$, respectively. Prove that the points $K, L, M$ are collinear.
|
First we will prove that $K A$ is tangent to $(\omega)$.
Indeed, it is a well-known fact that $F A=F B=F I$ and $E A=E C=E I$, so $F E$ is the perpendicular bisector of $A I$. It follows that $K A=K I$ and
$$
\angle K A F=\angle K I F=\angle F C B=\angle F E B=\angle F E A,
$$
so $K A$ is tangent to $(\omega)$. Similarly we can prove that $L B, M C$ are tangent to $(\omega)$ as well.
Let $A^{\prime}, B^{\prime}, C^{\prime}$ the intersections of $A I, B I, C I$ with $B C, C A, A B$ respectively. From Pascal's Theorem on the cyclic hexagon $A A C D E B$ we get $K, C^{\prime}, B^{\prime}$ collinear. Similarly $L, C^{\prime}, A^{\prime}$ collinear and $M, B^{\prime}, A^{\prime}$ collinear.
Then from Desargues' Theorem for $\triangle D E F, \triangle A^{\prime} B^{\prime} C^{\prime}$ which are perspective from the point $I$, we get that points $K, L, M$ of the intersection of their corresponding sides are collinear as wanted.
Remark (P.S.C.). After proving that $K A, L B, M C$ are tangent to ( $\omega$ ), we can argue as follows:
It readily follows that $\triangle K A F \sim \triangle K A E$ and so $\frac{K A}{K E}=\frac{K F}{K A}=\frac{A F}{A E}$, thus $\frac{K F}{K E}=\left(\frac{A F}{A E}\right)^{2}$. In a similar way we can find that $\frac{M E}{M D}=\left(\frac{C E}{C D}\right)^{2}$ and $\frac{L D}{L F}=\left(\frac{B D}{B F}\right)^{2}$. Multiplying we obtain $\frac{K F}{K E} \cdot \frac{M E}{M D} \cdot \frac{L D}{L F}=1$, so by the converse of Menelaus theorem applied in the triangle $D E F$ we get that the points $K, L, M$ are collinear.
|
{
"resource_path": "Balkan_MO/segmented/en-2015-BMO-type1.jsonl",
"problem_match": "\nProblem 2.",
"solution_match": "\nSolution."
}
|
3c83929b-30fa-5c9f-9884-62502da37077
| 604,399
|
A jury of 3366 film critics are judging the Oscars. Each critic makes a single vote for his favourite actor, and a single vote for his favourite actress. It turns out that for every integer $n \in\{1,2, \ldots, 100\}$ there is an actor or actress who has been voted for exactly $n$ times. Show that there are two critics who voted for the same actor and for the same actress.
|
Let us assume that every critic votes for a different pair of actor and actress. We'll arrive at a contradiction proving the required result. Indeed:
Call the vote of each critic, i.e his choice for the pair of an actor and an actress, as a double-vote, and call as a single-vote each one of the two choices he makes, i.e. the one for an actor and the other one for an actress. In this terminology, a double-vote corresponds to two single-votes.
For each $n=34,35, \ldots, 100$ let us pick out one actor or one actress who has been voted by exactly $n$ critics (i.e. appears in exactly $n$ single-votes) and call $S$ the set of these movie stars. Calling $a, b$ the number of men and women in $S$, we have $a+b=67$.
Now let $S_{1}$ be the set of double-votes, each having exactly one of its two corresponding singlevotes in $S$, and let $S_{2}$ be the set of double-votes with both its single-votes in $S$. If $s_{1}, s_{2}$ are the number of elements in $S_{1}, S_{2}$ respectively, we have that the number of all double-votes with at least one single-vote in $S$ is $s_{1}+s_{2}$, whereas the number of all double-votes with both single-votes in $S$ is $s_{2} \leq a b$.
Since all double-votes are distinct, there must exist at least $s_{1}+s_{2}$ critics. But the number of all single-votes in $S$ is $s_{1}+2 s_{2}=34+35+\cdots+100=4489$, and moreover $s \leq a b$. So there exist at least $s_{1}+s_{2}=s_{1}+2 s_{2}-s_{2} \geq 4489-a b$ critics.
Now notice that as $a+b=67$, the maximum value of $a b$ with $a, b$ integers is obtained for $\{a, b\}=$ $\{33,34\}$, so $a b \leq 33 \cdot 34=1122$. A quick proof of this is the following: $a b=\frac{(a+b)^{2}-(a-b)^{2}}{4}=$ $\frac{67^{2}-(a-b)^{2}}{4}$ which is maximized (for not equal integers $a, b$ as $a+b=67$ ) whenever $|a-b|=1$, thus for $\{a, b\}=\{33,34\}$.
Thus there exist at least $4489-1122=3367$ critics which is a contradiction and we are done.
Remark. We are going here to give some motivation about the choice of number 34, used in the above solution.
Let us assume that every critic votes for a different pair of actor and actress. One can again start by picking out one actor or one actress who has been voted by exactly $n$ critics for $n=k, k+1, \ldots, 100$. Then $a+b=100-k+1=101-k$ and the number of all single-votes is $s_{1}+2 s_{2}=k+k+1+\cdots+100=$ $5050-\frac{k(k-1)}{2}$, so there exist at least $s_{1}+s_{2}=s_{1}+2 s_{2}-s_{2} \geq 5050-\frac{k(k-1)}{2}-a b$ and
$$
a b=\frac{(a+b)^{2}-(a-b)^{2}}{4}=\frac{(101-k)^{2}-(a-b)^{2}}{4} \leq \frac{(101-k)^{2}-1}{4} .
$$
After all, the number of critics is at least
$$
5050-\frac{k(k-1)}{2}-\frac{(101-k)^{2}-1}{4}
$$
In order to arrive at a contradiction we have to choose $k$ such that
$$
5050-\frac{k(k-1)}{2}-\frac{(101-k)^{2}-1}{4} \geq 3367
$$
and solving the inequality with respect to $k$, the only value that makes the last one true is $k=34$.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
A jury of 3366 film critics are judging the Oscars. Each critic makes a single vote for his favourite actor, and a single vote for his favourite actress. It turns out that for every integer $n \in\{1,2, \ldots, 100\}$ there is an actor or actress who has been voted for exactly $n$ times. Show that there are two critics who voted for the same actor and for the same actress.
|
Let us assume that every critic votes for a different pair of actor and actress. We'll arrive at a contradiction proving the required result. Indeed:
Call the vote of each critic, i.e his choice for the pair of an actor and an actress, as a double-vote, and call as a single-vote each one of the two choices he makes, i.e. the one for an actor and the other one for an actress. In this terminology, a double-vote corresponds to two single-votes.
For each $n=34,35, \ldots, 100$ let us pick out one actor or one actress who has been voted by exactly $n$ critics (i.e. appears in exactly $n$ single-votes) and call $S$ the set of these movie stars. Calling $a, b$ the number of men and women in $S$, we have $a+b=67$.
Now let $S_{1}$ be the set of double-votes, each having exactly one of its two corresponding singlevotes in $S$, and let $S_{2}$ be the set of double-votes with both its single-votes in $S$. If $s_{1}, s_{2}$ are the number of elements in $S_{1}, S_{2}$ respectively, we have that the number of all double-votes with at least one single-vote in $S$ is $s_{1}+s_{2}$, whereas the number of all double-votes with both single-votes in $S$ is $s_{2} \leq a b$.
Since all double-votes are distinct, there must exist at least $s_{1}+s_{2}$ critics. But the number of all single-votes in $S$ is $s_{1}+2 s_{2}=34+35+\cdots+100=4489$, and moreover $s \leq a b$. So there exist at least $s_{1}+s_{2}=s_{1}+2 s_{2}-s_{2} \geq 4489-a b$ critics.
Now notice that as $a+b=67$, the maximum value of $a b$ with $a, b$ integers is obtained for $\{a, b\}=$ $\{33,34\}$, so $a b \leq 33 \cdot 34=1122$. A quick proof of this is the following: $a b=\frac{(a+b)^{2}-(a-b)^{2}}{4}=$ $\frac{67^{2}-(a-b)^{2}}{4}$ which is maximized (for not equal integers $a, b$ as $a+b=67$ ) whenever $|a-b|=1$, thus for $\{a, b\}=\{33,34\}$.
Thus there exist at least $4489-1122=3367$ critics which is a contradiction and we are done.
Remark. We are going here to give some motivation about the choice of number 34, used in the above solution.
Let us assume that every critic votes for a different pair of actor and actress. One can again start by picking out one actor or one actress who has been voted by exactly $n$ critics for $n=k, k+1, \ldots, 100$. Then $a+b=100-k+1=101-k$ and the number of all single-votes is $s_{1}+2 s_{2}=k+k+1+\cdots+100=$ $5050-\frac{k(k-1)}{2}$, so there exist at least $s_{1}+s_{2}=s_{1}+2 s_{2}-s_{2} \geq 5050-\frac{k(k-1)}{2}-a b$ and
$$
a b=\frac{(a+b)^{2}-(a-b)^{2}}{4}=\frac{(101-k)^{2}-(a-b)^{2}}{4} \leq \frac{(101-k)^{2}-1}{4} .
$$
After all, the number of critics is at least
$$
5050-\frac{k(k-1)}{2}-\frac{(101-k)^{2}-1}{4}
$$
In order to arrive at a contradiction we have to choose $k$ such that
$$
5050-\frac{k(k-1)}{2}-\frac{(101-k)^{2}-1}{4} \geq 3367
$$
and solving the inequality with respect to $k$, the only value that makes the last one true is $k=34$.
|
{
"resource_path": "Balkan_MO/segmented/en-2015-BMO-type1.jsonl",
"problem_match": "\nProblem 3.",
"solution_match": "\nSolution."
}
|
9de7f38a-3a1d-5f43-ac45-ccdb1b909d08
| 604,408
|
Prove that among any 20 consecutive positive integers there exists an integer $d$ such that for each positive integer $n$ we have the inequality
$$
n \sqrt{d}\{n \sqrt{d}\}>\frac{5}{2}
$$
where $\{x\}$ denotes the fractional part of the real number $x$. The fractional part of a real number $x$ is $x$ minus the greatest integer less than or equal to $x$.
|
Among the given numbers there is a number of the form $20 k+15=5(4 k+3)$. We shall prove that $d=5(4 k+3)$ satisfies the statement's condition. Since $d \equiv-1(\bmod 4)$, it follows that $d$ is not a perfect square, and thus for any $n \in \mathbb{N}$ there exists $a \in \mathbb{N}$ such that $a+1>n \sqrt{d}>a$, that is, $(a+1)^{2}>n^{2} d>a^{2}$. Actually, we are going to prove that $n^{2} d \geq a^{2}+5$. Indeed:
It is known that each positive integer of the form $4 s+3$ has a prime divisor of the same form. Let $p \mid 4 k+3$ and $p \equiv-1(\bmod 4)$. Because of the form of $p$, the numbers $a^{2}+1^{2}$ and $a^{2}+2^{2}$ are not divisible by $p$, and since $p \mid n^{2} d$, it follows that $n^{2} d \neq a^{2}+1, a^{2}+4$. On the other hand, $5 \mid n^{2} d$, and since $5 \nmid a^{2}+2, a^{2}+3$, we conclude $n^{2} d \neq a^{2}+2, a^{2}+3$. Since $n^{2} d>a^{2}$ we must have $n^{2} d \geq a^{2}+5$ as claimed. Therefore,
$$
n \sqrt{d}\{n \sqrt{d}\}=n \sqrt{d}(n \sqrt{d}-a) \geq a^{2}+5-a \sqrt{a^{2}+5}>a^{2}+5-\frac{a^{2}+\left(a^{2}+5\right)}{2}=\frac{5}{2},
$$
which was to be proved.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Prove that among any 20 consecutive positive integers there exists an integer $d$ such that for each positive integer $n$ we have the inequality
$$
n \sqrt{d}\{n \sqrt{d}\}>\frac{5}{2}
$$
where $\{x\}$ denotes the fractional part of the real number $x$. The fractional part of a real number $x$ is $x$ minus the greatest integer less than or equal to $x$.
|
Among the given numbers there is a number of the form $20 k+15=5(4 k+3)$. We shall prove that $d=5(4 k+3)$ satisfies the statement's condition. Since $d \equiv-1(\bmod 4)$, it follows that $d$ is not a perfect square, and thus for any $n \in \mathbb{N}$ there exists $a \in \mathbb{N}$ such that $a+1>n \sqrt{d}>a$, that is, $(a+1)^{2}>n^{2} d>a^{2}$. Actually, we are going to prove that $n^{2} d \geq a^{2}+5$. Indeed:
It is known that each positive integer of the form $4 s+3$ has a prime divisor of the same form. Let $p \mid 4 k+3$ and $p \equiv-1(\bmod 4)$. Because of the form of $p$, the numbers $a^{2}+1^{2}$ and $a^{2}+2^{2}$ are not divisible by $p$, and since $p \mid n^{2} d$, it follows that $n^{2} d \neq a^{2}+1, a^{2}+4$. On the other hand, $5 \mid n^{2} d$, and since $5 \nmid a^{2}+2, a^{2}+3$, we conclude $n^{2} d \neq a^{2}+2, a^{2}+3$. Since $n^{2} d>a^{2}$ we must have $n^{2} d \geq a^{2}+5$ as claimed. Therefore,
$$
n \sqrt{d}\{n \sqrt{d}\}=n \sqrt{d}(n \sqrt{d}-a) \geq a^{2}+5-a \sqrt{a^{2}+5}>a^{2}+5-\frac{a^{2}+\left(a^{2}+5\right)}{2}=\frac{5}{2},
$$
which was to be proved.
|
{
"resource_path": "Balkan_MO/segmented/en-2015-BMO-type1.jsonl",
"problem_match": "\nProblem 4.",
"solution_match": "\nSolution."
}
|
de1b3225-0afc-5bd8-a794-6020065dae9a
| 604,421
|
Find all injective functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for every real number $x$ and every positive integer $n$,
$$
\left|\sum_{i=1}^{n} i(f(x+i+1)-f(f(x+i)))\right|<2016
$$
|
From the condition of the problem we get
$$
\left|\sum_{i=1}^{n-1} i(f(x+i+1)-f(f(x+i)))\right|<2016
$$
Then
$$
\begin{aligned}
& |n(f(x+n+1)-f(f(x+n)))| \\
= & \left|\sum_{i=1}^{n} i(f(x+i+1)-f(f(x+i)))-\sum_{i=1}^{n-1} i(f(x+i+1)-f(f(x+i)))\right| \\
< & 2 \cdot 2016=4032
\end{aligned}
$$
implying
$$
|f(x+n+1)-f(f(x+n))|<\frac{4032}{n}
$$
for every real number $x$ and every positive integer $n$.
Let $y \in \mathbb{R}$ be arbitrary. Then there exists $x$ such that $y=x+n$. We obtain
$$
|f(y+1)-f(f(y))|<\frac{4032}{n}
$$
for every real number $y$ and every positive integer $n$. The last inequality holds for every positive integer $n$ from where $f(y+1)=f(f(y))$ for every $y \in \mathbb{R}$ and since the function $f$ is an injection, then $f(y)=y+1$. The function $f(y)=y+1$ satisfies the required condition.
|
f(y)=y+1
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find all injective functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for every real number $x$ and every positive integer $n$,
$$
\left|\sum_{i=1}^{n} i(f(x+i+1)-f(f(x+i)))\right|<2016
$$
|
From the condition of the problem we get
$$
\left|\sum_{i=1}^{n-1} i(f(x+i+1)-f(f(x+i)))\right|<2016
$$
Then
$$
\begin{aligned}
& |n(f(x+n+1)-f(f(x+n)))| \\
= & \left|\sum_{i=1}^{n} i(f(x+i+1)-f(f(x+i)))-\sum_{i=1}^{n-1} i(f(x+i+1)-f(f(x+i)))\right| \\
< & 2 \cdot 2016=4032
\end{aligned}
$$
implying
$$
|f(x+n+1)-f(f(x+n))|<\frac{4032}{n}
$$
for every real number $x$ and every positive integer $n$.
Let $y \in \mathbb{R}$ be arbitrary. Then there exists $x$ such that $y=x+n$. We obtain
$$
|f(y+1)-f(f(y))|<\frac{4032}{n}
$$
for every real number $y$ and every positive integer $n$. The last inequality holds for every positive integer $n$ from where $f(y+1)=f(f(y))$ for every $y \in \mathbb{R}$ and since the function $f$ is an injection, then $f(y)=y+1$. The function $f(y)=y+1$ satisfies the required condition.
|
{
"resource_path": "Balkan_MO/segmented/en-2016-BMO-type1.jsonl",
"problem_match": "# Problem 1.",
"solution_match": "\nSolution."
}
|
14db278a-6e10-5fb0-8af1-40e880016393
| 604,434
|
Let $A B C D$ be a cyclic quadrilateral with $A B<C D$. The diagonals intersect at the point $F$ and lines $A D$ and $B C$ intersect at the point $E$. Let $K$ and $L$ be the orthogonal projections of $F$ onto lines $A D$ and $B C$ respectively, and let $M, S$ and $T$ be the midpoints of $E F, C F$ and $D F$ respectively. Prove that the second intersection point of the circumcircles of triangles $M K T$ and $M L S$ lies on the segment $C D$.
|
Let $N$ be the midpoint of $C D$. We will prove that the circumcircles of the triangles $M K T$ and $M L S$ pass through $N$. (1)
First will prove that the circumcircle of $M L S$ passes through $N$.
Let $Q$ be the midpoint of $E C$. Note that the circumcircle of $M L S$ is the Euler circle (2) of the triangle $E F C$, so it passes also through $Q .\left({ }^{*}\right)(3)$
We will prove that
$$
\angle S L Q=\angle Q N S \quad \text { or } \quad \angle S L Q+\angle Q N S=180^{\circ}
$$
Indeed, since $F L C$ is right-angled and $L S$ is its median, we have that $S L=S C$ and
$$
\angle S L C=\angle S C L=\angle A C B
$$
In addition, since $N$ and $S$ are the midpoints of $D C$ and $F C$ we have that $S N \| F D$ and similarly, since $Q$ and $N$ are the midpoints of $E C$ and $C D$, so $Q N \| E D$.
It follows that the angles $\angle E D B$ and $\angle Q N S$ have parallel sides, and since $A B<C D$, they are acute, and as a result we have that
$$
\angle E D B=\angle Q N S \quad \text { or } \quad \angle E D B+\angle Q N S=180^{\circ}
$$
But, from the cyclic quadrilateral $A B C D$, we get that
$$
\angle E D B=\angle A C B
$$
Now, from (2),(3) and (4) we obtain immediately (1), so the quadrilateral $L N S Q$ is cyclic. Since from $\left(^{*}\right)$, its circumcircle passes also through $M$, we get that the points $M, L, Q, S, N$ are cocylic and this means that the circumcircle of $M L S$ passes through $N$.
Similarly, the circumcircle of $M K T$ passes also through $N$ and we have the desired.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C D$ be a cyclic quadrilateral with $A B<C D$. The diagonals intersect at the point $F$ and lines $A D$ and $B C$ intersect at the point $E$. Let $K$ and $L$ be the orthogonal projections of $F$ onto lines $A D$ and $B C$ respectively, and let $M, S$ and $T$ be the midpoints of $E F, C F$ and $D F$ respectively. Prove that the second intersection point of the circumcircles of triangles $M K T$ and $M L S$ lies on the segment $C D$.
|
Let $N$ be the midpoint of $C D$. We will prove that the circumcircles of the triangles $M K T$ and $M L S$ pass through $N$. (1)
First will prove that the circumcircle of $M L S$ passes through $N$.
Let $Q$ be the midpoint of $E C$. Note that the circumcircle of $M L S$ is the Euler circle (2) of the triangle $E F C$, so it passes also through $Q .\left({ }^{*}\right)(3)$
We will prove that
$$
\angle S L Q=\angle Q N S \quad \text { or } \quad \angle S L Q+\angle Q N S=180^{\circ}
$$
Indeed, since $F L C$ is right-angled and $L S$ is its median, we have that $S L=S C$ and
$$
\angle S L C=\angle S C L=\angle A C B
$$
In addition, since $N$ and $S$ are the midpoints of $D C$ and $F C$ we have that $S N \| F D$ and similarly, since $Q$ and $N$ are the midpoints of $E C$ and $C D$, so $Q N \| E D$.
It follows that the angles $\angle E D B$ and $\angle Q N S$ have parallel sides, and since $A B<C D$, they are acute, and as a result we have that
$$
\angle E D B=\angle Q N S \quad \text { or } \quad \angle E D B+\angle Q N S=180^{\circ}
$$
But, from the cyclic quadrilateral $A B C D$, we get that
$$
\angle E D B=\angle A C B
$$
Now, from (2),(3) and (4) we obtain immediately (1), so the quadrilateral $L N S Q$ is cyclic. Since from $\left(^{*}\right)$, its circumcircle passes also through $M$, we get that the points $M, L, Q, S, N$ are cocylic and this means that the circumcircle of $M L S$ passes through $N$.
Similarly, the circumcircle of $M K T$ passes also through $N$ and we have the desired.
|
{
"resource_path": "Balkan_MO/segmented/en-2016-BMO-type1.jsonl",
"problem_match": "# Problem 2.",
"solution_match": "\nSolution."
}
|
2562b049-0197-5221-a500-54b38da1d2a4
| 604,449
|
Find all monic polynomials $f$ with integer coefficients satisfying the following condition: there exists a positive integer $N$ such that $p$ divides $2(f(p)!)+1$ for every prime $p>N$ for which $f(p)$ is a positive integer.
Note: A monic polynomial has leading coefficient equal to 1.
|
If $f$ is a constant polynomial then it's obvious that the condition cannot hold for
$$
p \geq 5 \text { since } f(p)=1
$$
From the divisibility relation $p \mid 2(f(p))$ ! +1 we conclude that:
$$
f(p)<p, \text { for all primes } p>N \quad(*)
$$
In fact, if for some prime number $p$ we have $f(p) \geq p$, then $p \mid(f(p))$ ! and then $p \mid 1$, which is absurd.
Now suppose that $\operatorname{deg} f=m>1$. Then $f(x)=x^{m}+Q(x), \operatorname{deg} Q(x) \leq m-1$ and so $f(p)=$ $p^{m}+Q(p)$. Hence for some large enough prime number $p$ holds that $f(p)>p$, which contradicts $\left.{ }^{*}\right)$. Therefore we must have $\operatorname{deg} f(x)=1$ and $f(x)=x-a$, for some positive integer $a$. (3)
Thus the given condition becomes:
$$
p \mid 2(p-a)!+1
$$
But we have (using Wilsons theorem)
$$
\begin{gathered}
2(p-3)!\equiv-(p-3)!(p-2) \equiv-(p-2)!\equiv-1(\bmod p) \\
\Rightarrow p \mid 2(p-3)!+1
\end{gathered}
$$
From (1) and (2) we get
$$
\begin{aligned}
& (p-3)!\equiv(p-a)!(\bmod p) \\
& (p-3)!(-1)^{a}(a-1)!\equiv(p-a)!(-1)^{a}(a-1)!(\bmod p) \\
& (p-3)!(-1)^{a}(a-1)!\equiv 1(\bmod p)
\end{aligned}
$$
Since $-2(p-3)!\equiv 1(\bmod p)$, it follows that
$$
(-1)^{a}(a-1)!\equiv-2(\bmod p)
$$
Taking $p>(a-1)$ !, we conclude that $a=3$ and so $f(x)=x-3$, for all $x$.
The function $f(x)=x-3$ satisfies the required condition.
|
f(x)=x-3
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Find all monic polynomials $f$ with integer coefficients satisfying the following condition: there exists a positive integer $N$ such that $p$ divides $2(f(p)!)+1$ for every prime $p>N$ for which $f(p)$ is a positive integer.
Note: A monic polynomial has leading coefficient equal to 1.
|
If $f$ is a constant polynomial then it's obvious that the condition cannot hold for
$$
p \geq 5 \text { since } f(p)=1
$$
From the divisibility relation $p \mid 2(f(p))$ ! +1 we conclude that:
$$
f(p)<p, \text { for all primes } p>N \quad(*)
$$
In fact, if for some prime number $p$ we have $f(p) \geq p$, then $p \mid(f(p))$ ! and then $p \mid 1$, which is absurd.
Now suppose that $\operatorname{deg} f=m>1$. Then $f(x)=x^{m}+Q(x), \operatorname{deg} Q(x) \leq m-1$ and so $f(p)=$ $p^{m}+Q(p)$. Hence for some large enough prime number $p$ holds that $f(p)>p$, which contradicts $\left.{ }^{*}\right)$. Therefore we must have $\operatorname{deg} f(x)=1$ and $f(x)=x-a$, for some positive integer $a$. (3)
Thus the given condition becomes:
$$
p \mid 2(p-a)!+1
$$
But we have (using Wilsons theorem)
$$
\begin{gathered}
2(p-3)!\equiv-(p-3)!(p-2) \equiv-(p-2)!\equiv-1(\bmod p) \\
\Rightarrow p \mid 2(p-3)!+1
\end{gathered}
$$
From (1) and (2) we get
$$
\begin{aligned}
& (p-3)!\equiv(p-a)!(\bmod p) \\
& (p-3)!(-1)^{a}(a-1)!\equiv(p-a)!(-1)^{a}(a-1)!(\bmod p) \\
& (p-3)!(-1)^{a}(a-1)!\equiv 1(\bmod p)
\end{aligned}
$$
Since $-2(p-3)!\equiv 1(\bmod p)$, it follows that
$$
(-1)^{a}(a-1)!\equiv-2(\bmod p)
$$
Taking $p>(a-1)$ !, we conclude that $a=3$ and so $f(x)=x-3$, for all $x$.
The function $f(x)=x-3$ satisfies the required condition.
|
{
"resource_path": "Balkan_MO/segmented/en-2016-BMO-type1.jsonl",
"problem_match": "# Problem 3.",
"solution_match": "\nSolution."
}
|
ee013b75-6402-5090-b494-5e37b30d4dd5
| 604,461
|
The plane is divided into unit squares by two sets of parallel lines, forming an infinite grid. Each unit square is coloured with one of 1201 colours so that no rectangle with perimeter 100 contains two squares of the same colour. Show that no rectangle of size $1 \times 1201$ or $1201 \times 1$ contains two squares of the same colour.
Note: Any rectangle is assumed here to have sides contained in the lines of the grid.
|
Let the centers of the unit squares be the integer points in the plane, and denote each unit square by the coordinates of its center.
Consider the set $D$ of all unit squares $(x, y)$ such that $|x|+|y| \leq 24$. Any integer translate of $D$ is called a diamond.
Since any two unit squares that belong to the same diamond also belong to some rectangle of perimeter 100 , a diamond cannot contain two unit squares of the same colour. Since a diamond contains exactly $24^{2}+25^{2}=1201$ unit squares, a diamond must contain every colour exactly once.
Choose one colour, say, green, and let $a_{1}, a_{2}, \ldots$ be all green unit squares. Let $P_{i}$ be the diamond of center $a_{i}$. We will show that no unit square is covered by two $P$ 's and that every unit square is covered by some $P_{i}$.
Indeed, suppose first that $P_{i}$ and $P_{j}$ contain the same unit square $b$. Then their centers lie within the same rectangle of perimeter 100, a contradiction.
Let, on the other hand, $b$ be an arbitrary unit square. The diamond of center $b$ must contain some green unit square $a_{i}$. The diamond $P_{i}$ of center $a_{i}$ will then contain $b$.
Therefore, $P_{1}, P_{2}, \ldots$ form a covering of the plane in exactly one layer. It is easy to see, though, that, up to translation and reflection, there exists a unique such covering. (Indeed, consider two neighbouring diamonds. Unless they fit neatly, uncoverable spaces of two unit squares are created near the corners: see Fig. 1.)
Figure 1:
Without loss of generality, then, this covering is given by the diamonds of centers $(x, y)$ such that $24 x+25 y$ is divisible by 1201. (See Fig. 2 for an analogous covering with smaller diamonds.) It follows from this that no rectangle of size $1 \times 1201$ can contain two green unit squares, and analogous reasoning works for the remaining colours.
Figure 2:
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
The plane is divided into unit squares by two sets of parallel lines, forming an infinite grid. Each unit square is coloured with one of 1201 colours so that no rectangle with perimeter 100 contains two squares of the same colour. Show that no rectangle of size $1 \times 1201$ or $1201 \times 1$ contains two squares of the same colour.
Note: Any rectangle is assumed here to have sides contained in the lines of the grid.
|
Let the centers of the unit squares be the integer points in the plane, and denote each unit square by the coordinates of its center.
Consider the set $D$ of all unit squares $(x, y)$ such that $|x|+|y| \leq 24$. Any integer translate of $D$ is called a diamond.
Since any two unit squares that belong to the same diamond also belong to some rectangle of perimeter 100 , a diamond cannot contain two unit squares of the same colour. Since a diamond contains exactly $24^{2}+25^{2}=1201$ unit squares, a diamond must contain every colour exactly once.
Choose one colour, say, green, and let $a_{1}, a_{2}, \ldots$ be all green unit squares. Let $P_{i}$ be the diamond of center $a_{i}$. We will show that no unit square is covered by two $P$ 's and that every unit square is covered by some $P_{i}$.
Indeed, suppose first that $P_{i}$ and $P_{j}$ contain the same unit square $b$. Then their centers lie within the same rectangle of perimeter 100, a contradiction.
Let, on the other hand, $b$ be an arbitrary unit square. The diamond of center $b$ must contain some green unit square $a_{i}$. The diamond $P_{i}$ of center $a_{i}$ will then contain $b$.
Therefore, $P_{1}, P_{2}, \ldots$ form a covering of the plane in exactly one layer. It is easy to see, though, that, up to translation and reflection, there exists a unique such covering. (Indeed, consider two neighbouring diamonds. Unless they fit neatly, uncoverable spaces of two unit squares are created near the corners: see Fig. 1.)
Figure 1:
Without loss of generality, then, this covering is given by the diamonds of centers $(x, y)$ such that $24 x+25 y$ is divisible by 1201. (See Fig. 2 for an analogous covering with smaller diamonds.) It follows from this that no rectangle of size $1 \times 1201$ can contain two green unit squares, and analogous reasoning works for the remaining colours.
Figure 2:
|
{
"resource_path": "Balkan_MO/segmented/en-2016-BMO-type1.jsonl",
"problem_match": "# Problem 4.",
"solution_match": "\nSolution."
}
|
613e8d76-170e-5d70-9737-a5092f74af9d
| 604,474
|
Find all ordered pairs $(x, y)$ of positive integers such that:
$$
x^{3}+y^{3}=x^{2}+42 x y+y^{2} \text {. }
$$
|
Possible initial thoughts about this equation might include:
(i) I can factorise the sum of cubes on the left.
(ii) How can I use the 42?
(iii) The left is cubic and the right is quadratic, so if $x$ or $y$ is very large there will be no solutions.
The first two might lead us to rewrite the equation as $(x+y)\left(x^{2}-x y+y^{2}\right)=x^{2}-x y+y^{2}+43 x y$. The number $43=42+1$ is prime which is good news since we have $(x+y-1)\left(x^{2}-x y+y^{2}\right)=43 x y$.
Now we can employ some wishful thinking: if $x$ and $y$ happen to be coprime, then $\left(x^{2}-x y+y^{2}\right)$ has no factors in common with $x$ or $y$ so it must divide 43 . This feels like a significant step except for the fact that $x$ and $y$ may not be coprime.
This suggests writing $x=d X$ and $y=d Y$ where $d$ is the highest common factor of $x$ and $y$.
We end up with $(d X+d Y-1)\left(X^{2}-X Y+Y^{2}\right)=43 X Y$ so $X^{2}-X Y+Y^{2}$ equals 1 or 43.
The first of these readily gives $X=Y=1$. A neat way to deal with the second is to assume $Y \leq X$ so $43=X^{2}-X Y+Y^{2} \geq Y^{2}$. This gives six cases for $Y$ which can be checked in turn. Alternatively you can solve $X^{2}-X Y+Y^{2}=43$ for $X$ and fuss about the discriminant.
In the end the only solutions turn out to be $(x, y)=(22,22),(1,7)$ or $(7,1)$.
Another reasonable initial plan is to bound $x+y$ (using observation (iii) above) and then use modular arithmetic to eliminate cases until only the correct solutions remain. Working modulo 7 is particularly appealing since $7 \mid 42$ and the only cubes modulo 7 are 0,1 and -1 .
We have $x^{3}+y^{3}=(x+y)^{3}-3 x y(x+y)$ and also $x^{2}+42 x y+y^{2}=(x+y)^{2}+40 x y$ so the equation becomes $(x+y)^{3}=(x+y)^{2}+x y(3 x y+40)$. Now using $x y \leq\left(\frac{x+y}{2}\right)^{2}$ leads to $x+y \leq 44$.
This leaves a mere 484 cases to check! The modulo 7 magic is not really enough to cut this down to an attractive number, and although the approach can obviously be made to work, to call it laborious is an understatement.
Other possible approaches, such as substituting $u=x+y, v=x-y$, seem to lie somewhere between the two described above in terms of the amount of fortitude needed to pull them off.
|
(22,22),(1,7),(7,1)
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find all ordered pairs $(x, y)$ of positive integers such that:
$$
x^{3}+y^{3}=x^{2}+42 x y+y^{2} \text {. }
$$
|
Possible initial thoughts about this equation might include:
(i) I can factorise the sum of cubes on the left.
(ii) How can I use the 42?
(iii) The left is cubic and the right is quadratic, so if $x$ or $y$ is very large there will be no solutions.
The first two might lead us to rewrite the equation as $(x+y)\left(x^{2}-x y+y^{2}\right)=x^{2}-x y+y^{2}+43 x y$. The number $43=42+1$ is prime which is good news since we have $(x+y-1)\left(x^{2}-x y+y^{2}\right)=43 x y$.
Now we can employ some wishful thinking: if $x$ and $y$ happen to be coprime, then $\left(x^{2}-x y+y^{2}\right)$ has no factors in common with $x$ or $y$ so it must divide 43 . This feels like a significant step except for the fact that $x$ and $y$ may not be coprime.
This suggests writing $x=d X$ and $y=d Y$ where $d$ is the highest common factor of $x$ and $y$.
We end up with $(d X+d Y-1)\left(X^{2}-X Y+Y^{2}\right)=43 X Y$ so $X^{2}-X Y+Y^{2}$ equals 1 or 43.
The first of these readily gives $X=Y=1$. A neat way to deal with the second is to assume $Y \leq X$ so $43=X^{2}-X Y+Y^{2} \geq Y^{2}$. This gives six cases for $Y$ which can be checked in turn. Alternatively you can solve $X^{2}-X Y+Y^{2}=43$ for $X$ and fuss about the discriminant.
In the end the only solutions turn out to be $(x, y)=(22,22),(1,7)$ or $(7,1)$.
Another reasonable initial plan is to bound $x+y$ (using observation (iii) above) and then use modular arithmetic to eliminate cases until only the correct solutions remain. Working modulo 7 is particularly appealing since $7 \mid 42$ and the only cubes modulo 7 are 0,1 and -1 .
We have $x^{3}+y^{3}=(x+y)^{3}-3 x y(x+y)$ and also $x^{2}+42 x y+y^{2}=(x+y)^{2}+40 x y$ so the equation becomes $(x+y)^{3}=(x+y)^{2}+x y(3 x y+40)$. Now using $x y \leq\left(\frac{x+y}{2}\right)^{2}$ leads to $x+y \leq 44$.
This leaves a mere 484 cases to check! The modulo 7 magic is not really enough to cut this down to an attractive number, and although the approach can obviously be made to work, to call it laborious is an understatement.
Other possible approaches, such as substituting $u=x+y, v=x-y$, seem to lie somewhere between the two described above in terms of the amount of fortitude needed to pull them off.
|
{
"resource_path": "Balkan_MO/segmented/en-2017-BMO-type3.jsonl",
"problem_match": "\n1. ",
"solution_match": "\nSolution."
}
|
a427b534-58ac-5be2-a523-e9c701e6428d
| 604,488
|
Let $A B C$ be an acute triangle with with $A B<A C$ and let $\Gamma$ be its circumcircle. Let the tangents to $\Gamma$ at $B$ and $C$ be $t_{B}$ and $t_{C}$ respectively and let their point of intersection be $L$. The line through $B$ parallel to $A C$ intersects $t_{C}$ at $D$. The line through $C$ parallel to $A B$ intersects $t_{B}$ at $E$. The circumcircle of triangle $B D C$ meets the side $A C$ at $T$ where $T$ lies between $A$ and $C$. The circumcircle of triangle $B E C$ meets the line $A B$ at $S$ where $B$ lies between $A$ and $S$.
Prove that the lines $S T, B C$ and $A L$ are concurrent.
|
How we attack this problem depends on how much triangle geometry we can effortlessly recall - a good knowledge of some standard results helps a great deal.
We might instantly note that $A L$ is a symmedian of $A B C$, and so divides the line $B C$ in the ratio $c^{2}: b^{2}$. Now the plan is to show that $S T$ also divides $B C$ in this ratio.
Since we are working with ratios of distances, Menelaus' theorem may prove useful.
A key step is to notice (based on a careful diagram) that $A C$ is tangent to circle $C B S$. Once spotted this is easy to prove. The parallels give us $\angle B=\angle B C E$, and, since $B E$ is tangent to circle $A B C$, we have $\angle E B C=\angle A$ by the alternate segment theorem. Now angles in a triangle give $\angle C=\angle C E B$, and we have the converse to the alternate segment theorem. We obtain $A B$ is tangent to circle $C B D$ in the same way.
Now we have some tangencies and want some ratios.
Tangent-secant yields $A T . b=c^{2}$ and $c . A S=b^{2}$ or, equivalently, $b . C T=b^{2}-c^{2}$ and $c . B S=b^{2}-c^{2}$.
By Menelaus we know that $S T$ divides $B C$ in the ratio $A T . B S: A S . C T$ and it's all over bar the shouting.
If we are not lucky enough to have the stuff about the symmedian at our fingertips, we can still get essentially the same solution with a bit more work. We begin with the second half of the proof above, and establish that $S T$ divides $B C$ in the ratio $c^{2}: b^{2}$. Now we need to prove that $A L$ divides $B C$ in the same ratio.
The next (non-obvious!) step is to draw a parallel to $B C$ through $A$ as shown.
Now $\triangle A B C \sim \triangle B^{\prime} A B \sim \triangle C C^{\prime} A$ and the ratio $B^{\prime} A: A C^{\prime}$ which equals $B X: X C$ drops out as $c^{2}: b^{2}$ as required.
Clearly knowing the standard symmedian configuration and corresponding ratio is an enormous advantage.
Finally it is worth noting that, to the right sort of mind, the problem screams out for areal coordinates. These turn out to kill it fairly easily, not least because all three circles pass through at least two vertices of $\triangle A B C$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute triangle with with $A B<A C$ and let $\Gamma$ be its circumcircle. Let the tangents to $\Gamma$ at $B$ and $C$ be $t_{B}$ and $t_{C}$ respectively and let their point of intersection be $L$. The line through $B$ parallel to $A C$ intersects $t_{C}$ at $D$. The line through $C$ parallel to $A B$ intersects $t_{B}$ at $E$. The circumcircle of triangle $B D C$ meets the side $A C$ at $T$ where $T$ lies between $A$ and $C$. The circumcircle of triangle $B E C$ meets the line $A B$ at $S$ where $B$ lies between $A$ and $S$.
Prove that the lines $S T, B C$ and $A L$ are concurrent.
|
How we attack this problem depends on how much triangle geometry we can effortlessly recall - a good knowledge of some standard results helps a great deal.
We might instantly note that $A L$ is a symmedian of $A B C$, and so divides the line $B C$ in the ratio $c^{2}: b^{2}$. Now the plan is to show that $S T$ also divides $B C$ in this ratio.
Since we are working with ratios of distances, Menelaus' theorem may prove useful.
A key step is to notice (based on a careful diagram) that $A C$ is tangent to circle $C B S$. Once spotted this is easy to prove. The parallels give us $\angle B=\angle B C E$, and, since $B E$ is tangent to circle $A B C$, we have $\angle E B C=\angle A$ by the alternate segment theorem. Now angles in a triangle give $\angle C=\angle C E B$, and we have the converse to the alternate segment theorem. We obtain $A B$ is tangent to circle $C B D$ in the same way.
Now we have some tangencies and want some ratios.
Tangent-secant yields $A T . b=c^{2}$ and $c . A S=b^{2}$ or, equivalently, $b . C T=b^{2}-c^{2}$ and $c . B S=b^{2}-c^{2}$.
By Menelaus we know that $S T$ divides $B C$ in the ratio $A T . B S: A S . C T$ and it's all over bar the shouting.
If we are not lucky enough to have the stuff about the symmedian at our fingertips, we can still get essentially the same solution with a bit more work. We begin with the second half of the proof above, and establish that $S T$ divides $B C$ in the ratio $c^{2}: b^{2}$. Now we need to prove that $A L$ divides $B C$ in the same ratio.
The next (non-obvious!) step is to draw a parallel to $B C$ through $A$ as shown.
Now $\triangle A B C \sim \triangle B^{\prime} A B \sim \triangle C C^{\prime} A$ and the ratio $B^{\prime} A: A C^{\prime}$ which equals $B X: X C$ drops out as $c^{2}: b^{2}$ as required.
Clearly knowing the standard symmedian configuration and corresponding ratio is an enormous advantage.
Finally it is worth noting that, to the right sort of mind, the problem screams out for areal coordinates. These turn out to kill it fairly easily, not least because all three circles pass through at least two vertices of $\triangle A B C$.
|
{
"resource_path": "Balkan_MO/segmented/en-2017-BMO-type3.jsonl",
"problem_match": "\n2. ",
"solution_match": "\nSolution."
}
|
9c7cba3a-a820-596a-9776-75f0f389b914
| 604,502
|
Let $\mathbb{N}$ be the set of positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that:
$$
n+f(m) \text { divides } f(n)+n f(m)
$$
|
The striking thing about this problem is that the relation concerns divisibility rather than equality. How can we exploit this? We are given that $n+f(m) \mid f(n)+n f(m)$ but we can certainly add or subtract multiples of the left hand side from the right hand side and preserve the divisibility. This leads to a key idea:
'Eliminate one of the variables from the right hand side.'
Clearly $n+f(m) \mid f(n)+n f(m)-n(n+f(m))$ so for any $n, m$ we have
$$
n+f(m) \mid f(n)-n^{2}
$$
This feels like a strong condition: if we fix $n$ and let $f(m)$ go to infinity, then $f(n)-n^{2}$ will have arbitrarily large factors, which implies it must be zero.
We must be careful: this argument is fine, so long as the function $f$ takes arbitrarily large values. (We also need to check that $f(n)=n^{2}$ satisfies the original statement which it does.)
We are left with the case where $f$ takes only finitely many values.
In this case $f$ must take the same value infinitely often, so it is natural to focus on an infinite set $S \subset \mathbb{N}$ such that $f(s)=k$ for all $s \in S$. If $n, m \in S$ then the original statement gives $n+k \mid k+n k$ where $k$ is fixed and $n$ can be as large as we like.
Now we recycle our key idea and eliminate $n$ from the right.
$n+k \mid k+n k-k(n+k)$ so $n+k \mid k-k^{2}$ for arbitrarily large $n$. This means that $k-k^{2}=0$ so $k=1$, since it must be positive.
At this point we suspect that $f(n)=1$ for all $n$ is the only bounded solution, so we pick some $t$ such that $f(t)=L>1$ and try to get a contradiction.
In the original statement we can set $m=t$ and get $n+L \mid f(n)+n L$. Eliminating $L$ from the right gives us nothing new, so how can we proceed? Well, we have an infinite set $S$ such that $f$ is constantly 1 on $S$ so we can take $n \in S$ to obtain $n+L \mid 1+n L$
Using our key idea one more time and eliminating $n$ from the right, we get $n+L \mid 1-L^{2}$ for arbitrarily large $n$ which is impossible if $L>1$.
A rather different solution can be found by playing around with small values of $m$ and $n$.
As before it helps to establish $(\star)$ but now $(n, m)=(1,1)$ gives $1+f(1) \mid f(1)-1$.
The left is bigger than the right, so the right must be zero $-f(1)=1$.
Now try $(n, m)=(2,1)$ and obtain $2+f(2) \mid f(2)-4$. Subtracting the left from the right gives $2+f(2) \mid-6$. Since $f(2) \in \mathbb{N}$ the left is a factor of -6 which is bigger than 2 . This gives $f(2)=1$ or $f(2)=4$.
In the first case we can plug this back into the orginal statement to get $2+f(m) \mid 1+2 f(m)$. Now taking two copies of the left away from the right we have $2+f(m) \mid-3$.
Thus $2+f(m)$ must a factor of -3 which is bigger than 2 , so $f(m)=1$ for any $m$.
Before proceeding with the case $f(2)=4$ we take another look at our strong result $(\star)$. Setting $n=m$ gives $n+f(n) \mid f(n)-n^{2}$ so taking $f(n)-n^{2}$ away from $n+f(n)$ shows that
$$
n+f(n) \mid n+n^{2}
$$
Let see if we can use $(\star)$ and $(\dagger)$ to pin down the value of $f(3)$, using $f(2)=4$.
From $(\star)$ we have $3+4 \mid f(3)-9$ and from $(\dagger)$ we have $3+f(3) \mid 12$. The second of these shows $f(3)$ is 1,3 or 9 , but 1 and 3 are too small to work in the first relation.
Similarly, setting $(n, m)=(4,3)$ in $(\star)$ gives $4+9 \mid f(4)-16$ while $n=4$ in $(\dagger)$ gives $4+f(4) \mid 20$. The latter shows $f(4) \leq 16$ so $13 \mid 16-f(4)$. The only possible multiples of 13 are 0 and 13 , of which only the first one works. Thus $f(4)=16$.
Now we are ready to try induction. Assume $f(n-1)=(n-1)^{2}$ and use $(\star)$ and $(\dagger)$ to obtain $n+(n-1)^{2} \mid f(n)-n^{2}$ and $n+f(n) \mid n+n^{2}$. The latter implies $f(n) \leq n^{2}$ so the former becomes $n^{2}-n+1 \mid n^{2}-f(n)$. If $f(n) \neq n^{2}$ then $n^{2}-f(n)=1 \times\left(n^{2}-n+1\right)$ since any other multiple would be too large. However, putting $f(n)=n-1$ into $n+f(n) \mid n+n^{2}$ implies $2 n-1 \mid n(1+n)$. This is a contradiction since $2 n-1$ is coprime to $n$ and clearly cannot divide $1+n$.
for all $m, n \in \mathbb{N}$.
|
f(n)=n^2 \text{ or } f(n)=1
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $\mathbb{N}$ be the set of positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that:
$$
n+f(m) \text { divides } f(n)+n f(m)
$$
|
The striking thing about this problem is that the relation concerns divisibility rather than equality. How can we exploit this? We are given that $n+f(m) \mid f(n)+n f(m)$ but we can certainly add or subtract multiples of the left hand side from the right hand side and preserve the divisibility. This leads to a key idea:
'Eliminate one of the variables from the right hand side.'
Clearly $n+f(m) \mid f(n)+n f(m)-n(n+f(m))$ so for any $n, m$ we have
$$
n+f(m) \mid f(n)-n^{2}
$$
This feels like a strong condition: if we fix $n$ and let $f(m)$ go to infinity, then $f(n)-n^{2}$ will have arbitrarily large factors, which implies it must be zero.
We must be careful: this argument is fine, so long as the function $f$ takes arbitrarily large values. (We also need to check that $f(n)=n^{2}$ satisfies the original statement which it does.)
We are left with the case where $f$ takes only finitely many values.
In this case $f$ must take the same value infinitely often, so it is natural to focus on an infinite set $S \subset \mathbb{N}$ such that $f(s)=k$ for all $s \in S$. If $n, m \in S$ then the original statement gives $n+k \mid k+n k$ where $k$ is fixed and $n$ can be as large as we like.
Now we recycle our key idea and eliminate $n$ from the right.
$n+k \mid k+n k-k(n+k)$ so $n+k \mid k-k^{2}$ for arbitrarily large $n$. This means that $k-k^{2}=0$ so $k=1$, since it must be positive.
At this point we suspect that $f(n)=1$ for all $n$ is the only bounded solution, so we pick some $t$ such that $f(t)=L>1$ and try to get a contradiction.
In the original statement we can set $m=t$ and get $n+L \mid f(n)+n L$. Eliminating $L$ from the right gives us nothing new, so how can we proceed? Well, we have an infinite set $S$ such that $f$ is constantly 1 on $S$ so we can take $n \in S$ to obtain $n+L \mid 1+n L$
Using our key idea one more time and eliminating $n$ from the right, we get $n+L \mid 1-L^{2}$ for arbitrarily large $n$ which is impossible if $L>1$.
A rather different solution can be found by playing around with small values of $m$ and $n$.
As before it helps to establish $(\star)$ but now $(n, m)=(1,1)$ gives $1+f(1) \mid f(1)-1$.
The left is bigger than the right, so the right must be zero $-f(1)=1$.
Now try $(n, m)=(2,1)$ and obtain $2+f(2) \mid f(2)-4$. Subtracting the left from the right gives $2+f(2) \mid-6$. Since $f(2) \in \mathbb{N}$ the left is a factor of -6 which is bigger than 2 . This gives $f(2)=1$ or $f(2)=4$.
In the first case we can plug this back into the orginal statement to get $2+f(m) \mid 1+2 f(m)$. Now taking two copies of the left away from the right we have $2+f(m) \mid-3$.
Thus $2+f(m)$ must a factor of -3 which is bigger than 2 , so $f(m)=1$ for any $m$.
Before proceeding with the case $f(2)=4$ we take another look at our strong result $(\star)$. Setting $n=m$ gives $n+f(n) \mid f(n)-n^{2}$ so taking $f(n)-n^{2}$ away from $n+f(n)$ shows that
$$
n+f(n) \mid n+n^{2}
$$
Let see if we can use $(\star)$ and $(\dagger)$ to pin down the value of $f(3)$, using $f(2)=4$.
From $(\star)$ we have $3+4 \mid f(3)-9$ and from $(\dagger)$ we have $3+f(3) \mid 12$. The second of these shows $f(3)$ is 1,3 or 9 , but 1 and 3 are too small to work in the first relation.
Similarly, setting $(n, m)=(4,3)$ in $(\star)$ gives $4+9 \mid f(4)-16$ while $n=4$ in $(\dagger)$ gives $4+f(4) \mid 20$. The latter shows $f(4) \leq 16$ so $13 \mid 16-f(4)$. The only possible multiples of 13 are 0 and 13 , of which only the first one works. Thus $f(4)=16$.
Now we are ready to try induction. Assume $f(n-1)=(n-1)^{2}$ and use $(\star)$ and $(\dagger)$ to obtain $n+(n-1)^{2} \mid f(n)-n^{2}$ and $n+f(n) \mid n+n^{2}$. The latter implies $f(n) \leq n^{2}$ so the former becomes $n^{2}-n+1 \mid n^{2}-f(n)$. If $f(n) \neq n^{2}$ then $n^{2}-f(n)=1 \times\left(n^{2}-n+1\right)$ since any other multiple would be too large. However, putting $f(n)=n-1$ into $n+f(n) \mid n+n^{2}$ implies $2 n-1 \mid n(1+n)$. This is a contradiction since $2 n-1$ is coprime to $n$ and clearly cannot divide $1+n$.
for all $m, n \in \mathbb{N}$.
|
{
"resource_path": "Balkan_MO/segmented/en-2017-BMO-type3.jsonl",
"problem_match": "\n3. ",
"solution_match": "\nSolution."
}
|
1001d966-a6ed-532c-8aa8-db0c14e5bc28
| 604,515
|
There are $n>2$ students sitting at a round table. Initially each student has exactly one candy. At each step, each student chooses one of the following operations:
(a) Pass one candy to the student on their left or the student on their right.
(b) Divide all their candies into two, possibly empty, sets and pass one set to the student on their left and the other to the student on their right.
At each step the students perform their chosen operations simultaneously.
An arrangement of candies is legal if it can be obtained in a finite number of steps.
Find the number of legal arrangements.
(Two arrangements are different if there is a student who has different numbers of candies in each one.)
|
One possible initial reaction to this problem is that there is rather too much movement of caramels ${ }^{2}$ at each step to keep track of easily. This leads to the question: 'How little can I do in, say, two steps?' If every student passes all their caramels left on one step using (b), and all their caramels right on the next step, then no caramels move. (This is rather too little movement.) Let us see what a small change to this sequence can accomplish. We choose a student with at least one caramel. At the first step, she passes one caramel to the right and any others she has to the left. Every one else passes everything left. At the next step everybody passes everything right. The effect of this is that exactly one caramel has moved exactly two places to the right. Similarly, there is a double step which moves exactly one caramel to places to the left.
If we have not already done so, now is the time to start working through some small values of $n$.
The case $n=3$ yields a useful observation. Going two places (let's call this a double jump) to the left on a triangle is the same as going one place to the right. Indeed if $n$ is odd, say $2 k+1$, then $k$ double jumps to the left moves the caramel one place to the right and vice versa.
It is now clear that, if $n$ is odd, any arrangement of caramels is possible. We simply move them into position one at a time.
In the case $n=4$ it seems hard to get all the caramels into one place. Indeed, if we limit ourselves to double jumps, then we can only get $(1,1,1,1),(1,2,1,0),(2,2,0,0)$ and rotations of these arrangements. What can we say about these? Well it seems that students one and three always hold two candies between them. Having noticed this, it is not too hard to make a more general observation: if $n$ is even then a double jump cannot change the total number of caramels held by the odd numbered students. However, double jumps are not the only moves available to us. For example, it is possible to go from $(2,2,0,0)$ to $(3,1,0,0)$. A double jump now gives $(2,1,1,0)$ as well.
To squeeze maximum value out of the $n=4$ case, it is worth looking at the arrangements we have not yet managed to get to. They are (rotations of) $(4,0,0,0),(3,0,1,0)$ and $(2,0,2,0)$. What do these have in common? They are precisely the arrangements where the even numbered students hold all the caramels. Can we prove that these are illegal? Well, what can we say about an arrangement which precedes one of these elusive ones? This question leads to the last big idea in the solution to this problem. If after some step the even numbered students have all the caramels, they cannot have had any at all before the step, else they would have passed at least one caramel to an odd numbered student.
Turning this around gives a crucial lemma for even values of $n$. Let's call a caramel held by an odd numbered student an odd caramel and define even caramels similarly. Let's call an arrangement with at least one odd caramel and at least one even caramel balanced. If the arrangement is balanced before some step, then it will be balanced after the step. The initial position is balanced, so every legal position is balanced.
Finally we are on the home straight. We claim that every balanced position is legal. Using double jumps we can move to $\left(\ldots, \frac{n}{2}, \frac{n}{2}, 0, \ldots\right)$. Now we need to tinker with the numbers of odd and even caramels. There are lots of usable sequences. For example:
$$
\begin{gathered}
(\ldots, a, b, 0, \ldots) \\
(\ldots, a-1,1, b, \ldots) \\
(\ldots, a-1, b+1,0, \ldots)
\end{gathered}
$$
can be used to change the number of odd caramels provided $a-1 \geq 1$.
Once we have the correct number of odd and even caramels, they can be moved into place using double jumps.
It remains to observe that there are $\binom{2 n-1}{n}$ possible arrangements of caramels, and that if $n$ is even, then $2\left(\frac{\frac{3 n}{2}-1}{n}\right)$ of these are not balanced.
Another sensible approach is to think about which steps are reversible. It turns out that many are, including all those where the students all use option (b).
It is possible to argue that if $n$ is odd, then we can start with any position, move to $(\ldots, n, \ldots)$ reversibly, then move to the initial position reversibly. Playing the whole tape backwards shows all positions are legal.
If $n$ is even it is possible to start from any balanced position and reversibly move to $(\ldots, n-1,1, \ldots)$ and thence to the initial position so we are done.
[^0]
[^0]: ${ }^{2}$ The word 'candy' was a little too grating for my delicate British ears. I am grateful to the Italians for suggesting the more elegant alternative.
|
not found
|
Yes
|
Incomplete
|
math-word-problem
|
Combinatorics
|
There are $n>2$ students sitting at a round table. Initially each student has exactly one candy. At each step, each student chooses one of the following operations:
(a) Pass one candy to the student on their left or the student on their right.
(b) Divide all their candies into two, possibly empty, sets and pass one set to the student on their left and the other to the student on their right.
At each step the students perform their chosen operations simultaneously.
An arrangement of candies is legal if it can be obtained in a finite number of steps.
Find the number of legal arrangements.
(Two arrangements are different if there is a student who has different numbers of candies in each one.)
|
One possible initial reaction to this problem is that there is rather too much movement of caramels ${ }^{2}$ at each step to keep track of easily. This leads to the question: 'How little can I do in, say, two steps?' If every student passes all their caramels left on one step using (b), and all their caramels right on the next step, then no caramels move. (This is rather too little movement.) Let us see what a small change to this sequence can accomplish. We choose a student with at least one caramel. At the first step, she passes one caramel to the right and any others she has to the left. Every one else passes everything left. At the next step everybody passes everything right. The effect of this is that exactly one caramel has moved exactly two places to the right. Similarly, there is a double step which moves exactly one caramel to places to the left.
If we have not already done so, now is the time to start working through some small values of $n$.
The case $n=3$ yields a useful observation. Going two places (let's call this a double jump) to the left on a triangle is the same as going one place to the right. Indeed if $n$ is odd, say $2 k+1$, then $k$ double jumps to the left moves the caramel one place to the right and vice versa.
It is now clear that, if $n$ is odd, any arrangement of caramels is possible. We simply move them into position one at a time.
In the case $n=4$ it seems hard to get all the caramels into one place. Indeed, if we limit ourselves to double jumps, then we can only get $(1,1,1,1),(1,2,1,0),(2,2,0,0)$ and rotations of these arrangements. What can we say about these? Well it seems that students one and three always hold two candies between them. Having noticed this, it is not too hard to make a more general observation: if $n$ is even then a double jump cannot change the total number of caramels held by the odd numbered students. However, double jumps are not the only moves available to us. For example, it is possible to go from $(2,2,0,0)$ to $(3,1,0,0)$. A double jump now gives $(2,1,1,0)$ as well.
To squeeze maximum value out of the $n=4$ case, it is worth looking at the arrangements we have not yet managed to get to. They are (rotations of) $(4,0,0,0),(3,0,1,0)$ and $(2,0,2,0)$. What do these have in common? They are precisely the arrangements where the even numbered students hold all the caramels. Can we prove that these are illegal? Well, what can we say about an arrangement which precedes one of these elusive ones? This question leads to the last big idea in the solution to this problem. If after some step the even numbered students have all the caramels, they cannot have had any at all before the step, else they would have passed at least one caramel to an odd numbered student.
Turning this around gives a crucial lemma for even values of $n$. Let's call a caramel held by an odd numbered student an odd caramel and define even caramels similarly. Let's call an arrangement with at least one odd caramel and at least one even caramel balanced. If the arrangement is balanced before some step, then it will be balanced after the step. The initial position is balanced, so every legal position is balanced.
Finally we are on the home straight. We claim that every balanced position is legal. Using double jumps we can move to $\left(\ldots, \frac{n}{2}, \frac{n}{2}, 0, \ldots\right)$. Now we need to tinker with the numbers of odd and even caramels. There are lots of usable sequences. For example:
$$
\begin{gathered}
(\ldots, a, b, 0, \ldots) \\
(\ldots, a-1,1, b, \ldots) \\
(\ldots, a-1, b+1,0, \ldots)
\end{gathered}
$$
can be used to change the number of odd caramels provided $a-1 \geq 1$.
Once we have the correct number of odd and even caramels, they can be moved into place using double jumps.
It remains to observe that there are $\binom{2 n-1}{n}$ possible arrangements of caramels, and that if $n$ is even, then $2\left(\frac{\frac{3 n}{2}-1}{n}\right)$ of these are not balanced.
Another sensible approach is to think about which steps are reversible. It turns out that many are, including all those where the students all use option (b).
It is possible to argue that if $n$ is odd, then we can start with any position, move to $(\ldots, n, \ldots)$ reversibly, then move to the initial position reversibly. Playing the whole tape backwards shows all positions are legal.
If $n$ is even it is possible to start from any balanced position and reversibly move to $(\ldots, n-1,1, \ldots)$ and thence to the initial position so we are done.
[^0]
[^0]: ${ }^{2}$ The word 'candy' was a little too grating for my delicate British ears. I am grateful to the Italians for suggesting the more elegant alternative.
|
{
"resource_path": "Balkan_MO/segmented/en-2017-BMO-type3.jsonl",
"problem_match": "\n4. ",
"solution_match": "\nSolution."
}
|
126688ee-90a7-5f0a-80f8-da086c5c2141
| 604,526
|
A quadrilateral $A B C D$ is inscribed in a circle $k$, where $A B>C D$ and $A B$ is not parallel to $C D$. Point $M$ is the intersection of the diagonals $A C$ and $B D$ and the perpendicular from $M$ to $A B$ intersects the segment $A B$ at the point $E$. If $E M$ bisects the angle $C E D$, prove that $A B$ is a diameter of the circle $k$.
|
Let the line through $M$ parallel to $A B$ meet the segments $A D, D H, B C, C H$ at points $K, P, L, Q$, respectively. Triangle $H P Q$ is isosceles, so $M P=M Q$. Now from
$$
\frac{M P}{B H}=\frac{D M}{D B}=\frac{K M}{A B} \quad \text { and } \quad \frac{M Q}{A H}=\frac{C M}{C A}=\frac{M L}{A B}
$$
we obtain $A H / H B=K M / M L$.
Let the lines $A D$ and $B C$ meet at point $S$ and let the line $S M$ meet $A B$ at $H^{\prime}$. Then $A H^{\prime} / H^{\prime} B=K M / M L=A H / H B$, so $H^{\prime} \equiv H$, i.e. $S$ lies on the line $M H$.
The quadrilateral $A B C D$ is not a trapezoid, so $A H \neq B H$. Consider the point $A^{\prime}$ on the ray $H B$ such that $H A^{\prime}=H A$. Since $\varangle S A^{\prime} M=\varangle S A M=\varangle S B M$, quadrilateral $A^{\prime} B S M$ is cyclic and therefore $\varangle A B C=\varangle A^{\prime} B S=\varangle A^{\prime} M H=\varangle A M H=90^{\circ}-\varangle B A C$, which implies that $\varangle A C B=90^{\circ}$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
A quadrilateral $A B C D$ is inscribed in a circle $k$, where $A B>C D$ and $A B$ is not parallel to $C D$. Point $M$ is the intersection of the diagonals $A C$ and $B D$ and the perpendicular from $M$ to $A B$ intersects the segment $A B$ at the point $E$. If $E M$ bisects the angle $C E D$, prove that $A B$ is a diameter of the circle $k$.
|
Let the line through $M$ parallel to $A B$ meet the segments $A D, D H, B C, C H$ at points $K, P, L, Q$, respectively. Triangle $H P Q$ is isosceles, so $M P=M Q$. Now from
$$
\frac{M P}{B H}=\frac{D M}{D B}=\frac{K M}{A B} \quad \text { and } \quad \frac{M Q}{A H}=\frac{C M}{C A}=\frac{M L}{A B}
$$
we obtain $A H / H B=K M / M L$.
Let the lines $A D$ and $B C$ meet at point $S$ and let the line $S M$ meet $A B$ at $H^{\prime}$. Then $A H^{\prime} / H^{\prime} B=K M / M L=A H / H B$, so $H^{\prime} \equiv H$, i.e. $S$ lies on the line $M H$.
The quadrilateral $A B C D$ is not a trapezoid, so $A H \neq B H$. Consider the point $A^{\prime}$ on the ray $H B$ such that $H A^{\prime}=H A$. Since $\varangle S A^{\prime} M=\varangle S A M=\varangle S B M$, quadrilateral $A^{\prime} B S M$ is cyclic and therefore $\varangle A B C=\varangle A^{\prime} B S=\varangle A^{\prime} M H=\varangle A M H=90^{\circ}-\varangle B A C$, which implies that $\varangle A C B=90^{\circ}$.
|
{
"resource_path": "Balkan_MO/segmented/en-2018-BMO-type3.jsonl",
"problem_match": "\n1. ",
"solution_match": "\nSolution."
}
|
d890d45a-f52e-578b-98f3-207b038a0650
| 604,537
|
Let $q$ be a positive rational number. Two ants are initially at the same point $X$ in the plane. In the $n$-th minute $(n=1,2, \ldots)$ each of them chooses whether to walk due north, east, south or west and then walks the distance of $q^{n}$ metres. After a whole number of minutes, they are at the same point in the plane (not necessarily $X$ ), but have not taken exactly the same route within that time. Determine all possible values of $q$.
|
Answer: $q=1$.
Let $x_{A}^{(n)}$ (resp. $x_{B}^{(n)}$ ) be the $x$-coordinates of the first (resp. second) ant's position after $n$ minutes. Then $x_{A}^{(n)}-x_{A}^{(n-1)} \in\left\{q^{n},-q^{n}, 0\right\}$, and so $x_{A}^{(n)}, x_{B}^{(n)}$ are given by polynomials in $q$ with coefficients in $\{-1,0,1\}$. So if the ants meet after $n$ minutes, then
$$
0=x_{A}^{(n)}-x_{B}^{(n)}=P(q),
$$
where $P$ is a polynomial with degree at most $n$ and coefficients in $\{-2,-, 1,0,1,2\}$. Thus if $q=\frac{a}{b}(a, b \in \mathbb{N})$, we have $a \mid 2$ and $b \mid 2$, i.e. $q \in\left\{\frac{1}{2}, 1,2\right\}$.
It is clearly possible when $q=1$.
We argue that $q=\frac{1}{2}$ is not possible. Assume that the ants diverge for the first time after the $k$ th minute, for $k \geqslant 0$. Then
$$
\left|x_{B}^{(k+1)}-x_{A}^{(k+1)}\right|+\left|y_{B}^{(k+1)}-y_{A}^{(k+1)}\right|=2 q^{k} .
$$
But also $\left|x_{A}^{(\ell+1)}-x_{A}^{(\ell)}\right|+\left|y_{A}^{(\ell+1)}-y_{A}^{(\ell)}\right|=q^{\ell}$ for each $l \geqslant k+1$, and so
$$
\left|x_{A}^{(n)}-x_{A}^{(k+1)}\right|+\left|y_{A}^{(n)}-y_{A}^{(k+1)}\right| \leqslant q^{k+1}+q^{k+2}+\ldots+q^{n-1}
$$
and similarly for the second ant. Combining (1) and (2) with the triangle inequality, we obtain for any $n \geqslant k+1$
$$
\left|x_{B}^{(n)}-x_{A}^{(n)}\right|+\left|y_{B}^{(n)}-y_{A}^{(n)}\right| \geqslant 2 q^{k}-2\left(q^{k+1}+q^{k+2}+\ldots+q^{n-1}\right),
$$
which is strictly positive for $q=\frac{1}{2}$. So for any $n \geqslant k+1$, the ants cannot meet after $n$ minutes. Thus $q \neq \frac{1}{2}$.
Finally, we show that $q=2$ is also not possible. Suppose to the contrary that there is a pair of routes for $q=2$, meeting after $n$ minutes. Now consider rescaling the plane by a factor $2^{-n}$, and looking at the routes in the opposite direction. This would then be an example for $q=1 / 2$ and we have just shown that this is not possible.
|
q=1
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $q$ be a positive rational number. Two ants are initially at the same point $X$ in the plane. In the $n$-th minute $(n=1,2, \ldots)$ each of them chooses whether to walk due north, east, south or west and then walks the distance of $q^{n}$ metres. After a whole number of minutes, they are at the same point in the plane (not necessarily $X$ ), but have not taken exactly the same route within that time. Determine all possible values of $q$.
|
Answer: $q=1$.
Let $x_{A}^{(n)}$ (resp. $x_{B}^{(n)}$ ) be the $x$-coordinates of the first (resp. second) ant's position after $n$ minutes. Then $x_{A}^{(n)}-x_{A}^{(n-1)} \in\left\{q^{n},-q^{n}, 0\right\}$, and so $x_{A}^{(n)}, x_{B}^{(n)}$ are given by polynomials in $q$ with coefficients in $\{-1,0,1\}$. So if the ants meet after $n$ minutes, then
$$
0=x_{A}^{(n)}-x_{B}^{(n)}=P(q),
$$
where $P$ is a polynomial with degree at most $n$ and coefficients in $\{-2,-, 1,0,1,2\}$. Thus if $q=\frac{a}{b}(a, b \in \mathbb{N})$, we have $a \mid 2$ and $b \mid 2$, i.e. $q \in\left\{\frac{1}{2}, 1,2\right\}$.
It is clearly possible when $q=1$.
We argue that $q=\frac{1}{2}$ is not possible. Assume that the ants diverge for the first time after the $k$ th minute, for $k \geqslant 0$. Then
$$
\left|x_{B}^{(k+1)}-x_{A}^{(k+1)}\right|+\left|y_{B}^{(k+1)}-y_{A}^{(k+1)}\right|=2 q^{k} .
$$
But also $\left|x_{A}^{(\ell+1)}-x_{A}^{(\ell)}\right|+\left|y_{A}^{(\ell+1)}-y_{A}^{(\ell)}\right|=q^{\ell}$ for each $l \geqslant k+1$, and so
$$
\left|x_{A}^{(n)}-x_{A}^{(k+1)}\right|+\left|y_{A}^{(n)}-y_{A}^{(k+1)}\right| \leqslant q^{k+1}+q^{k+2}+\ldots+q^{n-1}
$$
and similarly for the second ant. Combining (1) and (2) with the triangle inequality, we obtain for any $n \geqslant k+1$
$$
\left|x_{B}^{(n)}-x_{A}^{(n)}\right|+\left|y_{B}^{(n)}-y_{A}^{(n)}\right| \geqslant 2 q^{k}-2\left(q^{k+1}+q^{k+2}+\ldots+q^{n-1}\right),
$$
which is strictly positive for $q=\frac{1}{2}$. So for any $n \geqslant k+1$, the ants cannot meet after $n$ minutes. Thus $q \neq \frac{1}{2}$.
Finally, we show that $q=2$ is also not possible. Suppose to the contrary that there is a pair of routes for $q=2$, meeting after $n$ minutes. Now consider rescaling the plane by a factor $2^{-n}$, and looking at the routes in the opposite direction. This would then be an example for $q=1 / 2$ and we have just shown that this is not possible.
|
{
"resource_path": "Balkan_MO/segmented/en-2018-BMO-type3.jsonl",
"problem_match": "\n2. ",
"solution_match": "\nSolution 1."
}
|
865bc88c-bcdf-5a35-8485-6e8f0e24f325
| 604,551
|
Let $q$ be a positive rational number. Two ants are initially at the same point $X$ in the plane. In the $n$-th minute $(n=1,2, \ldots)$ each of them chooses whether to walk due north, east, south or west and then walks the distance of $q^{n}$ metres. After a whole number of minutes, they are at the same point in the plane (not necessarily $X$ ), but have not taken exactly the same route within that time. Determine all possible values of $q$.
|
Consider the ants' positions $\alpha_{k}$ and $\beta_{k}$ after $k$ steps in the complex plane, assuming that their initial positions are at the origin and that all steps are parallel to one of the axes. We have $\alpha_{k+1}-\alpha_{k}=a_{k} q^{k}$ and $\beta_{k+1}-\beta_{k}=b_{k} q^{k}$ with $a_{k}, b_{k} \in\{1,-1, i,-i\}$.
If $\alpha_{n}=\beta_{n}$ for some $n>0$, then
$$
\sum_{k=0}^{n-1}\left(a_{k}-b_{k}\right) q^{k}=0, \quad \text { where } \quad a_{k}-b_{k} \in\{0, \pm 1 \pm i, \pm 2, \pm 2 i\}
$$
Note that the coefficient $a_{k}-b_{k}$ is always divisible by $1+i$ in Gaussian integers: indeed,
$$
c_{k}=\frac{a_{k}-b_{k}}{1+i} \in\{0, \pm 1, \pm i, \pm 1 \pm i\}
$$
Canceling $1+i$, we obtain $c_{0}+c_{1} q+\cdots+c_{n-1} q^{n-1}=0$. Therefore if $q=\frac{a}{b}(a, b \in \mathbb{N})$, we have $a \mid c_{0}$ and $b \mid c_{n-1}$ in Gaussian integers, which is only possible if $a=b=1$.
|
q=1
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Let $q$ be a positive rational number. Two ants are initially at the same point $X$ in the plane. In the $n$-th minute $(n=1,2, \ldots)$ each of them chooses whether to walk due north, east, south or west and then walks the distance of $q^{n}$ metres. After a whole number of minutes, they are at the same point in the plane (not necessarily $X$ ), but have not taken exactly the same route within that time. Determine all possible values of $q$.
|
Consider the ants' positions $\alpha_{k}$ and $\beta_{k}$ after $k$ steps in the complex plane, assuming that their initial positions are at the origin and that all steps are parallel to one of the axes. We have $\alpha_{k+1}-\alpha_{k}=a_{k} q^{k}$ and $\beta_{k+1}-\beta_{k}=b_{k} q^{k}$ with $a_{k}, b_{k} \in\{1,-1, i,-i\}$.
If $\alpha_{n}=\beta_{n}$ for some $n>0$, then
$$
\sum_{k=0}^{n-1}\left(a_{k}-b_{k}\right) q^{k}=0, \quad \text { where } \quad a_{k}-b_{k} \in\{0, \pm 1 \pm i, \pm 2, \pm 2 i\}
$$
Note that the coefficient $a_{k}-b_{k}$ is always divisible by $1+i$ in Gaussian integers: indeed,
$$
c_{k}=\frac{a_{k}-b_{k}}{1+i} \in\{0, \pm 1, \pm i, \pm 1 \pm i\}
$$
Canceling $1+i$, we obtain $c_{0}+c_{1} q+\cdots+c_{n-1} q^{n-1}=0$. Therefore if $q=\frac{a}{b}(a, b \in \mathbb{N})$, we have $a \mid c_{0}$ and $b \mid c_{n-1}$ in Gaussian integers, which is only possible if $a=b=1$.
|
{
"resource_path": "Balkan_MO/segmented/en-2018-BMO-type3.jsonl",
"problem_match": "\n2. ",
"solution_match": "\nSolution 2."
}
|
865bc88c-bcdf-5a35-8485-6e8f0e24f325
| 604,551
|
Alice and Bob play the following game: They start with two non-empty piles of coins. Taking turns, with Alice playing first, each player chooses a pile with an even number of coins and moves half of the coins of this pile to the other pile. The game ends if a player cannot move, in which case the other player wins.
Determine all pairs $(a, b)$ of positive integers such that if initially the two piles have $a$ and $b$ coins respectively, then Bob has a winning strategy.
|
By $v_{2}(n)$ we denote the largest nonnegative integer $r$ such that $2^{r} \mid n$.
A position $(a, b)$ (i.e. two piles of sizes $a$ and $b$ ) is said to be $k$-happy if $v_{2}(a)=v_{2}(b)=k$ for some integer $k \geqslant 0$, and $k$-unhappy if $\min \left\{v_{2}(a), v_{2}(b)\right\}=k<\max \left\{v_{2}(a), v_{2}(b)\right\}$. We shall prove that Bob has a winning strategy if and only if the initial position is $k$-happy for some even $k$.
- Given a 0-happy position, the player in turn is unable to play and loses.
- Given a $k$-happy position $(a, b)$ with $k \geqslant 1$, the player in turn will transform it into one of the positions $\left(a+\frac{1}{2} b, \frac{1}{2} b\right)$ and $\left(b+\frac{1}{2} a, \frac{1}{2} a\right)$, both of which are ( $\left.k-1\right)$-happy because $v_{2}\left(a+\frac{1}{2} b\right)=v_{2}\left(\frac{1}{2} b\right)=v_{2}\left(b+\frac{1}{2} a\right)=v_{2}\left(\frac{1}{2} a\right)=k-1$.
Therefore, if the starting position is $k$-happy, after $k$ moves they will get stuck at a 0 -happy position, so Bob will win if and only if $k$ is even.
- Given a $k$-unhappy position $(a, b)$ with $k$ odd and $v_{2}(a)=k<v_{2}(b)=\ell$, Alice can move to position $\left(\frac{1}{2} a, b+\frac{1}{2} a\right)$. Since $v_{2}\left(\frac{1}{2} a\right)=v_{2}\left(b+\frac{1}{2} a\right)=k-1$, this position is ( $k-1$ )-happy with $2 \mid k-1$, so Alice will win.
- Given a $k$-unhappy position $(a, b)$ with $k$ even and $v_{2}(a)=k<v_{2}(b)=\ell$, Alice must not play to position $\left(\frac{1}{2} a, b+\frac{1}{2} a\right)$, because the new position is ( $\left.k-1\right)$-happy and will lead to Bob's victory. Thus she must play to position $\left(a+\frac{1}{2} b, \frac{1}{2} b\right)$. We claim that this position is also $k$-unhappy. Indeed, if $\ell>k+1$, then $v_{2}\left(a+\frac{1}{2} b\right)=$ $k<v_{2}\left(\frac{1}{2} b\right)=\ell-1$, whereas if $\ell=k+1$, then $v_{2}\left(a+\frac{1}{2} b\right)>v_{2}\left(\frac{1}{2} b\right)=k$.
Therefore a $k$-unhappy position is winning for Alice if $k$ is odd, and drawing if $k$ is even.
|
proof
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Alice and Bob play the following game: They start with two non-empty piles of coins. Taking turns, with Alice playing first, each player chooses a pile with an even number of coins and moves half of the coins of this pile to the other pile. The game ends if a player cannot move, in which case the other player wins.
Determine all pairs $(a, b)$ of positive integers such that if initially the two piles have $a$ and $b$ coins respectively, then Bob has a winning strategy.
|
By $v_{2}(n)$ we denote the largest nonnegative integer $r$ such that $2^{r} \mid n$.
A position $(a, b)$ (i.e. two piles of sizes $a$ and $b$ ) is said to be $k$-happy if $v_{2}(a)=v_{2}(b)=k$ for some integer $k \geqslant 0$, and $k$-unhappy if $\min \left\{v_{2}(a), v_{2}(b)\right\}=k<\max \left\{v_{2}(a), v_{2}(b)\right\}$. We shall prove that Bob has a winning strategy if and only if the initial position is $k$-happy for some even $k$.
- Given a 0-happy position, the player in turn is unable to play and loses.
- Given a $k$-happy position $(a, b)$ with $k \geqslant 1$, the player in turn will transform it into one of the positions $\left(a+\frac{1}{2} b, \frac{1}{2} b\right)$ and $\left(b+\frac{1}{2} a, \frac{1}{2} a\right)$, both of which are ( $\left.k-1\right)$-happy because $v_{2}\left(a+\frac{1}{2} b\right)=v_{2}\left(\frac{1}{2} b\right)=v_{2}\left(b+\frac{1}{2} a\right)=v_{2}\left(\frac{1}{2} a\right)=k-1$.
Therefore, if the starting position is $k$-happy, after $k$ moves they will get stuck at a 0 -happy position, so Bob will win if and only if $k$ is even.
- Given a $k$-unhappy position $(a, b)$ with $k$ odd and $v_{2}(a)=k<v_{2}(b)=\ell$, Alice can move to position $\left(\frac{1}{2} a, b+\frac{1}{2} a\right)$. Since $v_{2}\left(\frac{1}{2} a\right)=v_{2}\left(b+\frac{1}{2} a\right)=k-1$, this position is ( $k-1$ )-happy with $2 \mid k-1$, so Alice will win.
- Given a $k$-unhappy position $(a, b)$ with $k$ even and $v_{2}(a)=k<v_{2}(b)=\ell$, Alice must not play to position $\left(\frac{1}{2} a, b+\frac{1}{2} a\right)$, because the new position is ( $\left.k-1\right)$-happy and will lead to Bob's victory. Thus she must play to position $\left(a+\frac{1}{2} b, \frac{1}{2} b\right)$. We claim that this position is also $k$-unhappy. Indeed, if $\ell>k+1$, then $v_{2}\left(a+\frac{1}{2} b\right)=$ $k<v_{2}\left(\frac{1}{2} b\right)=\ell-1$, whereas if $\ell=k+1$, then $v_{2}\left(a+\frac{1}{2} b\right)>v_{2}\left(\frac{1}{2} b\right)=k$.
Therefore a $k$-unhappy position is winning for Alice if $k$ is odd, and drawing if $k$ is even.
|
{
"resource_path": "Balkan_MO/segmented/en-2018-BMO-type3.jsonl",
"problem_match": "\n3. ",
"solution_match": "\nSolution."
}
|
c00a2050-1db7-5b63-920d-83adb481e740
| 604,574
|
Find all primes $p$ and $q$ such that $3 p^{q-1}+1$ divides $11^{p}+17^{p}$.
Time allowed: 4 hours and 30 minutes.
Each problem is worth 10 points.
|
Answer: $(p, q)=(3,3)$.
For $p=2$ it is directly checked that there are no solutions. Assume that $p>2$.
Observe that $N=11^{p}+17^{p} \equiv 4(\bmod 8)$, so $8 \nmid 3 p^{q-1}+1>4$. Consider an odd prime divisor $r$ of $3 p^{q-1}+1$. Obviously, $r \notin\{3,11,17\}$. There exists $b$ such that $17 b \equiv 1$ $(\bmod r)$. Then $r \mid b^{p} N \equiv a^{p}+1(\bmod r)$, where $a=11 b$. Thus $r \mid a^{2 p}-1$, but $r \nmid a^{p}-1$, which means that $\operatorname{ord}_{r}(a) \mid 2 p$ and $\operatorname{ord}_{r}(a) \nmid p$, i.e. $\operatorname{ord}_{r}(a) \in\{2,2 p\}$.
Note that if $\operatorname{ord}_{r}(a)=2$, then $r \mid a^{2}-1 \equiv\left(11^{2}-17^{2}\right) b^{2}(\bmod r)$, which gives $r=7$ as the only possibility. On the other hand, $\operatorname{ord}_{r}(a)=2 p$ implies $2 p \mid r-1$. Thus, all prime divisors of $3 p^{q-1}+1$ other than 2 or 7 are congruent to 1 modulo $2 p$, i.e.
$$
3 p^{q-1}+1=2^{\alpha} 7^{\beta} p_{1}^{\gamma_{1}} \cdots p_{k}^{\gamma_{k}}
$$
where $p_{i} \notin\{2,7\}$ are prime divisors with $p_{i} \equiv 1(\bmod 2 p)$.
We already know that $\alpha \leqslant 2$. Also, note that
$$
\frac{11^{p}+17^{p}}{28}=11^{p-1}-11^{p-2} 17+11^{p-3} 17^{2}-\cdots+17^{p-1} \equiv p \cdot 4^{p-1} \quad(\bmod 7)
$$
so $11^{p}+17^{p}$ is not divisible by $7^{2}$ and hence $\beta \leqslant 1$.
If $q=2$, then $(*)$ becomes $3 p+1=2^{\alpha} 7^{\beta} p_{1}^{\gamma_{1}} \cdots p_{k}^{\gamma_{k}}$, but $p_{i} \geqslant 2 p+1$, which is only possible if $\gamma_{i}=0$ for all $i$, i.e. $3 p+1=2^{\alpha} 7^{\beta} \in\{2,4,14,28\}$, which gives us no solutions.
Thus $q>2$, which implies $4 \mid 3 p^{q-1}+1$, i.e. $\alpha=2$. Now the right hand side of $(*)$ is congruent to 4 or 28 modulo $p$, which gives us $p=3$. Consequently $3^{q}+1 \mid 6244$, which is only possible for $q=3$. The pair $(p, q)=(3,3)$ is indeed a solution.
|
(p, q)=(3,3)
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Find all primes $p$ and $q$ such that $3 p^{q-1}+1$ divides $11^{p}+17^{p}$.
Time allowed: 4 hours and 30 minutes.
Each problem is worth 10 points.
|
Answer: $(p, q)=(3,3)$.
For $p=2$ it is directly checked that there are no solutions. Assume that $p>2$.
Observe that $N=11^{p}+17^{p} \equiv 4(\bmod 8)$, so $8 \nmid 3 p^{q-1}+1>4$. Consider an odd prime divisor $r$ of $3 p^{q-1}+1$. Obviously, $r \notin\{3,11,17\}$. There exists $b$ such that $17 b \equiv 1$ $(\bmod r)$. Then $r \mid b^{p} N \equiv a^{p}+1(\bmod r)$, where $a=11 b$. Thus $r \mid a^{2 p}-1$, but $r \nmid a^{p}-1$, which means that $\operatorname{ord}_{r}(a) \mid 2 p$ and $\operatorname{ord}_{r}(a) \nmid p$, i.e. $\operatorname{ord}_{r}(a) \in\{2,2 p\}$.
Note that if $\operatorname{ord}_{r}(a)=2$, then $r \mid a^{2}-1 \equiv\left(11^{2}-17^{2}\right) b^{2}(\bmod r)$, which gives $r=7$ as the only possibility. On the other hand, $\operatorname{ord}_{r}(a)=2 p$ implies $2 p \mid r-1$. Thus, all prime divisors of $3 p^{q-1}+1$ other than 2 or 7 are congruent to 1 modulo $2 p$, i.e.
$$
3 p^{q-1}+1=2^{\alpha} 7^{\beta} p_{1}^{\gamma_{1}} \cdots p_{k}^{\gamma_{k}}
$$
where $p_{i} \notin\{2,7\}$ are prime divisors with $p_{i} \equiv 1(\bmod 2 p)$.
We already know that $\alpha \leqslant 2$. Also, note that
$$
\frac{11^{p}+17^{p}}{28}=11^{p-1}-11^{p-2} 17+11^{p-3} 17^{2}-\cdots+17^{p-1} \equiv p \cdot 4^{p-1} \quad(\bmod 7)
$$
so $11^{p}+17^{p}$ is not divisible by $7^{2}$ and hence $\beta \leqslant 1$.
If $q=2$, then $(*)$ becomes $3 p+1=2^{\alpha} 7^{\beta} p_{1}^{\gamma_{1}} \cdots p_{k}^{\gamma_{k}}$, but $p_{i} \geqslant 2 p+1$, which is only possible if $\gamma_{i}=0$ for all $i$, i.e. $3 p+1=2^{\alpha} 7^{\beta} \in\{2,4,14,28\}$, which gives us no solutions.
Thus $q>2$, which implies $4 \mid 3 p^{q-1}+1$, i.e. $\alpha=2$. Now the right hand side of $(*)$ is congruent to 4 or 28 modulo $p$, which gives us $p=3$. Consequently $3^{q}+1 \mid 6244$, which is only possible for $q=3$. The pair $(p, q)=(3,3)$ is indeed a solution.
|
{
"resource_path": "Balkan_MO/segmented/en-2018-BMO-type3.jsonl",
"problem_match": "\n4. ",
"solution_match": "\nSolution."
}
|
fa69d6b3-8d46-5c04-83fe-27bca7d5c608
| 604,588
|
Let $\mathbb{P}$ be the set of all prime numbers. Find all functions $f: \mathbb{P} \rightarrow \mathbb{P}$ such that
$$
f(p)^{f(q)}+q^{p}=f(q)^{f(p)}+p^{q}
$$
holds for all $p, q \in \mathbb{P}$.
Proposed by Albania
|
Obviously, the identical function $f(p)=p$ for all $p \in \mathbb{P}$ is a solution. We will show that this is the only one.
First we will show that $f(2)=2$. Taking $q=2$ and $p$ any odd prime number, we have
$$
f(p)^{f(2)}+2^{p}=f(2)^{f(p)}+p^{2}
$$
Assume that $f(2) \neq 2$. It follows that $f(2)$ is odd and so $f(p)=2$ for any odd prime number $p$.
Taking any two different odd prime numbers $p, q$ we have
$$
2^{2}+q^{p}=2^{2}+p^{q} \Rightarrow p^{q}=q^{p} \Rightarrow p=q,
$$
contradiction. Hence, $f(2)=2$.
So for any odd prime number $p$ we have
$$
f(p)^{2}+2^{p}=2^{f(p)}+p^{2} .
$$
Copy this relation as
$$
2^{p}-p^{2}=2^{f(p)}-f(p)^{2}
$$
Let $T$ be the set of all positive integers greater than 2 , i.e. $T=\{3,4,5, \ldots\}$. The function $g: T \rightarrow \mathbb{Z}, g(n)=2^{n}-n^{2}$, is strictly increasing, i.e.
$$
g(n+1)-g(n)=2^{n}-2 n-1>0
$$
for all $n \in T$. We show this by induction. Indeed, for $n=3$ it is true, $2^{3}-2 \cdot 3-1>0$. Assume that $2^{k}-2 k-1>0$. It follows that for $n=k+1$ we have
$$
2^{k+1}-2(k+1)-1=\left(2^{k}-2 k-1\right)+\left(2^{k}-2\right)>0
$$
for any $k \geq 3$. Therefore, (2) is true for all $n \in T$.
As consequence, (1) holds if and only if $f(p)=p$ for all odd prime numbers $p$, as well as for $p=2$.
Therefore, the only function that satisfies the given relation is $f(p)=p$, for all $p \in \mathbb{P}$.
|
f(p)=p
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $\mathbb{P}$ be the set of all prime numbers. Find all functions $f: \mathbb{P} \rightarrow \mathbb{P}$ such that
$$
f(p)^{f(q)}+q^{p}=f(q)^{f(p)}+p^{q}
$$
holds for all $p, q \in \mathbb{P}$.
Proposed by Albania
|
Obviously, the identical function $f(p)=p$ for all $p \in \mathbb{P}$ is a solution. We will show that this is the only one.
First we will show that $f(2)=2$. Taking $q=2$ and $p$ any odd prime number, we have
$$
f(p)^{f(2)}+2^{p}=f(2)^{f(p)}+p^{2}
$$
Assume that $f(2) \neq 2$. It follows that $f(2)$ is odd and so $f(p)=2$ for any odd prime number $p$.
Taking any two different odd prime numbers $p, q$ we have
$$
2^{2}+q^{p}=2^{2}+p^{q} \Rightarrow p^{q}=q^{p} \Rightarrow p=q,
$$
contradiction. Hence, $f(2)=2$.
So for any odd prime number $p$ we have
$$
f(p)^{2}+2^{p}=2^{f(p)}+p^{2} .
$$
Copy this relation as
$$
2^{p}-p^{2}=2^{f(p)}-f(p)^{2}
$$
Let $T$ be the set of all positive integers greater than 2 , i.e. $T=\{3,4,5, \ldots\}$. The function $g: T \rightarrow \mathbb{Z}, g(n)=2^{n}-n^{2}$, is strictly increasing, i.e.
$$
g(n+1)-g(n)=2^{n}-2 n-1>0
$$
for all $n \in T$. We show this by induction. Indeed, for $n=3$ it is true, $2^{3}-2 \cdot 3-1>0$. Assume that $2^{k}-2 k-1>0$. It follows that for $n=k+1$ we have
$$
2^{k+1}-2(k+1)-1=\left(2^{k}-2 k-1\right)+\left(2^{k}-2\right)>0
$$
for any $k \geq 3$. Therefore, (2) is true for all $n \in T$.
As consequence, (1) holds if and only if $f(p)=p$ for all odd prime numbers $p$, as well as for $p=2$.
Therefore, the only function that satisfies the given relation is $f(p)=p$, for all $p \in \mathbb{P}$.
|
{
"resource_path": "Balkan_MO/segmented/en-2019-BMO-type1.jsonl",
"problem_match": "# Problem 1.",
"solution_match": "\nSolution."
}
|
50d68342-be3e-5654-a97c-5436c378a90a
| 604,604
|
Let $a, b, c$ be real numbers, such that $0 \leq a \leq b \leq c$ and $a+b+c=a b+b c+c a>0$. Prove that $\sqrt{b c}(a+1) \geq 2$. Find all triples $(a, b, c)$ for which equality holds.
## Proposed by Romania
|
Let $a+b+c=a b+b c+c a=k$. Since $(a+b+c)^{2} \geq 3(a b+b c+c a)$, we get that $k^{2} \geq 3 k$. Since $k>0$, we obtain that $k \geq 3$.
We have $b c \geq c a \geq a b$, so from the above relation we deduce that $b c \geq 1$.
By AM-GM, $b+c \geq 2 \sqrt{b c}$ and consequently $b+c \geq 2$. The equality holds iff $b=c$.
The constraint gives us
$$
a=\frac{b+c-b c}{b+c-1}=1-\frac{b c-1}{b+c-1} \geq 1-\frac{b c-1}{2 \sqrt{b c}-1}=\frac{\sqrt{b c}(2-\sqrt{b c})}{2 \sqrt{b c}-1}
$$
For $\sqrt{b c}=2$ condition $a \geq 0$ gives $\sqrt{b c}(a+1) \geq 2$ with equality iff $a=0$ and $b=c=2$. For $\sqrt{b c}<2$, taking into account the estimation for $a$, we get
$$
a \sqrt{b c} \geq \frac{b c(2-\sqrt{b c})}{2 \sqrt{b c}-1}=\frac{b c}{2 \sqrt{b c}-1}(2-\sqrt{b c}) .
$$
Since $\frac{b c}{2 \sqrt{b c}-1} \geq 1$, with equality for $b c=1$, we get $\sqrt{b c}(a+1) \geq 2$ with equality iff $a=b=c=1$.
For $\sqrt{b c}>2$ we have $\sqrt{b c}(a+1)>2(a+1) \geq 2$.
The proof is complete.
The equality holds iff $a=b=c=1$ or $a=0$ and $b=c=2$.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $a, b, c$ be real numbers, such that $0 \leq a \leq b \leq c$ and $a+b+c=a b+b c+c a>0$. Prove that $\sqrt{b c}(a+1) \geq 2$. Find all triples $(a, b, c)$ for which equality holds.
## Proposed by Romania
|
Let $a+b+c=a b+b c+c a=k$. Since $(a+b+c)^{2} \geq 3(a b+b c+c a)$, we get that $k^{2} \geq 3 k$. Since $k>0$, we obtain that $k \geq 3$.
We have $b c \geq c a \geq a b$, so from the above relation we deduce that $b c \geq 1$.
By AM-GM, $b+c \geq 2 \sqrt{b c}$ and consequently $b+c \geq 2$. The equality holds iff $b=c$.
The constraint gives us
$$
a=\frac{b+c-b c}{b+c-1}=1-\frac{b c-1}{b+c-1} \geq 1-\frac{b c-1}{2 \sqrt{b c}-1}=\frac{\sqrt{b c}(2-\sqrt{b c})}{2 \sqrt{b c}-1}
$$
For $\sqrt{b c}=2$ condition $a \geq 0$ gives $\sqrt{b c}(a+1) \geq 2$ with equality iff $a=0$ and $b=c=2$. For $\sqrt{b c}<2$, taking into account the estimation for $a$, we get
$$
a \sqrt{b c} \geq \frac{b c(2-\sqrt{b c})}{2 \sqrt{b c}-1}=\frac{b c}{2 \sqrt{b c}-1}(2-\sqrt{b c}) .
$$
Since $\frac{b c}{2 \sqrt{b c}-1} \geq 1$, with equality for $b c=1$, we get $\sqrt{b c}(a+1) \geq 2$ with equality iff $a=b=c=1$.
For $\sqrt{b c}>2$ we have $\sqrt{b c}(a+1)>2(a+1) \geq 2$.
The proof is complete.
The equality holds iff $a=b=c=1$ or $a=0$ and $b=c=2$.
|
{
"resource_path": "Balkan_MO/segmented/en-2019-BMO-type1.jsonl",
"problem_match": "# Problem 2.",
"solution_match": "\nSolution."
}
|
1b5e20f6-1fa4-5adc-b5cd-0ffc39682953
| 604,617
|
Let $A B C$ be an acute scalene triangle. Let $X$ and $Y$ be two distinct interior points of the segment $B C$ such that $\angle C A X=\angle Y A B$. Suppose that:
1) $K$ and $S$ are the feet of perpendiculars from $B$ to the lines $A X$ and $A Y$ respectively;
2) $T$ and $L$ are the feet of perpendiculars from $C$ to the lines $A X$ and $A Y$ respectively. Prove that $K L$ and $S T$ intersect on the line $B C$.
## Proposed by Greece
|
Denote $\phi=\widehat{X A B}=\widehat{Y A C}, \alpha=\widehat{C A X}=\widehat{B A Y}$. Then, because the quadrilaterals ABSK and ACTL are cyclic, we have
$$
\widehat{B S K}+\widehat{B A K}=180^{\circ}=\widehat{B S K}+\phi=\widehat{L A C}+\widehat{L T C}=\widehat{L T C}+\phi
$$
so, due to the 90-degree angles formed, we have $\widehat{K S L}=\widehat{K T L}$. Thus, KLST is cyclic.
Figure 1: G6
Consider $M$ to be the midpoint of $B C$ and $K^{\prime}$ to be the symmetric point of $K$ with respect to $M$. Then, $B K C K^{\prime}$ is a parallelogram, and so $B K \| C K^{\prime}$. But $B K \| C T$, because they are both perpendicular to $A X$. So, $K^{\prime}$ lies on $C T$ and, as $\widehat{K T K^{\prime}}=90$ and $M$ is the midpoint of $K K^{\prime}, M K=M T$. In a similar way, we have that $M S=M L$. Thus, the center of $(K L S T)$ is $M$.
Consider $D$ to be the foot of altitude from $A$ to $B C$. Then, $D$ belongs in both $(A B K S)$ and $(A C L T)$. So,
$$
\widehat{A D T}+\widehat{A C T}=180^{\circ}=\widehat{A B S}+\widehat{A D S}=\widehat{A D T}+90^{\circ}-\alpha=\widehat{A D S}+90^{\circ}-\alpha
$$
and $A D$ is the bisector of $\widehat{S D T}$.
Because $D M$ is perpendicular to $A D, D M$ is the external bisector of this angle, and, as $M S=M T$, it follows that $D M S T$ is cyclic. In a similar way, we have that $D M L K$ is also cyclic.
So, we have that $S T, K L$ and $D M$ are the radical axes of these three circles, $(K L S T)$, $(D M S T),(D M K L)$. These lines are, therefore, concurrent, and we have proved the desired result.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute scalene triangle. Let $X$ and $Y$ be two distinct interior points of the segment $B C$ such that $\angle C A X=\angle Y A B$. Suppose that:
1) $K$ and $S$ are the feet of perpendiculars from $B$ to the lines $A X$ and $A Y$ respectively;
2) $T$ and $L$ are the feet of perpendiculars from $C$ to the lines $A X$ and $A Y$ respectively. Prove that $K L$ and $S T$ intersect on the line $B C$.
## Proposed by Greece
|
Denote $\phi=\widehat{X A B}=\widehat{Y A C}, \alpha=\widehat{C A X}=\widehat{B A Y}$. Then, because the quadrilaterals ABSK and ACTL are cyclic, we have
$$
\widehat{B S K}+\widehat{B A K}=180^{\circ}=\widehat{B S K}+\phi=\widehat{L A C}+\widehat{L T C}=\widehat{L T C}+\phi
$$
so, due to the 90-degree angles formed, we have $\widehat{K S L}=\widehat{K T L}$. Thus, KLST is cyclic.
Figure 1: G6
Consider $M$ to be the midpoint of $B C$ and $K^{\prime}$ to be the symmetric point of $K$ with respect to $M$. Then, $B K C K^{\prime}$ is a parallelogram, and so $B K \| C K^{\prime}$. But $B K \| C T$, because they are both perpendicular to $A X$. So, $K^{\prime}$ lies on $C T$ and, as $\widehat{K T K^{\prime}}=90$ and $M$ is the midpoint of $K K^{\prime}, M K=M T$. In a similar way, we have that $M S=M L$. Thus, the center of $(K L S T)$ is $M$.
Consider $D$ to be the foot of altitude from $A$ to $B C$. Then, $D$ belongs in both $(A B K S)$ and $(A C L T)$. So,
$$
\widehat{A D T}+\widehat{A C T}=180^{\circ}=\widehat{A B S}+\widehat{A D S}=\widehat{A D T}+90^{\circ}-\alpha=\widehat{A D S}+90^{\circ}-\alpha
$$
and $A D$ is the bisector of $\widehat{S D T}$.
Because $D M$ is perpendicular to $A D, D M$ is the external bisector of this angle, and, as $M S=M T$, it follows that $D M S T$ is cyclic. In a similar way, we have that $D M L K$ is also cyclic.
So, we have that $S T, K L$ and $D M$ are the radical axes of these three circles, $(K L S T)$, $(D M S T),(D M K L)$. These lines are, therefore, concurrent, and we have proved the desired result.
|
{
"resource_path": "Balkan_MO/segmented/en-2019-BMO-type1.jsonl",
"problem_match": "# Problem 3.",
"solution_match": "\nSolution."
}
|
2c00dbbb-d31f-5fc0-a1eb-f5b20853f219
| 604,628
|
A grid consists of all points of the form $(m, n)$ where $m$ and $n$ are integers with $|m| \leqslant 2019$, $|n| \leqslant 2019$ and $|m|+|n|<4038$. We call the points $(m, n)$ of the grid with either $|m|=2019$ or $|n|=2019$ the boundary points. The four lines $x= \pm 2019$ and $y= \pm 2019$ are called boundary lines. Two points in the grid are called neighbours if the distance between them is equal to 1 .
Anna and Bob play a game on this grid.
Anna starts with a token at the point $(0,0)$. They take turns, with Bob playing first.
1) On each of his turns, Bob deletes at most two boundary points on each boundary line.
2) On each of her turns, Anna makes exactly three steps, where a step consists of moving her token from its current point to any neighbouring point which has not been deleted.
As soon as Anna places her token on some boundary point which has not been deleted, the game is over and Anna wins.
Does Anna have a winning strategy?
Proposed by Cyprus
|
Anna does not have a winning strategy. We will provide a winning strategy for Bob. It is enough to describe his strategy for the deletions on the line $y=2019$.
Bob starts by deleting $(0,2019)$ and $(-1,2019)$. Once Anna completes her turn, he deletes the next two available points on the left if Anna decreased her $x$-coordinate, the next two available points on the right if Anna increased her $x$-coordinate, and the next available point to the left and the next available point to the right if Anna did not change her $x$-coordinate. The only exception to the above rule is on the very first time Anna decreases $x$ by exactly 1 . In that turn, Bob deletes the next available point to the left and the next available point to the right.
Bob's strategy guarantees the following: If Anna makes a sequence of steps reaching $(-x, y)$ with $x>0$ and the exact opposite sequence of steps in the horizontal direction reaching $(x, y)$, then Bob deletes at least as many points to the left of $(0,2019)$ in the first sequence than points to the right of $(0,2019)$ in the second sequence.
So we may assume for contradiction that Anna wins by placing her token at $(k, 2019)$ for some $k>0$.
Define $\Delta=3 m-(2 x+y)$, where $m$ is the total number of points deleted by Bob to the right of $(0,2019)$, and $(x, y)$ is the position of Anna's token.
For each sequence of steps performed first by Anna and then by Bob, $\Delta$ does not decrease. This can be seen by looking at the following table exhibiting the changes in 3 m and $2 x+y$. We have excluded the cases where $2 x+y<0$.
| Turn | $(0,3)$ | $(1,2)$ | $(-1,2)$ | $(2,1)$ | $(0,1)$ | $(3,0)$ | $(1,0)$ | $(2,-1)$ | $(1,-2)$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $m$ | 1 | 2 | 0 (or 1$)$ | 2 | 1 | 2 | 2 | 2 | 2 |
| $3 m$ | 3 | 6 | 0 (or 3$)$ | 6 | 3 | 6 | 6 | 6 | 6 |
| $2 x+y$ | 3 | 4 | 0 | 5 | 1 | 6 | 2 | 3 | 0 |
The table also shows that, if in this sequence of turns Anna changes $y$ by +1 or -2 , then $\Delta$ is increased by 1 . Also, if Anna changes $y$ by +2 or -1 , then the first time this happens $\Delta$ is increased by 2 . (This also holds if her turn is $(0,-1)$ or $(-2,-1)$, which are not shown in the table.)
Since Anna wins by placing her token at $(k, 2019)$ we must have $m \leqslant k-1$ and $k \leqslant 2018$. So at that exact moment we have:
$$
\Delta=3 m-(2 k+2019)=k-2022 \leqslant-4 .
$$
So in her last turn she must have decreased $\Delta$ by at least 4 . So her last turn must have been $(1,2)$ or $(2,1)$, which give a decrease of 4 and 5 respectively. (It could not be $(3,0)$ because then she must have already won. Also she could not have done just one or two steps in her last turn since this is not enough for the required decrease in $\Delta$.)
If her last turn was $(1,2)$, then just before doing it we had $y=2017$ and $\Delta=0$. This means that in one of her turns the total change in $y$ was not $0 \bmod 3$. However, in that case we have seen that $\Delta>0$, a contradiction.
If her last turn was $(2,1)$, then just before doing it we had $y=2018$ and $\Delta=0$ or $\Delta=1$. So she must have made at least two turns with the change of $y$ being +1 or -2 or at least one step with the change of $y$ being +2 or -1 . In both cases, consulting the table, we get an increase of at least 2 in $\Delta$, a contradiction.
Note 1: If Anna is allowed to make at most three steps at each turn, then she actually has a winning strategy.
Note 2: If 2019 is replaced by $N>1$, then Bob has a winning strategy if and only if $3 \mid N$.
|
proof
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
A grid consists of all points of the form $(m, n)$ where $m$ and $n$ are integers with $|m| \leqslant 2019$, $|n| \leqslant 2019$ and $|m|+|n|<4038$. We call the points $(m, n)$ of the grid with either $|m|=2019$ or $|n|=2019$ the boundary points. The four lines $x= \pm 2019$ and $y= \pm 2019$ are called boundary lines. Two points in the grid are called neighbours if the distance between them is equal to 1 .
Anna and Bob play a game on this grid.
Anna starts with a token at the point $(0,0)$. They take turns, with Bob playing first.
1) On each of his turns, Bob deletes at most two boundary points on each boundary line.
2) On each of her turns, Anna makes exactly three steps, where a step consists of moving her token from its current point to any neighbouring point which has not been deleted.
As soon as Anna places her token on some boundary point which has not been deleted, the game is over and Anna wins.
Does Anna have a winning strategy?
Proposed by Cyprus
|
Anna does not have a winning strategy. We will provide a winning strategy for Bob. It is enough to describe his strategy for the deletions on the line $y=2019$.
Bob starts by deleting $(0,2019)$ and $(-1,2019)$. Once Anna completes her turn, he deletes the next two available points on the left if Anna decreased her $x$-coordinate, the next two available points on the right if Anna increased her $x$-coordinate, and the next available point to the left and the next available point to the right if Anna did not change her $x$-coordinate. The only exception to the above rule is on the very first time Anna decreases $x$ by exactly 1 . In that turn, Bob deletes the next available point to the left and the next available point to the right.
Bob's strategy guarantees the following: If Anna makes a sequence of steps reaching $(-x, y)$ with $x>0$ and the exact opposite sequence of steps in the horizontal direction reaching $(x, y)$, then Bob deletes at least as many points to the left of $(0,2019)$ in the first sequence than points to the right of $(0,2019)$ in the second sequence.
So we may assume for contradiction that Anna wins by placing her token at $(k, 2019)$ for some $k>0$.
Define $\Delta=3 m-(2 x+y)$, where $m$ is the total number of points deleted by Bob to the right of $(0,2019)$, and $(x, y)$ is the position of Anna's token.
For each sequence of steps performed first by Anna and then by Bob, $\Delta$ does not decrease. This can be seen by looking at the following table exhibiting the changes in 3 m and $2 x+y$. We have excluded the cases where $2 x+y<0$.
| Turn | $(0,3)$ | $(1,2)$ | $(-1,2)$ | $(2,1)$ | $(0,1)$ | $(3,0)$ | $(1,0)$ | $(2,-1)$ | $(1,-2)$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $m$ | 1 | 2 | 0 (or 1$)$ | 2 | 1 | 2 | 2 | 2 | 2 |
| $3 m$ | 3 | 6 | 0 (or 3$)$ | 6 | 3 | 6 | 6 | 6 | 6 |
| $2 x+y$ | 3 | 4 | 0 | 5 | 1 | 6 | 2 | 3 | 0 |
The table also shows that, if in this sequence of turns Anna changes $y$ by +1 or -2 , then $\Delta$ is increased by 1 . Also, if Anna changes $y$ by +2 or -1 , then the first time this happens $\Delta$ is increased by 2 . (This also holds if her turn is $(0,-1)$ or $(-2,-1)$, which are not shown in the table.)
Since Anna wins by placing her token at $(k, 2019)$ we must have $m \leqslant k-1$ and $k \leqslant 2018$. So at that exact moment we have:
$$
\Delta=3 m-(2 k+2019)=k-2022 \leqslant-4 .
$$
So in her last turn she must have decreased $\Delta$ by at least 4 . So her last turn must have been $(1,2)$ or $(2,1)$, which give a decrease of 4 and 5 respectively. (It could not be $(3,0)$ because then she must have already won. Also she could not have done just one or two steps in her last turn since this is not enough for the required decrease in $\Delta$.)
If her last turn was $(1,2)$, then just before doing it we had $y=2017$ and $\Delta=0$. This means that in one of her turns the total change in $y$ was not $0 \bmod 3$. However, in that case we have seen that $\Delta>0$, a contradiction.
If her last turn was $(2,1)$, then just before doing it we had $y=2018$ and $\Delta=0$ or $\Delta=1$. So she must have made at least two turns with the change of $y$ being +1 or -2 or at least one step with the change of $y$ being +2 or -1 . In both cases, consulting the table, we get an increase of at least 2 in $\Delta$, a contradiction.
Note 1: If Anna is allowed to make at most three steps at each turn, then she actually has a winning strategy.
Note 2: If 2019 is replaced by $N>1$, then Bob has a winning strategy if and only if $3 \mid N$.
|
{
"resource_path": "Balkan_MO/segmented/en-2019-BMO-type1.jsonl",
"problem_match": "# Problem 4.",
"solution_match": "\nSolution."
}
|
dc94142b-b3fb-5193-9fde-57231db0086c
| 604,641
|
Let $A B C$ be an acute triangle with $A B=A C$, let $D$ be the midpoint of the side $A C$, and let $\gamma$ be the circumcircle of the triangle $A B D$. The tangent of $\gamma$ at $A$ crosses the line $B C$ at $E$. Let $O$ be the circumcentre of the triangle $A B E$. Prove that the midpoint of the segment $A O$ lies on $\gamma$.
|
. We will first prove that $C$ is the midpoint of the segment $B E$. From the angle equalities
- $\angle B C D=\angle A C B=\angle C B A=\angle E B A$
- $\angle B D C=\angle B A D+\angle D B A=\angle B A D+\angle D A E=\angle B A E$
we can conclude that the triangles $\triangle A B E$ and $\triangle D C B$ are similar.
Thus, $B E / B C=A B / C D=2$, which implies that $C$ is indeed the midpoint of the segment $B E$.
We will now prove that $A E$ is tangent to the circle $A C O$. From the angle equalities
- $\angle O A E=90^{\circ}-\angle E B A$
- $\angle O C A=\angle O C B-\angle A C B=90^{\circ}-\angle C B A=90^{\circ}-\angle E B A$
we can conclude that $\angle O A E=\angle O C A$, which implies that $A E$ is indeed tangent to the circle $A C O$.
Finally, let $\Gamma$ be the image of $\gamma$ under the homothethy of center $A$ and factor 2 . Clearly, $\Gamma$ is also tangent to $A E$ at $A$ and passes through $C$, so $\Gamma$ must coincide with the circle $A C O$, which obviously passes through $O$. Thus, $\gamma$ passes through the midpoint of the segment $A O$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute triangle with $A B=A C$, let $D$ be the midpoint of the side $A C$, and let $\gamma$ be the circumcircle of the triangle $A B D$. The tangent of $\gamma$ at $A$ crosses the line $B C$ at $E$. Let $O$ be the circumcentre of the triangle $A B E$. Prove that the midpoint of the segment $A O$ lies on $\gamma$.
|
. We will first prove that $C$ is the midpoint of the segment $B E$. From the angle equalities
- $\angle B C D=\angle A C B=\angle C B A=\angle E B A$
- $\angle B D C=\angle B A D+\angle D B A=\angle B A D+\angle D A E=\angle B A E$
we can conclude that the triangles $\triangle A B E$ and $\triangle D C B$ are similar.
Thus, $B E / B C=A B / C D=2$, which implies that $C$ is indeed the midpoint of the segment $B E$.
We will now prove that $A E$ is tangent to the circle $A C O$. From the angle equalities
- $\angle O A E=90^{\circ}-\angle E B A$
- $\angle O C A=\angle O C B-\angle A C B=90^{\circ}-\angle C B A=90^{\circ}-\angle E B A$
we can conclude that $\angle O A E=\angle O C A$, which implies that $A E$ is indeed tangent to the circle $A C O$.
Finally, let $\Gamma$ be the image of $\gamma$ under the homothethy of center $A$ and factor 2 . Clearly, $\Gamma$ is also tangent to $A E$ at $A$ and passes through $C$, so $\Gamma$ must coincide with the circle $A C O$, which obviously passes through $O$. Thus, $\gamma$ passes through the midpoint of the segment $A O$.
|
{
"resource_path": "Balkan_MO/segmented/en-2020-BMO-type1.jsonl",
"problem_match": "\nProblem 1.",
"solution_match": "\nSolution 1"
}
|
b44d9ca9-d550-5c53-9bd3-466717b20c72
| 604,656
|
Let $A B C$ be an acute triangle with $A B=A C$, let $D$ be the midpoint of the side $A C$, and let $\gamma$ be the circumcircle of the triangle $A B D$. The tangent of $\gamma$ at $A$ crosses the line $B C$ at $E$. Let $O$ be the circumcentre of the triangle $A B E$. Prove that the midpoint of the segment $A O$ lies on $\gamma$.
|
. Like in the previous solution, we will first prove that $C$ is the midpoint of the segment $B E$. Let $F$ be the midpoint of the segment $A B$. Because $\angle E A C=\angle A B D=\angle F C D$, we have that $C F \| A E$. This implies that $C F$ is a midline in triangle $\triangle B A E$, so $C$ is indeed the midpoint of the segment $B E$.
Let $L$ be the midpoint of the segment $A O$. Because $L D$ is a midline in triangle $\triangle A O C$, so $L D \| O C$, which means that $\angle A L D=\angle A O C$. From the angle equalities
- $\angle A L D=\angle A O C=\angle B O C+\angle A O B=\angle B A E+2 \angle B E A$
- $\angle A B D=\angle C E A=\angle B C A-\angle B E A=\angle A B E-\angle B E A$
- $\angle A L D+\angle A B D=\angle B A E+2 \angle B E A+\angle A B E-\angle B E A=180^{\circ}$
we obtain that the quadrilateral $A B D L$ is cyclic, thus $L$ lies on $\gamma$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute triangle with $A B=A C$, let $D$ be the midpoint of the side $A C$, and let $\gamma$ be the circumcircle of the triangle $A B D$. The tangent of $\gamma$ at $A$ crosses the line $B C$ at $E$. Let $O$ be the circumcentre of the triangle $A B E$. Prove that the midpoint of the segment $A O$ lies on $\gamma$.
|
. Like in the previous solution, we will first prove that $C$ is the midpoint of the segment $B E$. Let $F$ be the midpoint of the segment $A B$. Because $\angle E A C=\angle A B D=\angle F C D$, we have that $C F \| A E$. This implies that $C F$ is a midline in triangle $\triangle B A E$, so $C$ is indeed the midpoint of the segment $B E$.
Let $L$ be the midpoint of the segment $A O$. Because $L D$ is a midline in triangle $\triangle A O C$, so $L D \| O C$, which means that $\angle A L D=\angle A O C$. From the angle equalities
- $\angle A L D=\angle A O C=\angle B O C+\angle A O B=\angle B A E+2 \angle B E A$
- $\angle A B D=\angle C E A=\angle B C A-\angle B E A=\angle A B E-\angle B E A$
- $\angle A L D+\angle A B D=\angle B A E+2 \angle B E A+\angle A B E-\angle B E A=180^{\circ}$
we obtain that the quadrilateral $A B D L$ is cyclic, thus $L$ lies on $\gamma$.
|
{
"resource_path": "Balkan_MO/segmented/en-2020-BMO-type1.jsonl",
"problem_match": "\nProblem 1.",
"solution_match": "\nSolution 2"
}
|
b44d9ca9-d550-5c53-9bd3-466717b20c72
| 604,656
|
Let $A B C$ be an acute triangle with $A B=A C$, let $D$ be the midpoint of the side $A C$, and let $\gamma$ be the circumcircle of the triangle $A B D$. The tangent of $\gamma$ at $A$ crosses the line $B C$ at $E$. Let $O$ be the circumcentre of the triangle $A B E$. Prove that the midpoint of the segment $A O$ lies on $\gamma$.
|
. Like in the previous solutions, establish that $C$ is the midpoint of the segment $B E$.
Let $S$ be the intersection between $B D$ and $A E$. We will first show that $S$ also lies on $C O$. Because
$$
\angle B D C=\angle B A D+\angle D B A=\angle B A D+\angle D A E=\angle B A E
$$
(just like in solution 1), we obtain that the triangle $\triangle S B E$ is isosceles, so $C O$ passes through $S$, because it is the perpendicular bisector of the segment $B E$.
Because $\angle B O C=\angle B A S$, we obtain that the quadrilateral $A S O B$ is cyclic, so $\angle B A L=\angle B S O$. Denote by $L$ the intersection between $A O$ and $\gamma$, then $\angle L D S=\angle B A L$. Combining these two equalities leads to $\angle B S O=\angle L D S$, so $L D \| S O$.
This means that $L D$ is midline in triangle $\triangle A O C$, so $L$, which lies on $\gamma$, is the midpoint of the segment $A O$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute triangle with $A B=A C$, let $D$ be the midpoint of the side $A C$, and let $\gamma$ be the circumcircle of the triangle $A B D$. The tangent of $\gamma$ at $A$ crosses the line $B C$ at $E$. Let $O$ be the circumcentre of the triangle $A B E$. Prove that the midpoint of the segment $A O$ lies on $\gamma$.
|
. Like in the previous solutions, establish that $C$ is the midpoint of the segment $B E$.
Let $S$ be the intersection between $B D$ and $A E$. We will first show that $S$ also lies on $C O$. Because
$$
\angle B D C=\angle B A D+\angle D B A=\angle B A D+\angle D A E=\angle B A E
$$
(just like in solution 1), we obtain that the triangle $\triangle S B E$ is isosceles, so $C O$ passes through $S$, because it is the perpendicular bisector of the segment $B E$.
Because $\angle B O C=\angle B A S$, we obtain that the quadrilateral $A S O B$ is cyclic, so $\angle B A L=\angle B S O$. Denote by $L$ the intersection between $A O$ and $\gamma$, then $\angle L D S=\angle B A L$. Combining these two equalities leads to $\angle B S O=\angle L D S$, so $L D \| S O$.
This means that $L D$ is midline in triangle $\triangle A O C$, so $L$, which lies on $\gamma$, is the midpoint of the segment $A O$.
|
{
"resource_path": "Balkan_MO/segmented/en-2020-BMO-type1.jsonl",
"problem_match": "\nProblem 1.",
"solution_match": "\nSolution 3"
}
|
b44d9ca9-d550-5c53-9bd3-466717b20c72
| 604,656
|
Let $A B C$ be an acute triangle with $A B=A C$, let $D$ be the midpoint of the side $A C$, and let $\gamma$ be the circumcircle of the triangle $A B D$. The tangent of $\gamma$ at $A$ crosses the line $B C$ at $E$. Let $O$ be the circumcentre of the triangle $A B E$. Prove that the midpoint of the segment $A O$ lies on $\gamma$.
|
. Like in the previous solutions, establish that $C$ is the midpoint of the segment $B E$.
Let $F$ be the midpoint of the segment $A B$. Then both $F$ and $C$ lie on the circle of diameter $B O$. Because the quadrilateral $B F D C$ is cyclic, it means that $D$ also lies on that circle.
Let $K$ be the midpoint of $B O$. Then, $A K$ must be the perpendicular bisector of the segment $B C$, so $A K \| O C$, which implies that $\angle K A D=\angle D C O$. However, $\angle D C O=\angle K B D$, because $D B C O$ is cyclic. From the two equalities we obtain that $\angle K A D=\angle K B D$, so $K$ lies on $\gamma$. Furthermore, $A K$ is the bisector of $\angle B A D$, so $K$ is in fact the midpoint of the $\operatorname{arc} B D$.
Now consider a reflection across $O F$. Clearly, $B$ maps to $A$. Because $O F$ is the perpendicular bisector of the segment $A B, \gamma$ maps to itself through this reflection. Thus, $K$, the intersection between $O B$ and $\gamma$, will map to $L$, the intersection between $O A$ and $\gamma$. This implies that $L$ is the midpoint of the segment $A O$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute triangle with $A B=A C$, let $D$ be the midpoint of the side $A C$, and let $\gamma$ be the circumcircle of the triangle $A B D$. The tangent of $\gamma$ at $A$ crosses the line $B C$ at $E$. Let $O$ be the circumcentre of the triangle $A B E$. Prove that the midpoint of the segment $A O$ lies on $\gamma$.
|
. Like in the previous solutions, establish that $C$ is the midpoint of the segment $B E$.
Let $F$ be the midpoint of the segment $A B$. Then both $F$ and $C$ lie on the circle of diameter $B O$. Because the quadrilateral $B F D C$ is cyclic, it means that $D$ also lies on that circle.
Let $K$ be the midpoint of $B O$. Then, $A K$ must be the perpendicular bisector of the segment $B C$, so $A K \| O C$, which implies that $\angle K A D=\angle D C O$. However, $\angle D C O=\angle K B D$, because $D B C O$ is cyclic. From the two equalities we obtain that $\angle K A D=\angle K B D$, so $K$ lies on $\gamma$. Furthermore, $A K$ is the bisector of $\angle B A D$, so $K$ is in fact the midpoint of the $\operatorname{arc} B D$.
Now consider a reflection across $O F$. Clearly, $B$ maps to $A$. Because $O F$ is the perpendicular bisector of the segment $A B, \gamma$ maps to itself through this reflection. Thus, $K$, the intersection between $O B$ and $\gamma$, will map to $L$, the intersection between $O A$ and $\gamma$. This implies that $L$ is the midpoint of the segment $A O$.
|
{
"resource_path": "Balkan_MO/segmented/en-2020-BMO-type1.jsonl",
"problem_match": "\nProblem 1.",
"solution_match": "\nSolution 4"
}
|
b44d9ca9-d550-5c53-9bd3-466717b20c72
| 604,656
|
Denote $\mathbb{Z}_{>0}=\{1,2,3, \ldots\}$ the set of all positive integers. Determine all functions $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that, for each positive integer $n$,
i) $\sum_{k=1}^{n} f(k)$ is a perfect square, and
ii) $f(n)$ divides $n^{3}$.
|
Induct on $n$ to show that $f(n)=n^{3}$ for all positive integers $n$. It is readily checked that this $f$ satisfies the conditions in the statement. The base case, $n=1$, is clear.
Let $n \geqslant 2$ and assume that $f(m)=m^{3}$ for all positive integers $m<n$. Then $\sum_{k=1}^{n-1} f(k)=\frac{n^{2}(n-1)^{2}}{4}$, and reference to the first condition in the statement yields
$f(n)=\sum_{k=1}^{n} f(k)-\sum_{k=1}^{n-1} f(k)=\left(\frac{n(n-1)}{2}+k\right)^{2}-\frac{n^{2}(n-1)^{2}}{4}=k\left(n^{2}-n+k\right)$,
for some positive integer $k$.
The divisibility condition in the statement implies $k\left(n^{2}-n+k\right) \leqslant n^{3}$, which is equivalent to $(n-k)\left(n^{2}+k\right) \geqslant 0$, showing that $k \leqslant n$.
On the other hand, $n^{2}-n+k$ must also divide $n^{3}$. But, if $k<n$, then
$$
n<\frac{n^{3}}{n^{2}-1} \leqslant \frac{n^{3}}{n^{2}-n+k} \leqslant \frac{n^{3}}{n^{2}-n+1}<\frac{n^{3}+1}{n^{2}-n+1}=n+1
$$
therefore $\frac{n^{3}}{n^{2}-n+k}$ cannot be an integer.
Consequently, $k=n$, so $f(n)=n^{3}$. This completes induction and concludes the proof.
|
f(n)=n^{3}
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
Denote $\mathbb{Z}_{>0}=\{1,2,3, \ldots\}$ the set of all positive integers. Determine all functions $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that, for each positive integer $n$,
i) $\sum_{k=1}^{n} f(k)$ is a perfect square, and
ii) $f(n)$ divides $n^{3}$.
|
Induct on $n$ to show that $f(n)=n^{3}$ for all positive integers $n$. It is readily checked that this $f$ satisfies the conditions in the statement. The base case, $n=1$, is clear.
Let $n \geqslant 2$ and assume that $f(m)=m^{3}$ for all positive integers $m<n$. Then $\sum_{k=1}^{n-1} f(k)=\frac{n^{2}(n-1)^{2}}{4}$, and reference to the first condition in the statement yields
$f(n)=\sum_{k=1}^{n} f(k)-\sum_{k=1}^{n-1} f(k)=\left(\frac{n(n-1)}{2}+k\right)^{2}-\frac{n^{2}(n-1)^{2}}{4}=k\left(n^{2}-n+k\right)$,
for some positive integer $k$.
The divisibility condition in the statement implies $k\left(n^{2}-n+k\right) \leqslant n^{3}$, which is equivalent to $(n-k)\left(n^{2}+k\right) \geqslant 0$, showing that $k \leqslant n$.
On the other hand, $n^{2}-n+k$ must also divide $n^{3}$. But, if $k<n$, then
$$
n<\frac{n^{3}}{n^{2}-1} \leqslant \frac{n^{3}}{n^{2}-n+k} \leqslant \frac{n^{3}}{n^{2}-n+1}<\frac{n^{3}+1}{n^{2}-n+1}=n+1
$$
therefore $\frac{n^{3}}{n^{2}-n+k}$ cannot be an integer.
Consequently, $k=n$, so $f(n)=n^{3}$. This completes induction and concludes the proof.
|
{
"resource_path": "Balkan_MO/segmented/en-2020-BMO-type1.jsonl",
"problem_match": "## 2020 BMO, Problem 2",
"solution_match": "\nSolution."
}
|
ba3b682b-f7bb-5060-9e8f-8a4a0fb0983d
| 604,700
|
empty
|
Let $F(n)=f(1)+f(2)+\ldots+f(n)$. We use the following two observations:
Lemma $1 F(n) \leq\left(\frac{n(n+1)}{2}\right)^{2}$
Proof: Since $f(i) \mid i^{3}$, for all $i$ we have $f(i) \leq i^{3}$, and adding all up we get
$$
f(1)+f(2)+\ldots+f(n) \leq 1^{3}+2^{3}+\ldots+n^{3}=\left(\frac{n(n+1)}{2}\right)^{2}
$$
Lemma $2 F(n) \geq n^{2}$
Proof: Note that $F(n)$ is injective and increasing since $f(i)>0, \forall i$. Since $F(n)$ is a perfect square for all $n$ the desired result is obtained.
Lemma $3 f(p)=p^{3}$ for all $p$ prime.
Proof: Since $f(p) \mid p^{3}$, the only possible values for $f(p)$ are $1, p, p^{2}, p^{3}$. We show that $f(p)$ can not be 1 or $p$ or $p^{2}$.
Case 1: Suppose $f(p)=1$. This implies $F(p-1)$ and $F(p)$ are two consecutive numbers, grater than 1 and perfect squares. This is impossible, contradiction.
Case 2: Suppose $f(p)=p$. Let $F(p-1)=a^{2}$ and $F(p)=b^{2}$. Hence we have $p=(a-b)(a+b)$, so $a^{2}$ has to be $\left(\frac{(p-1)}{2}\right)^{2}$ and $b^{2}$ has to be $\left(\frac{(p+1)}{2}\right)^{2}$. But by Lemma 2 we know $F(p-1)=a^{2} \geq(p-1)^{2}$, contradiction.
Case 3: Suppose $f(p)=p^{2}$. Again we have $p^{2}=(a-b)(a+b)$ and since $a, b>0$ we have to have $a-b=1$ and $a+b=p^{2}$. This gives $F(p-1)=\left(\frac{p^{2}-1}{2}\right)^{2}$. But from Lemma 2, we know $F(p-1) \leq\left(\frac{p^{2}-p}{2}\right)^{2}$, hence we get a contradiction again and $f(p)$ can not be $p^{2}$.
To finish the proof, we need to show that $f(n)=n^{3}$ for all nonprime values as well. Let $p>n$ be a prime. We know $f(p)=p^{3}$ and $f(p)=F(p)-F(p-1)=(a-b)(a+b)$. By a reasoning similar to Case 2 above, we can not have $a-b=1$ and $a+b=p^{3}$ so we have to have $a-b=p$ and $a+b=p^{2}$. This gives us $F(p-1)=f(1)+f(2)+\ldots+f(p-1)=\left(\frac{p^{2}-p}{2}\right)$, so we have equality in Lemma 1 . We know this only happens if $f(i)=i^{3}$ for all $i \leq p-1$. This concludes the proof.
|
proof
|
Incomplete
|
Yes
|
math-word-problem
|
Other
|
empty
|
Let $F(n)=f(1)+f(2)+\ldots+f(n)$. We use the following two observations:
Lemma $1 F(n) \leq\left(\frac{n(n+1)}{2}\right)^{2}$
Proof: Since $f(i) \mid i^{3}$, for all $i$ we have $f(i) \leq i^{3}$, and adding all up we get
$$
f(1)+f(2)+\ldots+f(n) \leq 1^{3}+2^{3}+\ldots+n^{3}=\left(\frac{n(n+1)}{2}\right)^{2}
$$
Lemma $2 F(n) \geq n^{2}$
Proof: Note that $F(n)$ is injective and increasing since $f(i)>0, \forall i$. Since $F(n)$ is a perfect square for all $n$ the desired result is obtained.
Lemma $3 f(p)=p^{3}$ for all $p$ prime.
Proof: Since $f(p) \mid p^{3}$, the only possible values for $f(p)$ are $1, p, p^{2}, p^{3}$. We show that $f(p)$ can not be 1 or $p$ or $p^{2}$.
Case 1: Suppose $f(p)=1$. This implies $F(p-1)$ and $F(p)$ are two consecutive numbers, grater than 1 and perfect squares. This is impossible, contradiction.
Case 2: Suppose $f(p)=p$. Let $F(p-1)=a^{2}$ and $F(p)=b^{2}$. Hence we have $p=(a-b)(a+b)$, so $a^{2}$ has to be $\left(\frac{(p-1)}{2}\right)^{2}$ and $b^{2}$ has to be $\left(\frac{(p+1)}{2}\right)^{2}$. But by Lemma 2 we know $F(p-1)=a^{2} \geq(p-1)^{2}$, contradiction.
Case 3: Suppose $f(p)=p^{2}$. Again we have $p^{2}=(a-b)(a+b)$ and since $a, b>0$ we have to have $a-b=1$ and $a+b=p^{2}$. This gives $F(p-1)=\left(\frac{p^{2}-1}{2}\right)^{2}$. But from Lemma 2, we know $F(p-1) \leq\left(\frac{p^{2}-p}{2}\right)^{2}$, hence we get a contradiction again and $f(p)$ can not be $p^{2}$.
To finish the proof, we need to show that $f(n)=n^{3}$ for all nonprime values as well. Let $p>n$ be a prime. We know $f(p)=p^{3}$ and $f(p)=F(p)-F(p-1)=(a-b)(a+b)$. By a reasoning similar to Case 2 above, we can not have $a-b=1$ and $a+b=p^{3}$ so we have to have $a-b=p$ and $a+b=p^{2}$. This gives us $F(p-1)=f(1)+f(2)+\ldots+f(p-1)=\left(\frac{p^{2}-p}{2}\right)$, so we have equality in Lemma 1 . We know this only happens if $f(i)=i^{3}$ for all $i \leq p-1$. This concludes the proof.
|
{
"resource_path": "Balkan_MO/segmented/en-2020-BMO-type1.jsonl",
"problem_match": "## 2020 BMO, Problem 2",
"solution_match": "# Solution 2"
}
|
3b9c1a14-0e38-5ad0-a0aa-5704eeb28d1d
| 604,716
|
Let $k$ be a positive integer. Determine the least integer $n \geqslant k+1$ for which the game below can be played indefinitely:
Consider $n$ boxes, labelled $b_{1}, b_{2}, \ldots, b_{n}$. For each index $i$, box $b_{i}$ contains initially exactly $i$ coins. At each step, the following three substeps are performed in order:
(1) Choose $k+1$ boxes;
(2) Of these $k+1$ boxes, choose $k$ and remove at least half of the coins from each, and add to the remaining box, if labelled $b_{i}$, a number of $i$ coins.
(3) If one of the boxes is left empty, the game ends; otherwise, go to the next step.
|
The required minimum is $n=2^{k}+k-1$.
In this case the game can be played indefinitely by choosing the last $k+1$ boxes, $b_{2^{k}-1}, b_{2^{k}}, \ldots, b_{2^{k}+k-1}$, at each step: At step $r$, if box $b_{2^{k}+i-1}$ has exactly $m_{i}$ coins, then $\left\lceil m_{i} / 2\right\rceil$ coins are removed from that box, unless $i \equiv r-1(\bmod k+1)$, in which case $2^{k}+i-1$ coins are added. Thus, after step $r$ has been performed, box $b_{2^{k}+i-1}$ contains exactly $\left\lfloor m_{i} / 2\right\rfloor$ coins, unless $i \equiv r-1(\bmod k+1)$, in which case it contains exactly $m_{i}+2^{k}+i-1$ coins. This game goes on indefinitely, since each time a box is supplied, at least $2^{k}-1$ coins are added, so it will then contain at least $2^{k}$ coins, good enough to survive the $k$ steps to its next supply.
We now show that no smaller value of $n$ works. So, let $n \leqslant 2^{k}+k-2$ and suppose, if possible, that a game can be played indefinitely. Notice that a box currently containing exactly $m$ coins survives at most $w=\left\lfloor\log _{2} m\right\rfloor$ withdrawals; this $w$ will be referred to as the weight of that box. The sum of the weigths of all boxes will referred to as the total weight. The argument hinges on the lemma below, proved at the end of the solution.
Lemma. Performing a step does not increase the total weight. Moreover, supplying one of the first $2^{k}-2$ boxes strictly decreases the total weight.
Since the total weight cannot strictly decrease indefinitely, $n>2^{k}-2$, and from some stage on none of the first $2^{k}-2$ boxes is ever supplied. Recall that each step involves a $(k+1)$-box choice. Since $n \leqslant 2^{k}+k-2$, from that stage on, each step involves a withdrawal from at least one of the first $2^{k}-2$ boxes. This cannot go on indefinitely, so the game must eventually come to an end, contradicting the assumption.
Consequently, a game that can be played indefinitely requires $n \geqslant 2^{k}+$ $k-1$.
Proof of the Lemma. Since a withdrawal from a box decreases its weight by at least 1 , it is sufficient to show that supplying a box increases its weight by at most $k$; and if the latter is amongst the first $2^{k}-2$ boxes, then its weight increases by at most $k-1$. Let the box to be supplied be $b_{i}$ and let it currently contain exactly $m_{i}$ coins, to proceed by case analysis:
If $m_{i}=1$, the weight increases by $\left\lfloor\log _{2}(i+1)\right\rfloor \leqslant\left\lfloor\log _{2}\left(2^{k}+k-1\right)\right\rfloor \leqslant$ $\left\lfloor\log _{2}\left(2^{k+1}-2\right)\right\rfloor \leqslant k$; and if, in addition, $i \leqslant 2^{k}-2$, then the weight increases by $\left\lfloor\log _{2}(i+1)\right\rfloor \leqslant\left\lfloor\log _{2}\left(2^{k}-1\right)\right\rfloor=k-1$.
If $m_{i}=2$, then the weight increases by $\left\lfloor\log _{2}(i+2)\right\rfloor-\left\lfloor\log _{2} 2\right\rfloor \leqslant$ $\left\lfloor\log _{2}\left(2^{k}+k\right)\right\rfloor-1 \leqslant k-1$.
If $m_{i} \geqslant 3$, then the weight increases by
$$
\begin{aligned}
\left\lfloor\log _{2}\left(i+m_{i}\right)\right\rfloor-\left\lfloor\log _{2} m_{i}\right\rfloor & \leqslant\left\lfloor\log _{2}\left(i+m_{i}\right)-\log _{2} m_{i}\right\rfloor+1 \\
& \leqslant\left\lfloor\log _{2}\left(1+\frac{2^{k}+k-2}{3}\right)\right\rfloor+1 \leqslant k,
\end{aligned}
$$
since $1+\frac{1}{3}\left(2^{k}+k-2\right)=\frac{1}{3}\left(2^{k}+k+1\right)<\frac{1}{3}\left(2^{k}+2^{k+1}\right)=2^{k}$.
Finally, let $i \leqslant 2^{k}-2$ to consider the subcases $m_{i}=3$ and $m_{i} \geqslant 4$. In the former subcase, the weight increases by
$$
\left\lfloor\log _{2}(i+3)\right\rfloor-\left\lfloor\log _{2} 3\right\rfloor \leqslant\left\lfloor\log _{2}\left(2^{k}+1\right)\right\rfloor-1=k-1,
$$
and in the latter by
$$
\begin{aligned}
\left\lfloor\log _{2}\left(i+m_{i}\right)\right\rfloor-\left\lfloor\log _{2} m_{i}\right\rfloor & \leqslant\left\lfloor\log _{2}\left(i+m_{i}\right)-\log _{2} m_{i}\right\rfloor+1 \\
& \leqslant\left\lfloor\log _{2}\left(1+\frac{2^{k}-2}{4}\right)\right\rfloor+1 \leqslant k-1,
\end{aligned}
$$
since $1+\frac{1}{4}\left(2^{k}-2\right)=\frac{1}{4}\left(2^{k}+2\right)<2^{k-2}+1$. This ends the proof and completes the solution.
|
2^{k}+k-1
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Let $k$ be a positive integer. Determine the least integer $n \geqslant k+1$ for which the game below can be played indefinitely:
Consider $n$ boxes, labelled $b_{1}, b_{2}, \ldots, b_{n}$. For each index $i$, box $b_{i}$ contains initially exactly $i$ coins. At each step, the following three substeps are performed in order:
(1) Choose $k+1$ boxes;
(2) Of these $k+1$ boxes, choose $k$ and remove at least half of the coins from each, and add to the remaining box, if labelled $b_{i}$, a number of $i$ coins.
(3) If one of the boxes is left empty, the game ends; otherwise, go to the next step.
|
The required minimum is $n=2^{k}+k-1$.
In this case the game can be played indefinitely by choosing the last $k+1$ boxes, $b_{2^{k}-1}, b_{2^{k}}, \ldots, b_{2^{k}+k-1}$, at each step: At step $r$, if box $b_{2^{k}+i-1}$ has exactly $m_{i}$ coins, then $\left\lceil m_{i} / 2\right\rceil$ coins are removed from that box, unless $i \equiv r-1(\bmod k+1)$, in which case $2^{k}+i-1$ coins are added. Thus, after step $r$ has been performed, box $b_{2^{k}+i-1}$ contains exactly $\left\lfloor m_{i} / 2\right\rfloor$ coins, unless $i \equiv r-1(\bmod k+1)$, in which case it contains exactly $m_{i}+2^{k}+i-1$ coins. This game goes on indefinitely, since each time a box is supplied, at least $2^{k}-1$ coins are added, so it will then contain at least $2^{k}$ coins, good enough to survive the $k$ steps to its next supply.
We now show that no smaller value of $n$ works. So, let $n \leqslant 2^{k}+k-2$ and suppose, if possible, that a game can be played indefinitely. Notice that a box currently containing exactly $m$ coins survives at most $w=\left\lfloor\log _{2} m\right\rfloor$ withdrawals; this $w$ will be referred to as the weight of that box. The sum of the weigths of all boxes will referred to as the total weight. The argument hinges on the lemma below, proved at the end of the solution.
Lemma. Performing a step does not increase the total weight. Moreover, supplying one of the first $2^{k}-2$ boxes strictly decreases the total weight.
Since the total weight cannot strictly decrease indefinitely, $n>2^{k}-2$, and from some stage on none of the first $2^{k}-2$ boxes is ever supplied. Recall that each step involves a $(k+1)$-box choice. Since $n \leqslant 2^{k}+k-2$, from that stage on, each step involves a withdrawal from at least one of the first $2^{k}-2$ boxes. This cannot go on indefinitely, so the game must eventually come to an end, contradicting the assumption.
Consequently, a game that can be played indefinitely requires $n \geqslant 2^{k}+$ $k-1$.
Proof of the Lemma. Since a withdrawal from a box decreases its weight by at least 1 , it is sufficient to show that supplying a box increases its weight by at most $k$; and if the latter is amongst the first $2^{k}-2$ boxes, then its weight increases by at most $k-1$. Let the box to be supplied be $b_{i}$ and let it currently contain exactly $m_{i}$ coins, to proceed by case analysis:
If $m_{i}=1$, the weight increases by $\left\lfloor\log _{2}(i+1)\right\rfloor \leqslant\left\lfloor\log _{2}\left(2^{k}+k-1\right)\right\rfloor \leqslant$ $\left\lfloor\log _{2}\left(2^{k+1}-2\right)\right\rfloor \leqslant k$; and if, in addition, $i \leqslant 2^{k}-2$, then the weight increases by $\left\lfloor\log _{2}(i+1)\right\rfloor \leqslant\left\lfloor\log _{2}\left(2^{k}-1\right)\right\rfloor=k-1$.
If $m_{i}=2$, then the weight increases by $\left\lfloor\log _{2}(i+2)\right\rfloor-\left\lfloor\log _{2} 2\right\rfloor \leqslant$ $\left\lfloor\log _{2}\left(2^{k}+k\right)\right\rfloor-1 \leqslant k-1$.
If $m_{i} \geqslant 3$, then the weight increases by
$$
\begin{aligned}
\left\lfloor\log _{2}\left(i+m_{i}\right)\right\rfloor-\left\lfloor\log _{2} m_{i}\right\rfloor & \leqslant\left\lfloor\log _{2}\left(i+m_{i}\right)-\log _{2} m_{i}\right\rfloor+1 \\
& \leqslant\left\lfloor\log _{2}\left(1+\frac{2^{k}+k-2}{3}\right)\right\rfloor+1 \leqslant k,
\end{aligned}
$$
since $1+\frac{1}{3}\left(2^{k}+k-2\right)=\frac{1}{3}\left(2^{k}+k+1\right)<\frac{1}{3}\left(2^{k}+2^{k+1}\right)=2^{k}$.
Finally, let $i \leqslant 2^{k}-2$ to consider the subcases $m_{i}=3$ and $m_{i} \geqslant 4$. In the former subcase, the weight increases by
$$
\left\lfloor\log _{2}(i+3)\right\rfloor-\left\lfloor\log _{2} 3\right\rfloor \leqslant\left\lfloor\log _{2}\left(2^{k}+1\right)\right\rfloor-1=k-1,
$$
and in the latter by
$$
\begin{aligned}
\left\lfloor\log _{2}\left(i+m_{i}\right)\right\rfloor-\left\lfloor\log _{2} m_{i}\right\rfloor & \leqslant\left\lfloor\log _{2}\left(i+m_{i}\right)-\log _{2} m_{i}\right\rfloor+1 \\
& \leqslant\left\lfloor\log _{2}\left(1+\frac{2^{k}-2}{4}\right)\right\rfloor+1 \leqslant k-1,
\end{aligned}
$$
since $1+\frac{1}{4}\left(2^{k}-2\right)=\frac{1}{4}\left(2^{k}+2\right)<2^{k-2}+1$. This ends the proof and completes the solution.
|
{
"resource_path": "Balkan_MO/segmented/en-2020-BMO-type1.jsonl",
"problem_match": "## 2020 BMO, Problem 3",
"solution_match": "\nSolution."
}
|
21fcc46f-d1ea-5c74-a214-14209ff3879d
| 604,727
|
Let $a_{1}=2$ and, for every positive integer $n$, let $a_{n+1}$ be the smallest integer strictly greater than $a_{n}$ that has more positive divisors than $a_{n}$. Prove that $2 a_{n+1}=3 a_{n}$ only for finitely many indices $n$.
|
Begin with a mere remark on the terms of the sequence under consideration.
Lemma 1. Each $a_{n}$ is minimal amongst all positive integers having the same number of positive divisors as $a_{n}$.
Proof. Suppose, if possible, that for some $n$, some positive integer $b<a_{n}$ has as many positive divisors as $a_{n}$. Then $a_{m}<b \leqslant a_{m+1}$ for some $m<n$, and the definition of the sequence forces $b=a_{m+1}$. Since $b<a_{n}$, it follows that $m+1<n$, which is a contradiction, as $a_{m+1}$ should have less positive divisors than $a_{n}$.
Let $p_{1}<p_{2}<\cdots<p_{n}<\cdots$ be the strictly increasing sequence of prime numbers, and write canonical factorisations into primes in the form $N=\prod_{i \geqslant 1} p_{i}^{e_{i}}$, where $e_{i} \geqslant 0$ for all $i$, and $e_{i}=0$ for all but finitely many indices $i$; in this notation, the number of positive divisors of $N$ is $\tau(N)=$ $\prod_{i \geqslant 1}\left(e_{i}+1\right)$.
Lemma 2. The exponents in the canonical factorisation of each $a_{n}$ into primes form a non-strictly decreasing sequence.
Proof. Indeed, if $e_{i}<e_{j}$ for some $i<j$ in the canonical decomposition of $a_{n}$ into primes, then swapping the two exponents yields a smaller integer with the same number of positive divisors, contradicting Lemma 1.
We are now in a position to prove the required result. For convenience, a term $a_{n}$ satisfying $3 a_{n}=2 a_{n+1}$ will be referred to as a special term of the sequence.
Suppose now, if possible, that the sequence has infinitely many special terms, so the latter form a strictly increasing, and hence unbounded, subsequence. To reach a contradiction, it is sufficient to show that:
(1) The exponents of the primes in the factorisation of special terms have a common upper bound $e$; and
(2) For all large enough primes $p$, no special term is divisible by $p$.
Refer to Lemma 2 to write $a_{n}=\prod_{i \geqslant 1} p_{i}^{e_{i}(n)}$, where $e_{i}(n) \geqslant e_{i+1}(n)$ for all $i$.
Statement (2) is a straightforward consequence of (1) and Lemma 1. Suppose, if possible, that some special term $a_{n}$ is divisible by a prime $p_{i}>2^{e+1}$, where $e$ is the integer provided by (1). Then $e \geqslant e_{i}(n)>0$, so $2^{e_{1}(n) e_{i}(n)+e_{i}(n)} a_{n} / p_{i}^{e_{i}(n)}$ is a positive integer with the same number of positive divisors as $a_{n}$, but smaller than $a_{n}$. This contradicts Lemma 1. Consequently, no special term is divisible by a prime exceeding $2^{e+1}$.
To prove (1), it is sufficient to show that, as $a_{n}$ runs through the special terms, the exponents $e_{1}(n)$ are bounded from above. Then, Lemma 2 shows that such an upper bound $e$ suits all primes.
Consider a large enough special $a_{n}$. The condition $\tau\left(a_{n}\right)<\tau\left(a_{n+1}\right)$ is then equivalent to $\left(e_{1}(n)+1\right)\left(e_{2}(n)+1\right)<e_{1}(n)\left(e_{2}(n)+2\right)$. Alternatively, but equivalently, $e_{1}(n) \geqslant e_{2}(n)+2$. The latter implies that $a_{n}$ is divisible by 8 , for either $e_{1}(n) \geqslant 3$ or $a_{n}$ is a large enough power of 2 .
Next, note that $9 a_{n} / 8$ is an integer strictly between $a_{n}$ and $a_{n+1}$, so $\tau\left(9 a_{n} / 8\right) \leqslant \tau\left(a_{n}\right)$, which is equivalent to
$$
\left(e_{1}(n)-2\right)\left(e_{2}(n)+3\right) \leqslant\left(e_{1}(n)+1\right)\left(e_{2}(n)+1\right),
$$
so $2 e_{1}(n) \leqslant 3 e_{2}(n)+7$. This shows that $a_{n}$ is divisible by 3 , for otherwise, letting $a_{n}$ run through the special terms, 3 would be an upper bound for all but finitely many $e_{1}(n)$, and the special terms would therefore form a bounded sequence.
Thus, $4 a_{n} / 3$ is another integer strictly between $a_{n}$ and $a_{n+1}$. As before, $\tau\left(4 a_{n} / 3\right) \leqslant \tau\left(a_{n}\right)$. Alternatively, but equivalently,
$$
\left(e_{1}(n)+3\right) e_{2}(n) \leqslant\left(e_{1}(n)+1\right)\left(e_{2}(n)+1\right),
$$
so $2 e_{2}(n)-1 \leqslant e_{1}(n)$. Combine this with the inequality in the previous paragraph to write $4 e_{2}(n)-2 \leqslant 2 e_{1}(n) \leqslant 3 e_{2}(n)+7$ and infer that $e_{2}(n) \leqslant 9$. Consequently, $2 e_{1}(n) \leqslant 3 e_{2}(n)+7 \leqslant 34$, showing that $e=17$ is suitable for (1) to hold. This establishes (1) and completes the solution.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $a_{1}=2$ and, for every positive integer $n$, let $a_{n+1}$ be the smallest integer strictly greater than $a_{n}$ that has more positive divisors than $a_{n}$. Prove that $2 a_{n+1}=3 a_{n}$ only for finitely many indices $n$.
|
Begin with a mere remark on the terms of the sequence under consideration.
Lemma 1. Each $a_{n}$ is minimal amongst all positive integers having the same number of positive divisors as $a_{n}$.
Proof. Suppose, if possible, that for some $n$, some positive integer $b<a_{n}$ has as many positive divisors as $a_{n}$. Then $a_{m}<b \leqslant a_{m+1}$ for some $m<n$, and the definition of the sequence forces $b=a_{m+1}$. Since $b<a_{n}$, it follows that $m+1<n$, which is a contradiction, as $a_{m+1}$ should have less positive divisors than $a_{n}$.
Let $p_{1}<p_{2}<\cdots<p_{n}<\cdots$ be the strictly increasing sequence of prime numbers, and write canonical factorisations into primes in the form $N=\prod_{i \geqslant 1} p_{i}^{e_{i}}$, where $e_{i} \geqslant 0$ for all $i$, and $e_{i}=0$ for all but finitely many indices $i$; in this notation, the number of positive divisors of $N$ is $\tau(N)=$ $\prod_{i \geqslant 1}\left(e_{i}+1\right)$.
Lemma 2. The exponents in the canonical factorisation of each $a_{n}$ into primes form a non-strictly decreasing sequence.
Proof. Indeed, if $e_{i}<e_{j}$ for some $i<j$ in the canonical decomposition of $a_{n}$ into primes, then swapping the two exponents yields a smaller integer with the same number of positive divisors, contradicting Lemma 1.
We are now in a position to prove the required result. For convenience, a term $a_{n}$ satisfying $3 a_{n}=2 a_{n+1}$ will be referred to as a special term of the sequence.
Suppose now, if possible, that the sequence has infinitely many special terms, so the latter form a strictly increasing, and hence unbounded, subsequence. To reach a contradiction, it is sufficient to show that:
(1) The exponents of the primes in the factorisation of special terms have a common upper bound $e$; and
(2) For all large enough primes $p$, no special term is divisible by $p$.
Refer to Lemma 2 to write $a_{n}=\prod_{i \geqslant 1} p_{i}^{e_{i}(n)}$, where $e_{i}(n) \geqslant e_{i+1}(n)$ for all $i$.
Statement (2) is a straightforward consequence of (1) and Lemma 1. Suppose, if possible, that some special term $a_{n}$ is divisible by a prime $p_{i}>2^{e+1}$, where $e$ is the integer provided by (1). Then $e \geqslant e_{i}(n)>0$, so $2^{e_{1}(n) e_{i}(n)+e_{i}(n)} a_{n} / p_{i}^{e_{i}(n)}$ is a positive integer with the same number of positive divisors as $a_{n}$, but smaller than $a_{n}$. This contradicts Lemma 1. Consequently, no special term is divisible by a prime exceeding $2^{e+1}$.
To prove (1), it is sufficient to show that, as $a_{n}$ runs through the special terms, the exponents $e_{1}(n)$ are bounded from above. Then, Lemma 2 shows that such an upper bound $e$ suits all primes.
Consider a large enough special $a_{n}$. The condition $\tau\left(a_{n}\right)<\tau\left(a_{n+1}\right)$ is then equivalent to $\left(e_{1}(n)+1\right)\left(e_{2}(n)+1\right)<e_{1}(n)\left(e_{2}(n)+2\right)$. Alternatively, but equivalently, $e_{1}(n) \geqslant e_{2}(n)+2$. The latter implies that $a_{n}$ is divisible by 8 , for either $e_{1}(n) \geqslant 3$ or $a_{n}$ is a large enough power of 2 .
Next, note that $9 a_{n} / 8$ is an integer strictly between $a_{n}$ and $a_{n+1}$, so $\tau\left(9 a_{n} / 8\right) \leqslant \tau\left(a_{n}\right)$, which is equivalent to
$$
\left(e_{1}(n)-2\right)\left(e_{2}(n)+3\right) \leqslant\left(e_{1}(n)+1\right)\left(e_{2}(n)+1\right),
$$
so $2 e_{1}(n) \leqslant 3 e_{2}(n)+7$. This shows that $a_{n}$ is divisible by 3 , for otherwise, letting $a_{n}$ run through the special terms, 3 would be an upper bound for all but finitely many $e_{1}(n)$, and the special terms would therefore form a bounded sequence.
Thus, $4 a_{n} / 3$ is another integer strictly between $a_{n}$ and $a_{n+1}$. As before, $\tau\left(4 a_{n} / 3\right) \leqslant \tau\left(a_{n}\right)$. Alternatively, but equivalently,
$$
\left(e_{1}(n)+3\right) e_{2}(n) \leqslant\left(e_{1}(n)+1\right)\left(e_{2}(n)+1\right),
$$
so $2 e_{2}(n)-1 \leqslant e_{1}(n)$. Combine this with the inequality in the previous paragraph to write $4 e_{2}(n)-2 \leqslant 2 e_{1}(n) \leqslant 3 e_{2}(n)+7$ and infer that $e_{2}(n) \leqslant 9$. Consequently, $2 e_{1}(n) \leqslant 3 e_{2}(n)+7 \leqslant 34$, showing that $e=17$ is suitable for (1) to hold. This establishes (1) and completes the solution.
|
{
"resource_path": "Balkan_MO/segmented/en-2020-BMO-type1.jsonl",
"problem_match": "## 2020 BMO, Problem 4",
"solution_match": "\nSolution."
}
|
17f3c23d-30bf-5bf4-add0-149a7077a44b
| 604,740
|
Let $A B C$ be a triangle with $A B<A C$. Let $\omega$ be a circle passing through $B, C$ and assume that $A$ is inside $\omega$. Suppose $X, Y$ lie on $\omega$ such that $\angle B X A=\angle A Y C$. Suppose also that $X$ and $C$ lie on opposite sides of the line $A B$ and that $Y$ and $B$ lie on opposite sides of the line $A C$.
Show that, as $X, Y$ vary on $\omega$, the line $X Y$ passes through a fixed point.
|
. Extend $X A$ and $Y A$ to meet $\omega$ again at $X^{\prime}$ and $Y^{\prime}$ respectively. We then have that:
$$
\angle Y^{\prime} Y C=\angle A Y C=\angle B X A=\angle B X X^{\prime} .
$$
so $B C X^{\prime} Y^{\prime}$ is an isosceles trapezium and hence $X^{\prime} Y^{\prime} \| B C$.
Let $\ell$ be the line through $A$ parallel to $B C$ and let $\ell$ intersect $\omega$ at $P, Q$ with $P$ on the opposite side of $A B$ to $C$. As $X^{\prime} Y^{\prime}\|B C\| P Q$ then
$$
\angle X A P=\angle X X^{\prime} Y^{\prime}=\angle X Y Y^{\prime}=\angle X Y A
$$
which shows that $\ell$ is tangent to the circumcircle of triangle $A X Y$. Let $X Y$ intersect $P Q$ at $Z$. By power of a point we have that
$$
Z A^{2}=Z X \cdot Z Y=Z P \cdot Z Q
$$
As $P, Q$ are independent of the positions of $X, Y$, this shows that $Z$ is fixed and hence $X Y$ passes through a fixed point.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle with $A B<A C$. Let $\omega$ be a circle passing through $B, C$ and assume that $A$ is inside $\omega$. Suppose $X, Y$ lie on $\omega$ such that $\angle B X A=\angle A Y C$. Suppose also that $X$ and $C$ lie on opposite sides of the line $A B$ and that $Y$ and $B$ lie on opposite sides of the line $A C$.
Show that, as $X, Y$ vary on $\omega$, the line $X Y$ passes through a fixed point.
|
. Extend $X A$ and $Y A$ to meet $\omega$ again at $X^{\prime}$ and $Y^{\prime}$ respectively. We then have that:
$$
\angle Y^{\prime} Y C=\angle A Y C=\angle B X A=\angle B X X^{\prime} .
$$
so $B C X^{\prime} Y^{\prime}$ is an isosceles trapezium and hence $X^{\prime} Y^{\prime} \| B C$.
Let $\ell$ be the line through $A$ parallel to $B C$ and let $\ell$ intersect $\omega$ at $P, Q$ with $P$ on the opposite side of $A B$ to $C$. As $X^{\prime} Y^{\prime}\|B C\| P Q$ then
$$
\angle X A P=\angle X X^{\prime} Y^{\prime}=\angle X Y Y^{\prime}=\angle X Y A
$$
which shows that $\ell$ is tangent to the circumcircle of triangle $A X Y$. Let $X Y$ intersect $P Q$ at $Z$. By power of a point we have that
$$
Z A^{2}=Z X \cdot Z Y=Z P \cdot Z Q
$$
As $P, Q$ are independent of the positions of $X, Y$, this shows that $Z$ is fixed and hence $X Y$ passes through a fixed point.
|
{
"resource_path": "Balkan_MO/segmented/en-2021-BMO-type1.jsonl",
"problem_match": "## BMO 2021 - Problem 1",
"solution_match": "\nSolution 1"
}
|
323e6cc8-3aa0-5b14-96d5-52936f05b76d
| 604,756
|
Let $A B C$ be a triangle with $A B<A C$. Let $\omega$ be a circle passing through $B, C$ and assume that $A$ is inside $\omega$. Suppose $X, Y$ lie on $\omega$ such that $\angle B X A=\angle A Y C$. Suppose also that $X$ and $C$ lie on opposite sides of the line $A B$ and that $Y$ and $B$ lie on opposite sides of the line $A C$.
Show that, as $X, Y$ vary on $\omega$, the line $X Y$ passes through a fixed point.
|
. Let $B^{\prime}$ and $C^{\prime}$ be the points of intersection of the lines $A B$ and $A C$ with $\omega$ respectively and let $\omega_{1}$ be the circumcircle of the triangle $A B^{\prime} C^{\prime}$. Let $\varepsilon$ be the tangent to $\omega_{1}$ at the point $A$. Because $A B<A C$ the lines $B^{\prime} C^{\prime}$ and $\varepsilon$ intersects at a point $Z$ which is fixed and independent of $X$ and $Y$.
We have
$$
\angle Z A C^{\prime}=\angle C^{\prime} B^{\prime} A=\angle C^{\prime} B^{\prime} B=\angle C^{\prime} C B .
$$
Therefore, $\varepsilon \| B C$.
Let $X^{\prime}, Y^{\prime}$ be the points of intersection of the lines $X A, Y A$ with $\omega$ respecively. From the hypothesis we have $\angle B X X^{\prime}=\angle Y^{\prime} Y C$. Therefore
$$
\widehat{B X^{\prime}}=\widehat{Y^{\prime} C} \Longrightarrow \widehat{B C}+\widehat{C X^{\prime}}=\widehat{Y^{\prime} B}+\widehat{B C} \Longrightarrow \widehat{C X^{\prime}}=\widehat{Y^{\prime} B}
$$
and so $X^{\prime} Y^{\prime}\|B C\| \varepsilon$. Thus
$$
\angle X A Z=\angle X X^{\prime} Y^{\prime}=\angle X Y Y^{\prime}=\angle X Y A .
$$
From the last equality we have that $\varepsilon$ is also tangent to the circmucircle $\omega_{2}$ of the triangle $X A Y$.
Consider now the radical centre of the circles $\omega, \omega_{1}, \omega_{2}$. This is the point of intersection of the radical axes $B^{\prime} C^{\prime}\left(\right.$ of $\omega$ and $\left.\omega_{1}\right), \varepsilon\left(\right.$ of $\omega_{1}$ and $\left.\omega_{2}\right)$ and $X Y$ (of $\omega$ and $\omega_{2}$ ).
This must be point $Z$ and therefore the variable line $X Y$ passes through the fixed point $Z$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle with $A B<A C$. Let $\omega$ be a circle passing through $B, C$ and assume that $A$ is inside $\omega$. Suppose $X, Y$ lie on $\omega$ such that $\angle B X A=\angle A Y C$. Suppose also that $X$ and $C$ lie on opposite sides of the line $A B$ and that $Y$ and $B$ lie on opposite sides of the line $A C$.
Show that, as $X, Y$ vary on $\omega$, the line $X Y$ passes through a fixed point.
|
. Let $B^{\prime}$ and $C^{\prime}$ be the points of intersection of the lines $A B$ and $A C$ with $\omega$ respectively and let $\omega_{1}$ be the circumcircle of the triangle $A B^{\prime} C^{\prime}$. Let $\varepsilon$ be the tangent to $\omega_{1}$ at the point $A$. Because $A B<A C$ the lines $B^{\prime} C^{\prime}$ and $\varepsilon$ intersects at a point $Z$ which is fixed and independent of $X$ and $Y$.
We have
$$
\angle Z A C^{\prime}=\angle C^{\prime} B^{\prime} A=\angle C^{\prime} B^{\prime} B=\angle C^{\prime} C B .
$$
Therefore, $\varepsilon \| B C$.
Let $X^{\prime}, Y^{\prime}$ be the points of intersection of the lines $X A, Y A$ with $\omega$ respecively. From the hypothesis we have $\angle B X X^{\prime}=\angle Y^{\prime} Y C$. Therefore
$$
\widehat{B X^{\prime}}=\widehat{Y^{\prime} C} \Longrightarrow \widehat{B C}+\widehat{C X^{\prime}}=\widehat{Y^{\prime} B}+\widehat{B C} \Longrightarrow \widehat{C X^{\prime}}=\widehat{Y^{\prime} B}
$$
and so $X^{\prime} Y^{\prime}\|B C\| \varepsilon$. Thus
$$
\angle X A Z=\angle X X^{\prime} Y^{\prime}=\angle X Y Y^{\prime}=\angle X Y A .
$$
From the last equality we have that $\varepsilon$ is also tangent to the circmucircle $\omega_{2}$ of the triangle $X A Y$.
Consider now the radical centre of the circles $\omega, \omega_{1}, \omega_{2}$. This is the point of intersection of the radical axes $B^{\prime} C^{\prime}\left(\right.$ of $\omega$ and $\left.\omega_{1}\right), \varepsilon\left(\right.$ of $\omega_{1}$ and $\left.\omega_{2}\right)$ and $X Y$ (of $\omega$ and $\omega_{2}$ ).
This must be point $Z$ and therefore the variable line $X Y$ passes through the fixed point $Z$.
|
{
"resource_path": "Balkan_MO/segmented/en-2021-BMO-type1.jsonl",
"problem_match": "## BMO 2021 - Problem 1",
"solution_match": "\nSolution 2"
}
|
323e6cc8-3aa0-5b14-96d5-52936f05b76d
| 604,756
|
Find all functions $f:(0,+\infty) \rightarrow(0,+\infty)$ such that
$$
f(x+f(x)+f(y))=2 f(x)+y
$$
holds for all $x, y \in(0,+\infty)$.
|
. We will show that $f(x)=x$ for every $x \in \mathbb{R}^{+}$. It is easy to check that this function satisfies the equation.
We write $P(x, y)$ for the assertion that $f(x+f(x)+f(y))=2 f(x)+y$.
We first show that $f$ is injective. So assume $f(a)=f(b)$. Now $P(1, a)$ and $P(1, b)$ show that
$$
2 f(1)+a=f(1+f(1)+f(a))=f(1+f(1)+f(b))=2 f(1)+b
$$
and therefore $a=b$.
Let $A=\left\{x \in \mathbb{R}^{+}: f(x)=x\right\}$. It is enough to show that $A=\mathbb{R}^{+}$.
$P(x, x)$ shows that $x+2 f(x) \in A$ for every $x \in \mathbb{R}^{+}$. Now $P(x, x+2 f(x))$ gives that
$$
f(2 x+3 f(x))=x+4 f(x)
$$
for every $x \in \mathbb{R}^{+}$. Therefore $P(x, 2 x+3 f(x))$ gives that $2 x+5 f(x) \in A$ for every $x \in \mathbb{R}^{+}$. Suppose $x, y \in \mathbb{R}^{+}$such that $x, 2 x+y \in A$. Then $P(x, y)$ gives that
$$
f(2 x+f(y))=f(x+f(x)+f(y))=2 f(x)+y=2 x+y=f(2 x+y)
$$
and by the injectivity of $f$ we have that $2 x+f(y)=2 x+y$. We conlude that $y \in A$ as well. Now since $x+2 f(x) \in A$ and $2 x+5 f(x)=2(x+2 f(x))+f(x) \in A$ we deduce that $f(x) \in A$ for every $x \in \mathbb{R}^{+}$. I.e. $f(f(x))=f(x)$ for every $x \in \mathbb{R}^{+}$.
By injectivity of $f$ we now conclude that $f(x)=x$ for every $x \in \mathbb{R}^{+}$.
|
f(x)=x
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find all functions $f:(0,+\infty) \rightarrow(0,+\infty)$ such that
$$
f(x+f(x)+f(y))=2 f(x)+y
$$
holds for all $x, y \in(0,+\infty)$.
|
. We will show that $f(x)=x$ for every $x \in \mathbb{R}^{+}$. It is easy to check that this function satisfies the equation.
We write $P(x, y)$ for the assertion that $f(x+f(x)+f(y))=2 f(x)+y$.
We first show that $f$ is injective. So assume $f(a)=f(b)$. Now $P(1, a)$ and $P(1, b)$ show that
$$
2 f(1)+a=f(1+f(1)+f(a))=f(1+f(1)+f(b))=2 f(1)+b
$$
and therefore $a=b$.
Let $A=\left\{x \in \mathbb{R}^{+}: f(x)=x\right\}$. It is enough to show that $A=\mathbb{R}^{+}$.
$P(x, x)$ shows that $x+2 f(x) \in A$ for every $x \in \mathbb{R}^{+}$. Now $P(x, x+2 f(x))$ gives that
$$
f(2 x+3 f(x))=x+4 f(x)
$$
for every $x \in \mathbb{R}^{+}$. Therefore $P(x, 2 x+3 f(x))$ gives that $2 x+5 f(x) \in A$ for every $x \in \mathbb{R}^{+}$. Suppose $x, y \in \mathbb{R}^{+}$such that $x, 2 x+y \in A$. Then $P(x, y)$ gives that
$$
f(2 x+f(y))=f(x+f(x)+f(y))=2 f(x)+y=2 x+y=f(2 x+y)
$$
and by the injectivity of $f$ we have that $2 x+f(y)=2 x+y$. We conlude that $y \in A$ as well. Now since $x+2 f(x) \in A$ and $2 x+5 f(x)=2(x+2 f(x))+f(x) \in A$ we deduce that $f(x) \in A$ for every $x \in \mathbb{R}^{+}$. I.e. $f(f(x))=f(x)$ for every $x \in \mathbb{R}^{+}$.
By injectivity of $f$ we now conclude that $f(x)=x$ for every $x \in \mathbb{R}^{+}$.
|
{
"resource_path": "Balkan_MO/segmented/en-2021-BMO-type1.jsonl",
"problem_match": "## BMO 2021 - Problem 2",
"solution_match": "\nSolution 1"
}
|
41954388-c82e-518c-9068-42f48dc13063
| 604,779
|
Find all functions $f:(0,+\infty) \rightarrow(0,+\infty)$ such that
$$
f(x+f(x)+f(y))=2 f(x)+y
$$
holds for all $x, y \in(0,+\infty)$.
|
. As in Solution 1, $f$ is injective. Furthermore, letting $m=2 f(1)$ we have that the image of $f$ contains $(m, \infty)$. Indeed, if $t>m$, say $t=m+y$ for some $y>0$, then $P(1, y)$ shows that $f(1+f(1)+f(y))=t$.
Let $a, b \in \mathbb{R}$. We will show that $f(a)-a=f(b)-b$. Define $c=2 f(a)-2 f(b)$ and $d=a+f(a)-b-f(b)$. It is enough to show that $c=d$. By interchanging the roles of $a$ and $b$ in necessary, we may assume that $d \geqslant 0$.
From $P(a, y)$ and $P(b, y)$, after subtraction, we get
$$
f(a+f(a)+f(y))-f(b+f(b)+f(y))=2 f(a)-2 f(b)=c .
$$
so for any $t>m$ (picking $y$ such that $f(y)=t$ in (1)) we get
$$
f(a+f(a)+t)-f(b+f(b)+t)=2 f(a)-2 f(b)=c .
$$
Now for any $z>m+b+f(b)$, taking $t=z-b-f(b)$ in (2) we get
$$
f(z+d)-f(z)=c
$$
Now for any $x>m+b+f(b)$ from (3) we get that
$$
2 f(x+d)+y=2 f(x)+y+2 c
$$
Also, for any $x$ large enough, $(x>\max \{m+b+f(b), m+b+f(b)+c-d\}$ will do), by repeated application of (3), we have
$$
\begin{aligned}
f(x+d+f(x+d)+f(y)) & =f(x+f(x+d)+y)+c \\
& =f(x+f(x)+y+c)+c \\
& =f(x+f(x)+y+c-d)+2 c .
\end{aligned}
$$
(In the first equality we applied (3) with $z=x+f(x+d)+y>x>m+b+f(b)$, in the second with $z=x>m+b+f(b)$ and in the third with $z=x+f(x)+y-c+d>x+c-d>m+b+f(b)$.
In particular, now $P(x+d, y)$ implies that
$$
f(x+f(x)+y+c-d)=2 f(x)+y=f(x+f(x)+y)
$$
for every large enough $x$. By injectivity of $f$ we deduce that $x+f(x)+y+c-d=x+f(x)+y$ and therefore $c=d$ as required.
It now follows that $f(x)=x+k$ for every $x \in \mathbb{R}^{+}$and some fixed constant $k$. Substituting in the initial equation we get $k=0$.
|
f(x)=x
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find all functions $f:(0,+\infty) \rightarrow(0,+\infty)$ such that
$$
f(x+f(x)+f(y))=2 f(x)+y
$$
holds for all $x, y \in(0,+\infty)$.
|
. As in Solution 1, $f$ is injective. Furthermore, letting $m=2 f(1)$ we have that the image of $f$ contains $(m, \infty)$. Indeed, if $t>m$, say $t=m+y$ for some $y>0$, then $P(1, y)$ shows that $f(1+f(1)+f(y))=t$.
Let $a, b \in \mathbb{R}$. We will show that $f(a)-a=f(b)-b$. Define $c=2 f(a)-2 f(b)$ and $d=a+f(a)-b-f(b)$. It is enough to show that $c=d$. By interchanging the roles of $a$ and $b$ in necessary, we may assume that $d \geqslant 0$.
From $P(a, y)$ and $P(b, y)$, after subtraction, we get
$$
f(a+f(a)+f(y))-f(b+f(b)+f(y))=2 f(a)-2 f(b)=c .
$$
so for any $t>m$ (picking $y$ such that $f(y)=t$ in (1)) we get
$$
f(a+f(a)+t)-f(b+f(b)+t)=2 f(a)-2 f(b)=c .
$$
Now for any $z>m+b+f(b)$, taking $t=z-b-f(b)$ in (2) we get
$$
f(z+d)-f(z)=c
$$
Now for any $x>m+b+f(b)$ from (3) we get that
$$
2 f(x+d)+y=2 f(x)+y+2 c
$$
Also, for any $x$ large enough, $(x>\max \{m+b+f(b), m+b+f(b)+c-d\}$ will do), by repeated application of (3), we have
$$
\begin{aligned}
f(x+d+f(x+d)+f(y)) & =f(x+f(x+d)+y)+c \\
& =f(x+f(x)+y+c)+c \\
& =f(x+f(x)+y+c-d)+2 c .
\end{aligned}
$$
(In the first equality we applied (3) with $z=x+f(x+d)+y>x>m+b+f(b)$, in the second with $z=x>m+b+f(b)$ and in the third with $z=x+f(x)+y-c+d>x+c-d>m+b+f(b)$.
In particular, now $P(x+d, y)$ implies that
$$
f(x+f(x)+y+c-d)=2 f(x)+y=f(x+f(x)+y)
$$
for every large enough $x$. By injectivity of $f$ we deduce that $x+f(x)+y+c-d=x+f(x)+y$ and therefore $c=d$ as required.
It now follows that $f(x)=x+k$ for every $x \in \mathbb{R}^{+}$and some fixed constant $k$. Substituting in the initial equation we get $k=0$.
|
{
"resource_path": "Balkan_MO/segmented/en-2021-BMO-type1.jsonl",
"problem_match": "## BMO 2021 - Problem 2",
"solution_match": "\nSolution 2"
}
|
41954388-c82e-518c-9068-42f48dc13063
| 604,779
|
Let $a, b$ and $c$ be positive integers satisfying the equation
$$
(a, b)+[a, b]=2021^{c} .
$$
If $|a-b|$ is a prime number, prove that the number $(a+b)^{2}+4$ is composite.
Here, $(a, b)$ denotes the greatest common divisor of $a$ and $b$, and $[a, b]$ denotes the least common multiple of $a$ and $b$.
|
We write $p=|a-b|$ and assume for contradiction that $q=(a+b)^{2}+4$ is a prime number.
Since $(a, b) \mid[a, b]$, we have that $(a, b) \mid 2021^{c}$. As $(a, b)$ also divides $p=|a-b|$, it follows that $(a, b) \in\{1,43,47\}$. We will consider all 3 cases separately:
(1) If $(a, b)=1$, then $1+a b=2021^{c}$, and therefore
$$
q=(a+b)^{2}+4=(a-b)^{2}+4(1+a b)=p^{2}+4 \cdot 2021^{c} .
$$
(a) Suppose $c$ is even. Since $q \equiv 1 \bmod 4$, it can be represented uniquely (up to order) as a sum of two (non-negative) squares. But (1) gives potentially two such representations so in order to have uniqueness we must have $p=2$. But then $4 \mid q$ a contradiction.
(b) If $c$ is odd then $a b=2021^{c}-1 \equiv 1 \bmod 3$. Thus $a \equiv b \bmod 3$ implying that $p=|a-b| \equiv 0 \bmod 3$. Therefore $p=3$. Without loss of generality $b=a+3$. Then $2021^{c}=a b+1=a^{2}+3 a+1$ and so
$$
(2 a+3)^{2}=4 a^{2}+12 a+9=4 \cdot 2021^{c}+5
$$
So 5 is a quadratic residue modulo 47, a contradiction as
$$
\left(\frac{5}{47}\right)=\left(\frac{47}{5}\right)=\left(\frac{2}{5}\right)=-1 .
$$
(2) If $(a, b)=43$, then $p=|a-b|=43$ and we may assume that $a=43 k$ and $b=43(k+1)$, for some $k \in \mathbb{N}$. Then $2021^{c}=43+43 k(k+1)$ giving that
$$
(2 k+1)^{2}=4 k^{2}+4 k+4-3=4 \cdot 43^{c-1} \cdot 47-3 .
$$
So -3 is a quadratic residue modulo 47 , a contradiction as
$$
\left(\frac{-3}{47}\right)=\left(\frac{-1}{47}\right)\left(\frac{3}{47}\right)=\left(\frac{47}{3}\right)=\left(\frac{2}{3}\right)=-1
$$
(3) If $(a, b)=47$ then analogously there is a $k \in \mathbb{N}$ such that
$$
(2 k+1)^{2}=4 \cdot 43^{c} \cdot 47^{c-1}-3 .
$$
If $c>1$ then we get a contradiction in exactly the same way as in (2). If $c=1$ then $(2 k+1)^{2}=169$ giving $k=6$. This implies that $a+b=47 \cdot 6+47 \cdot 7=47 \cdot 13 \equiv 1 \bmod 5$. Thus $q=(a+b)^{2}+4 \equiv 0 \bmod 5$, a contradiction.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $a, b$ and $c$ be positive integers satisfying the equation
$$
(a, b)+[a, b]=2021^{c} .
$$
If $|a-b|$ is a prime number, prove that the number $(a+b)^{2}+4$ is composite.
Here, $(a, b)$ denotes the greatest common divisor of $a$ and $b$, and $[a, b]$ denotes the least common multiple of $a$ and $b$.
|
We write $p=|a-b|$ and assume for contradiction that $q=(a+b)^{2}+4$ is a prime number.
Since $(a, b) \mid[a, b]$, we have that $(a, b) \mid 2021^{c}$. As $(a, b)$ also divides $p=|a-b|$, it follows that $(a, b) \in\{1,43,47\}$. We will consider all 3 cases separately:
(1) If $(a, b)=1$, then $1+a b=2021^{c}$, and therefore
$$
q=(a+b)^{2}+4=(a-b)^{2}+4(1+a b)=p^{2}+4 \cdot 2021^{c} .
$$
(a) Suppose $c$ is even. Since $q \equiv 1 \bmod 4$, it can be represented uniquely (up to order) as a sum of two (non-negative) squares. But (1) gives potentially two such representations so in order to have uniqueness we must have $p=2$. But then $4 \mid q$ a contradiction.
(b) If $c$ is odd then $a b=2021^{c}-1 \equiv 1 \bmod 3$. Thus $a \equiv b \bmod 3$ implying that $p=|a-b| \equiv 0 \bmod 3$. Therefore $p=3$. Without loss of generality $b=a+3$. Then $2021^{c}=a b+1=a^{2}+3 a+1$ and so
$$
(2 a+3)^{2}=4 a^{2}+12 a+9=4 \cdot 2021^{c}+5
$$
So 5 is a quadratic residue modulo 47, a contradiction as
$$
\left(\frac{5}{47}\right)=\left(\frac{47}{5}\right)=\left(\frac{2}{5}\right)=-1 .
$$
(2) If $(a, b)=43$, then $p=|a-b|=43$ and we may assume that $a=43 k$ and $b=43(k+1)$, for some $k \in \mathbb{N}$. Then $2021^{c}=43+43 k(k+1)$ giving that
$$
(2 k+1)^{2}=4 k^{2}+4 k+4-3=4 \cdot 43^{c-1} \cdot 47-3 .
$$
So -3 is a quadratic residue modulo 47 , a contradiction as
$$
\left(\frac{-3}{47}\right)=\left(\frac{-1}{47}\right)\left(\frac{3}{47}\right)=\left(\frac{47}{3}\right)=\left(\frac{2}{3}\right)=-1
$$
(3) If $(a, b)=47$ then analogously there is a $k \in \mathbb{N}$ such that
$$
(2 k+1)^{2}=4 \cdot 43^{c} \cdot 47^{c-1}-3 .
$$
If $c>1$ then we get a contradiction in exactly the same way as in (2). If $c=1$ then $(2 k+1)^{2}=169$ giving $k=6$. This implies that $a+b=47 \cdot 6+47 \cdot 7=47 \cdot 13 \equiv 1 \bmod 5$. Thus $q=(a+b)^{2}+4 \equiv 0 \bmod 5$, a contradiction.
|
{
"resource_path": "Balkan_MO/segmented/en-2021-BMO-type1.jsonl",
"problem_match": "## BMO 2021 - Problem 3",
"solution_match": "\nSolution."
}
|
c9b8ab81-62f4-5c71-83bc-a5084935c369
| 604,804
|
Angel has a warehouse, which initially contains 100 piles of 100 pieces of rubbish each. Each morning, Angel performs exactly one of the following moves:
(a) He clears every piece of rubbish from a single pile.
(b) He clears one piece of rubbish from each pile.
However, every evening, a demon sneaks into the warehouse and performs exactly one of the following moves:
(a) He adds one piece of rubbish to each non-empty pile.
(b) He creates a new pile with one piece of rubbish.
What is the first morning when Angel can guarantee to have cleared all the rubbish from the warehouse?
|
. We will show that he can do so by the morning of day 199 but not earlier.If we have $n$ piles with at least two pieces of rubbish and $m$ piles with exactly one piece of rubbish, then we define the value of the pile to be
$$
V= \begin{cases}n & m=0 \\ n+\frac{1}{2} & m=1 \\ n+1 & m \geqslant 2\end{cases}
$$
We also denote this position by $(n, m)$. Implicitly we will also write $k$ for the number of piles with exactly two pieces of rubbish.
Angel's strategy is the following:
(i) From position $(0, m)$ remove one piece from each pile to go position $(0,0)$. The game ends.
(ii) From position $(n, 0)$, where $n \geqslant 1$, remove one pile to go to position $(n-1,0)$. Either the game ends, or the demon can move to position $(n-1,0)$ or $(n-1,1)$. In any case $V$ reduces by at least $1 / 2$.
(iii) From position $(n, 1)$, where $n \geqslant 1$, remove one pile with at least two pieces to go to position $(n-1,1)$. The demon can move to position $(n, 0)$ or $(n-1,2)$. In any case $V$ reduces by (at least) $1 / 2$.
(iv) From position $(n, m)$, where $n \geqslant 1$ and $m \geqslant 2$, remove one piece from each pile to go to position $(n-k, k)$. The demon can move to position $(n, 0)$ or $(n-k, k+1)$. In any case $V$ reduces by at least $1 / 2$. (The value of position $(n-k, k+1)$ is $n+\frac{1}{2}$ if $k=0$, and $n-k+1 \leqslant n$ if $k \geqslant 1$.)
So during every day if the game does not end then $V$ is decreased by at least $1 / 2$. So after 198 days if the game did not already end we will have $V \leqslant 1$ and we will be in one of positions $(0, m),(1,0)$. The game can then end on the morning of day 199.
We will now provide a strategy for demon which guarantees that at the end of each day $V$ has decreased by at most $1 / 2$ and furthermore at the end of the day $m \leqslant 1$.
(i) If Angel moves from $(n, 0)$ to $(n-1,0)$ (by removing a pile) then create a new pile with one piece to move to $(n-1,1)$. Then $V$ decreases by $1 / 2$ and and $m=1 \leqslant 1$
(ii) If Angel moves from $(n, 0)$ to $(n-k, k)$ (by removing one piece from each pile) then add one piece back to each pile to move to $(n, 0)$. Then $V$ stays the same and $m=0 \leqslant 1$.
(iii) If Angels moves from $(n, 1)$ to $(n-1,1)$ or $(n, 0)$ (by removing a pile) then add one piece to each pile to move to $(n, 0)$. Then $V$ decreases by $1 / 2$ and $m=0 \leqslant 1$.
(iv) If Angel moves from $(n, 1)$ to $(n-k, k)$ (by removing a piece from each pile) then add one piece to each pile to move to $(n, 0)$. Then $V$ decreases by $1 / 2$ and $m=0 \leqslant 1$.
Since after every move of demon we have $m \leqslant 1$, in order for Angel to finish the game in the next morning we must have $n=1, m=0$ or $n=0, m=1$ and therefore we must have $V \leqslant 1$. But now inductively the demon can guarantee that by the end of day $N$, where $N \leqslant 198$ the game has not yet finished and that $V \geqslant 100-N / 2$.
|
199
|
Yes
|
Yes
|
math-word-problem
|
Logic and Puzzles
|
Angel has a warehouse, which initially contains 100 piles of 100 pieces of rubbish each. Each morning, Angel performs exactly one of the following moves:
(a) He clears every piece of rubbish from a single pile.
(b) He clears one piece of rubbish from each pile.
However, every evening, a demon sneaks into the warehouse and performs exactly one of the following moves:
(a) He adds one piece of rubbish to each non-empty pile.
(b) He creates a new pile with one piece of rubbish.
What is the first morning when Angel can guarantee to have cleared all the rubbish from the warehouse?
|
. We will show that he can do so by the morning of day 199 but not earlier.If we have $n$ piles with at least two pieces of rubbish and $m$ piles with exactly one piece of rubbish, then we define the value of the pile to be
$$
V= \begin{cases}n & m=0 \\ n+\frac{1}{2} & m=1 \\ n+1 & m \geqslant 2\end{cases}
$$
We also denote this position by $(n, m)$. Implicitly we will also write $k$ for the number of piles with exactly two pieces of rubbish.
Angel's strategy is the following:
(i) From position $(0, m)$ remove one piece from each pile to go position $(0,0)$. The game ends.
(ii) From position $(n, 0)$, where $n \geqslant 1$, remove one pile to go to position $(n-1,0)$. Either the game ends, or the demon can move to position $(n-1,0)$ or $(n-1,1)$. In any case $V$ reduces by at least $1 / 2$.
(iii) From position $(n, 1)$, where $n \geqslant 1$, remove one pile with at least two pieces to go to position $(n-1,1)$. The demon can move to position $(n, 0)$ or $(n-1,2)$. In any case $V$ reduces by (at least) $1 / 2$.
(iv) From position $(n, m)$, where $n \geqslant 1$ and $m \geqslant 2$, remove one piece from each pile to go to position $(n-k, k)$. The demon can move to position $(n, 0)$ or $(n-k, k+1)$. In any case $V$ reduces by at least $1 / 2$. (The value of position $(n-k, k+1)$ is $n+\frac{1}{2}$ if $k=0$, and $n-k+1 \leqslant n$ if $k \geqslant 1$.)
So during every day if the game does not end then $V$ is decreased by at least $1 / 2$. So after 198 days if the game did not already end we will have $V \leqslant 1$ and we will be in one of positions $(0, m),(1,0)$. The game can then end on the morning of day 199.
We will now provide a strategy for demon which guarantees that at the end of each day $V$ has decreased by at most $1 / 2$ and furthermore at the end of the day $m \leqslant 1$.
(i) If Angel moves from $(n, 0)$ to $(n-1,0)$ (by removing a pile) then create a new pile with one piece to move to $(n-1,1)$. Then $V$ decreases by $1 / 2$ and and $m=1 \leqslant 1$
(ii) If Angel moves from $(n, 0)$ to $(n-k, k)$ (by removing one piece from each pile) then add one piece back to each pile to move to $(n, 0)$. Then $V$ stays the same and $m=0 \leqslant 1$.
(iii) If Angels moves from $(n, 1)$ to $(n-1,1)$ or $(n, 0)$ (by removing a pile) then add one piece to each pile to move to $(n, 0)$. Then $V$ decreases by $1 / 2$ and $m=0 \leqslant 1$.
(iv) If Angel moves from $(n, 1)$ to $(n-k, k)$ (by removing a piece from each pile) then add one piece to each pile to move to $(n, 0)$. Then $V$ decreases by $1 / 2$ and $m=0 \leqslant 1$.
Since after every move of demon we have $m \leqslant 1$, in order for Angel to finish the game in the next morning we must have $n=1, m=0$ or $n=0, m=1$ and therefore we must have $V \leqslant 1$. But now inductively the demon can guarantee that by the end of day $N$, where $N \leqslant 198$ the game has not yet finished and that $V \geqslant 100-N / 2$.
|
{
"resource_path": "Balkan_MO/segmented/en-2021-BMO-type1.jsonl",
"problem_match": "## BMO 2021 - Problem 4",
"solution_match": "\nSolution 1"
}
|
bbae61a6-833f-5992-ae9e-196273a0a96d
| 604,819
|
Angel has a warehouse, which initially contains 100 piles of 100 pieces of rubbish each. Each morning, Angel performs exactly one of the following moves:
(a) He clears every piece of rubbish from a single pile.
(b) He clears one piece of rubbish from each pile.
However, every evening, a demon sneaks into the warehouse and performs exactly one of the following moves:
(a) He adds one piece of rubbish to each non-empty pile.
(b) He creates a new pile with one piece of rubbish.
What is the first morning when Angel can guarantee to have cleared all the rubbish from the warehouse?
|
.
Define Angel's score $S_{A}$ to be $S_{A}=2 n+m-1$. The Angel can clear the rubbish in at most $\max \left\{S_{A}, 1\right\}$ days. The proof is by induction on $(n, m)$ in lexicographic order.
Angel's strategy is the same as in Solution 1 and in each of cases (ii)-(iv) one needs to check that $S_{A}$ reduces by at least 1 in each day. (Case (i) is trivial as the game ends in one day.)
Now define demon's score $S_{D}$ to be $S_{D}=2 n-1$ if $m=0$ and $S_{D}=2 n$ if $m \geqslant 1$. The claim is the if $(n, m) \neq(0,0)$, then the demon can ensure that Angel requires $S_{D}$ days to clear the rubbish.
Again, demon's strategy is the same as in the Solution by PSC and in each of cases (i)-(iv) one needs to check that $S_{D}$ reduced by at most 1 in each day.
|
not found
|
Yes
|
Yes
|
math-word-problem
|
Logic and Puzzles
|
Angel has a warehouse, which initially contains 100 piles of 100 pieces of rubbish each. Each morning, Angel performs exactly one of the following moves:
(a) He clears every piece of rubbish from a single pile.
(b) He clears one piece of rubbish from each pile.
However, every evening, a demon sneaks into the warehouse and performs exactly one of the following moves:
(a) He adds one piece of rubbish to each non-empty pile.
(b) He creates a new pile with one piece of rubbish.
What is the first morning when Angel can guarantee to have cleared all the rubbish from the warehouse?
|
.
Define Angel's score $S_{A}$ to be $S_{A}=2 n+m-1$. The Angel can clear the rubbish in at most $\max \left\{S_{A}, 1\right\}$ days. The proof is by induction on $(n, m)$ in lexicographic order.
Angel's strategy is the same as in Solution 1 and in each of cases (ii)-(iv) one needs to check that $S_{A}$ reduces by at least 1 in each day. (Case (i) is trivial as the game ends in one day.)
Now define demon's score $S_{D}$ to be $S_{D}=2 n-1$ if $m=0$ and $S_{D}=2 n$ if $m \geqslant 1$. The claim is the if $(n, m) \neq(0,0)$, then the demon can ensure that Angel requires $S_{D}$ days to clear the rubbish.
Again, demon's strategy is the same as in the Solution by PSC and in each of cases (i)-(iv) one needs to check that $S_{D}$ reduced by at most 1 in each day.
|
{
"resource_path": "Balkan_MO/segmented/en-2021-BMO-type1.jsonl",
"problem_match": "## BMO 2021 - Problem 4",
"solution_match": "# Solution 2"
}
|
bbae61a6-833f-5992-ae9e-196273a0a96d
| 604,819
|
Problem. Let $A B C$ be an acute triangle such that $C A \neq C B$ with circumcircle $\omega$ and circumcentre $O$. Let $t_{A}$ and $t_{B}$ be the tangents to $\omega$ at $A$ and $B$ respectively, which meet at $X$. Let $Y$ be the foot of the perpendicular from $O$ onto the line segment $C X$. The line through $C$ parallel to line $A B$ meets $t_{A}$ at $Z$. Prove that the line $Y Z$ passes through the midpoint of the line segment $A C$.
|
. Firstly observe that $O A X B$ is cyclic, with diameter $O X$, and $Y$ also lies on this circle since $O Y \perp X C$. Hence:
$$
\angle A Z C=\angle X A B=\angle A B X=\angle A Y X
$$
and so $C Y A Z$ is cyclic.
Let $M$ be the intersection of $Y Z$ and $A C$ and let $C Y$ intersect $\omega$ again at $W$. Using the new cyclic relation we get $\angle C Y Z=\angle C A Z$ and then using that $Z A$ is tangent to $\omega$ we get $\angle C A Z=\angle C W A$, so $\angle C Y M=\angle C W A$. Therefore the triangles $C W A$ and $C Y M$ are similar. But $C W$ is a chord of $\omega$, and $Y$ is the foot of the perpendicular from $O$, hence $Y$ is the midpoint of $C W$. It follows from the similarity relation that $M$ is the midpoint of $A C$, as required.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Problem. Let $A B C$ be an acute triangle such that $C A \neq C B$ with circumcircle $\omega$ and circumcentre $O$. Let $t_{A}$ and $t_{B}$ be the tangents to $\omega$ at $A$ and $B$ respectively, which meet at $X$. Let $Y$ be the foot of the perpendicular from $O$ onto the line segment $C X$. The line through $C$ parallel to line $A B$ meets $t_{A}$ at $Z$. Prove that the line $Y Z$ passes through the midpoint of the line segment $A C$.
|
. Firstly observe that $O A X B$ is cyclic, with diameter $O X$, and $Y$ also lies on this circle since $O Y \perp X C$. Hence:
$$
\angle A Z C=\angle X A B=\angle A B X=\angle A Y X
$$
and so $C Y A Z$ is cyclic.
Let $M$ be the intersection of $Y Z$ and $A C$ and let $C Y$ intersect $\omega$ again at $W$. Using the new cyclic relation we get $\angle C Y Z=\angle C A Z$ and then using that $Z A$ is tangent to $\omega$ we get $\angle C A Z=\angle C W A$, so $\angle C Y M=\angle C W A$. Therefore the triangles $C W A$ and $C Y M$ are similar. But $C W$ is a chord of $\omega$, and $Y$ is the foot of the perpendicular from $O$, hence $Y$ is the midpoint of $C W$. It follows from the similarity relation that $M$ is the midpoint of $A C$, as required.
|
{
"resource_path": "Balkan_MO/segmented/en-2022-BMO-type1.jsonl",
"problem_match": "# Problem 1",
"solution_match": "\nSolution 1"
}
|
d0d1dab7-85d9-56ab-aa41-17a423c90108
| 604,845
|
Problem. Let $A B C$ be an acute triangle such that $C A \neq C B$ with circumcircle $\omega$ and circumcentre $O$. Let $t_{A}$ and $t_{B}$ be the tangents to $\omega$ at $A$ and $B$ respectively, which meet at $X$. Let $Y$ be the foot of the perpendicular from $O$ onto the line segment $C X$. The line through $C$ parallel to line $A B$ meets $t_{A}$ at $Z$. Prove that the line $Y Z$ passes through the midpoint of the line segment $A C$.
|
. Let $M$ be the midpoint of $A C$. We have $\angle C A Z=\angle C B A$ and $\angle Z C A=$ $\angle B A C$ so the triangles $C A Z$ and $A B C$ are similar. The line $C Y X$ is the $C$-symmedian of triangle $A B C$, and $Z M$ is the corresponding median in triangle $C A Z$, hence by isogonality $\angle A Z M=\angle A C Y$. So
$$
\angle Z M A=180^{\circ}-\angle A Z M-\angle M A Z=180^{\circ}-\angle A C Y-\angle C B A
$$
Now observe $\angle O M C=\angle O Y C=90^{\circ}$, so $C M Y O$ is cyclic. Thus:
$$
\angle C Y M=\angle C O M=\frac{1}{2} \angle C O A=\angle C B A
$$
This shows that
$$
\angle Y M C=180^{\circ}-\angle M C Y-\angle C Y M=180^{\circ}-\angle A C Y-\angle C B A
$$
Combining this with (1) we get that $\angle Y M C=\angle Z M A$ and as $A, C, M$ are collinear, it follows that $Z, M, Y$ are collinear as required.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Problem. Let $A B C$ be an acute triangle such that $C A \neq C B$ with circumcircle $\omega$ and circumcentre $O$. Let $t_{A}$ and $t_{B}$ be the tangents to $\omega$ at $A$ and $B$ respectively, which meet at $X$. Let $Y$ be the foot of the perpendicular from $O$ onto the line segment $C X$. The line through $C$ parallel to line $A B$ meets $t_{A}$ at $Z$. Prove that the line $Y Z$ passes through the midpoint of the line segment $A C$.
|
. Let $M$ be the midpoint of $A C$. We have $\angle C A Z=\angle C B A$ and $\angle Z C A=$ $\angle B A C$ so the triangles $C A Z$ and $A B C$ are similar. The line $C Y X$ is the $C$-symmedian of triangle $A B C$, and $Z M$ is the corresponding median in triangle $C A Z$, hence by isogonality $\angle A Z M=\angle A C Y$. So
$$
\angle Z M A=180^{\circ}-\angle A Z M-\angle M A Z=180^{\circ}-\angle A C Y-\angle C B A
$$
Now observe $\angle O M C=\angle O Y C=90^{\circ}$, so $C M Y O$ is cyclic. Thus:
$$
\angle C Y M=\angle C O M=\frac{1}{2} \angle C O A=\angle C B A
$$
This shows that
$$
\angle Y M C=180^{\circ}-\angle M C Y-\angle C Y M=180^{\circ}-\angle A C Y-\angle C B A
$$
Combining this with (1) we get that $\angle Y M C=\angle Z M A$ and as $A, C, M$ are collinear, it follows that $Z, M, Y$ are collinear as required.
|
{
"resource_path": "Balkan_MO/segmented/en-2022-BMO-type1.jsonl",
"problem_match": "# Problem 1",
"solution_match": "\nSolution 2"
}
|
d0d1dab7-85d9-56ab-aa41-17a423c90108
| 604,845
|
Problem. Let $A B C$ be an acute triangle such that $C A \neq C B$ with circumcircle $\omega$ and circumcentre $O$. Let $t_{A}$ and $t_{B}$ be the tangents to $\omega$ at $A$ and $B$ respectively, which meet at $X$. Let $Y$ be the foot of the perpendicular from $O$ onto the line segment $C X$. The line through $C$ parallel to line $A B$ meets $t_{A}$ at $Z$. Prove that the line $Y Z$ passes through the midpoint of the line segment $A C$.
|
. As in Solution 2 we have that $C X$ is the $A$-symmedian of triangle $A B C$ and that triangle $A B C$ is similar to triangle $C A Z$.
Let $f$ be the spiral similarity which maps $A C$ onto $A B$ and let $g$ be the reflection on the perpendicular bisector of $A B$. Note that $f$ is a rotation about $A$ by an angle of $\angle C A B$ (clockwise in our figure) followed by a homothety centered at $A$ by a factor of $A B / A C$. By the similarity of triangles $A B C$ and $C A Z$ we have that $g(f(Z))=C$, so actually $f(Z)$ is the other point of intersection, say $C^{\prime}$, of $C Z$ with $\omega$.
As in Solution 1 we have that $C Y A Z$ is cyclic. Therefore, letting $W$ be the other point of intersection of $C Y$ with $\omega$, we have $\angle W A B=\angle W C B=\angle C A Y$. We also have $\angle A C Y=$ $\angle A B W$. It follows that $f(Y)=W$.
Let $W^{\prime}=g(W)$. Then $W^{\prime} \in \omega$ and since $C W$ is the $A$-symmedian, then $C W^{\prime}$ passes through the midpoint $N$ of $A B$. Now $C W^{\prime}$ and $C^{\prime} W$ intersect on the perpendicular bisector of $A B$ and therefore they intersect on $N$. It follows that $N=A B \cap C^{\prime} W=A f(C) \cap f(Z) f(Y)$ is the image of $M=A C \cap Z Y$ under $f$. Since $N$ is the midpoint of $A B$, then $M$ is the midpoint of $A C$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Problem. Let $A B C$ be an acute triangle such that $C A \neq C B$ with circumcircle $\omega$ and circumcentre $O$. Let $t_{A}$ and $t_{B}$ be the tangents to $\omega$ at $A$ and $B$ respectively, which meet at $X$. Let $Y$ be the foot of the perpendicular from $O$ onto the line segment $C X$. The line through $C$ parallel to line $A B$ meets $t_{A}$ at $Z$. Prove that the line $Y Z$ passes through the midpoint of the line segment $A C$.
|
. As in Solution 2 we have that $C X$ is the $A$-symmedian of triangle $A B C$ and that triangle $A B C$ is similar to triangle $C A Z$.
Let $f$ be the spiral similarity which maps $A C$ onto $A B$ and let $g$ be the reflection on the perpendicular bisector of $A B$. Note that $f$ is a rotation about $A$ by an angle of $\angle C A B$ (clockwise in our figure) followed by a homothety centered at $A$ by a factor of $A B / A C$. By the similarity of triangles $A B C$ and $C A Z$ we have that $g(f(Z))=C$, so actually $f(Z)$ is the other point of intersection, say $C^{\prime}$, of $C Z$ with $\omega$.
As in Solution 1 we have that $C Y A Z$ is cyclic. Therefore, letting $W$ be the other point of intersection of $C Y$ with $\omega$, we have $\angle W A B=\angle W C B=\angle C A Y$. We also have $\angle A C Y=$ $\angle A B W$. It follows that $f(Y)=W$.
Let $W^{\prime}=g(W)$. Then $W^{\prime} \in \omega$ and since $C W$ is the $A$-symmedian, then $C W^{\prime}$ passes through the midpoint $N$ of $A B$. Now $C W^{\prime}$ and $C^{\prime} W$ intersect on the perpendicular bisector of $A B$ and therefore they intersect on $N$. It follows that $N=A B \cap C^{\prime} W=A f(C) \cap f(Z) f(Y)$ is the image of $M=A C \cap Z Y$ under $f$. Since $N$ is the midpoint of $A B$, then $M$ is the midpoint of $A C$.
|
{
"resource_path": "Balkan_MO/segmented/en-2022-BMO-type1.jsonl",
"problem_match": "# Problem 1",
"solution_match": "\nSolution 3"
}
|
d0d1dab7-85d9-56ab-aa41-17a423c90108
| 604,845
|
Problem. Let $A B C$ be an acute triangle such that $C A \neq C B$ with circumcircle $\omega$ and circumcentre $O$. Let $t_{A}$ and $t_{B}$ be the tangents to $\omega$ at $A$ and $B$ respectively, which meet at $X$. Let $Y$ be the foot of the perpendicular from $O$ onto the line segment $C X$. The line through $C$ parallel to line $A B$ meets $t_{A}$ at $Z$. Prove that the line $Y Z$ passes through the midpoint of the line segment $A C$.
|
. Let $E=A B \cap C X$ and $F=A W \cap C Z$. We have $(C, W ; X, E)=-1$. Projecting from the line $C X$ onto the line $C Z$ from $A$ we get that $(C, F ; Z, \infty)=-1$. Thus $Z$ is the midpoint of $C F$. Since also $Y$ is the midpoint of $C W$, we get that $Z Y$ bisects $C A$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Problem. Let $A B C$ be an acute triangle such that $C A \neq C B$ with circumcircle $\omega$ and circumcentre $O$. Let $t_{A}$ and $t_{B}$ be the tangents to $\omega$ at $A$ and $B$ respectively, which meet at $X$. Let $Y$ be the foot of the perpendicular from $O$ onto the line segment $C X$. The line through $C$ parallel to line $A B$ meets $t_{A}$ at $Z$. Prove that the line $Y Z$ passes through the midpoint of the line segment $A C$.
|
. Let $E=A B \cap C X$ and $F=A W \cap C Z$. We have $(C, W ; X, E)=-1$. Projecting from the line $C X$ onto the line $C Z$ from $A$ we get that $(C, F ; Z, \infty)=-1$. Thus $Z$ is the midpoint of $C F$. Since also $Y$ is the midpoint of $C W$, we get that $Z Y$ bisects $C A$.
|
{
"resource_path": "Balkan_MO/segmented/en-2022-BMO-type1.jsonl",
"problem_match": "# Problem 1",
"solution_match": "\nSolution 4"
}
|
d0d1dab7-85d9-56ab-aa41-17a423c90108
| 604,845
|
Problem. Let $a, b$ and $n$ be positive integers with $a>b$ such that all of the following hold:
(i) $a^{2021}$ divides $n$,
(ii) $b^{2021}$ divides $n$,
(iii) 2022 divides $a-b$.
Prove that there is a subset $T$ of the set of positive divisors of the number $n$ such that the sum of the elements of $T$ is divisible by 2022 but not divisible by $2022^{2}$.
|
If $1011 \mid a$, then $1011^{2021} \mid n$ and we can take $T=\left\{1011,1011^{2}\right\}$. So we can assume that $3 \nmid a$ or $337 \nmid a$.
We continue with the following claim:
Claim. If $k$ is a positive integer, then $a^{k} b^{2021-k} \mid n$.
Proof of the Claim. We have that $n^{2021}=n^{k} \cdot n^{2021-k}$ is divisible by $a^{2021 k} \cdot b^{2021(2021-k)}$ and taking the 2021-root we get the desired result.
Back to the problem, we will prove that the set $T=\left\{a^{k} b^{2021-k}: k \geqslant 0\right\}$ consisting of 2022 divisors of $n$, has the desired property. The sum of its elements is equal to
$$
S=\sum_{k=0}^{2021} a^{k} b^{2021-k} \equiv \sum_{k=0}^{2021} a^{2021} \equiv 0 \bmod 2022
$$
On the other hand, the last sum is equal to $\frac{a^{2022}-b^{2022}}{a-b}$.
If $3 \nmid a$, we will prove that $S$ is not divisible by 9 . Indeed if $3 \nmid a$ then we also have $3 \nmid b$. So if $3^{t} \| a-b$ then, since $3^{1} \| 2022$, by the Lifting the Exponent Lemma, we have that $3^{t+1} \| a^{2022}-b^{2022}$. This implies that $S$ is not divisible by 9 , therefore, $2022^{2}$ doesn't divide $S$.
If $3 \mid a$, then we have $337 \nmid a$ and a similar argument shows that $337^{2} \nmid S$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Problem. Let $a, b$ and $n$ be positive integers with $a>b$ such that all of the following hold:
(i) $a^{2021}$ divides $n$,
(ii) $b^{2021}$ divides $n$,
(iii) 2022 divides $a-b$.
Prove that there is a subset $T$ of the set of positive divisors of the number $n$ such that the sum of the elements of $T$ is divisible by 2022 but not divisible by $2022^{2}$.
|
If $1011 \mid a$, then $1011^{2021} \mid n$ and we can take $T=\left\{1011,1011^{2}\right\}$. So we can assume that $3 \nmid a$ or $337 \nmid a$.
We continue with the following claim:
Claim. If $k$ is a positive integer, then $a^{k} b^{2021-k} \mid n$.
Proof of the Claim. We have that $n^{2021}=n^{k} \cdot n^{2021-k}$ is divisible by $a^{2021 k} \cdot b^{2021(2021-k)}$ and taking the 2021-root we get the desired result.
Back to the problem, we will prove that the set $T=\left\{a^{k} b^{2021-k}: k \geqslant 0\right\}$ consisting of 2022 divisors of $n$, has the desired property. The sum of its elements is equal to
$$
S=\sum_{k=0}^{2021} a^{k} b^{2021-k} \equiv \sum_{k=0}^{2021} a^{2021} \equiv 0 \bmod 2022
$$
On the other hand, the last sum is equal to $\frac{a^{2022}-b^{2022}}{a-b}$.
If $3 \nmid a$, we will prove that $S$ is not divisible by 9 . Indeed if $3 \nmid a$ then we also have $3 \nmid b$. So if $3^{t} \| a-b$ then, since $3^{1} \| 2022$, by the Lifting the Exponent Lemma, we have that $3^{t+1} \| a^{2022}-b^{2022}$. This implies that $S$ is not divisible by 9 , therefore, $2022^{2}$ doesn't divide $S$.
If $3 \mid a$, then we have $337 \nmid a$ and a similar argument shows that $337^{2} \nmid S$.
|
{
"resource_path": "Balkan_MO/segmented/en-2022-BMO-type1.jsonl",
"problem_match": "# Problem 2",
"solution_match": "\nSolution."
}
|
e8749cc7-f2ad-5639-a45e-ab0c82e60044
| 604,887
|
Problem. Find all functions $f:(0, \infty) \rightarrow(0, \infty)$ such that
$$
f\left(y(f(x))^{3}+x\right)=x^{3} f(y)+f(x)
$$
for all $x, y>0$.
|
. Setting $y=\frac{t}{f(x)^{3}}$ we get
$$
f(x+t)=x^{3} f\left(\frac{t}{f(x)^{3}}\right)+f(x)
$$
for every $x, t>0$.
From (1) it is immediate that $f$ is increasing.
Claim. $f(1)=1$
Proof of Claim. Let $c=f(1)$. If $c<1$, taking $x=1$ and $y=\frac{1}{1-c^{3}}$ we have $y-y c^{3}=1$, so $y f(1)^{3}+1=y$ and $f\left(y f(1)^{3}+1\right)=f(y)=1^{3} f(y)$. Thus $f(1)=0$, a contradiction. Assume now for contradiction that $c>1$. We claim that
$$
f\left(1+c^{3}+\cdots+c^{3 n}\right)=(n+1) c
$$
for every $n \in \mathbb{N}$. We proceed by induction, the case $n=0$ being trivial. The inductive step follows easily by taking $x=1, t=c^{3}+c^{6}+\cdots+c^{3(k+1)}$ in (1).
Now taking $x=1+c^{3}+\cdots+c^{3 n-3}, t=c^{3 n}$ in (1) we get
$$
(n+1) c=f\left(1+c^{3}+\cdots+c^{3 n}\right)=\left(1+c^{3}+\cdots+c^{3 n-3}\right)^{3} f\left(\frac{c^{3 n}}{(c n)^{3}}\right)+n c
$$
giving
$$
f\left(\frac{c^{3 n-3}}{n^{3}}\right)=\frac{c}{\left(1+c^{3}+\cdots+c^{3 n}\right)^{3}}<c=f(1) \Longrightarrow \frac{c^{3 n-3}}{n^{3}}<1 .
$$
But this leads to a contradiction if $n$ is large enough.
Now for $x=1$ we get $f(y+1)=f(y)+1$ and since $f(1)=1$ inductively we get $f(n)=n$ for every $n \in \mathbb{N}$. For $m, n \in \mathbb{N}$, setting $x=n, y=q=m / n$ we get
$$
m n^{2}+n=f\left(q n^{3}+n\right)=f\left(y f(x)^{3}+x\right)=x^{3} f(y)+f(x)=n^{3} f(q)+n \Longrightarrow f(q)=q .
$$
Since $f$ is strictly increasing with $f(q)=q$ for every $q \in \mathbb{Q}^{>0}$ we deduce that $f(x)=x$ for every $x>0$. It is easily checked that this satisfies the functional equation.
|
f(x)=x
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Problem. Find all functions $f:(0, \infty) \rightarrow(0, \infty)$ such that
$$
f\left(y(f(x))^{3}+x\right)=x^{3} f(y)+f(x)
$$
for all $x, y>0$.
|
. Setting $y=\frac{t}{f(x)^{3}}$ we get
$$
f(x+t)=x^{3} f\left(\frac{t}{f(x)^{3}}\right)+f(x)
$$
for every $x, t>0$.
From (1) it is immediate that $f$ is increasing.
Claim. $f(1)=1$
Proof of Claim. Let $c=f(1)$. If $c<1$, taking $x=1$ and $y=\frac{1}{1-c^{3}}$ we have $y-y c^{3}=1$, so $y f(1)^{3}+1=y$ and $f\left(y f(1)^{3}+1\right)=f(y)=1^{3} f(y)$. Thus $f(1)=0$, a contradiction. Assume now for contradiction that $c>1$. We claim that
$$
f\left(1+c^{3}+\cdots+c^{3 n}\right)=(n+1) c
$$
for every $n \in \mathbb{N}$. We proceed by induction, the case $n=0$ being trivial. The inductive step follows easily by taking $x=1, t=c^{3}+c^{6}+\cdots+c^{3(k+1)}$ in (1).
Now taking $x=1+c^{3}+\cdots+c^{3 n-3}, t=c^{3 n}$ in (1) we get
$$
(n+1) c=f\left(1+c^{3}+\cdots+c^{3 n}\right)=\left(1+c^{3}+\cdots+c^{3 n-3}\right)^{3} f\left(\frac{c^{3 n}}{(c n)^{3}}\right)+n c
$$
giving
$$
f\left(\frac{c^{3 n-3}}{n^{3}}\right)=\frac{c}{\left(1+c^{3}+\cdots+c^{3 n}\right)^{3}}<c=f(1) \Longrightarrow \frac{c^{3 n-3}}{n^{3}}<1 .
$$
But this leads to a contradiction if $n$ is large enough.
Now for $x=1$ we get $f(y+1)=f(y)+1$ and since $f(1)=1$ inductively we get $f(n)=n$ for every $n \in \mathbb{N}$. For $m, n \in \mathbb{N}$, setting $x=n, y=q=m / n$ we get
$$
m n^{2}+n=f\left(q n^{3}+n\right)=f\left(y f(x)^{3}+x\right)=x^{3} f(y)+f(x)=n^{3} f(q)+n \Longrightarrow f(q)=q .
$$
Since $f$ is strictly increasing with $f(q)=q$ for every $q \in \mathbb{Q}^{>0}$ we deduce that $f(x)=x$ for every $x>0$. It is easily checked that this satisfies the functional equation.
|
{
"resource_path": "Balkan_MO/segmented/en-2022-BMO-type1.jsonl",
"problem_match": "# Problem 3",
"solution_match": "\nSolution 1"
}
|
1d8b9eca-2f95-5ed8-b541-906cc371dc5c
| 604,899
|
Problem. Find all functions $f:(0, \infty) \rightarrow(0, \infty)$ such that
$$
f\left(y(f(x))^{3}+x\right)=x^{3} f(y)+f(x)
$$
for all $x, y>0$.
|
. We can also derive a contradiction in the case $c>1$ as follows:
Since $f$ is strictly increasing then
$$
f(y)+f(1)=f\left(y f(1)^{3}+1\right)>f\left(y f(1)^{3}\right) \Longrightarrow f\left(c^{3} y\right)<f(y)+c
$$
for every $y>0$. So by induction we get $f\left(c^{3 n}\right)<(n+1) c$ for every $n \in \mathbb{N}$. Setting $x=c^{3 n}$ and $t=c^{3 n+3}-c^{3 n}$ in (1) we get
$$
(n+2) c>f\left(c^{3 n+3}\right)>f\left(c^{3 n+3}\right)-f\left(c^{3 n}\right)=c^{9 n} f\left(\frac{c^{3 n+3}-c^{3 n}}{f\left(c^{3 n}\right)^{3}}\right)>c^{9 n} f\left(\frac{c^{3 n+3}-c^{3 n}}{c^{3}(n+1)^{3}}\right)
$$
But
$$
\frac{c^{3 n+3}-c^{3 n}}{c^{3}(n+1)^{3}}=\frac{c^{3 n}}{(n+1)^{3}} \cdot \frac{c^{3}-1}{c^{3}}>1
$$
for $c$ large enough. So $(n+2) c>c^{9 n+1}$ which leads to a contradiction if $n$ is large enough.
|
proof
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Problem. Find all functions $f:(0, \infty) \rightarrow(0, \infty)$ such that
$$
f\left(y(f(x))^{3}+x\right)=x^{3} f(y)+f(x)
$$
for all $x, y>0$.
|
. We can also derive a contradiction in the case $c>1$ as follows:
Since $f$ is strictly increasing then
$$
f(y)+f(1)=f\left(y f(1)^{3}+1\right)>f\left(y f(1)^{3}\right) \Longrightarrow f\left(c^{3} y\right)<f(y)+c
$$
for every $y>0$. So by induction we get $f\left(c^{3 n}\right)<(n+1) c$ for every $n \in \mathbb{N}$. Setting $x=c^{3 n}$ and $t=c^{3 n+3}-c^{3 n}$ in (1) we get
$$
(n+2) c>f\left(c^{3 n+3}\right)>f\left(c^{3 n+3}\right)-f\left(c^{3 n}\right)=c^{9 n} f\left(\frac{c^{3 n+3}-c^{3 n}}{f\left(c^{3 n}\right)^{3}}\right)>c^{9 n} f\left(\frac{c^{3 n+3}-c^{3 n}}{c^{3}(n+1)^{3}}\right)
$$
But
$$
\frac{c^{3 n+3}-c^{3 n}}{c^{3}(n+1)^{3}}=\frac{c^{3 n}}{(n+1)^{3}} \cdot \frac{c^{3}-1}{c^{3}}>1
$$
for $c$ large enough. So $(n+2) c>c^{9 n+1}$ which leads to a contradiction if $n$ is large enough.
|
{
"resource_path": "Balkan_MO/segmented/en-2022-BMO-type1.jsonl",
"problem_match": "# Problem 3",
"solution_match": "\nSolution 2"
}
|
1d8b9eca-2f95-5ed8-b541-906cc371dc5c
| 604,899
|
Problem. Consider an $n \times n$ grid consisting of $n^{2}$ unit cells, where $n \geqslant 3$ is a given odd positive integer. First, Dionysus colours each cell either red or blue. It is known that a frog can hop from one cell to another if and only if these cells have the same colour and share at least one vertex. Then, Xanthias views the colouring and next places $k$ frogs on the cells so that each of the $n^{2}$ cells can be reached by a frog in a finite number (possibly zero) of hops. Find the least value of $k$ for which this is always possible regardless of the colouring chosen by Dionysus.
Note. Dionysus and Xanthias are characters from the play of Aristophanes 'frogs'. Dionysus is the known god of wine and Xanthias is his witty slave.
|
. Let $G$ be the graph whose vertices are all $(n+1)^{2}$ vertices of the grid and where two vertices are adjacent if and only if they are adjacent in the grid and moreover the two cells in either side of the corresponding edge have different colours.
The connnected components of $G$, excluding the isolated vertices, are precisely the boundaries between pairs of monochromatic regions each of which can be covered by a single frog. Each time we add one of these components in the grid, it creates exactly one new monochromatic region. So the number of frogs required is one more than the number of such components of $G$.
It is easy to check that every corner vertex of the grid has degree 0 , every boundary vertex of the grid has degree 0 or 1 and every 'internal' vertex of the grid has degree 0,2 or 4 . It is also easy to see that every component of $G$ which is not an isolated vertex must contain at least four vertices unless it is the boundary of a single corner of the grid, in which case it contains only three vertices.
Writing $N$ for the number of components which are not isolated vertices, we see that in total they contain at least $4 N-4$ vertices. (As at most four of them contain 3 vertices and all others contain 4 vertices.) Since we also have at least 4 components which are isolated vertices, then $4 N=(4 N-4)+4 \leqslant(n+1)^{2}$. Thus $N \leqslant \frac{(n+1)^{2}}{4}$ and therefore the minimal number of frogs required is $\frac{(n+1)^{2}}{4}+1$.
This bound for $n=2 m+1$ is achieved by putting coordinates $(x, y)$ with $x, y \in\{0,1, \ldots, 2 m\}$ in the cells and colouring red all cells both of whose coordinates are even, and blue all other cells. An example for $n=9$ is shown above.
|
\frac{(n+1)^{2}}{4}+1
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Problem. Consider an $n \times n$ grid consisting of $n^{2}$ unit cells, where $n \geqslant 3$ is a given odd positive integer. First, Dionysus colours each cell either red or blue. It is known that a frog can hop from one cell to another if and only if these cells have the same colour and share at least one vertex. Then, Xanthias views the colouring and next places $k$ frogs on the cells so that each of the $n^{2}$ cells can be reached by a frog in a finite number (possibly zero) of hops. Find the least value of $k$ for which this is always possible regardless of the colouring chosen by Dionysus.
Note. Dionysus and Xanthias are characters from the play of Aristophanes 'frogs'. Dionysus is the known god of wine and Xanthias is his witty slave.
|
. Let $G$ be the graph whose vertices are all $(n+1)^{2}$ vertices of the grid and where two vertices are adjacent if and only if they are adjacent in the grid and moreover the two cells in either side of the corresponding edge have different colours.
The connnected components of $G$, excluding the isolated vertices, are precisely the boundaries between pairs of monochromatic regions each of which can be covered by a single frog. Each time we add one of these components in the grid, it creates exactly one new monochromatic region. So the number of frogs required is one more than the number of such components of $G$.
It is easy to check that every corner vertex of the grid has degree 0 , every boundary vertex of the grid has degree 0 or 1 and every 'internal' vertex of the grid has degree 0,2 or 4 . It is also easy to see that every component of $G$ which is not an isolated vertex must contain at least four vertices unless it is the boundary of a single corner of the grid, in which case it contains only three vertices.
Writing $N$ for the number of components which are not isolated vertices, we see that in total they contain at least $4 N-4$ vertices. (As at most four of them contain 3 vertices and all others contain 4 vertices.) Since we also have at least 4 components which are isolated vertices, then $4 N=(4 N-4)+4 \leqslant(n+1)^{2}$. Thus $N \leqslant \frac{(n+1)^{2}}{4}$ and therefore the minimal number of frogs required is $\frac{(n+1)^{2}}{4}+1$.
This bound for $n=2 m+1$ is achieved by putting coordinates $(x, y)$ with $x, y \in\{0,1, \ldots, 2 m\}$ in the cells and colouring red all cells both of whose coordinates are even, and blue all other cells. An example for $n=9$ is shown above.
|
{
"resource_path": "Balkan_MO/segmented/en-2022-BMO-type1.jsonl",
"problem_match": "# Problem 4",
"solution_match": "\nSolution 1"
}
|
b85b77fe-2996-59e8-8f01-d8b24287296b
| 604,919
|
Problem. Consider an $n \times n$ grid consisting of $n^{2}$ unit cells, where $n \geqslant 3$ is a given odd positive integer. First, Dionysus colours each cell either red or blue. It is known that a frog can hop from one cell to another if and only if these cells have the same colour and share at least one vertex. Then, Xanthias views the colouring and next places $k$ frogs on the cells so that each of the $n^{2}$ cells can be reached by a frog in a finite number (possibly zero) of hops. Find the least value of $k$ for which this is always possible regardless of the colouring chosen by Dionysus.
Note. Dionysus and Xanthias are characters from the play of Aristophanes 'frogs'. Dionysus is the known god of wine and Xanthias is his witty slave.
|
. Consider an $n \times m$ grid with $n, m \geqslant 3$ being odd. We say that a column is of 'Type A' if, when partitioned into its monochromatic pieces, the first and last piece have the same colour with each one containing at least two cells. Otherwise we say that that it is of 'Type B'.
It is enough to show that the number $F$ of frogs required satisfies the inequality
$$
F \leqslant \frac{(m+1)(n+1)}{4}+1-C
$$
where $C$ is the number of boundary columns of Type $A$.
We will proceed by induction but we first need a preliminary result.
Claim. Consider two neighbouring columns of height $n$ which when taken alone need $k$ and $\ell$ frogs respectively. Let $k+t$ be the number of frogs required when both columns are taken together. (It is allowed for $t$ to be negative.) Then the maximum value of $t$ is given by the following table according to the types of the two columns:
| Column 1 | Column 2 | $t$ |
| :---: | :---: | :---: |
| $A$ | $A$ | $\min \left\{\frac{\ell+1}{2}, \frac{n-k}{2}\right\}$ |
| $A$ | $B$ | $\min \left\{\frac{\ell+1}{2}, \frac{n-k+2}{2}\right\}$ |
| $B$ | $A$ | $\min \left\{\frac{\ell-1}{2}, \frac{n-k}{2}\right\}$ |
| $B$ | $B$ | $\min \left\{\frac{\ell}{2}, \frac{n-k+1}{2}\right\}$ |
Proof of Claim. Note that for every two consecutive monochromatic regions of the second column, one can be covered by a frog from the first column. This is because there is a cell in the first column which neighbours both of them and a from can jump from it to the region of the corresponding colour. So the new frogs needed is at most $\frac{\ell+1}{2}$. Furthermore, if we have equality, then $\ell$ must be odd so its top and bottom cell have the same colour. Furthermore the neighbouring cells in the first column must be of opposite colour, so the first column is of Type $A$. If the first column is of Type $B$ and the second column is of Type $A$, then even $\frac{\ell}{2}$ cannot be achieved. If it could, then $\ell$ ought to be even but this contradicts the fact that the second column is of type $A$.
We draw the $k-1$ horizontal lines separating the first column into monochromatics regions and suppose that those they have heights $h_{1}, h_{2}, \ldots, h_{k}$. Note that the cells touching these lines in the second column do not need any frog as a frog from the first column can jump to them. So the remaining cells are partitioned in columns of heights $h_{1}-1, h_{2}-2, \ldots, h_{k-1}-$ $2, h_{k}-1$ all of whose cells to the left are the same colour. Now in each one of them we will need at most $\frac{h_{1}}{2}, \frac{h_{2}-1}{2}, \ldots, \frac{h_{k-1}-1}{2}, \frac{h_{k}}{2}$ frogs. Their sum is $\frac{n-k+2}{2}$ so we need at most that many frogs. Equality holds only if $h_{1}, h_{k}$ are even and the other $h_{i}$ 's are odd. In that case,
since their sum is equal to $n$ which is odd we must have that $k$ is odd. So the first column must be of Type $A$. Furthermore, if the second column is of Type $A$, then the first and last monochromatic regions need at most $\frac{h_{1}-1}{2}$ and $\frac{h_{k}-1}{2}$ new frogs respectively. So the total number of new frogs needed is at most $\frac{n-k}{2}$.
Suppose now that $m=3$ and the middle column needs $k$ frogs. So depending on the type of the three columns we need at most the following number of frogs:
| Column 1 | Column 2 | Column 3 | Number of Frogs |
| :---: | :---: | :---: | :---: |
| $A$ | $A$ | $A$ | $k+\frac{n-k}{2}+\frac{n-k}{2}=n$ |
| $A$ | $A$ | $B$ | $k+\frac{n-k}{2}+\frac{n-k+2}{2}=n+1$ |
| $A$ | $B$ | $A$ | $k+\frac{n-k}{2}+\frac{n-k}{2}=n$ |
| $A$ | $B$ | $B$ | $k+\frac{n-k}{2}+\frac{n-k+1}{2}=n-\frac{1}{2}$ |
| $B$ | $A$ | $A$ | $k+\frac{n-k+2}{2}+\frac{n-k}{2}=n+1$ |
| $B$ | $A$ | $B$ | $k+\frac{n-k+2}{2}+\frac{n-k+2}{2}=n+2$ |
| $B$ | $B$ | $A$ | $k+\frac{n-k+1}{2}+\frac{n-k}{2}=n+\frac{1}{2}$ |
| $B$ | $B$ | $B$ | $k+\frac{n-k+1}{2}+\frac{n-k+1}{2}=n+1$ |
This proves (1) for the case $m=3$ as the claim is $F \leqslant n+2-C$ and it can be checked that this is satisfied in all cases.
Suppose now by induction that the result is true for $m$ and we are trying to prove it for $m+2$. We attach two columns at the end of the table. We need to show that we need additionally at most $\frac{n+1}{2}+C_{\text {old }}-C_{\text {new }}$ number of frogs.
Suppose they need $k$ and $\ell$ frogs respectively. So depending on the type of these two columns with the previous one we need at most the following additional number of frogs:
| Column 1 | Column 2 | Column 3 | Number of Frogs |
| :---: | :---: | :---: | :---: |
| $A$ | $A$ | $A$ | $\frac{k+1}{2}+\frac{n-k}{2}=\frac{n+1}{2}$ |
| $A$ | $A$ | $B$ | $\frac{k+1}{2}+\frac{n-k+2}{2}=\frac{n+3}{2}$ |
| $A$ | $B$ | $A$ | $\frac{k+1}{2}+\frac{n-k}{2}=\frac{n+1}{2}$ |
| $A$ | $B$ | $B$ | $\frac{k+1}{2}+\frac{n-k+1}{2}=\frac{n+2}{2}$ |
| $B$ | $A$ | $A$ | $\frac{k-1}{2}+\frac{n-k}{2}=\frac{n-1}{2}$ |
| $B$ | $A$ | $B$ | $\frac{k-1}{2}+\frac{n-k+2}{2}=\frac{n+1}{2}$ |
| $B$ | $B$ | $A$ | $\frac{k}{2}+\frac{n-k}{2}=\frac{n}{2}$ |
| $B$ | $B$ | $B$ | $\frac{k}{2}+\frac{n-k+1}{2}=\frac{n+1}{2}$ |
It can now be checked (using also that $n$ is odd) that this completes the inductive step.
|
not found
|
Yes
|
Incomplete
|
math-word-problem
|
Combinatorics
|
Problem. Consider an $n \times n$ grid consisting of $n^{2}$ unit cells, where $n \geqslant 3$ is a given odd positive integer. First, Dionysus colours each cell either red or blue. It is known that a frog can hop from one cell to another if and only if these cells have the same colour and share at least one vertex. Then, Xanthias views the colouring and next places $k$ frogs on the cells so that each of the $n^{2}$ cells can be reached by a frog in a finite number (possibly zero) of hops. Find the least value of $k$ for which this is always possible regardless of the colouring chosen by Dionysus.
Note. Dionysus and Xanthias are characters from the play of Aristophanes 'frogs'. Dionysus is the known god of wine and Xanthias is his witty slave.
|
. Consider an $n \times m$ grid with $n, m \geqslant 3$ being odd. We say that a column is of 'Type A' if, when partitioned into its monochromatic pieces, the first and last piece have the same colour with each one containing at least two cells. Otherwise we say that that it is of 'Type B'.
It is enough to show that the number $F$ of frogs required satisfies the inequality
$$
F \leqslant \frac{(m+1)(n+1)}{4}+1-C
$$
where $C$ is the number of boundary columns of Type $A$.
We will proceed by induction but we first need a preliminary result.
Claim. Consider two neighbouring columns of height $n$ which when taken alone need $k$ and $\ell$ frogs respectively. Let $k+t$ be the number of frogs required when both columns are taken together. (It is allowed for $t$ to be negative.) Then the maximum value of $t$ is given by the following table according to the types of the two columns:
| Column 1 | Column 2 | $t$ |
| :---: | :---: | :---: |
| $A$ | $A$ | $\min \left\{\frac{\ell+1}{2}, \frac{n-k}{2}\right\}$ |
| $A$ | $B$ | $\min \left\{\frac{\ell+1}{2}, \frac{n-k+2}{2}\right\}$ |
| $B$ | $A$ | $\min \left\{\frac{\ell-1}{2}, \frac{n-k}{2}\right\}$ |
| $B$ | $B$ | $\min \left\{\frac{\ell}{2}, \frac{n-k+1}{2}\right\}$ |
Proof of Claim. Note that for every two consecutive monochromatic regions of the second column, one can be covered by a frog from the first column. This is because there is a cell in the first column which neighbours both of them and a from can jump from it to the region of the corresponding colour. So the new frogs needed is at most $\frac{\ell+1}{2}$. Furthermore, if we have equality, then $\ell$ must be odd so its top and bottom cell have the same colour. Furthermore the neighbouring cells in the first column must be of opposite colour, so the first column is of Type $A$. If the first column is of Type $B$ and the second column is of Type $A$, then even $\frac{\ell}{2}$ cannot be achieved. If it could, then $\ell$ ought to be even but this contradicts the fact that the second column is of type $A$.
We draw the $k-1$ horizontal lines separating the first column into monochromatics regions and suppose that those they have heights $h_{1}, h_{2}, \ldots, h_{k}$. Note that the cells touching these lines in the second column do not need any frog as a frog from the first column can jump to them. So the remaining cells are partitioned in columns of heights $h_{1}-1, h_{2}-2, \ldots, h_{k-1}-$ $2, h_{k}-1$ all of whose cells to the left are the same colour. Now in each one of them we will need at most $\frac{h_{1}}{2}, \frac{h_{2}-1}{2}, \ldots, \frac{h_{k-1}-1}{2}, \frac{h_{k}}{2}$ frogs. Their sum is $\frac{n-k+2}{2}$ so we need at most that many frogs. Equality holds only if $h_{1}, h_{k}$ are even and the other $h_{i}$ 's are odd. In that case,
since their sum is equal to $n$ which is odd we must have that $k$ is odd. So the first column must be of Type $A$. Furthermore, if the second column is of Type $A$, then the first and last monochromatic regions need at most $\frac{h_{1}-1}{2}$ and $\frac{h_{k}-1}{2}$ new frogs respectively. So the total number of new frogs needed is at most $\frac{n-k}{2}$.
Suppose now that $m=3$ and the middle column needs $k$ frogs. So depending on the type of the three columns we need at most the following number of frogs:
| Column 1 | Column 2 | Column 3 | Number of Frogs |
| :---: | :---: | :---: | :---: |
| $A$ | $A$ | $A$ | $k+\frac{n-k}{2}+\frac{n-k}{2}=n$ |
| $A$ | $A$ | $B$ | $k+\frac{n-k}{2}+\frac{n-k+2}{2}=n+1$ |
| $A$ | $B$ | $A$ | $k+\frac{n-k}{2}+\frac{n-k}{2}=n$ |
| $A$ | $B$ | $B$ | $k+\frac{n-k}{2}+\frac{n-k+1}{2}=n-\frac{1}{2}$ |
| $B$ | $A$ | $A$ | $k+\frac{n-k+2}{2}+\frac{n-k}{2}=n+1$ |
| $B$ | $A$ | $B$ | $k+\frac{n-k+2}{2}+\frac{n-k+2}{2}=n+2$ |
| $B$ | $B$ | $A$ | $k+\frac{n-k+1}{2}+\frac{n-k}{2}=n+\frac{1}{2}$ |
| $B$ | $B$ | $B$ | $k+\frac{n-k+1}{2}+\frac{n-k+1}{2}=n+1$ |
This proves (1) for the case $m=3$ as the claim is $F \leqslant n+2-C$ and it can be checked that this is satisfied in all cases.
Suppose now by induction that the result is true for $m$ and we are trying to prove it for $m+2$. We attach two columns at the end of the table. We need to show that we need additionally at most $\frac{n+1}{2}+C_{\text {old }}-C_{\text {new }}$ number of frogs.
Suppose they need $k$ and $\ell$ frogs respectively. So depending on the type of these two columns with the previous one we need at most the following additional number of frogs:
| Column 1 | Column 2 | Column 3 | Number of Frogs |
| :---: | :---: | :---: | :---: |
| $A$ | $A$ | $A$ | $\frac{k+1}{2}+\frac{n-k}{2}=\frac{n+1}{2}$ |
| $A$ | $A$ | $B$ | $\frac{k+1}{2}+\frac{n-k+2}{2}=\frac{n+3}{2}$ |
| $A$ | $B$ | $A$ | $\frac{k+1}{2}+\frac{n-k}{2}=\frac{n+1}{2}$ |
| $A$ | $B$ | $B$ | $\frac{k+1}{2}+\frac{n-k+1}{2}=\frac{n+2}{2}$ |
| $B$ | $A$ | $A$ | $\frac{k-1}{2}+\frac{n-k}{2}=\frac{n-1}{2}$ |
| $B$ | $A$ | $B$ | $\frac{k-1}{2}+\frac{n-k+2}{2}=\frac{n+1}{2}$ |
| $B$ | $B$ | $A$ | $\frac{k}{2}+\frac{n-k}{2}=\frac{n}{2}$ |
| $B$ | $B$ | $B$ | $\frac{k}{2}+\frac{n-k+1}{2}=\frac{n+1}{2}$ |
It can now be checked (using also that $n$ is odd) that this completes the inductive step.
|
{
"resource_path": "Balkan_MO/segmented/en-2022-BMO-type1.jsonl",
"problem_match": "# Problem 4",
"solution_match": "\nSolution 2"
}
|
b85b77fe-2996-59e8-8f01-d8b24287296b
| 604,919
|
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x, y \in \mathbb{R}$,
$$
x f(x+f(y))=(y-x) f(f(x))
$$
|
Answer: For any real $c, f(x)=c-x$ for all $x \in \mathbb{R}$ and $f(x)=0$ for all $x \in \mathbb{R}$.
Let $P(x, y)$ denote the proposition that $x$ and $y$ satisfy the given equation. $P(0,1)$ gives us $f(f(0))=0$.
From $P(x, x)$ we get that $x f(x+f(x))=0$ for all $x \in \mathbb{R}$, which together with $f(f(0))=$ 0 gives us $f(x+f(x))=0$ for all $x$.
Now let $t$ be any real number such that $f(t)=0$. If $y$ is any number, we have from $P(t-f(y), y)$ the equality
$$
(y+f(y)-t) f(f(t-f(y)))=0
$$
for all $y$ and all $t$ such that $f(t)=0$. So, by taking $y=f(0)$ we obtain
$$
(f(0)-t) f(f(t))=0 \quad \text { and hence } \quad(f(0)-t) f(0)=0
$$
Recall that as $t$ with $f(t)=0$ we can take $x+f(x)$ for any real number $x$. If for all reals $x$ we have $x+f(x)=f(0)$, then $f(x)$ must be of the form $f(x)=c-x$ for some real $c$. It is straightforward that all functions of this form are indeed solutions.
Otherwise we can find some $a \neq 0$ so that $a+f(a) \neq f(0)$. If $t=a+f(a)$ in (A1-1), then $f(0)$ must be equal to 0 . Now $P(x, 0)$ gives us $f(f(x))=-f(x)$ for all real $x$. From here $P(x, x+f(x))$ gives us $x f(x)=-f(x)^{2}$ for all real $x$, which means for every $x$ either $f(x)=0$ or $f(x)=-x$.
Let us assume that in this case there is some $b \neq 0$ so that $f(b)=-b$. For any $y$ we get from $P(b, y)$ and $f(f(b))=-f(b)=b$ the equality $b f(b+f(y))=(y-b) b$, which gives us $f(b+f(y))=y-b$ for all $y$. If $y \neq b$, then the right hand side in the previous equality is not zero, so we must have $f(b+f(y))=-b-f(y)$, which means that $-b-f(y)=y-b$, or that $f(y)=-y$ for all real $y$. But we already covered this solution (take $c=0$ above). If there is no such $b$, then $f(x)=0$ for all $x$, which gives us the final solution.
Thus, all such functions are of the form $c-x$ for real $c$ or the zero function.
|
proof
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x, y \in \mathbb{R}$,
$$
x f(x+f(y))=(y-x) f(f(x))
$$
|
Answer: For any real $c, f(x)=c-x$ for all $x \in \mathbb{R}$ and $f(x)=0$ for all $x \in \mathbb{R}$.
Let $P(x, y)$ denote the proposition that $x$ and $y$ satisfy the given equation. $P(0,1)$ gives us $f(f(0))=0$.
From $P(x, x)$ we get that $x f(x+f(x))=0$ for all $x \in \mathbb{R}$, which together with $f(f(0))=$ 0 gives us $f(x+f(x))=0$ for all $x$.
Now let $t$ be any real number such that $f(t)=0$. If $y$ is any number, we have from $P(t-f(y), y)$ the equality
$$
(y+f(y)-t) f(f(t-f(y)))=0
$$
for all $y$ and all $t$ such that $f(t)=0$. So, by taking $y=f(0)$ we obtain
$$
(f(0)-t) f(f(t))=0 \quad \text { and hence } \quad(f(0)-t) f(0)=0
$$
Recall that as $t$ with $f(t)=0$ we can take $x+f(x)$ for any real number $x$. If for all reals $x$ we have $x+f(x)=f(0)$, then $f(x)$ must be of the form $f(x)=c-x$ for some real $c$. It is straightforward that all functions of this form are indeed solutions.
Otherwise we can find some $a \neq 0$ so that $a+f(a) \neq f(0)$. If $t=a+f(a)$ in (A1-1), then $f(0)$ must be equal to 0 . Now $P(x, 0)$ gives us $f(f(x))=-f(x)$ for all real $x$. From here $P(x, x+f(x))$ gives us $x f(x)=-f(x)^{2}$ for all real $x$, which means for every $x$ either $f(x)=0$ or $f(x)=-x$.
Let us assume that in this case there is some $b \neq 0$ so that $f(b)=-b$. For any $y$ we get from $P(b, y)$ and $f(f(b))=-f(b)=b$ the equality $b f(b+f(y))=(y-b) b$, which gives us $f(b+f(y))=y-b$ for all $y$. If $y \neq b$, then the right hand side in the previous equality is not zero, so we must have $f(b+f(y))=-b-f(y)$, which means that $-b-f(y)=y-b$, or that $f(y)=-y$ for all real $y$. But we already covered this solution (take $c=0$ above). If there is no such $b$, then $f(x)=0$ for all $x$, which gives us the final solution.
Thus, all such functions are of the form $c-x$ for real $c$ or the zero function.
|
{
"resource_path": "Balkan_MO/segmented/en-2023-BMO-type1.jsonl",
"problem_match": "\nProblem 1.",
"solution_match": "\nSolution."
}
|
62e924b4-6d70-563a-9c8e-191a2c7a2396
| 604,945
|
In triangle $A B C$, the incircle touches sides $B C, C A, A B$ at $D, E, F$ respectively. Assume there exists a point $X$ on the line $E F$ such that
$$
\angle X B C=\angle X C B=45^{\circ} .
$$
Let $M$ be the midpoint of the arc $B C$ on the circumcircle of $A B C$ not containing $A$. Prove that the line $M D$ passes through $E$ or $F$.
|
. We first state a well-known lemma.
Lemma: In triangle $A B C$, let $D, E, F$ be the points of tangency of the incircle to the sides $B C, C A, A B$ and let $I$ be the incenter. Then the intersection of $E F$ and $B I$ lies on the circle of diameter $B C$.
Proof: Let $S$ be the intersection of $E F$ and $B I . \angle B I C=\angle E F C$, hence $S$ lies on the circumcircle of $F I C$ in which $I C$ is a diameter. Thus $\angle I S C=90^{\circ}$, hence $\angle B S C=90^{\circ}$.
Returning to the problem, let $I$ be the incenter. The lemma implies that the two intersection points of $E F$ with the circle of diameter $B C$ are precisely the intersection points of $E F$ with $B I$ and $C I$. We have $\angle B X C=90^{\circ}$, therefore either $B X$ or $C X$ is an internal angle bisector, which means either $\angle B=90^{\circ}$ or $\angle C=90^{\circ}$.
Assume, without loss of generality, that $\angle B=90^{\circ}$. Then we have $\angle A M C=90^{\circ}$, so $M$ is the second intersection point of $A I$ with the circle of diameter $A C$, thus the lemma implies that $M$ lies on $D F$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
In triangle $A B C$, the incircle touches sides $B C, C A, A B$ at $D, E, F$ respectively. Assume there exists a point $X$ on the line $E F$ such that
$$
\angle X B C=\angle X C B=45^{\circ} .
$$
Let $M$ be the midpoint of the arc $B C$ on the circumcircle of $A B C$ not containing $A$. Prove that the line $M D$ passes through $E$ or $F$.
|
. We first state a well-known lemma.
Lemma: In triangle $A B C$, let $D, E, F$ be the points of tangency of the incircle to the sides $B C, C A, A B$ and let $I$ be the incenter. Then the intersection of $E F$ and $B I$ lies on the circle of diameter $B C$.
Proof: Let $S$ be the intersection of $E F$ and $B I . \angle B I C=\angle E F C$, hence $S$ lies on the circumcircle of $F I C$ in which $I C$ is a diameter. Thus $\angle I S C=90^{\circ}$, hence $\angle B S C=90^{\circ}$.
Returning to the problem, let $I$ be the incenter. The lemma implies that the two intersection points of $E F$ with the circle of diameter $B C$ are precisely the intersection points of $E F$ with $B I$ and $C I$. We have $\angle B X C=90^{\circ}$, therefore either $B X$ or $C X$ is an internal angle bisector, which means either $\angle B=90^{\circ}$ or $\angle C=90^{\circ}$.
Assume, without loss of generality, that $\angle B=90^{\circ}$. Then we have $\angle A M C=90^{\circ}$, so $M$ is the second intersection point of $A I$ with the circle of diameter $A C$, thus the lemma implies that $M$ lies on $D F$.
|
{
"resource_path": "Balkan_MO/segmented/en-2023-BMO-type1.jsonl",
"problem_match": "\nProblem 2.",
"solution_match": "\nSolution 1"
}
|
22bb783e-be47-5aaa-9760-57cffe9db953
| 604,953
|
In triangle $A B C$, the incircle touches sides $B C, C A, A B$ at $D, E, F$ respectively. Assume there exists a point $X$ on the line $E F$ such that
$$
\angle X B C=\angle X C B=45^{\circ} .
$$
Let $M$ be the midpoint of the arc $B C$ on the circumcircle of $A B C$ not containing $A$. Prove that the line $M D$ passes through $E$ or $F$.
|
.
Let $I$ be the incenter of $\triangle A B C$ and let $K$ be the foot of the perpendicular from $D$ to $E F$. We begin by proving that $B K X C$ is cyclic, which can be done in two ways:
First Way. Note that $\angle K F D=90^{\circ}-\frac{\angle C}{2}$ and $\angle K E D=90^{\circ}-\frac{\angle B}{2}$, so by using $K D \perp E F$, we have $\frac{F K}{E D}=\frac{\tan \frac{\angle C}{2}}{\tan \frac{\angle B}{2}}$. Similarly, since $\angle I B D=\frac{\angle B}{2}$ and $\angle I C D=\frac{\angle C}{2}$, by using $I D \perp B C$, we have $\frac{B F}{E C}=\frac{B D}{D C}=\frac{\tan \frac{\angle C}{2}}{\tan \frac{\angle B}{2}}$. Then, since $\angle B F K=90^{\circ}+\frac{\angle A}{2}=$ $\angle K E C$, we conclude that $\triangle B F K$ and $\triangle C E K$ are similar, so $\angle F K B=\angle C K E$ which shows line $E F$ is the external-angle bisector of $\angle B K C$. Therefore, $X$ lies on both the perpendicular bisector of the segment $B C$ and the external angle bisector of $\angle B K C$ (and these lines are distinct) thus it lies on the circumcirle of $\triangle B K C$ (in particular the midpoint of arc $B K C$ ).
Second Way. Let $T$ be the intersection of $E F$ and $B C$, and $N$ be the midpoint of the segment $B C$. It is well-known that $(T, D ; B, C)$ is harmonic and $T B \cdot T C=T D \cdot T N$. On the other hand, since $X B=X C$, we have $\angle X N D=90^{\circ}=\angle X K D$, so $X K D N$ is cyclic and $T D \cdot T N=T K \cdot T X$. Therefore, we have $T K \cdot T X=T B \cdot T C$, which implies $B K X C$ is cyclic.
Now we will show that either $\angle B=90^{\circ}$ or $\angle C=90^{\circ}$. Note that $\angle B K C=\angle B X C=$ $90^{\circ}=\angle F K D=\angle E K D$ and $\angle F K B=\angle E K C$. Then we have
$$
\angle F K B=\angle B K D=\angle D K C=\angle C K E=45^{\circ} .
$$
Hence $B K$ bisects $\angle F K D$, but $B$ also lies on the perpendicular bisector of $D F$. Therefore, either $F K D B$ is cyclic or $K F=K D$ while the former implies that $\angle B=180^{\circ}-\angle F K D=$ $90^{\circ}$. In the latter case, we have $K B \perp F D$, which gives $90^{\circ}-\frac{\angle C}{2}=\angle K F D=90^{\circ}-$ $\angle F K B=45^{\circ}$ and so $\angle C=90^{\circ}$ as desired.
We consider, without loss of generality, the case where $\angle B=90^{\circ}$. Observing that $A, I, M$ are collinear we get:
$$
\angle C D I=90^{\circ}=\angle C B A=\angle C M A=\angle C M I
$$
Hence $M D I C$ is cyclic so:
$$
\angle M D C=\angle M I C=180^{\circ}-\angle C I A=180^{\circ}-\left(90^{\circ}+\frac{\angle B}{2}\right)=45^{\circ}
$$
We also have $\angle F D B=90^{\circ}-\frac{\angle B}{2}=45^{\circ}$ so $\angle F D B=\angle M D C$ and thus $M, D, F$ are collinear as required.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
In triangle $A B C$, the incircle touches sides $B C, C A, A B$ at $D, E, F$ respectively. Assume there exists a point $X$ on the line $E F$ such that
$$
\angle X B C=\angle X C B=45^{\circ} .
$$
Let $M$ be the midpoint of the arc $B C$ on the circumcircle of $A B C$ not containing $A$. Prove that the line $M D$ passes through $E$ or $F$.
|
.
Let $I$ be the incenter of $\triangle A B C$ and let $K$ be the foot of the perpendicular from $D$ to $E F$. We begin by proving that $B K X C$ is cyclic, which can be done in two ways:
First Way. Note that $\angle K F D=90^{\circ}-\frac{\angle C}{2}$ and $\angle K E D=90^{\circ}-\frac{\angle B}{2}$, so by using $K D \perp E F$, we have $\frac{F K}{E D}=\frac{\tan \frac{\angle C}{2}}{\tan \frac{\angle B}{2}}$. Similarly, since $\angle I B D=\frac{\angle B}{2}$ and $\angle I C D=\frac{\angle C}{2}$, by using $I D \perp B C$, we have $\frac{B F}{E C}=\frac{B D}{D C}=\frac{\tan \frac{\angle C}{2}}{\tan \frac{\angle B}{2}}$. Then, since $\angle B F K=90^{\circ}+\frac{\angle A}{2}=$ $\angle K E C$, we conclude that $\triangle B F K$ and $\triangle C E K$ are similar, so $\angle F K B=\angle C K E$ which shows line $E F$ is the external-angle bisector of $\angle B K C$. Therefore, $X$ lies on both the perpendicular bisector of the segment $B C$ and the external angle bisector of $\angle B K C$ (and these lines are distinct) thus it lies on the circumcirle of $\triangle B K C$ (in particular the midpoint of arc $B K C$ ).
Second Way. Let $T$ be the intersection of $E F$ and $B C$, and $N$ be the midpoint of the segment $B C$. It is well-known that $(T, D ; B, C)$ is harmonic and $T B \cdot T C=T D \cdot T N$. On the other hand, since $X B=X C$, we have $\angle X N D=90^{\circ}=\angle X K D$, so $X K D N$ is cyclic and $T D \cdot T N=T K \cdot T X$. Therefore, we have $T K \cdot T X=T B \cdot T C$, which implies $B K X C$ is cyclic.
Now we will show that either $\angle B=90^{\circ}$ or $\angle C=90^{\circ}$. Note that $\angle B K C=\angle B X C=$ $90^{\circ}=\angle F K D=\angle E K D$ and $\angle F K B=\angle E K C$. Then we have
$$
\angle F K B=\angle B K D=\angle D K C=\angle C K E=45^{\circ} .
$$
Hence $B K$ bisects $\angle F K D$, but $B$ also lies on the perpendicular bisector of $D F$. Therefore, either $F K D B$ is cyclic or $K F=K D$ while the former implies that $\angle B=180^{\circ}-\angle F K D=$ $90^{\circ}$. In the latter case, we have $K B \perp F D$, which gives $90^{\circ}-\frac{\angle C}{2}=\angle K F D=90^{\circ}-$ $\angle F K B=45^{\circ}$ and so $\angle C=90^{\circ}$ as desired.
We consider, without loss of generality, the case where $\angle B=90^{\circ}$. Observing that $A, I, M$ are collinear we get:
$$
\angle C D I=90^{\circ}=\angle C B A=\angle C M A=\angle C M I
$$
Hence $M D I C$ is cyclic so:
$$
\angle M D C=\angle M I C=180^{\circ}-\angle C I A=180^{\circ}-\left(90^{\circ}+\frac{\angle B}{2}\right)=45^{\circ}
$$
We also have $\angle F D B=90^{\circ}-\frac{\angle B}{2}=45^{\circ}$ so $\angle F D B=\angle M D C$ and thus $M, D, F$ are collinear as required.
|
{
"resource_path": "Balkan_MO/segmented/en-2023-BMO-type1.jsonl",
"problem_match": "\nProblem 2.",
"solution_match": "# Solution 2"
}
|
22bb783e-be47-5aaa-9760-57cffe9db953
| 604,953
|
For each positive integer $n$, denote by $\omega(n)$ the number of distinct prime divisors of $n$ (for example, $\omega(1)=0$ and $\omega(12)=2$ ). Find all polynomials $P(x)$ with integer coefficients, such that whenever $n$ is a positive integer satisfying $\omega(n)>2023^{2023}$, then $P(n)$ is also a positive integer with
$$
\omega(n) \geq \omega(P(n)) .
$$
|
Answer: All polynomials of the form $f(x)=x^{m}$ for some $m \in \mathbb{Z}^{+}$and $f(x)=c$ for some $c \in \mathbb{Z}^{+}$with $\omega(c) \leq 2023^{2023}+1$.
First of all we prove the following (well-known) Lemma. Lemma: Let $f(x)$ be a nonconstant polynomial with integer coefficients. Then, the number of primes $p$ such that $p \mid f(n)$ for some $n$ is infinite.
Proof: If $f(0)=0$, then the Lemma is obvious. Otherwise, define the polynomial
$$
g(x)=\frac{f(x f(0))}{f(0)}
$$
which has integer coefficients. Observe that $g(0)=1$ and if $g$ satisfies the property of the Lemma, then so does $f$. So, we need to prove that there are infinitely many primes $p$ such that $p \mid g(n)$ for some $n$. Suppose, for the sake of contradiction that the number of such primes is finite, and let those be $p_{1}, \ldots, p_{k}$. Then, set $n=N p_{1} \cdots p_{k}$ for some large $N$, such that $|g(n)|>1$. It is evident that $g(n)$ has a prime divisor, but it is none of the $p_{i}$ 's. This is a contradiction and therefore the result follows.
Let $M=2023{ }^{2023}+1$. Observe that constant polynomials $f(x)=c$ with $c \in \mathbb{N}$ such that $\omega(c) \leq M$ satisfy the conditions of the problem. On the other hand, if $f(x)=c$ with $\omega(c)>M$, we can choose some $n$ such that $\omega(n)=M$ to see that the condition of the problem is not satisfied. Next, we look for non-constant polynomials that satisfy the conditions of the problem. Let $f(x)=x^{m} g(x)$, where $m \geq 0$ and $g(x)$ is a polynomial with $g(0) \neq 0$. We claim that $g$ is a constant polynomial. Indeed, if it is not, then (due to the Lemma) there exist pairwise distinct primes $q_{1}, \ldots, q_{M+1}$ and non-zero integers $n_{1}, \ldots, n_{M+1}$ such that $q_{i}>|g(0)|$ and $q_{i} \mid g\left(n_{i}\right)$ for $i=1,2, \ldots, M+1$. Set $n=p_{1} p_{2} \cdots p_{M}$, where $p_{1}, \ldots, p_{M}$ are distinct primes such that
$$
p_{1} \equiv n_{i} \quad\left(\bmod q_{i}\right), \forall i=1,2, \ldots, M+1
$$
and
$$
p_{j} \equiv 1 \quad\left(\bmod q_{i}\right), \forall i=1,2, \ldots, M+1, \forall j=2,3, \ldots, M
$$
Observe that since $q_{i}>|g(0)|$, it is impossible to have $q_{i} \mid n_{i}$, so the existence of such primes is guaranteed by the Chinese Remainder Theorem and the Dirichlet's Theorem. Now, for every $i=1,2, \ldots, M+1$ we can see that $n=p_{1} \cdots p_{M} \equiv n_{i}\left(\bmod q_{i}\right)$, which means that
$$
g(n) \equiv g\left(n_{i}\right) \equiv 0 \quad\left(\bmod q_{i}\right) \forall i=1,2, \ldots, M+1
$$
Thus, $\omega(f(n)) \geq \omega(g(n)) \geq M+1>M=\omega(n)$, which gives the desired contradiction. Therefore, $f(x)=c x^{m}$, for some $m \geq 1$ (since $f$ was non-constant). If $c<0$, take some $n$ with $\omega(n)=M$ to see that $f(n)$ is negative and so, does not satisfy the conditions of the problem. If $c>1$, choose some $n$ with $\omega(n)=M$ and $\operatorname{gcd}(n, c)=1$ to observe that $f$ cannot satisfy the conditions of the problem. This means that $f(x)=x^{m}$ (which is for sure a solution to the problem) for some $m \geq 1$ and $f(x)=c$ for some $c \in \mathbb{Z}^{+}$with $\omega(c) \leq M$ are the only polynomials that satisfy the conditions of the problem.
|
proof
|
Yes
|
Yes
|
math-word-problem
|
Number Theory
|
For each positive integer $n$, denote by $\omega(n)$ the number of distinct prime divisors of $n$ (for example, $\omega(1)=0$ and $\omega(12)=2$ ). Find all polynomials $P(x)$ with integer coefficients, such that whenever $n$ is a positive integer satisfying $\omega(n)>2023^{2023}$, then $P(n)$ is also a positive integer with
$$
\omega(n) \geq \omega(P(n)) .
$$
|
Answer: All polynomials of the form $f(x)=x^{m}$ for some $m \in \mathbb{Z}^{+}$and $f(x)=c$ for some $c \in \mathbb{Z}^{+}$with $\omega(c) \leq 2023^{2023}+1$.
First of all we prove the following (well-known) Lemma. Lemma: Let $f(x)$ be a nonconstant polynomial with integer coefficients. Then, the number of primes $p$ such that $p \mid f(n)$ for some $n$ is infinite.
Proof: If $f(0)=0$, then the Lemma is obvious. Otherwise, define the polynomial
$$
g(x)=\frac{f(x f(0))}{f(0)}
$$
which has integer coefficients. Observe that $g(0)=1$ and if $g$ satisfies the property of the Lemma, then so does $f$. So, we need to prove that there are infinitely many primes $p$ such that $p \mid g(n)$ for some $n$. Suppose, for the sake of contradiction that the number of such primes is finite, and let those be $p_{1}, \ldots, p_{k}$. Then, set $n=N p_{1} \cdots p_{k}$ for some large $N$, such that $|g(n)|>1$. It is evident that $g(n)$ has a prime divisor, but it is none of the $p_{i}$ 's. This is a contradiction and therefore the result follows.
Let $M=2023{ }^{2023}+1$. Observe that constant polynomials $f(x)=c$ with $c \in \mathbb{N}$ such that $\omega(c) \leq M$ satisfy the conditions of the problem. On the other hand, if $f(x)=c$ with $\omega(c)>M$, we can choose some $n$ such that $\omega(n)=M$ to see that the condition of the problem is not satisfied. Next, we look for non-constant polynomials that satisfy the conditions of the problem. Let $f(x)=x^{m} g(x)$, where $m \geq 0$ and $g(x)$ is a polynomial with $g(0) \neq 0$. We claim that $g$ is a constant polynomial. Indeed, if it is not, then (due to the Lemma) there exist pairwise distinct primes $q_{1}, \ldots, q_{M+1}$ and non-zero integers $n_{1}, \ldots, n_{M+1}$ such that $q_{i}>|g(0)|$ and $q_{i} \mid g\left(n_{i}\right)$ for $i=1,2, \ldots, M+1$. Set $n=p_{1} p_{2} \cdots p_{M}$, where $p_{1}, \ldots, p_{M}$ are distinct primes such that
$$
p_{1} \equiv n_{i} \quad\left(\bmod q_{i}\right), \forall i=1,2, \ldots, M+1
$$
and
$$
p_{j} \equiv 1 \quad\left(\bmod q_{i}\right), \forall i=1,2, \ldots, M+1, \forall j=2,3, \ldots, M
$$
Observe that since $q_{i}>|g(0)|$, it is impossible to have $q_{i} \mid n_{i}$, so the existence of such primes is guaranteed by the Chinese Remainder Theorem and the Dirichlet's Theorem. Now, for every $i=1,2, \ldots, M+1$ we can see that $n=p_{1} \cdots p_{M} \equiv n_{i}\left(\bmod q_{i}\right)$, which means that
$$
g(n) \equiv g\left(n_{i}\right) \equiv 0 \quad\left(\bmod q_{i}\right) \forall i=1,2, \ldots, M+1
$$
Thus, $\omega(f(n)) \geq \omega(g(n)) \geq M+1>M=\omega(n)$, which gives the desired contradiction. Therefore, $f(x)=c x^{m}$, for some $m \geq 1$ (since $f$ was non-constant). If $c<0$, take some $n$ with $\omega(n)=M$ to see that $f(n)$ is negative and so, does not satisfy the conditions of the problem. If $c>1$, choose some $n$ with $\omega(n)=M$ and $\operatorname{gcd}(n, c)=1$ to observe that $f$ cannot satisfy the conditions of the problem. This means that $f(x)=x^{m}$ (which is for sure a solution to the problem) for some $m \geq 1$ and $f(x)=c$ for some $c \in \mathbb{Z}^{+}$with $\omega(c) \leq M$ are the only polynomials that satisfy the conditions of the problem.
|
{
"resource_path": "Balkan_MO/segmented/en-2023-BMO-type1.jsonl",
"problem_match": "\nProblem 3.",
"solution_match": "\nSolution."
}
|
1c941603-0ac9-5552-a454-31d564025b5a
| 604,972
|
Find the greatest integer $k \leq 2023$ for which the following holds: whenever Alice colours exactly $k$ numbers of the set $\{1,2, \ldots, 2023\}$ in red, Bob can colour some of the remaining uncoloured numbers in blue, such that the sum of the red numbers is the same as the sum of the blue numbers.
|
Answer: 592.
For $k \geq 593$, Alice can color the greatest 593 numbers $1431,1432, \ldots, 2023$ and any other $(k-593)$ numbers so that their sum $s$ would satisfy
$$
s \geq \frac{2023 \cdot 2024}{2}-\frac{1430 \cdot 1431}{2}>\frac{1}{2} \cdot\left(\frac{2023 \cdot 2024}{2}\right),
$$
thus anyhow Bob chooses his numbers, the sum of his numbers will be less than Alice's numbers' sum.
We now show that $k=592$ satisfies the condition. Let $s$ be the sum of Alice's 592 numbers; note that $s<\frac{1}{2} \cdot\left(\frac{2023 \cdot 2024}{2}\right)$. Below is a strategy for Bob to find some of the remaining 1431 numbers so that their sum is
$$
s_{0}=\min \left\{s, \frac{2023.2024}{2}-2 s\right\} \leqslant \frac{1}{3} \cdot\left(\frac{2023 \cdot 2024}{2}\right)
$$
(Clearly, if Bob finds some numbers whose sum is $\frac{2023.2024}{2}-2 s$, then the sum of remaining numbers will be $s$ ).
Case 1. $s_{0} \geq 2024$. Let $s_{0}=2024 a+b$, where $0 \leqslant b \leqslant 2023$. Bob finds two of the remaining numbers with sum $b$ or $2024+b$, then he finds $a$ (or $a-1$ ) pairs among the remaining numbers with sum 2024 . Note that $a \leq 337$ since $s_{0} \leq \frac{1}{3} \cdot\left(\frac{2023 \cdot 2024}{2}\right)$.
The $\left\lfloor\frac{b-1}{2}\right\rfloor$ pairs
$$
(1, b-1),(2, b-2), \ldots,\left(\left\lfloor\frac{b-1}{2}\right\rfloor, b-\left\lfloor\frac{b-1}{2}\right\rfloor\right),
$$
have sum of their components equal to $b$ and the $\left\lfloor\frac{2023+b}{2}\right\rfloor-b$ pairs
$$
(2023, b+1),(2022, b+2), \ldots,\left(2024+b-\left\lfloor\frac{2023+b}{2}\right\rfloor,\left\lfloor\frac{2023+b}{2}\right\rfloor\right)
$$
have sum of their components equal to $2024+b$. The total number of these pairs is
$$
\left\lfloor\frac{2023+b}{2}\right\rfloor-b+\left\lfloor\frac{b-1}{2}\right\rfloor \geq \frac{2022+b}{2}+\frac{b-2}{2}-b=\frac{2020}{2}=1010>592
$$
hence some of these pairs have no red-colored components, so Bob can choose one of these pairs and color those two numbers in blue. Thus 594 numbers are colored so far.
Further, the 1011 pairs
$$
(1,2023),(2,2022), \ldots,(1011,1013)
$$
have sum of the components equal to 2024. Among these, at least $1011-594=417>$ $337 \geq a$ pairs have no components colored, so Bob can choose $a$ (or $a-1$ ) uncolored pairs and color them all blue to achieve a collection of blue numbers with their sum equal to $s_{0}$.
Case 2. $s_{0} \leq 2023$. Note that $s \geq 1+2+\ldots+592>2023$, thus we have $s_{0}=$ $\frac{2023 \cdot 2024}{2}-2 s$, i.e. $s=\frac{2023 \cdot 2024}{4}-\frac{s_{0}}{2}$.
If $s_{0}>2 \cdot 593$, at least one of the 593 pairs
$$
\left(1, s_{0}-1\right),\left(2, s_{0}-2\right), \ldots,\left(593, s_{0}-593\right)
$$
have no red-colored components, so Bob can choose these two numbers and immediately achieve the sum of $s_{0}$. And if $s_{0} \leq 2 \cdot 593$, then
$s=\frac{2023 \cdot 2024}{4}-\frac{s_{0}}{2} \geq(1432+1433+\ldots+2023)-593=839+(1434+1435+\ldots+2023)$,
hence Alice cannot have colored any of the numbers $1,2, \ldots, 838$. Then Bob can easily choose one or two of these numbers having the sum of $s_{0}$.
|
592
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
Find the greatest integer $k \leq 2023$ for which the following holds: whenever Alice colours exactly $k$ numbers of the set $\{1,2, \ldots, 2023\}$ in red, Bob can colour some of the remaining uncoloured numbers in blue, such that the sum of the red numbers is the same as the sum of the blue numbers.
|
Answer: 592.
For $k \geq 593$, Alice can color the greatest 593 numbers $1431,1432, \ldots, 2023$ and any other $(k-593)$ numbers so that their sum $s$ would satisfy
$$
s \geq \frac{2023 \cdot 2024}{2}-\frac{1430 \cdot 1431}{2}>\frac{1}{2} \cdot\left(\frac{2023 \cdot 2024}{2}\right),
$$
thus anyhow Bob chooses his numbers, the sum of his numbers will be less than Alice's numbers' sum.
We now show that $k=592$ satisfies the condition. Let $s$ be the sum of Alice's 592 numbers; note that $s<\frac{1}{2} \cdot\left(\frac{2023 \cdot 2024}{2}\right)$. Below is a strategy for Bob to find some of the remaining 1431 numbers so that their sum is
$$
s_{0}=\min \left\{s, \frac{2023.2024}{2}-2 s\right\} \leqslant \frac{1}{3} \cdot\left(\frac{2023 \cdot 2024}{2}\right)
$$
(Clearly, if Bob finds some numbers whose sum is $\frac{2023.2024}{2}-2 s$, then the sum of remaining numbers will be $s$ ).
Case 1. $s_{0} \geq 2024$. Let $s_{0}=2024 a+b$, where $0 \leqslant b \leqslant 2023$. Bob finds two of the remaining numbers with sum $b$ or $2024+b$, then he finds $a$ (or $a-1$ ) pairs among the remaining numbers with sum 2024 . Note that $a \leq 337$ since $s_{0} \leq \frac{1}{3} \cdot\left(\frac{2023 \cdot 2024}{2}\right)$.
The $\left\lfloor\frac{b-1}{2}\right\rfloor$ pairs
$$
(1, b-1),(2, b-2), \ldots,\left(\left\lfloor\frac{b-1}{2}\right\rfloor, b-\left\lfloor\frac{b-1}{2}\right\rfloor\right),
$$
have sum of their components equal to $b$ and the $\left\lfloor\frac{2023+b}{2}\right\rfloor-b$ pairs
$$
(2023, b+1),(2022, b+2), \ldots,\left(2024+b-\left\lfloor\frac{2023+b}{2}\right\rfloor,\left\lfloor\frac{2023+b}{2}\right\rfloor\right)
$$
have sum of their components equal to $2024+b$. The total number of these pairs is
$$
\left\lfloor\frac{2023+b}{2}\right\rfloor-b+\left\lfloor\frac{b-1}{2}\right\rfloor \geq \frac{2022+b}{2}+\frac{b-2}{2}-b=\frac{2020}{2}=1010>592
$$
hence some of these pairs have no red-colored components, so Bob can choose one of these pairs and color those two numbers in blue. Thus 594 numbers are colored so far.
Further, the 1011 pairs
$$
(1,2023),(2,2022), \ldots,(1011,1013)
$$
have sum of the components equal to 2024. Among these, at least $1011-594=417>$ $337 \geq a$ pairs have no components colored, so Bob can choose $a$ (or $a-1$ ) uncolored pairs and color them all blue to achieve a collection of blue numbers with their sum equal to $s_{0}$.
Case 2. $s_{0} \leq 2023$. Note that $s \geq 1+2+\ldots+592>2023$, thus we have $s_{0}=$ $\frac{2023 \cdot 2024}{2}-2 s$, i.e. $s=\frac{2023 \cdot 2024}{4}-\frac{s_{0}}{2}$.
If $s_{0}>2 \cdot 593$, at least one of the 593 pairs
$$
\left(1, s_{0}-1\right),\left(2, s_{0}-2\right), \ldots,\left(593, s_{0}-593\right)
$$
have no red-colored components, so Bob can choose these two numbers and immediately achieve the sum of $s_{0}$. And if $s_{0} \leq 2 \cdot 593$, then
$s=\frac{2023 \cdot 2024}{4}-\frac{s_{0}}{2} \geq(1432+1433+\ldots+2023)-593=839+(1434+1435+\ldots+2023)$,
hence Alice cannot have colored any of the numbers $1,2, \ldots, 838$. Then Bob can easily choose one or two of these numbers having the sum of $s_{0}$.
|
{
"resource_path": "Balkan_MO/segmented/en-2023-BMO-type1.jsonl",
"problem_match": "\nProblem 4.",
"solution_match": "\nSolution."
}
|
3d6962fe-c852-5854-8352-f3b3c572a632
| 604,984
|
Let $A B C$ be an acute-angled triangle with $A C>A B$ and let $D$ be the foot of the $A$-angle bisector on $B C$. The reflections of lines $A B$ and $A C$ in line $B C$ meet $A C$ and $A B$ at points $E$ and $F$ respectively. A line through $D$ meets $A C$ and $A B$ at $G$ and $H$ respectively such that $G$ lies strictly between $A$ and $C$ while $H$ lies strictly between $B$ and $F$. Prove that the circumcircles of $\triangle E D G$ and $\triangle F D H$ are tangent to each other.
|
. Let $X$ and $Y$ lie on the tangent to the circumcircle of $\triangle E D G$ on the opposite side to $D$ as shown in the figure below. Regarding diagram dependency, the acute condition with $A C>A B$ ensures $E$ lies on extension of $C A$ beyond $A$, and $F$ lies on extension of $A B$ beyond $B$. The condition on $\ell$ means the points lie in the orders $E, A, G, C$ and $A, B, H, F$.
Using the alternate segment theorem, the condition that $\odot E D G$ and $\odot F D H$ are tangent at $D$ can be rewritten as
$$
\Varangle H F D=\Varangle Y D H .
$$
But using the same theorem, we get $\Varangle Y D H=\Varangle X D G=\Varangle D E G$. So we can remove $G, H$ from the figure, and it is sufficient to prove that $\Varangle D E A=\Varangle D F B$.
The reflection property means that $A D$ and $B D$ are external angle bisectors in $\triangle E A B$ and hence $D$
is the $E$-excentre of this triangle. Thus $D E$ (internally) bisects $\Varangle B E A$, giving
$$
\Varangle D E A=\Varangle D E B .
$$
Now observe that the pairs of lines $(B E, C E)$ and $(B F, C F)$ are reflections in $B C$ thus $E, F$ are reflections in $B C$. Also $D$ is its own reflection in $B C$. Hence $\Varangle D E B=\Varangle D F B$ and so
$$
\Varangle D E A=\Varangle D E B=\Varangle D F B,
$$
as required.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute-angled triangle with $A C>A B$ and let $D$ be the foot of the $A$-angle bisector on $B C$. The reflections of lines $A B$ and $A C$ in line $B C$ meet $A C$ and $A B$ at points $E$ and $F$ respectively. A line through $D$ meets $A C$ and $A B$ at $G$ and $H$ respectively such that $G$ lies strictly between $A$ and $C$ while $H$ lies strictly between $B$ and $F$. Prove that the circumcircles of $\triangle E D G$ and $\triangle F D H$ are tangent to each other.
|
. Let $X$ and $Y$ lie on the tangent to the circumcircle of $\triangle E D G$ on the opposite side to $D$ as shown in the figure below. Regarding diagram dependency, the acute condition with $A C>A B$ ensures $E$ lies on extension of $C A$ beyond $A$, and $F$ lies on extension of $A B$ beyond $B$. The condition on $\ell$ means the points lie in the orders $E, A, G, C$ and $A, B, H, F$.
Using the alternate segment theorem, the condition that $\odot E D G$ and $\odot F D H$ are tangent at $D$ can be rewritten as
$$
\Varangle H F D=\Varangle Y D H .
$$
But using the same theorem, we get $\Varangle Y D H=\Varangle X D G=\Varangle D E G$. So we can remove $G, H$ from the figure, and it is sufficient to prove that $\Varangle D E A=\Varangle D F B$.
The reflection property means that $A D$ and $B D$ are external angle bisectors in $\triangle E A B$ and hence $D$
is the $E$-excentre of this triangle. Thus $D E$ (internally) bisects $\Varangle B E A$, giving
$$
\Varangle D E A=\Varangle D E B .
$$
Now observe that the pairs of lines $(B E, C E)$ and $(B F, C F)$ are reflections in $B C$ thus $E, F$ are reflections in $B C$. Also $D$ is its own reflection in $B C$. Hence $\Varangle D E B=\Varangle D F B$ and so
$$
\Varangle D E A=\Varangle D E B=\Varangle D F B,
$$
as required.
|
{
"resource_path": "Balkan_MO/segmented/en-2024-BMO-type1.jsonl",
"problem_match": "\nProblem 1.",
"solution_match": "\nSolution 1"
}
|
d70b96ff-607b-5cca-9e18-9bcb23679048
| 604,993
|
Let $n \geq k \geq 3$ be integers. Show that for every integer sequence $1 \leq a_{1}<a_{2}<\ldots<$ $a_{k} \leq n$ one can choose non-negative integers $b_{1}, b_{2}, \ldots, b_{k}$, satisfying the following conditions:
(i) $0 \leq b_{i} \leq n$ for each $1 \leq i \leq k$,
(ii) all the positive $b_{i}$ are distinct,
(iii) the sums $a_{i}+b_{i}, 1 \leq i \leq k$, form a permutation of the first $k$ terms of a non-constant arithmetic progression.
|
. Let the resulting progression be $A n s:=\left\{a_{k}-(k-1), a_{k}-(k-2), \ldots, a_{k}\right\}$ and $a_{t}$ be the largest number not belonging to $A n s$. Clearly the set $A n s \backslash\left\{a_{1}, a_{2}, \ldots, a_{k}\right\}$ has cardinality $t$; let its members be $c_{1}>c_{2}>\cdots>c_{t}$. Define $b_{j}:=c_{j}-a_{j}$ for $1 \leq j \leq t$ or zero otherwise. Since $\left\{c_{j}\right\}$ is decreasing and $\left\{a_{j}\right\}$ is increasing, all $b_{j}$ are distinct and clearly $b_{1}<n$. After we add $b_{j}$ to $a_{j}$ we get a permutation of Ans as desired.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Let $n \geq k \geq 3$ be integers. Show that for every integer sequence $1 \leq a_{1}<a_{2}<\ldots<$ $a_{k} \leq n$ one can choose non-negative integers $b_{1}, b_{2}, \ldots, b_{k}$, satisfying the following conditions:
(i) $0 \leq b_{i} \leq n$ for each $1 \leq i \leq k$,
(ii) all the positive $b_{i}$ are distinct,
(iii) the sums $a_{i}+b_{i}, 1 \leq i \leq k$, form a permutation of the first $k$ terms of a non-constant arithmetic progression.
|
. Let the resulting progression be $A n s:=\left\{a_{k}-(k-1), a_{k}-(k-2), \ldots, a_{k}\right\}$ and $a_{t}$ be the largest number not belonging to $A n s$. Clearly the set $A n s \backslash\left\{a_{1}, a_{2}, \ldots, a_{k}\right\}$ has cardinality $t$; let its members be $c_{1}>c_{2}>\cdots>c_{t}$. Define $b_{j}:=c_{j}-a_{j}$ for $1 \leq j \leq t$ or zero otherwise. Since $\left\{c_{j}\right\}$ is decreasing and $\left\{a_{j}\right\}$ is increasing, all $b_{j}$ are distinct and clearly $b_{1}<n$. After we add $b_{j}$ to $a_{j}$ we get a permutation of Ans as desired.
|
{
"resource_path": "Balkan_MO/segmented/en-2024-BMO-type1.jsonl",
"problem_match": "\nProblem 2.",
"solution_match": "\nSolution 1"
}
|
8820a8f3-b899-5a15-9bb2-0e08ec9d94eb
| 605,005
|
Let $n \geq k \geq 3$ be integers. Show that for every integer sequence $1 \leq a_{1}<a_{2}<\ldots<$ $a_{k} \leq n$ one can choose non-negative integers $b_{1}, b_{2}, \ldots, b_{k}$, satisfying the following conditions:
(i) $0 \leq b_{i} \leq n$ for each $1 \leq i \leq k$,
(ii) all the positive $b_{i}$ are distinct,
(iii) the sums $a_{i}+b_{i}, 1 \leq i \leq k$, form a permutation of the first $k$ terms of a non-constant arithmetic progression.
|
. Let the resulting progression be $A n s:=\left\{a_{k}-(k-1), a_{k}-(k-2), \ldots, a_{k}\right\}$.
We proceed with the following reduction. Let $\delta$ be the smallest $b$ we used before (in the beginning it is $n$ ). While $a_{1} \notin A n s$ we map $a_{1}$ to the largest element $q$ of $A n s \backslash\left\{a_{1}, a_{2}, \ldots, a_{k}\right\}$ and put $\delta_{\text {new }}:=b_{1}:=q-a_{1}$. Now we rearrange the sequence of $a$-s. We do not touch $\operatorname{Ans} \cap\left\{a_{1}, a_{2}, \ldots, a_{k}\right\}$ so every $b$ is defined at most once (in the end undefined $b$-s become zeros). Also $b<\delta$ and $\delta$ decreases at each step, because $q$ decreases and $a_{1}$ grows, and hence all nonzero $b$-s are distinct.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Let $n \geq k \geq 3$ be integers. Show that for every integer sequence $1 \leq a_{1}<a_{2}<\ldots<$ $a_{k} \leq n$ one can choose non-negative integers $b_{1}, b_{2}, \ldots, b_{k}$, satisfying the following conditions:
(i) $0 \leq b_{i} \leq n$ for each $1 \leq i \leq k$,
(ii) all the positive $b_{i}$ are distinct,
(iii) the sums $a_{i}+b_{i}, 1 \leq i \leq k$, form a permutation of the first $k$ terms of a non-constant arithmetic progression.
|
. Let the resulting progression be $A n s:=\left\{a_{k}-(k-1), a_{k}-(k-2), \ldots, a_{k}\right\}$.
We proceed with the following reduction. Let $\delta$ be the smallest $b$ we used before (in the beginning it is $n$ ). While $a_{1} \notin A n s$ we map $a_{1}$ to the largest element $q$ of $A n s \backslash\left\{a_{1}, a_{2}, \ldots, a_{k}\right\}$ and put $\delta_{\text {new }}:=b_{1}:=q-a_{1}$. Now we rearrange the sequence of $a$-s. We do not touch $\operatorname{Ans} \cap\left\{a_{1}, a_{2}, \ldots, a_{k}\right\}$ so every $b$ is defined at most once (in the end undefined $b$-s become zeros). Also $b<\delta$ and $\delta$ decreases at each step, because $q$ decreases and $a_{1}$ grows, and hence all nonzero $b$-s are distinct.
|
{
"resource_path": "Balkan_MO/segmented/en-2024-BMO-type1.jsonl",
"problem_match": "\nProblem 2.",
"solution_match": "\nSolution 2"
}
|
8820a8f3-b899-5a15-9bb2-0e08ec9d94eb
| 605,005
|
Let $a$ and $b$ be distinct positive integers such that $3^{a}+2$ is divisible by $3^{b}+2$. Prove that $a>b^{2}$ 。
|
. Obviously we have $a>b$. Let $a=b q+r$, where $0 \leq r<b$. Then
$$
3^{a} \equiv 3^{b q+r} \equiv(-2)^{q} \cdot 3^{r} \equiv-2 \quad\left(\bmod 3^{b}+2\right)
$$
So $3^{b}+2$ divides $A=(-2)^{q} .3^{r}+2$ and it follows that
$$
\left|(-2)^{q} \cdot 3^{r}+2\right| \geq 3^{b}+2 \text { or }(-2)^{q} \cdot 3^{r}+2=0
$$
We make case distinction:
1. $(-2)^{q} .3^{r}+2=0$. Then $q=1$ and $r=0$ or $a=b$, a contradiction.
2. $q$ is even. Then
$$
A=2^{q} \cdot 3^{r}+2=\left(3^{b}+2\right) \cdot k
$$
Consider both sides of the last equation modulo $3^{r}$. Since $b>r$ :
$$
2 \equiv 2^{q} \cdot 3^{r}+2=\left(3^{b}+2\right) k \equiv 2 k \quad\left(\bmod 3^{r}\right)
$$
so it follows that $3^{r} \mid k-1$. If $k=1$ then $2^{q} .3^{r}=3^{b}$, a contradiction. So $k \geq 3^{r}+1$, and therefore:
$$
A=2^{q} .3^{r}+2=\left(3^{b}+2\right) k \geq\left(3^{b}+2\right)\left(3^{r}+1\right)>3^{b} .3^{r}+2
$$
It follows that
$$
2^{q} .3^{r}>3^{b} .3^{r} \text {, i.e. } 2^{q}>3^{b} \text {, which implies } 3^{b^{2}}<2^{b q}<3^{b q} \leq 3^{b q+r}=3^{a} \text {. }
$$
Consequently $a>b^{2}$.
3. If $q$ is odd. Then
$$
2^{q} \cdot 3^{r}-2=\left(3^{b}+2\right) k
$$
Considering both sides of the last equation modulo $3^{r}$, and since $b>r$, we get: $k+1$ is divisible by $3^{r}$ and therefore $k \geq 3^{r}-1$. Thus $r>0$ because $k>0$, and:
$$
\begin{array}{r}
2^{q} .3^{r}-2=\left(3^{b}+2\right) k \geq\left(3^{b}+2\right)\left(3^{r}-1\right), \text { and therefore } \\
2^{q} .3^{r}>\left(3^{b}+2\right)\left(3^{r}-1\right)>3^{b}\left(3^{r}-1\right)>3^{b} \frac{3^{r}}{2}, \text { which shows } \\
2^{q+1}>3^{b} .
\end{array}
$$
But for $q>1$ we have $2^{q+1}<3^{q}$, which combined with the above inequality, implies that $3^{b^{2}}<2^{(q+1) b}<3^{q b} \leq 3^{a}$, q.e.d. Finally, If $q=1$ then $2^{q} .3^{r}-2=\left(3^{b}+2\right) k$ and consequently $2.3^{r}-2 \geq 3^{b}+2 \geq 3^{r+1}+2>2.3^{r}-2$, a contradiction.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $a$ and $b$ be distinct positive integers such that $3^{a}+2$ is divisible by $3^{b}+2$. Prove that $a>b^{2}$ 。
|
. Obviously we have $a>b$. Let $a=b q+r$, where $0 \leq r<b$. Then
$$
3^{a} \equiv 3^{b q+r} \equiv(-2)^{q} \cdot 3^{r} \equiv-2 \quad\left(\bmod 3^{b}+2\right)
$$
So $3^{b}+2$ divides $A=(-2)^{q} .3^{r}+2$ and it follows that
$$
\left|(-2)^{q} \cdot 3^{r}+2\right| \geq 3^{b}+2 \text { or }(-2)^{q} \cdot 3^{r}+2=0
$$
We make case distinction:
1. $(-2)^{q} .3^{r}+2=0$. Then $q=1$ and $r=0$ or $a=b$, a contradiction.
2. $q$ is even. Then
$$
A=2^{q} \cdot 3^{r}+2=\left(3^{b}+2\right) \cdot k
$$
Consider both sides of the last equation modulo $3^{r}$. Since $b>r$ :
$$
2 \equiv 2^{q} \cdot 3^{r}+2=\left(3^{b}+2\right) k \equiv 2 k \quad\left(\bmod 3^{r}\right)
$$
so it follows that $3^{r} \mid k-1$. If $k=1$ then $2^{q} .3^{r}=3^{b}$, a contradiction. So $k \geq 3^{r}+1$, and therefore:
$$
A=2^{q} .3^{r}+2=\left(3^{b}+2\right) k \geq\left(3^{b}+2\right)\left(3^{r}+1\right)>3^{b} .3^{r}+2
$$
It follows that
$$
2^{q} .3^{r}>3^{b} .3^{r} \text {, i.e. } 2^{q}>3^{b} \text {, which implies } 3^{b^{2}}<2^{b q}<3^{b q} \leq 3^{b q+r}=3^{a} \text {. }
$$
Consequently $a>b^{2}$.
3. If $q$ is odd. Then
$$
2^{q} \cdot 3^{r}-2=\left(3^{b}+2\right) k
$$
Considering both sides of the last equation modulo $3^{r}$, and since $b>r$, we get: $k+1$ is divisible by $3^{r}$ and therefore $k \geq 3^{r}-1$. Thus $r>0$ because $k>0$, and:
$$
\begin{array}{r}
2^{q} .3^{r}-2=\left(3^{b}+2\right) k \geq\left(3^{b}+2\right)\left(3^{r}-1\right), \text { and therefore } \\
2^{q} .3^{r}>\left(3^{b}+2\right)\left(3^{r}-1\right)>3^{b}\left(3^{r}-1\right)>3^{b} \frac{3^{r}}{2}, \text { which shows } \\
2^{q+1}>3^{b} .
\end{array}
$$
But for $q>1$ we have $2^{q+1}<3^{q}$, which combined with the above inequality, implies that $3^{b^{2}}<2^{(q+1) b}<3^{q b} \leq 3^{a}$, q.e.d. Finally, If $q=1$ then $2^{q} .3^{r}-2=\left(3^{b}+2\right) k$ and consequently $2.3^{r}-2 \geq 3^{b}+2 \geq 3^{r+1}+2>2.3^{r}-2$, a contradiction.
|
{
"resource_path": "Balkan_MO/segmented/en-2024-BMO-type1.jsonl",
"problem_match": "\nProblem 3.",
"solution_match": "\nSolution 1"
}
|
72a043a3-5f0e-5dc8-9a80-40bb5d686c27
| 605,025
|
Let $a$ and $b$ be distinct positive integers such that $3^{a}+2$ is divisible by $3^{b}+2$. Prove that $a>b^{2}$ 。
|
. $D=a-b$, and we shall show $D>b^{2}-b$. We have $3^{b}+2 \mid 3^{a}+2$, so $3^{b}+2 \mid 3^{D}-1$. Let $D=b q+r$ where $r<b$. First suppose that $r \neq 0$. We have
$$
1 \equiv 3^{D} \equiv 3^{b q+r} \equiv(-2)^{q+1} 3^{r-b} \quad\left(\bmod 3^{b}+2\right) \Longrightarrow 3^{b-r} \equiv(-2)^{q+1} \quad\left(\bmod 3^{b}+2\right)
$$
Therefore
$$
3^{b}+2 \leq\left|(-2)^{q+1}-3^{b-r}\right| \leq 2^{q+1}+3^{b-r} \leq 2^{q+1}+3^{b-1}
$$
Hence
$$
2 \times 3^{b-1}+2 \leq 2^{q+1} \Longrightarrow 3^{b-1}<2^{q} \Longrightarrow \frac{\log 3}{\log 2}(b-1)<q
$$
Which yields $D=b q+r>b q>\frac{\log 3}{\log 2} b(b-1) \geq b^{2}-b$ as desired. Now for the case $r=0,(-2)^{q} \equiv 1$ $\left(\bmod 3^{b}+2\right)$ and so
$$
3^{b}+2 \leq\left|(-2)^{q}-1\right| \leq 2^{q}+1 \Longrightarrow 3^{b-1}<3^{b}<2^{q} \Longrightarrow \frac{\log 3}{\log 2}(b-1)<q
$$
and analogous to the previous case, $D=b q+r=b q>\frac{\log 3}{\log 2} b(b-1) \geq b^{2}-b$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $a$ and $b$ be distinct positive integers such that $3^{a}+2$ is divisible by $3^{b}+2$. Prove that $a>b^{2}$ 。
|
. $D=a-b$, and we shall show $D>b^{2}-b$. We have $3^{b}+2 \mid 3^{a}+2$, so $3^{b}+2 \mid 3^{D}-1$. Let $D=b q+r$ where $r<b$. First suppose that $r \neq 0$. We have
$$
1 \equiv 3^{D} \equiv 3^{b q+r} \equiv(-2)^{q+1} 3^{r-b} \quad\left(\bmod 3^{b}+2\right) \Longrightarrow 3^{b-r} \equiv(-2)^{q+1} \quad\left(\bmod 3^{b}+2\right)
$$
Therefore
$$
3^{b}+2 \leq\left|(-2)^{q+1}-3^{b-r}\right| \leq 2^{q+1}+3^{b-r} \leq 2^{q+1}+3^{b-1}
$$
Hence
$$
2 \times 3^{b-1}+2 \leq 2^{q+1} \Longrightarrow 3^{b-1}<2^{q} \Longrightarrow \frac{\log 3}{\log 2}(b-1)<q
$$
Which yields $D=b q+r>b q>\frac{\log 3}{\log 2} b(b-1) \geq b^{2}-b$ as desired. Now for the case $r=0,(-2)^{q} \equiv 1$ $\left(\bmod 3^{b}+2\right)$ and so
$$
3^{b}+2 \leq\left|(-2)^{q}-1\right| \leq 2^{q}+1 \Longrightarrow 3^{b-1}<3^{b}<2^{q} \Longrightarrow \frac{\log 3}{\log 2}(b-1)<q
$$
and analogous to the previous case, $D=b q+r=b q>\frac{\log 3}{\log 2} b(b-1) \geq b^{2}-b$.
|
{
"resource_path": "Balkan_MO/segmented/en-2024-BMO-type1.jsonl",
"problem_match": "\nProblem 3.",
"solution_match": "\nSolution 2"
}
|
72a043a3-5f0e-5dc8-9a80-40bb5d686c27
| 605,025
|
Let $\mathbb{R}^{+}=(0, \infty)$ be the set of all positive real numbers. Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ and polynomials $P(x)$ with non-negative real coefficients such that $P(0)=0$ which satisfy the equality
$$
f(f(x)+P(y))=f(x-y)+2 y
$$
for all real numbers $x>y>0$.
|
. Assume that $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$and the polynomial $P$ with non-negative coefficients and $P(0)=0$ satisfy the conditions of the problem. For positive reals with $x>y$, we shall write $Q(x, y)$ for the relation:
$$
f(f(x)+P(y))=f(x-y)+2 y
$$
1. Step 1. $f(x) \geq x$. Assume that this is not true. Since $P(0)=0$ and $P$ is with non-negative coefficients, $P(x)+x$ is surjective on positive reals. If $f(x)<x$ for some positive real $x$, then setting $y$ such that $y+P(y)=x-f(x)$ (where obviously $y<x$ ), we shall get $f(x)+P(y)=x-y$ and by $Q(x, y), f(f(x)+P(y))=f(x-y)+2 y$, we get $2 y=0$, a contradiction.
2. Step 2. $P(x)=c x$ for some non-negative real $c$. We will show $\operatorname{deg} P \leq 1$ and together with $P(0)=0$ the result will follow. Assume the contrary. Hence there exists a positive $l$ such that $P(x) \geq 2 x$ for all $x \geq l$. By Step 1 we get
$$
\forall x>y \geq l: f(x-y)+2 y=f(f(x)+P(y)) \geq f(x)+P(y) \geq f(x)+2 y
$$
and therefore $f(x-y) \geq f(x)$. We get $f(y) \geq f(2 y) \geq \cdots \geq f(n y) \geq n y$ for all positive integers $n$, which is a contradiction.
3. Step 3. If $c \neq 0$, then $f\left(f(x)+2 z+c^{2}\right)=f(x+1)+2(z-1)+2 c$ for $z>1$. Indeed by $Q(f(x+z)+c z, c)$, we get
$$
f\left(f(f(x+z)+c z)+c^{2}\right)=f(f(x+z)+c z-c)+2 c=f(x+1)+2(z-1)+2 c
$$
On the other hand by $Q(x+z, z)$, we have:
$$
f(x)+2 z+c^{2}=f(f(x+z)+P(z))+c^{2}=f(f(x+z)+c z)+c^{2} .
$$
Substituting in the LHS of $Q(f(x+z)+c z, c)$, we get $f\left(f(x)+2 z+c^{2}\right)=f(x+1)+2(z-1)+2 c$.
4. Step 4. There is $x_{0}$, such that $f(x)$ is linear on $\left(x_{0}, \infty\right)$. If $c \neq 0$, then by Step 3 , fixing $x=1$, we get $f\left(f(1)+2 z+c^{2}\right)=f(2)+2(z-1)+2 c$ which implies that $f$ is linear for $z>f(1)+2+c^{2}$. As for the case $c=0$, consider $y, z \in(0, \infty)$. Pick $x>\max (y, z)$, then by $Q(x, x-y)$ and $Q(x, x-z)$ we get:
$$
f(y)+2(x-y)=f(f(x))=f(z)+2(x-z)
$$
which proves that $f(y)-2 y=f(z)-2 z$ and there fore $f$ is linear on $(0, \infty)$.
5. Step 5. $P(y)=y$ and $f(x)=x$ on $\left(x_{0}, \infty\right)$. By Step 4 , let $f(x)=a x+b$ on $\left(x_{0}, \infty\right)$. Since $f$ takes only positive values, $a \geq 0$. If $a=0$, then by $Q(x+y, y)$ for $y>x_{0}$ we get:
$$
2 y+f(x)=f(f(x+y)+P(y))=f(b+c y)
$$
Since the LHS is not constant, we conclude $c \neq 0$, but then for $y>x_{0} / c$, we get that the RHS equals $b$ which is a contradiction.
Hence $a>0$. Now for $x>x_{0}$ and $x>\left(x_{0}-b\right) / a$ large enough by $P(x+y, y)$ we get:
$a x+b+2 y=f(x)+2 y=f(f(x+y)+P(y))=f(a x+a y+b+c y)=a(a x+a y+b+c y)+b$.
Comparing the coefficients before $x$, we see $a^{2}=a$ and since $a \neq 0, a=1$. Now $2 b=b$ and thus $b=0$. Finally, equalising the coefficients before $y$, we conclude $2=1+c$ and therefore $c=1$.
Now we know that $f(x)=x$ on $\left(x_{0}, \infty\right)$ and $P(y)=y$. Let $y>x_{0}$. Then by $Q(x+y, x)$ we conclude:
$$
f(x)+2 y=f(f(x+y)+P(y))=f(x+y+y)=x+2 y .
$$
Therefore $f(x)=x$ for every $x$. Conversely, it is straightforward that $f(x)=x$ and $P(y)=y$ do indeed satisfy the conditions of the problem.
|
f(x)=x \text{ and } P(y)=y
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $\mathbb{R}^{+}=(0, \infty)$ be the set of all positive real numbers. Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ and polynomials $P(x)$ with non-negative real coefficients such that $P(0)=0$ which satisfy the equality
$$
f(f(x)+P(y))=f(x-y)+2 y
$$
for all real numbers $x>y>0$.
|
. Assume that $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$and the polynomial $P$ with non-negative coefficients and $P(0)=0$ satisfy the conditions of the problem. For positive reals with $x>y$, we shall write $Q(x, y)$ for the relation:
$$
f(f(x)+P(y))=f(x-y)+2 y
$$
1. Step 1. $f(x) \geq x$. Assume that this is not true. Since $P(0)=0$ and $P$ is with non-negative coefficients, $P(x)+x$ is surjective on positive reals. If $f(x)<x$ for some positive real $x$, then setting $y$ such that $y+P(y)=x-f(x)$ (where obviously $y<x$ ), we shall get $f(x)+P(y)=x-y$ and by $Q(x, y), f(f(x)+P(y))=f(x-y)+2 y$, we get $2 y=0$, a contradiction.
2. Step 2. $P(x)=c x$ for some non-negative real $c$. We will show $\operatorname{deg} P \leq 1$ and together with $P(0)=0$ the result will follow. Assume the contrary. Hence there exists a positive $l$ such that $P(x) \geq 2 x$ for all $x \geq l$. By Step 1 we get
$$
\forall x>y \geq l: f(x-y)+2 y=f(f(x)+P(y)) \geq f(x)+P(y) \geq f(x)+2 y
$$
and therefore $f(x-y) \geq f(x)$. We get $f(y) \geq f(2 y) \geq \cdots \geq f(n y) \geq n y$ for all positive integers $n$, which is a contradiction.
3. Step 3. If $c \neq 0$, then $f\left(f(x)+2 z+c^{2}\right)=f(x+1)+2(z-1)+2 c$ for $z>1$. Indeed by $Q(f(x+z)+c z, c)$, we get
$$
f\left(f(f(x+z)+c z)+c^{2}\right)=f(f(x+z)+c z-c)+2 c=f(x+1)+2(z-1)+2 c
$$
On the other hand by $Q(x+z, z)$, we have:
$$
f(x)+2 z+c^{2}=f(f(x+z)+P(z))+c^{2}=f(f(x+z)+c z)+c^{2} .
$$
Substituting in the LHS of $Q(f(x+z)+c z, c)$, we get $f\left(f(x)+2 z+c^{2}\right)=f(x+1)+2(z-1)+2 c$.
4. Step 4. There is $x_{0}$, such that $f(x)$ is linear on $\left(x_{0}, \infty\right)$. If $c \neq 0$, then by Step 3 , fixing $x=1$, we get $f\left(f(1)+2 z+c^{2}\right)=f(2)+2(z-1)+2 c$ which implies that $f$ is linear for $z>f(1)+2+c^{2}$. As for the case $c=0$, consider $y, z \in(0, \infty)$. Pick $x>\max (y, z)$, then by $Q(x, x-y)$ and $Q(x, x-z)$ we get:
$$
f(y)+2(x-y)=f(f(x))=f(z)+2(x-z)
$$
which proves that $f(y)-2 y=f(z)-2 z$ and there fore $f$ is linear on $(0, \infty)$.
5. Step 5. $P(y)=y$ and $f(x)=x$ on $\left(x_{0}, \infty\right)$. By Step 4 , let $f(x)=a x+b$ on $\left(x_{0}, \infty\right)$. Since $f$ takes only positive values, $a \geq 0$. If $a=0$, then by $Q(x+y, y)$ for $y>x_{0}$ we get:
$$
2 y+f(x)=f(f(x+y)+P(y))=f(b+c y)
$$
Since the LHS is not constant, we conclude $c \neq 0$, but then for $y>x_{0} / c$, we get that the RHS equals $b$ which is a contradiction.
Hence $a>0$. Now for $x>x_{0}$ and $x>\left(x_{0}-b\right) / a$ large enough by $P(x+y, y)$ we get:
$a x+b+2 y=f(x)+2 y=f(f(x+y)+P(y))=f(a x+a y+b+c y)=a(a x+a y+b+c y)+b$.
Comparing the coefficients before $x$, we see $a^{2}=a$ and since $a \neq 0, a=1$. Now $2 b=b$ and thus $b=0$. Finally, equalising the coefficients before $y$, we conclude $2=1+c$ and therefore $c=1$.
Now we know that $f(x)=x$ on $\left(x_{0}, \infty\right)$ and $P(y)=y$. Let $y>x_{0}$. Then by $Q(x+y, x)$ we conclude:
$$
f(x)+2 y=f(f(x+y)+P(y))=f(x+y+y)=x+2 y .
$$
Therefore $f(x)=x$ for every $x$. Conversely, it is straightforward that $f(x)=x$ and $P(y)=y$ do indeed satisfy the conditions of the problem.
|
{
"resource_path": "Balkan_MO/segmented/en-2024-BMO-type1.jsonl",
"problem_match": "\nProblem 4.",
"solution_match": "\nSolution 1"
}
|
010d1bc6-5b57-56e7-8f48-ac10ac4a51bb
| 605,048
|
Let $\mathbb{R}^{+}=(0, \infty)$ be the set of all positive real numbers. Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ and polynomials $P(x)$ with non-negative real coefficients such that $P(0)=0$ which satisfy the equality
$$
f(f(x)+P(y))=f(x-y)+2 y
$$
for all real numbers $x>y>0$.
|
. Assume that the function $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$and the polynomial with non-negative coefficients $P(y)=y P_{1}(y)$ satisfy the given equation. Fix $x=x_{0}>0$ and note that:
$$
f\left(f\left(x_{0}+y\right)+P(y)\right)=f\left(x_{0}+y-y\right)+2 y=f\left(x_{0}\right)+2 y
$$
Assume that $g=0$. Then $f(f(x+y))=f(x)+2 y$ for $x, y>0$. Let $x>0$ and $z>0$. Pick $y>0$. Then:
$$
2 y+f(x+z)=f(f(x+y+z))=f(f(x+z+y))=f(x)+2(z+y) .
$$
Therefore $f(x+z)=f(x)+2 z$ for any $x>0$ and $z>0$. Setting $c=f(1)$, we see that $f(z+1)=c+2 z$ for all positive $z$. Therefore if $x, y>1$ we have that $f(x+y)=c+2(x+y-1)>1$. This shows that:
$$
f(f(x+y))=c+2(f(x+y)-1)=3 c+4(x+y)-4
$$
On the other hand $f(x)+2 y=c+2 x+2 y$. Therefore the equality $f(f(x+y))=f(x)+2 y$ is not universally satisfied.
From now on, we assume that $g \neq 0$. Therefore $P$ is strictly increasing with $P(0)=0, \lim _{y \rightarrow \infty} P(y)=$ $\infty$, i.e. $g$ is bijective on $[0, \infty)$ and $P(0)=0$.
Let $x>0, y>0$ and set $u=f(x+y), v=P(y)$. From above, we have $u>0$ and $v>0$. Therefore:
$$
f(f(u+v)+P(v))=f(u)+2 v=f(f(x+y))+2 P(y)
$$
On the other hand $f(u+v)=f(f(x+y)+P(y))=f(x)+2 y$. Therefore we obtain that:
$$
f(f(x)+2 y+P(P(y)))=f(f(x+y))+2 P(y)
$$
Since $g$ is bijective from $(0, \infty)$ to $(0, \infty)$ for any $z>0$ there is $t$ such that $P(t)=z$. Applying this observation to $z=P(P(y))+2 y$ and setting $x^{\prime}=x+t$, we obtain that:
$f(f(x+t+y))+2 P(y)=f\left(f\left(x^{\prime}+y\right)\right)+2 P(y)=f\left(f\left(x^{\prime}\right)+P(P(y))+2 y\right)=f(f(x+t)+P(t))=f(x)+2 t$.
Thus if we denote $h(y)=P(P(y))+2 y$, then $t=P^{(-1)}(h(y))$ and the above equality can be rewritten as:
$f\left(f\left(x+P^{(-1)}(h(y))+y\right)\right)=f(x)+2 P^{(-1)}(h(y))-2 P(y)=f(x)+2 P^{(-1)}(h(y))+2 y-2 y-2 P(y)$.
Let $s(y)=P^{(-1)}(h(y))+y$ and note that since $h$ is continuous and monotone increasing, $g$ is continuous and monotone increasing, then so are $P^{(-1)}$ and consequently $P^{(-1)} \circ h$ and $s$. It is also clear, that $\lim _{y \rightarrow 0} s(y)=0$ and $\lim _{y \rightarrow \infty} s(y)=\infty$. Therefore $s$ is continuously bijective from $[0, \infty)$ to $[0, \infty)$ with $s(0)=0$.
Thus we have:
$$
f(f(x+s(y)))=f(x)+2 s(y)-2 y-2 P(y)
$$
and using that $s$ is invertible, we obtain:
$$
f(f(x+y))=f(x)+2 y-2 s^{(-1)}(y)-2 P\left(s^{(-1)}(y)\right)
$$
Now fix $x_{0}$, then for any $x>x_{0}$ and any $y>0$ we have:
$$
\begin{aligned}
f(x)+2 y-2 s^{(-1)}(y)-2 P\left(s^{(-1)}(y)\right) & =f(f(x+y))=f\left(f\left(x_{0}+x+y-x_{0}\right)\right) \\
& =f\left(x_{0}\right)+2\left(x+y-x_{0}\right)-2 s^{(-1)}\left(x+y-x_{0}\right)-2 P\left(s^{(-1)}\left(x+y-x_{0}\right)\right) .
\end{aligned}
$$
Setting $y=x_{0}$, we get:
$$
f(x)+2 x_{0}-2 s^{(-1)}\left(x_{0}\right)-2 P\left(s^{(-1)}\left(x_{0}\right)\right)=f\left(x_{0}\right)+2 x-2 s^{(-1)}(x)-2 P\left(s^{(-1)}(x)\right) .
$$
Since this equality is valid for any $x>x_{0}$ we actually have that:
$$
f(x)-2 x+2 s^{(-1)}(x)+2 P\left(s^{(-1)}(x)\right)=c \text { for some fixed constant } c \in \mathbb{R} \text { and all } x \in \mathbb{R}^{+} .
$$
Let $\phi(x)=-x+2 s^{(-1)}(x)+2 P\left(s^{(-1)}(x)\right)$. Then $f(x)=x-\phi(x)+c$ and since $\phi$ is a sum of continuous functions that are continuous at 0 . Therefore $f$ is continuous and can be extended to a continuous function on $[0, \infty)$. Back in the original equation we fix $x>0$ and let $y$ tend to 0 . Using the continuity of $f$ and $g$ on $[0, \infty)$ and $P(0)=0$ we obtain:
$$
f(f(x))=\lim _{y \rightarrow 0+} f(f(x)+P(y))=\lim _{y \rightarrow 0+}(f(x-y)+P(y))=f(x)+P(0)=f(x) .
$$
Finally, fixing $x=1$ and varying $y>0$, we obtain:
$$
f(f(1+y)+P(y))=f(1)+2 y
$$
It follows that $f$ takes every value on $(f(1), \infty)$. Therefore for any $y \in(f(1), \infty)$ there is $z$ such that $f(z)=y$. Using that $f(f(z))=f(z)$ we conclude that $f(y)=y$ for all $y \in(f(1), \infty)$.
Now fix $x$ and take $y>f(1)$. Hence
$$
f(x)+2 y=f(f(x+y)+P(y))=f(x+y+P(y))=x+y+P(y)
$$
We conclude $f(x)-x=P(y)-y$ for every $x$ an $y>f(1)$. In particular $f\left(x_{1}\right)-x_{1}=f\left(x_{2}\right)-x_{2}$ for all $x_{1}, x_{2} \in(0, \infty)$ and since $f(x)=x$ for $x \in(f(1), \infty)$, we get $f(x)=x$ on $(0, \infty)$.
Finally, $x+2 y=f(x)+2 y=f(f(x+y)+P(y))=f(x+y)+P(y)=x+y+P(y)$, which shows that $P(y)=y$ for every $y \in(0, \infty)$.
It is also straightforward to check that $f(x)=x$ and $P(y)=y$ satisfy the equality:
$$
f(f(x+y)+P(y))=f(x+2 y)=x+2 y=f(x)+2 y
$$
|
f(x)=x \text{ and } P(y)=y
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $\mathbb{R}^{+}=(0, \infty)$ be the set of all positive real numbers. Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ and polynomials $P(x)$ with non-negative real coefficients such that $P(0)=0$ which satisfy the equality
$$
f(f(x)+P(y))=f(x-y)+2 y
$$
for all real numbers $x>y>0$.
|
. Assume that the function $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$and the polynomial with non-negative coefficients $P(y)=y P_{1}(y)$ satisfy the given equation. Fix $x=x_{0}>0$ and note that:
$$
f\left(f\left(x_{0}+y\right)+P(y)\right)=f\left(x_{0}+y-y\right)+2 y=f\left(x_{0}\right)+2 y
$$
Assume that $g=0$. Then $f(f(x+y))=f(x)+2 y$ for $x, y>0$. Let $x>0$ and $z>0$. Pick $y>0$. Then:
$$
2 y+f(x+z)=f(f(x+y+z))=f(f(x+z+y))=f(x)+2(z+y) .
$$
Therefore $f(x+z)=f(x)+2 z$ for any $x>0$ and $z>0$. Setting $c=f(1)$, we see that $f(z+1)=c+2 z$ for all positive $z$. Therefore if $x, y>1$ we have that $f(x+y)=c+2(x+y-1)>1$. This shows that:
$$
f(f(x+y))=c+2(f(x+y)-1)=3 c+4(x+y)-4
$$
On the other hand $f(x)+2 y=c+2 x+2 y$. Therefore the equality $f(f(x+y))=f(x)+2 y$ is not universally satisfied.
From now on, we assume that $g \neq 0$. Therefore $P$ is strictly increasing with $P(0)=0, \lim _{y \rightarrow \infty} P(y)=$ $\infty$, i.e. $g$ is bijective on $[0, \infty)$ and $P(0)=0$.
Let $x>0, y>0$ and set $u=f(x+y), v=P(y)$. From above, we have $u>0$ and $v>0$. Therefore:
$$
f(f(u+v)+P(v))=f(u)+2 v=f(f(x+y))+2 P(y)
$$
On the other hand $f(u+v)=f(f(x+y)+P(y))=f(x)+2 y$. Therefore we obtain that:
$$
f(f(x)+2 y+P(P(y)))=f(f(x+y))+2 P(y)
$$
Since $g$ is bijective from $(0, \infty)$ to $(0, \infty)$ for any $z>0$ there is $t$ such that $P(t)=z$. Applying this observation to $z=P(P(y))+2 y$ and setting $x^{\prime}=x+t$, we obtain that:
$f(f(x+t+y))+2 P(y)=f\left(f\left(x^{\prime}+y\right)\right)+2 P(y)=f\left(f\left(x^{\prime}\right)+P(P(y))+2 y\right)=f(f(x+t)+P(t))=f(x)+2 t$.
Thus if we denote $h(y)=P(P(y))+2 y$, then $t=P^{(-1)}(h(y))$ and the above equality can be rewritten as:
$f\left(f\left(x+P^{(-1)}(h(y))+y\right)\right)=f(x)+2 P^{(-1)}(h(y))-2 P(y)=f(x)+2 P^{(-1)}(h(y))+2 y-2 y-2 P(y)$.
Let $s(y)=P^{(-1)}(h(y))+y$ and note that since $h$ is continuous and monotone increasing, $g$ is continuous and monotone increasing, then so are $P^{(-1)}$ and consequently $P^{(-1)} \circ h$ and $s$. It is also clear, that $\lim _{y \rightarrow 0} s(y)=0$ and $\lim _{y \rightarrow \infty} s(y)=\infty$. Therefore $s$ is continuously bijective from $[0, \infty)$ to $[0, \infty)$ with $s(0)=0$.
Thus we have:
$$
f(f(x+s(y)))=f(x)+2 s(y)-2 y-2 P(y)
$$
and using that $s$ is invertible, we obtain:
$$
f(f(x+y))=f(x)+2 y-2 s^{(-1)}(y)-2 P\left(s^{(-1)}(y)\right)
$$
Now fix $x_{0}$, then for any $x>x_{0}$ and any $y>0$ we have:
$$
\begin{aligned}
f(x)+2 y-2 s^{(-1)}(y)-2 P\left(s^{(-1)}(y)\right) & =f(f(x+y))=f\left(f\left(x_{0}+x+y-x_{0}\right)\right) \\
& =f\left(x_{0}\right)+2\left(x+y-x_{0}\right)-2 s^{(-1)}\left(x+y-x_{0}\right)-2 P\left(s^{(-1)}\left(x+y-x_{0}\right)\right) .
\end{aligned}
$$
Setting $y=x_{0}$, we get:
$$
f(x)+2 x_{0}-2 s^{(-1)}\left(x_{0}\right)-2 P\left(s^{(-1)}\left(x_{0}\right)\right)=f\left(x_{0}\right)+2 x-2 s^{(-1)}(x)-2 P\left(s^{(-1)}(x)\right) .
$$
Since this equality is valid for any $x>x_{0}$ we actually have that:
$$
f(x)-2 x+2 s^{(-1)}(x)+2 P\left(s^{(-1)}(x)\right)=c \text { for some fixed constant } c \in \mathbb{R} \text { and all } x \in \mathbb{R}^{+} .
$$
Let $\phi(x)=-x+2 s^{(-1)}(x)+2 P\left(s^{(-1)}(x)\right)$. Then $f(x)=x-\phi(x)+c$ and since $\phi$ is a sum of continuous functions that are continuous at 0 . Therefore $f$ is continuous and can be extended to a continuous function on $[0, \infty)$. Back in the original equation we fix $x>0$ and let $y$ tend to 0 . Using the continuity of $f$ and $g$ on $[0, \infty)$ and $P(0)=0$ we obtain:
$$
f(f(x))=\lim _{y \rightarrow 0+} f(f(x)+P(y))=\lim _{y \rightarrow 0+}(f(x-y)+P(y))=f(x)+P(0)=f(x) .
$$
Finally, fixing $x=1$ and varying $y>0$, we obtain:
$$
f(f(1+y)+P(y))=f(1)+2 y
$$
It follows that $f$ takes every value on $(f(1), \infty)$. Therefore for any $y \in(f(1), \infty)$ there is $z$ such that $f(z)=y$. Using that $f(f(z))=f(z)$ we conclude that $f(y)=y$ for all $y \in(f(1), \infty)$.
Now fix $x$ and take $y>f(1)$. Hence
$$
f(x)+2 y=f(f(x+y)+P(y))=f(x+y+P(y))=x+y+P(y)
$$
We conclude $f(x)-x=P(y)-y$ for every $x$ an $y>f(1)$. In particular $f\left(x_{1}\right)-x_{1}=f\left(x_{2}\right)-x_{2}$ for all $x_{1}, x_{2} \in(0, \infty)$ and since $f(x)=x$ for $x \in(f(1), \infty)$, we get $f(x)=x$ on $(0, \infty)$.
Finally, $x+2 y=f(x)+2 y=f(f(x+y)+P(y))=f(x+y)+P(y)=x+y+P(y)$, which shows that $P(y)=y$ for every $y \in(0, \infty)$.
It is also straightforward to check that $f(x)=x$ and $P(y)=y$ satisfy the equality:
$$
f(f(x+y)+P(y))=f(x+2 y)=x+2 y=f(x)+2 y
$$
|
{
"resource_path": "Balkan_MO/segmented/en-2024-BMO-type1.jsonl",
"problem_match": "\nProblem 4.",
"solution_match": "\nSolution 2"
}
|
010d1bc6-5b57-56e7-8f48-ac10ac4a51bb
| 605,048
|
Let $a, b, c$ be positive real numbers. Prove that
$$
\sqrt{a^{3} b+a^{3} c}+\sqrt{b^{3} c+b^{3} a}+\sqrt{c^{3} a+c^{3} b} \geq \frac{4}{3}(a b+b c+c a)
$$
|
W.L.O.G. $a \geq b \geq c$.
$$
\begin{gathered}
a \geq b \geq c \Rightarrow a b \geq a c \geq b c \Rightarrow a b+a c \geq a b+b c \geq a c+b c \Rightarrow \sqrt{a b+a c} \geq \sqrt{b c+b a} \geq \sqrt{a c+b c} \\
\sqrt{a^{3} b+a^{3} c}+\sqrt{b^{3} c+b^{3} a}+\sqrt{c^{3} a+c^{3} b}=a \sqrt{a b+a c}+b \sqrt{b c+b a}+c \sqrt{c a+c b} \geq \\
\frac{(a+b+c)}{3}(\sqrt{a b+a c}+\sqrt{b c+b a}+\sqrt{c a+c b})=\frac{(a+b+c)}{3}(\sqrt{a(b+c)}+\sqrt{b(c+a)}+\sqrt{c(a+b)}) \geq \\
\frac{(a+b+c)}{3}\left(\frac{2 a(b+c)}{a+b+c}+\frac{2 b(c+a)}{b+c+a}+\frac{c(a+b)}{c+a+b}\right)=\frac{(a+b+c)}{3} \frac{4(a b+b c+c a)}{a+b+c}=\frac{4}{3}(a b+b c+c a)
\end{gathered}
$$
(1) Chebyshev's inequality
(2) $\mathrm{GM} \geq \mathrm{HM}$
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $a, b, c$ be positive real numbers. Prove that
$$
\sqrt{a^{3} b+a^{3} c}+\sqrt{b^{3} c+b^{3} a}+\sqrt{c^{3} a+c^{3} b} \geq \frac{4}{3}(a b+b c+c a)
$$
|
W.L.O.G. $a \geq b \geq c$.
$$
\begin{gathered}
a \geq b \geq c \Rightarrow a b \geq a c \geq b c \Rightarrow a b+a c \geq a b+b c \geq a c+b c \Rightarrow \sqrt{a b+a c} \geq \sqrt{b c+b a} \geq \sqrt{a c+b c} \\
\sqrt{a^{3} b+a^{3} c}+\sqrt{b^{3} c+b^{3} a}+\sqrt{c^{3} a+c^{3} b}=a \sqrt{a b+a c}+b \sqrt{b c+b a}+c \sqrt{c a+c b} \geq \\
\frac{(a+b+c)}{3}(\sqrt{a b+a c}+\sqrt{b c+b a}+\sqrt{c a+c b})=\frac{(a+b+c)}{3}(\sqrt{a(b+c)}+\sqrt{b(c+a)}+\sqrt{c(a+b)}) \geq \\
\frac{(a+b+c)}{3}\left(\frac{2 a(b+c)}{a+b+c}+\frac{2 b(c+a)}{b+c+a}+\frac{c(a+b)}{c+a+b}\right)=\frac{(a+b+c)}{3} \frac{4(a b+b c+c a)}{a+b+c}=\frac{4}{3}(a b+b c+c a)
\end{gathered}
$$
(1) Chebyshev's inequality
(2) $\mathrm{GM} \geq \mathrm{HM}$
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"problem_match": "\nA1.",
"solution_match": "\nSolution."
}
|
8a6cfa24-98e9-5786-838e-4d000734ec8f
| 605,069
|
For all $x, y, z>0$ satisfying $\frac{x}{y z}+\frac{y}{z x}+\frac{z}{x y} \leq x+y+z$, prove that
$$
\frac{1}{x^{2}+y+z}+\frac{1}{y^{2}+z+x}+\frac{1}{z^{2}+x+y} \leq 1
$$
|
By Cauchy-Schwarz inequality, we have
$$
\left(x^{2}+y+z\right)\left(y^{2}+y z^{2}+z x^{2}\right) \geq(x y+y z+z x)^{2}
$$
and hence we obtain that
$$
\frac{1}{x^{2}+y+z}+\frac{1}{y^{2}+z+x}+\frac{1}{z^{2}+x+y} \leq \frac{2\left(x y^{2}+y z^{2}+z x^{2}\right)+x^{2}+y^{2}+z^{2}}{(x y+y z+z x)^{2}}
$$
Using the condition $\frac{x}{y z}+\frac{y}{z x}+\frac{z}{x y} \leq x+y+z$, we also have
$$
x^{2}+y^{2}+z^{2} \leq x y z(x+y+z)
$$
and hence
$$
2\left(x^{2}+y^{2}+z^{2}\right)+x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2} \leq(x y+y z+z x)^{2} .
$$
Finally, by AM-GM
$$
x^{2}+z^{2} x^{2} \geq 2 z x^{2}
$$
which yields that
$$
2\left(x^{2}+y^{2}+z^{2}\right)+x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2} \geq 2\left(x y^{2}+y z^{2}+z x^{2}\right)+x^{2}+y^{2}+z^{2}
$$
Using (1), (2) and (3), we are done.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
For all $x, y, z>0$ satisfying $\frac{x}{y z}+\frac{y}{z x}+\frac{z}{x y} \leq x+y+z$, prove that
$$
\frac{1}{x^{2}+y+z}+\frac{1}{y^{2}+z+x}+\frac{1}{z^{2}+x+y} \leq 1
$$
|
By Cauchy-Schwarz inequality, we have
$$
\left(x^{2}+y+z\right)\left(y^{2}+y z^{2}+z x^{2}\right) \geq(x y+y z+z x)^{2}
$$
and hence we obtain that
$$
\frac{1}{x^{2}+y+z}+\frac{1}{y^{2}+z+x}+\frac{1}{z^{2}+x+y} \leq \frac{2\left(x y^{2}+y z^{2}+z x^{2}\right)+x^{2}+y^{2}+z^{2}}{(x y+y z+z x)^{2}}
$$
Using the condition $\frac{x}{y z}+\frac{y}{z x}+\frac{z}{x y} \leq x+y+z$, we also have
$$
x^{2}+y^{2}+z^{2} \leq x y z(x+y+z)
$$
and hence
$$
2\left(x^{2}+y^{2}+z^{2}\right)+x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2} \leq(x y+y z+z x)^{2} .
$$
Finally, by AM-GM
$$
x^{2}+z^{2} x^{2} \geq 2 z x^{2}
$$
which yields that
$$
2\left(x^{2}+y^{2}+z^{2}\right)+x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2} \geq 2\left(x y^{2}+y z^{2}+z x^{2}\right)+x^{2}+y^{2}+z^{2}
$$
Using (1), (2) and (3), we are done.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"problem_match": "\nA2.",
"solution_match": "\n## Solution."
}
|
f7bd6a9c-8c53-5cce-a307-3e67c842e84d
| 605,082
|
Find all monotonic functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying the condition: for every real number $x$ and every natural number $n$
$$
\left|\sum_{i=1}^{n} i\left(f(x+i+1)-f^{2}(x+i)\right)\right|<C
$$
where $C>0$ is independent of $x$ and $f^{2}(x)=f(f(x))$.
|
From the condition of the problem we get $\left|\sum_{i=1}^{n-1} i\left(f(x+i+1)-f^{2}(x+i)\right)\right|<C$. Then $\left|n\left(f(x+n+1)-f^{2}(x+n)\right)\right|=\left|\sum_{i=1}^{n} i\left(f(x+i+1)-f^{2}(x+i)\right)-\sum_{i=1}^{n-1} i\left(f(x+i+1)-f^{2}(x+i)\right)\right|<2 C$ implying $\left|f(x+n+1)-f^{2}(x+n)\right|<\frac{2 C}{n}$ for every real number $x$ and every natural number $n$. Let $y \in \mathbb{R}$ be arbitrary. Then there exists $x$ such that $y=x+n$. We obtain $\left|f(y+1)-f^{2}(y)\right|<\frac{2 C}{n}$ for every real number $y$ and every natural number $n$. The last inequality holds for every natural number $n$ from where $f(y+1)=f^{2}(y)$ for every $y \in \mathbb{R}$. The function $f$ is monotonic which implies that it is an injection and the latter implies $f(y)=y+1$.
|
f(y)=y+1
|
Yes
|
Yes
|
proof
|
Algebra
|
Find all monotonic functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying the condition: for every real number $x$ and every natural number $n$
$$
\left|\sum_{i=1}^{n} i\left(f(x+i+1)-f^{2}(x+i)\right)\right|<C
$$
where $C>0$ is independent of $x$ and $f^{2}(x)=f(f(x))$.
|
From the condition of the problem we get $\left|\sum_{i=1}^{n-1} i\left(f(x+i+1)-f^{2}(x+i)\right)\right|<C$. Then $\left|n\left(f(x+n+1)-f^{2}(x+n)\right)\right|=\left|\sum_{i=1}^{n} i\left(f(x+i+1)-f^{2}(x+i)\right)-\sum_{i=1}^{n-1} i\left(f(x+i+1)-f^{2}(x+i)\right)\right|<2 C$ implying $\left|f(x+n+1)-f^{2}(x+n)\right|<\frac{2 C}{n}$ for every real number $x$ and every natural number $n$. Let $y \in \mathbb{R}$ be arbitrary. Then there exists $x$ such that $y=x+n$. We obtain $\left|f(y+1)-f^{2}(y)\right|<\frac{2 C}{n}$ for every real number $y$ and every natural number $n$. The last inequality holds for every natural number $n$ from where $f(y+1)=f^{2}(y)$ for every $y \in \mathbb{R}$. The function $f$ is monotonic which implies that it is an injection and the latter implies $f(y)=y+1$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"problem_match": "\n## A3.",
"solution_match": "\nSolution."
}
|
bd5d0a0a-45e5-5439-8fee-386b96dcc29a
| 605,092
|
The positive real numbers $a, b, c$ satisfy the equality $a+b+c=1$. For every natural number $n$ find the minimal possible value of the expression
$$
E=\frac{a^{-n}+b}{1-a}+\frac{b^{-n}+c}{1-b}+\frac{c^{-n}+a}{1-c}
$$
|
We transform the first term of the expression $E$ in the following way:
$$
\begin{aligned}
\frac{a^{-n}+b}{1-a}=\frac{1+a^{n} b}{a^{n}(b+c)}=\frac{a^{n+1}+a^{n} b+1-a^{n+1}}{a^{n}(b+c)} & =\frac{a^{n}(a+b)+(1-a)\left(1+a+a^{2}+\ldots+a^{n}\right)}{a^{n}(b+c)} \\
\frac{a^{n}(a+b)}{a^{n}(b+c)}+\frac{(b+c)\left(1+a+a^{2}+\ldots+a^{n}\right)}{a^{n}(b+c)} & =\frac{a+b}{b+c}+1+\frac{1}{a}+\frac{1}{a^{2}}+\ldots+\frac{1}{a^{n}}
\end{aligned}
$$
Analogously, we obtain
$$
\begin{aligned}
& \frac{b^{-n}+c}{1-b}=\frac{b+c}{c+a}+1+\frac{1}{b}+\frac{1}{b^{2}}+\ldots+\frac{1}{b^{n}} \\
& \frac{c^{-n}+a}{1-c}=\frac{c+a}{a+b}+1+\frac{1}{c}+\frac{1}{c^{2}}+\ldots+\frac{1}{c^{n}}
\end{aligned}
$$
The expression $E$ can be written in the form
$$
E=\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}+3+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right)+\ldots+\left(\frac{1}{a^{n}}+\frac{1}{b^{n}}+\frac{1}{c^{n}}\right)
$$
By virtue of the inequalities $\quad \frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b} \geq 3 \sqrt{\frac{a+b}{b+c} \cdot \frac{b+c}{c+a} \cdot \frac{c+a}{a+b}}=3$
and $\quad m_{k}=\left(\frac{a^{k}+b^{k}+c^{k}}{3}\right)^{\frac{1}{k}} \leq m_{1}=\frac{a+b+c}{3}=\frac{1}{3}$ for every $k=-1,-2,-3, \ldots,-n$, we have $\quad \frac{1}{a^{m}}+\frac{1}{b^{m}}+\frac{1}{c^{m}} \geq 3^{m+1}$ for every $m=1,2, \ldots, m$ and
$$
E=3+3+3^{2}+\ldots+3^{n+1}=2+\frac{3^{n+2}-1}{2}=\frac{3^{n+2}+3}{3}
$$
For $a=b=c=\frac{1}{3}$ we obtain $E=\frac{3^{n+2}+3}{3}$. So, $\min E=\frac{3^{n+2}+3}{3}$.
Remark(PSC): The original solution received from the PSC contained some typos, the correct result is $\min E=\frac{3^{n+2}+3}{2}$.
|
\frac{3^{n+2}+3}{2}
|
Yes
|
Incomplete
|
math-word-problem
|
Inequalities
|
The positive real numbers $a, b, c$ satisfy the equality $a+b+c=1$. For every natural number $n$ find the minimal possible value of the expression
$$
E=\frac{a^{-n}+b}{1-a}+\frac{b^{-n}+c}{1-b}+\frac{c^{-n}+a}{1-c}
$$
|
We transform the first term of the expression $E$ in the following way:
$$
\begin{aligned}
\frac{a^{-n}+b}{1-a}=\frac{1+a^{n} b}{a^{n}(b+c)}=\frac{a^{n+1}+a^{n} b+1-a^{n+1}}{a^{n}(b+c)} & =\frac{a^{n}(a+b)+(1-a)\left(1+a+a^{2}+\ldots+a^{n}\right)}{a^{n}(b+c)} \\
\frac{a^{n}(a+b)}{a^{n}(b+c)}+\frac{(b+c)\left(1+a+a^{2}+\ldots+a^{n}\right)}{a^{n}(b+c)} & =\frac{a+b}{b+c}+1+\frac{1}{a}+\frac{1}{a^{2}}+\ldots+\frac{1}{a^{n}}
\end{aligned}
$$
Analogously, we obtain
$$
\begin{aligned}
& \frac{b^{-n}+c}{1-b}=\frac{b+c}{c+a}+1+\frac{1}{b}+\frac{1}{b^{2}}+\ldots+\frac{1}{b^{n}} \\
& \frac{c^{-n}+a}{1-c}=\frac{c+a}{a+b}+1+\frac{1}{c}+\frac{1}{c^{2}}+\ldots+\frac{1}{c^{n}}
\end{aligned}
$$
The expression $E$ can be written in the form
$$
E=\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}+3+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right)+\ldots+\left(\frac{1}{a^{n}}+\frac{1}{b^{n}}+\frac{1}{c^{n}}\right)
$$
By virtue of the inequalities $\quad \frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b} \geq 3 \sqrt{\frac{a+b}{b+c} \cdot \frac{b+c}{c+a} \cdot \frac{c+a}{a+b}}=3$
and $\quad m_{k}=\left(\frac{a^{k}+b^{k}+c^{k}}{3}\right)^{\frac{1}{k}} \leq m_{1}=\frac{a+b+c}{3}=\frac{1}{3}$ for every $k=-1,-2,-3, \ldots,-n$, we have $\quad \frac{1}{a^{m}}+\frac{1}{b^{m}}+\frac{1}{c^{m}} \geq 3^{m+1}$ for every $m=1,2, \ldots, m$ and
$$
E=3+3+3^{2}+\ldots+3^{n+1}=2+\frac{3^{n+2}-1}{2}=\frac{3^{n+2}+3}{3}
$$
For $a=b=c=\frac{1}{3}$ we obtain $E=\frac{3^{n+2}+3}{3}$. So, $\min E=\frac{3^{n+2}+3}{3}$.
Remark(PSC): The original solution received from the PSC contained some typos, the correct result is $\min E=\frac{3^{n+2}+3}{2}$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"problem_match": "\n## A4.",
"solution_match": "\nSolution."
}
|
41324078-dea6-5847-a176-b38dc973410a
| 605,106
|
Let $a, b, c$ and $d$ be real numbers such that $a+b+c+d=2$ and $a b+b c+c d+d a+a c+b d=0$.
Find the minimum value and the maximum value of the product $a b c d$.
|
Let's find the minimum first.
$$
a^{2}+b^{2}+c^{2}+d^{2}=(a+b+c+d)^{2}-2(a b+b c+c d+d a+a c+b d)=4
$$
By AM-GM, $4=a^{2}+b^{2}+c^{2}+d^{2} \geq 4 \sqrt{|a b c d|} \Rightarrow 1 \geq|a b c d| \Rightarrow a b c d \geq-1$.
Note that if $a=b=c=1$ and $d=-1$, then $a b c d=-1$.
We'll find the maximum. We search for $a b c d>0$.
Obviously, the numbers $a, b, c$ and $d$ can not be all positive or all negative.
WLOG $a, b>0$ and $c, d<0$. Denote $-c=x,-d=y$.
We have $a, b, x, y>0, a+b-x-y=2$ and $a^{2}+b^{2}+x^{2}+y^{2}=4$. We need to find $\max (a b x y)$. We get: $x+y=a+b-2$ and $x^{2}+y^{2}=4-\left(a^{2}+b^{2}\right)$. Since $(x+y)^{2} \leq 2\left(x^{2}+y^{2}\right)$, then $2\left(a^{2}+b^{2}\right)+(a+b-2)^{2} \leq 8 ;$ on the other hand, $(a+b)^{2} \leq 2\left(a^{2}+b^{2}\right) \Rightarrow(a+b)^{2}+(a+b-2)^{2} \leq 8$.
Let $a+b=2 s \Rightarrow 2 s^{2}-2 s-1 \leq 0 \Rightarrow s \leq \frac{\sqrt{3}+1}{2}=k$.
But $a b \leq s^{2} \Rightarrow a b \leq k^{2}$.
Now $a+b=x+y+2$ and $a^{2}+b^{2}=4-\left(x^{2}+y^{2}\right)$. Since $(a+b)^{2} \leq 2\left(a^{2}+b^{2}\right)$, then $2\left(x^{2}+y^{2}\right)+(x+y+2)^{2} \leq 8$; on the other hand, $(x+y)^{2} \leq 2\left(x^{2}+y^{2}\right) \Rightarrow$
$\Rightarrow(x+y)^{2}+(x+y+2)^{2} \leq 8$. Let $x+y=2 q \Rightarrow 2 q^{2}+2 q-1 \leq 0 \Rightarrow q \leq \frac{\sqrt{3}-1}{2}=\frac{1}{2 k}$.
But $x y \leq q^{2} \Rightarrow x y \leq \frac{1}{4 k^{2}}$.
In conclusion, $a b x y \leq k^{2} \cdot \frac{1}{4 k^{2}}=\frac{1}{4} \Rightarrow a b c d \leq \frac{1}{4}$.
Note that if $a=b=k$ and $c=d=-\frac{1}{2 k}$, then $a b c d=\frac{1}{4}$
In conclusion, $\min (a b c d)=-1$ and $\max (a b c d)=\frac{1}{4}$.
|
\min (a b c d)=-1 \text{ and } \max (a b c d)=\frac{1}{4}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Let $a, b, c$ and $d$ be real numbers such that $a+b+c+d=2$ and $a b+b c+c d+d a+a c+b d=0$.
Find the minimum value and the maximum value of the product $a b c d$.
|
Let's find the minimum first.
$$
a^{2}+b^{2}+c^{2}+d^{2}=(a+b+c+d)^{2}-2(a b+b c+c d+d a+a c+b d)=4
$$
By AM-GM, $4=a^{2}+b^{2}+c^{2}+d^{2} \geq 4 \sqrt{|a b c d|} \Rightarrow 1 \geq|a b c d| \Rightarrow a b c d \geq-1$.
Note that if $a=b=c=1$ and $d=-1$, then $a b c d=-1$.
We'll find the maximum. We search for $a b c d>0$.
Obviously, the numbers $a, b, c$ and $d$ can not be all positive or all negative.
WLOG $a, b>0$ and $c, d<0$. Denote $-c=x,-d=y$.
We have $a, b, x, y>0, a+b-x-y=2$ and $a^{2}+b^{2}+x^{2}+y^{2}=4$. We need to find $\max (a b x y)$. We get: $x+y=a+b-2$ and $x^{2}+y^{2}=4-\left(a^{2}+b^{2}\right)$. Since $(x+y)^{2} \leq 2\left(x^{2}+y^{2}\right)$, then $2\left(a^{2}+b^{2}\right)+(a+b-2)^{2} \leq 8 ;$ on the other hand, $(a+b)^{2} \leq 2\left(a^{2}+b^{2}\right) \Rightarrow(a+b)^{2}+(a+b-2)^{2} \leq 8$.
Let $a+b=2 s \Rightarrow 2 s^{2}-2 s-1 \leq 0 \Rightarrow s \leq \frac{\sqrt{3}+1}{2}=k$.
But $a b \leq s^{2} \Rightarrow a b \leq k^{2}$.
Now $a+b=x+y+2$ and $a^{2}+b^{2}=4-\left(x^{2}+y^{2}\right)$. Since $(a+b)^{2} \leq 2\left(a^{2}+b^{2}\right)$, then $2\left(x^{2}+y^{2}\right)+(x+y+2)^{2} \leq 8$; on the other hand, $(x+y)^{2} \leq 2\left(x^{2}+y^{2}\right) \Rightarrow$
$\Rightarrow(x+y)^{2}+(x+y+2)^{2} \leq 8$. Let $x+y=2 q \Rightarrow 2 q^{2}+2 q-1 \leq 0 \Rightarrow q \leq \frac{\sqrt{3}-1}{2}=\frac{1}{2 k}$.
But $x y \leq q^{2} \Rightarrow x y \leq \frac{1}{4 k^{2}}$.
In conclusion, $a b x y \leq k^{2} \cdot \frac{1}{4 k^{2}}=\frac{1}{4} \Rightarrow a b c d \leq \frac{1}{4}$.
Note that if $a=b=k$ and $c=d=-\frac{1}{2 k}$, then $a b c d=\frac{1}{4}$
In conclusion, $\min (a b c d)=-1$ and $\max (a b c d)=\frac{1}{4}$.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"problem_match": "\n## A5.",
"solution_match": "\nSolution."
}
|
c3ddc0e0-b748-5225-a870-e28325144864
| 605,116
|
Prove that there is no function from positive real numbers to itself, $f:(0,+\infty) \rightarrow(0,+\infty)$ such that:
$$
f(f(x)+y)=f(x)+3 x+y f(y) \quad \text {,for every } \quad x, y \in(0,+\infty)
$$
|
First we prove that $f(x) \geq x$ for all $x>0$.
Indeed, if there is an $a>0$ with $f(a)<a$ then from the initial for $x=a$ and $y=a-f(a)>0$ we get that $3 a+(a-f(a)) f(a-f(a))=0$. This is absurd since $3 a+(a-f(a)) f(a-f(a))>0$.
So we have that
$$
f(x) \geq x \quad \text {,for all } \quad x>0
$$
Then using (1) we have that $f(x)+3 x+y f(x)=f(f(x)+y) \geq f(x)+y$, so
$$
3 x+y f(y) \geq y \quad, \text { all } \quad x, y>0
$$
Suppose that $y f(y)<y$ for a $y>0$ and let $-y f(y)=b>0$ then for $x=\frac{b}{4}$ we get $\frac{3 b}{4}-b \geq 0$ so $b \leq 0$, absurd. So we have that $y f(y) \geq y$, for all $y>0$, and so
$$
f(y) \geq 1 \quad \text {,for all } \quad y>0
$$
Substituting $y$ with $f(y)$ at the initial we have that
$$
f(x)+3 x+f(y) f(f(y))=f(f(x)+f(y))
$$
and changing the roles of $x, y$ we have that:
$$
f(y)+3 y+f(x) f(f(x))=f(f(x)+f(x))
$$
So we have $f(x) f(f(x))-f(x)-3 x=f(y) f(f(y))-f(y)-3 y$, which means that the function $f(x) f(f(x))-$ $f(x)-3 x$ is constant. That means that there exist a constant $c$ such that
$$
f(x) f(f(x))=f(x)+3 x+c \quad \text {,for all } \quad x>0
$$
So we can write (6) in the form $f(x)(f(f(x))-1)=3 x+c$ and since $f(f(x))>1$ we have that $3 x+c \geq 0$, for all $x>0$. if $c<0$ then for $x=-\frac{c}{4}>0$ we get that $c>0$ which is absurd. So $c \geq 0$.
We write (6) in the form
$$
f(f(x))=1+\frac{3 x}{f(x)}+\frac{c}{f(x)}
$$
Since $c \geq 0$ then from (7) with the help of (1) and (3) we have that $f(f(x)) \leq 4+c$.
But from (1) we have that $f(f(x)) \geq f(x) \geq x$, for all $x \geq 0$, and so
$$
4+c \geq f(f(x)) \geq x \quad \text {,for all } \quad x>0
$$
Taking $x=5+c$ we get that the last one cannot hold. So there is no such a function.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
Prove that there is no function from positive real numbers to itself, $f:(0,+\infty) \rightarrow(0,+\infty)$ such that:
$$
f(f(x)+y)=f(x)+3 x+y f(y) \quad \text {,for every } \quad x, y \in(0,+\infty)
$$
|
First we prove that $f(x) \geq x$ for all $x>0$.
Indeed, if there is an $a>0$ with $f(a)<a$ then from the initial for $x=a$ and $y=a-f(a)>0$ we get that $3 a+(a-f(a)) f(a-f(a))=0$. This is absurd since $3 a+(a-f(a)) f(a-f(a))>0$.
So we have that
$$
f(x) \geq x \quad \text {,for all } \quad x>0
$$
Then using (1) we have that $f(x)+3 x+y f(x)=f(f(x)+y) \geq f(x)+y$, so
$$
3 x+y f(y) \geq y \quad, \text { all } \quad x, y>0
$$
Suppose that $y f(y)<y$ for a $y>0$ and let $-y f(y)=b>0$ then for $x=\frac{b}{4}$ we get $\frac{3 b}{4}-b \geq 0$ so $b \leq 0$, absurd. So we have that $y f(y) \geq y$, for all $y>0$, and so
$$
f(y) \geq 1 \quad \text {,for all } \quad y>0
$$
Substituting $y$ with $f(y)$ at the initial we have that
$$
f(x)+3 x+f(y) f(f(y))=f(f(x)+f(y))
$$
and changing the roles of $x, y$ we have that:
$$
f(y)+3 y+f(x) f(f(x))=f(f(x)+f(x))
$$
So we have $f(x) f(f(x))-f(x)-3 x=f(y) f(f(y))-f(y)-3 y$, which means that the function $f(x) f(f(x))-$ $f(x)-3 x$ is constant. That means that there exist a constant $c$ such that
$$
f(x) f(f(x))=f(x)+3 x+c \quad \text {,for all } \quad x>0
$$
So we can write (6) in the form $f(x)(f(f(x))-1)=3 x+c$ and since $f(f(x))>1$ we have that $3 x+c \geq 0$, for all $x>0$. if $c<0$ then for $x=-\frac{c}{4}>0$ we get that $c>0$ which is absurd. So $c \geq 0$.
We write (6) in the form
$$
f(f(x))=1+\frac{3 x}{f(x)}+\frac{c}{f(x)}
$$
Since $c \geq 0$ then from (7) with the help of (1) and (3) we have that $f(f(x)) \leq 4+c$.
But from (1) we have that $f(f(x)) \geq f(x) \geq x$, for all $x \geq 0$, and so
$$
4+c \geq f(f(x)) \geq x \quad \text {,for all } \quad x>0
$$
Taking $x=5+c$ we get that the last one cannot hold. So there is no such a function.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"problem_match": "\n## A6.",
"solution_match": "\nSolution."
}
|
7ad1058b-b125-578d-bbf2-c05bbf15e5a0
| 605,128
|
Find all integers $n \geq 2$ for which there exist the real numbers $a_{k}, 1 \leq k \leq n$, which are satisfying the following conditions:
$$
\sum_{k=1}^{n} a_{k}=0, \sum_{k=1}^{n} a_{k}^{2}=1 \text { and } \sqrt{n} \cdot\left(\sum_{k=1}^{n} a_{k}^{3}\right)=2(b \sqrt{n}-1), \text { where } b=\max _{1 \leq k \leq n}\left\{a_{k}\right\}
$$
|
We have: $\left(a_{k}+\frac{1}{\sqrt{n}}\right)^{2}\left(a_{k}-b\right) \leq 0 \Rightarrow\left(a_{k}^{2}+\frac{2}{\sqrt{n}} \cdot a_{k}+\frac{1}{n}\right)\left(a_{k}-b\right) \leq 0 \Rightarrow$
$$
a_{k}^{3} \leq\left(b-\frac{2}{\sqrt{n}}\right) \cdot a_{k}^{2}+\left(\frac{2 b}{\sqrt{n}}-\frac{1}{n}\right) \cdot a_{k}+\frac{b}{n} \forall \in\{1,2, \cdot, n\} \cdot(k)
$$
Adding up the inequalities ( $k$ ) we get
$$
\begin{gathered}
\sum_{k=1}^{n} a_{k}^{3} \leq\left(b-\frac{2}{\sqrt{n}}\right) \cdot\left(\sum_{k=1}^{n} a_{k}^{2}\right)+\left(\frac{2 b}{\sqrt{n}}-\frac{1}{n}\right) \cdot\left(\sum_{k=1}^{n} a_{k}\right)+b \leftrightarrow \\
\sum_{k=1}^{n} a_{k}^{3} \leq b-\frac{2}{\sqrt{n}}+b \leftrightarrow \sqrt{n} \cdot\left(\sum_{k=1}^{n} a_{k}^{3}\right) \leq 2(b \sqrt{n}-1) .
\end{gathered}
$$
But according to hypothesis,
$$
\sqrt{n} \cdot\left(\sum_{k=1}^{n} a_{k}^{3}\right)=2(b \sqrt{n}-1) .
$$
Hence is necessarily that
$$
\begin{gathered}
a_{k}^{3}=\left(b-\frac{2}{\sqrt{n}}\right) \cdot a_{k}^{2}+\left(\frac{2 b}{\sqrt{n}}-\frac{1}{n}\right) \cdot a_{k}+\frac{b}{n} \forall k \in\{1,2, \cdots, n\} \leftrightarrow \\
\left(a_{k}+\frac{1}{\sqrt{n}}\right)^{2}\left(a_{k}-b\right)=0 \forall k \in\{1,2, \cdots, n\} \leftrightarrow a_{k} \in\left\{-\frac{1}{\sqrt{n}}, b\right\} \forall k \in\{1,2, \cdots, n\}
\end{gathered}
$$
We'll prove that $b>0$. Indeed, if $b<0$ then $0=\sum_{k=1}^{n} a_{k} \leq n b<0$, which is absurd. If $b=0$, since $\sum_{k=1}^{n} a_{k}=0$, then $a_{k}=0 \forall k \in\{1,2, \cdots, n\} \Rightarrow 1=\sum_{k=1}^{n} a_{k}^{2}=0$, which is absurd.
In conclusion $b>0$.
If $a_{k}=-\frac{1}{\sqrt{n}} \forall k \in\{1,2, \cdots, n\}$ then $\sum_{k=1}^{n} a_{k}=-\sqrt{n}<0$, which is absurd and similarly if
$a_{k}=b \forall k \in\{1,2, \cdots, n\}$ then $\sum_{k=1}^{n} a_{k}=n b>0$, which is absurd. Hence $\exists m \in\{1,2, \cdots, n-1\}$
such that among the numbers $a_{k}$ we have $n-m$ equal to $-\frac{1}{\sqrt{n}}$ and $m$ equal to $b$. We get $\left\{\begin{array}{l}-\frac{n-m}{\sqrt{n}}+m b=0 \\ \frac{n-m}{n}+m b^{2}=1\end{array}\right.$.
From here, $b=\frac{n-m}{m \sqrt{n}} \Rightarrow \frac{n-m}{n}+\frac{(n-m)^{2}}{m n}=1 \Rightarrow$
$\Rightarrow n-m=m \Rightarrow m=\frac{n}{2}$. Hence $n$ is even.
Conversely, for any even integer $n \geq 2$ we get that there exist the real numbers $a_{k}, 1 \leq k \leq n$, such that
$$
\sum_{k=1}^{n} a_{k}=0, \sum_{k=1}^{n} a_{k}^{2}=1 \text { and } \sqrt{n} \cdot\left(\sum_{k=1}^{n} a_{k}^{3}\right)=2(b \sqrt{n}-1), \text { where } b=\max _{1 \leq k \leq n}\left\{a_{k}\right\}
$$
(We may choose for example $a_{1}=\cdots=a_{\frac{n}{2}}=-\frac{1}{\sqrt{n}}$ and $a_{\frac{n}{2}+1}=\cdots=a_{n}=\frac{1}{\sqrt{n}}$ ).
|
n \text{ is even}
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find all integers $n \geq 2$ for which there exist the real numbers $a_{k}, 1 \leq k \leq n$, which are satisfying the following conditions:
$$
\sum_{k=1}^{n} a_{k}=0, \sum_{k=1}^{n} a_{k}^{2}=1 \text { and } \sqrt{n} \cdot\left(\sum_{k=1}^{n} a_{k}^{3}\right)=2(b \sqrt{n}-1), \text { where } b=\max _{1 \leq k \leq n}\left\{a_{k}\right\}
$$
|
We have: $\left(a_{k}+\frac{1}{\sqrt{n}}\right)^{2}\left(a_{k}-b\right) \leq 0 \Rightarrow\left(a_{k}^{2}+\frac{2}{\sqrt{n}} \cdot a_{k}+\frac{1}{n}\right)\left(a_{k}-b\right) \leq 0 \Rightarrow$
$$
a_{k}^{3} \leq\left(b-\frac{2}{\sqrt{n}}\right) \cdot a_{k}^{2}+\left(\frac{2 b}{\sqrt{n}}-\frac{1}{n}\right) \cdot a_{k}+\frac{b}{n} \forall \in\{1,2, \cdot, n\} \cdot(k)
$$
Adding up the inequalities ( $k$ ) we get
$$
\begin{gathered}
\sum_{k=1}^{n} a_{k}^{3} \leq\left(b-\frac{2}{\sqrt{n}}\right) \cdot\left(\sum_{k=1}^{n} a_{k}^{2}\right)+\left(\frac{2 b}{\sqrt{n}}-\frac{1}{n}\right) \cdot\left(\sum_{k=1}^{n} a_{k}\right)+b \leftrightarrow \\
\sum_{k=1}^{n} a_{k}^{3} \leq b-\frac{2}{\sqrt{n}}+b \leftrightarrow \sqrt{n} \cdot\left(\sum_{k=1}^{n} a_{k}^{3}\right) \leq 2(b \sqrt{n}-1) .
\end{gathered}
$$
But according to hypothesis,
$$
\sqrt{n} \cdot\left(\sum_{k=1}^{n} a_{k}^{3}\right)=2(b \sqrt{n}-1) .
$$
Hence is necessarily that
$$
\begin{gathered}
a_{k}^{3}=\left(b-\frac{2}{\sqrt{n}}\right) \cdot a_{k}^{2}+\left(\frac{2 b}{\sqrt{n}}-\frac{1}{n}\right) \cdot a_{k}+\frac{b}{n} \forall k \in\{1,2, \cdots, n\} \leftrightarrow \\
\left(a_{k}+\frac{1}{\sqrt{n}}\right)^{2}\left(a_{k}-b\right)=0 \forall k \in\{1,2, \cdots, n\} \leftrightarrow a_{k} \in\left\{-\frac{1}{\sqrt{n}}, b\right\} \forall k \in\{1,2, \cdots, n\}
\end{gathered}
$$
We'll prove that $b>0$. Indeed, if $b<0$ then $0=\sum_{k=1}^{n} a_{k} \leq n b<0$, which is absurd. If $b=0$, since $\sum_{k=1}^{n} a_{k}=0$, then $a_{k}=0 \forall k \in\{1,2, \cdots, n\} \Rightarrow 1=\sum_{k=1}^{n} a_{k}^{2}=0$, which is absurd.
In conclusion $b>0$.
If $a_{k}=-\frac{1}{\sqrt{n}} \forall k \in\{1,2, \cdots, n\}$ then $\sum_{k=1}^{n} a_{k}=-\sqrt{n}<0$, which is absurd and similarly if
$a_{k}=b \forall k \in\{1,2, \cdots, n\}$ then $\sum_{k=1}^{n} a_{k}=n b>0$, which is absurd. Hence $\exists m \in\{1,2, \cdots, n-1\}$
such that among the numbers $a_{k}$ we have $n-m$ equal to $-\frac{1}{\sqrt{n}}$ and $m$ equal to $b$. We get $\left\{\begin{array}{l}-\frac{n-m}{\sqrt{n}}+m b=0 \\ \frac{n-m}{n}+m b^{2}=1\end{array}\right.$.
From here, $b=\frac{n-m}{m \sqrt{n}} \Rightarrow \frac{n-m}{n}+\frac{(n-m)^{2}}{m n}=1 \Rightarrow$
$\Rightarrow n-m=m \Rightarrow m=\frac{n}{2}$. Hence $n$ is even.
Conversely, for any even integer $n \geq 2$ we get that there exist the real numbers $a_{k}, 1 \leq k \leq n$, such that
$$
\sum_{k=1}^{n} a_{k}=0, \sum_{k=1}^{n} a_{k}^{2}=1 \text { and } \sqrt{n} \cdot\left(\sum_{k=1}^{n} a_{k}^{3}\right)=2(b \sqrt{n}-1), \text { where } b=\max _{1 \leq k \leq n}\left\{a_{k}\right\}
$$
(We may choose for example $a_{1}=\cdots=a_{\frac{n}{2}}=-\frac{1}{\sqrt{n}}$ and $a_{\frac{n}{2}+1}=\cdots=a_{n}=\frac{1}{\sqrt{n}}$ ).
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"problem_match": "\nA7.",
"solution_match": "\nSolution."
}
|
e2d31732-3db8-50dd-92b0-8fe652b247fa
| 605,137
|
Find all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ for which $f(g(n))-g(f(n))$ is independent on $n$ for any $g: \mathbb{Z} \rightarrow \mathbb{Z}$.
|
First observe that if $f(n)=n$, then $f(g(n))-g(f(n))=0$. Therefore the identity function satisfies the problem condition.
If there is $n_{0}$ with $f\left(n_{0}\right) \neq n_{0}$, consider the characteristic function $g$ that is defined as $g\left(f\left(n_{0}\right)\right)=1$ and $g(n)=0$ for $n \neq f\left(n_{0}\right)$. In this case, let $n \neq f\left(n_{0}\right)$ be arbirtrary. One has $f(g(n))-g(f(n))=$ $f\left(g\left(n_{0}\right)\right)-g\left(f\left(n_{0}\right)\right) \Rightarrow f(0)-g(f(n))=f(0)-g\left(f\left(n_{0}\right)\right) \Rightarrow g(f(n))=1 \Rightarrow f(n)=f\left(n_{0}\right)$.
Now, consider the very similar function $g$ that is defined as $g\left(f\left(n_{0}\right)\right)=a$ and $g(n)=b$ for $n \neq f\left(n_{0}\right)$, where $a, b$ are integers with $a \neq b \neq f\left(n_{0}\right)$. We have chosen $a, b$ so as to ensure that $f(a)=f\left(n_{0}\right)=f(b)$. Now, we find that $f\left(g\left(f\left(n_{0}\right)\right)\right)-g\left(f\left(f\left(n_{0}\right)\right)\right)=f\left(g\left(n_{0}\right)\right)-g\left(f\left(n_{0}\right)\right) \Rightarrow f(a)-g\left(f\left(f\left(n_{0}\right)\right)\right)=f(b)-$ $g\left(f\left(n_{0}\right)\right) \Rightarrow g\left(f\left(f\left(n_{0}\right)\right)\right)=g\left(f\left(n_{0}\right)\right)=a \Rightarrow f\left(f\left(n_{0}\right)\right)=f\left(n_{0}\right)$.
In summary, $f(n)=f\left(n_{0}\right)$ for any $n \neq f\left(n_{0}\right)$ and $f\left(f\left(n_{0}\right)\right)=f\left(n_{0}\right)$. Therefore $f$ is a constant function. Let us now see that a constant function indeed satisfies the problem condition: If $f(n)=c$ for all $n, f(g(n))-g(f(n))=c-g(c)$ is independent of $n$.
The answers are the identity function $f(n)=n$ and the constant functions $f(n)=c$.
## Combinatorics
|
proof
|
Yes
|
Yes
|
math-word-problem
|
Algebra
|
Find all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ for which $f(g(n))-g(f(n))$ is independent on $n$ for any $g: \mathbb{Z} \rightarrow \mathbb{Z}$.
|
First observe that if $f(n)=n$, then $f(g(n))-g(f(n))=0$. Therefore the identity function satisfies the problem condition.
If there is $n_{0}$ with $f\left(n_{0}\right) \neq n_{0}$, consider the characteristic function $g$ that is defined as $g\left(f\left(n_{0}\right)\right)=1$ and $g(n)=0$ for $n \neq f\left(n_{0}\right)$. In this case, let $n \neq f\left(n_{0}\right)$ be arbirtrary. One has $f(g(n))-g(f(n))=$ $f\left(g\left(n_{0}\right)\right)-g\left(f\left(n_{0}\right)\right) \Rightarrow f(0)-g(f(n))=f(0)-g\left(f\left(n_{0}\right)\right) \Rightarrow g(f(n))=1 \Rightarrow f(n)=f\left(n_{0}\right)$.
Now, consider the very similar function $g$ that is defined as $g\left(f\left(n_{0}\right)\right)=a$ and $g(n)=b$ for $n \neq f\left(n_{0}\right)$, where $a, b$ are integers with $a \neq b \neq f\left(n_{0}\right)$. We have chosen $a, b$ so as to ensure that $f(a)=f\left(n_{0}\right)=f(b)$. Now, we find that $f\left(g\left(f\left(n_{0}\right)\right)\right)-g\left(f\left(f\left(n_{0}\right)\right)\right)=f\left(g\left(n_{0}\right)\right)-g\left(f\left(n_{0}\right)\right) \Rightarrow f(a)-g\left(f\left(f\left(n_{0}\right)\right)\right)=f(b)-$ $g\left(f\left(n_{0}\right)\right) \Rightarrow g\left(f\left(f\left(n_{0}\right)\right)\right)=g\left(f\left(n_{0}\right)\right)=a \Rightarrow f\left(f\left(n_{0}\right)\right)=f\left(n_{0}\right)$.
In summary, $f(n)=f\left(n_{0}\right)$ for any $n \neq f\left(n_{0}\right)$ and $f\left(f\left(n_{0}\right)\right)=f\left(n_{0}\right)$. Therefore $f$ is a constant function. Let us now see that a constant function indeed satisfies the problem condition: If $f(n)=c$ for all $n, f(g(n))-g(f(n))=c-g(c)$ is independent of $n$.
The answers are the identity function $f(n)=n$ and the constant functions $f(n)=c$.
## Combinatorics
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"problem_match": "\n## A8.",
"solution_match": "\nSolution."
}
|
d86e0426-3421-57d4-9b28-e5fbd27db0a1
| 605,146
|
Let positive integers $K$ and $d$ be given. Prove that there exists a positive integer $n$ and a sequence of $K$ positive integers $b_{1}, b_{2}, \ldots, b_{K}$ such that the number $n$ is a $d$-digit palindrome in all number bases $b_{1}, b_{2}, \ldots, b_{K}$.
|
Let a positive integer $d$ be given. We shall prove that, for each large enough $n$, the number $(n!)^{d-1}$ is a $d$-digit palindrome in all number bases $\frac{n!}{i}-1$ for $1 \leqslant i \leqslant n$. In particular, we shall prove that the digit expansion of $(n!)^{d-1}$ in the base $\frac{n!}{i}-1$ is
$$
\left\langle i^{d-1}\binom{d-1}{d-1}, i^{d-1}\binom{d-1}{d-2}, i^{d-1}\binom{d-1}{d-3}, \ldots, i^{d-1}\binom{d-1}{1}, i^{d-1}\binom{d-1}{0}\right\rangle_{\frac{n!}{i}-1}
$$
We first show that, for each large enough $n$, all these digits are smaller than the considered base, that is, they are indeed digits in that base. It is enough to check this assertion for $i=n$, that is, to show the inequality $n^{d-1}\binom{d-1}{j}<(n-1)$ ! -1 . However, since for a fixed $d$ the right-hand side clearly grows faster than the left-hand side, this is indeed true for all large enough $n$.
Everything that is left is to evaluate:
$$
\begin{aligned}
\sum_{j=0}^{d-1} i^{d-1}\binom{d-1}{j}\left(\frac{n!}{i}-1\right)^{j} & =i^{d-1} \sum_{j=0}^{d-1}\binom{d-1}{j}\left(\frac{n!}{i}-1\right)^{j}=i^{d-1} \sum_{j=0}^{d-1}\binom{d-1}{j}\left(\frac{n!}{i}-1\right)^{j} \\
& =i^{d-1}\left(\frac{n!}{i}-1+1\right)^{d-1}=(n!)^{d-1}
\end{aligned}
$$
which completes the proof.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let positive integers $K$ and $d$ be given. Prove that there exists a positive integer $n$ and a sequence of $K$ positive integers $b_{1}, b_{2}, \ldots, b_{K}$ such that the number $n$ is a $d$-digit palindrome in all number bases $b_{1}, b_{2}, \ldots, b_{K}$.
|
Let a positive integer $d$ be given. We shall prove that, for each large enough $n$, the number $(n!)^{d-1}$ is a $d$-digit palindrome in all number bases $\frac{n!}{i}-1$ for $1 \leqslant i \leqslant n$. In particular, we shall prove that the digit expansion of $(n!)^{d-1}$ in the base $\frac{n!}{i}-1$ is
$$
\left\langle i^{d-1}\binom{d-1}{d-1}, i^{d-1}\binom{d-1}{d-2}, i^{d-1}\binom{d-1}{d-3}, \ldots, i^{d-1}\binom{d-1}{1}, i^{d-1}\binom{d-1}{0}\right\rangle_{\frac{n!}{i}-1}
$$
We first show that, for each large enough $n$, all these digits are smaller than the considered base, that is, they are indeed digits in that base. It is enough to check this assertion for $i=n$, that is, to show the inequality $n^{d-1}\binom{d-1}{j}<(n-1)$ ! -1 . However, since for a fixed $d$ the right-hand side clearly grows faster than the left-hand side, this is indeed true for all large enough $n$.
Everything that is left is to evaluate:
$$
\begin{aligned}
\sum_{j=0}^{d-1} i^{d-1}\binom{d-1}{j}\left(\frac{n!}{i}-1\right)^{j} & =i^{d-1} \sum_{j=0}^{d-1}\binom{d-1}{j}\left(\frac{n!}{i}-1\right)^{j}=i^{d-1} \sum_{j=0}^{d-1}\binom{d-1}{j}\left(\frac{n!}{i}-1\right)^{j} \\
& =i^{d-1}\left(\frac{n!}{i}-1+1\right)^{d-1}=(n!)^{d-1}
\end{aligned}
$$
which completes the proof.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"problem_match": "\n## C1.",
"solution_match": "\nSolution."
}
|
bf3a52f1-a449-58dd-a3d4-06030b8d8e51
| 605,155
|
There are 2016 costumers who entered a shop on a particular day. Every customer entered the shop exactly once. (i.e. the customer entered the shop, stayed there for some time and then left the shop without returning back.)
Find the maximal $k$ such that the following holds:
There are $k$ customers such that either all of them were in the shop at a specific time instance or no two of them were both in the shop at any time instance.
|
We show that the maximal $k$ is 45 .
First we show that no larger $k$ can be achieved: We break the day at 45 disjoint time intervals and assume that at each time interval there were exactly 45 costumers who stayed in the shop only during that time interval (except in the last interval in which there were only 36 customers). We observe that there are no 46 people with the required property.
Now we show that $k=45$ can be achieved: Suppose that customers $C_{1}, C_{2}, \ldots, C_{2016}$ visited the shop in this order. (If two or more customers entered the shop at exactly the same time then we break ties arbitrarily.)
We define groups $A_{1}, A_{2}, \ldots$ of customers as follows: Starting with $C_{1}$ and proceeding in order, we place customer $C_{j}$ into the group $A_{i}$ where $i$ is the smallest index such that $A_{i}$ contains no customer $C_{j^{\prime}}$ with $j^{\prime}<j$ and such that $C_{j^{\prime}}$ was inside the shop once $C_{j}$ entered it.
Clearly no two customers who are in the same group were inside the shop at the exact same time. So we may assume that every $A_{i}$ has at most 45 customers. Since $44 \cdot 45<2016$, by the pigeonhole principle there must be at least 45 (non-empty) groups.
Let $C_{j}$ be a person in group $A_{45}$ and suppose that $C_{j}$ entered the shop at time $t_{j}$. Since we placed $C_{j}$ in group $A_{45}$ this means that for each $i<45$, there is a $j_{i}<j$ such that $C_{j_{i}} \in A_{i}$ and $C_{j_{i}}$ is still inside the shop at time $t_{j}$.
Thus we have found a specific time instance, namely $t_{j}$, during which at least 45 customers were all inside the shop.
Note: Instead of asking for the maximal $k$, an easier version is the following:
Show that there are 45 customers such that either all of them were in the shop at a specific time instance or no two of them were both in the shop at any time instance.
|
45
|
Yes
|
Yes
|
math-word-problem
|
Combinatorics
|
There are 2016 costumers who entered a shop on a particular day. Every customer entered the shop exactly once. (i.e. the customer entered the shop, stayed there for some time and then left the shop without returning back.)
Find the maximal $k$ such that the following holds:
There are $k$ customers such that either all of them were in the shop at a specific time instance or no two of them were both in the shop at any time instance.
|
We show that the maximal $k$ is 45 .
First we show that no larger $k$ can be achieved: We break the day at 45 disjoint time intervals and assume that at each time interval there were exactly 45 costumers who stayed in the shop only during that time interval (except in the last interval in which there were only 36 customers). We observe that there are no 46 people with the required property.
Now we show that $k=45$ can be achieved: Suppose that customers $C_{1}, C_{2}, \ldots, C_{2016}$ visited the shop in this order. (If two or more customers entered the shop at exactly the same time then we break ties arbitrarily.)
We define groups $A_{1}, A_{2}, \ldots$ of customers as follows: Starting with $C_{1}$ and proceeding in order, we place customer $C_{j}$ into the group $A_{i}$ where $i$ is the smallest index such that $A_{i}$ contains no customer $C_{j^{\prime}}$ with $j^{\prime}<j$ and such that $C_{j^{\prime}}$ was inside the shop once $C_{j}$ entered it.
Clearly no two customers who are in the same group were inside the shop at the exact same time. So we may assume that every $A_{i}$ has at most 45 customers. Since $44 \cdot 45<2016$, by the pigeonhole principle there must be at least 45 (non-empty) groups.
Let $C_{j}$ be a person in group $A_{45}$ and suppose that $C_{j}$ entered the shop at time $t_{j}$. Since we placed $C_{j}$ in group $A_{45}$ this means that for each $i<45$, there is a $j_{i}<j$ such that $C_{j_{i}} \in A_{i}$ and $C_{j_{i}}$ is still inside the shop at time $t_{j}$.
Thus we have found a specific time instance, namely $t_{j}$, during which at least 45 customers were all inside the shop.
Note: Instead of asking for the maximal $k$, an easier version is the following:
Show that there are 45 customers such that either all of them were in the shop at a specific time instance or no two of them were both in the shop at any time instance.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"problem_match": "\n## C2.",
"solution_match": "\nSolution."
}
|
ebcb4d04-2cb8-5da6-a385-f39daf4d2621
| 605,167
|
The plane is divided into unit squares by means of two sets of parallel lines. The unit squares are coloured in 1201 colours so that no rectangle of perimeter 100 contains two squares of the same colour. Show that no rectangle of size $1 \times 1201$ contains two squares of the same colour.
|
Let the centers of the unit squares be the integer points in the plane, and denote each unit square by the coordinates of its center.
Consider the set $D$ of all unit squares $(x, y)$ such that $|x|+|y| \leq 24$. Any translate of $D$ is called a diamond.
Since any two unit squares that belong to the same diamond also belong to some rectangle of perimeter 100, a diamond cannot contain two unit squares of the same colour. Since a diamond contains exactly $24^{2}+25^{2}=1201$ unit squares, a diamond must contain every colour exactly once.
Choose one colour, say, green, and let $a_{1}, a_{2}, \ldots$ be all green unit squares. Let $P_{i}$ be the diamond of center $a_{i}$. We will show that no unit square is covered by two $P$ 's and that every unit square is covered by some $P_{i}$.
Indeed, suppose first that $P_{i}$ and $P_{j}$ contain the same unit square $b$. Then their centers lie within the same rectangle of perimeter 100, a contradiction.
Let, on the other hand, $b$ be an arbitrary unit square. The diamond of center $b$ must contain some green unit square $a_{i}$. The diamond $P_{i}$ of center $a_{i}$ will then contain $b$.
Therefore, $P_{1}, P_{2}, \ldots$ form a covering of the plane in exactly one layer. It is easy to see, though, that, up to translation and reflection, there exists a unique such covering. (Indeed, consider two neighbouring diamonds. Unless they fit neatly, uncoverable spaces of two unit squares are created near the corners: see Fig. 1.)
Figure 1:
Without loss of generality, then, this covering is given by the diamonds of centers $(x, y)$ such that $24 x+25 y$ is divisible by 1201. (See Fig. 2 for an analogous covering with smaller diamonds.) It follows from this that no rectangle of size $1 \times 1201$ can contain two green unit squares, and analogous reasoning works for the remaining colours.
Figure 2:
Remark(PSC): The number of the unit squares in a diamond can be evaluated alternatively with the formula
$$
2 \times(1+3+5+\ldots+47)+49=2 \times 24^{2}+49=1201
$$
## Geometry
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
The plane is divided into unit squares by means of two sets of parallel lines. The unit squares are coloured in 1201 colours so that no rectangle of perimeter 100 contains two squares of the same colour. Show that no rectangle of size $1 \times 1201$ contains two squares of the same colour.
|
Let the centers of the unit squares be the integer points in the plane, and denote each unit square by the coordinates of its center.
Consider the set $D$ of all unit squares $(x, y)$ such that $|x|+|y| \leq 24$. Any translate of $D$ is called a diamond.
Since any two unit squares that belong to the same diamond also belong to some rectangle of perimeter 100, a diamond cannot contain two unit squares of the same colour. Since a diamond contains exactly $24^{2}+25^{2}=1201$ unit squares, a diamond must contain every colour exactly once.
Choose one colour, say, green, and let $a_{1}, a_{2}, \ldots$ be all green unit squares. Let $P_{i}$ be the diamond of center $a_{i}$. We will show that no unit square is covered by two $P$ 's and that every unit square is covered by some $P_{i}$.
Indeed, suppose first that $P_{i}$ and $P_{j}$ contain the same unit square $b$. Then their centers lie within the same rectangle of perimeter 100, a contradiction.
Let, on the other hand, $b$ be an arbitrary unit square. The diamond of center $b$ must contain some green unit square $a_{i}$. The diamond $P_{i}$ of center $a_{i}$ will then contain $b$.
Therefore, $P_{1}, P_{2}, \ldots$ form a covering of the plane in exactly one layer. It is easy to see, though, that, up to translation and reflection, there exists a unique such covering. (Indeed, consider two neighbouring diamonds. Unless they fit neatly, uncoverable spaces of two unit squares are created near the corners: see Fig. 1.)
Figure 1:
Without loss of generality, then, this covering is given by the diamonds of centers $(x, y)$ such that $24 x+25 y$ is divisible by 1201. (See Fig. 2 for an analogous covering with smaller diamonds.) It follows from this that no rectangle of size $1 \times 1201$ can contain two green unit squares, and analogous reasoning works for the remaining colours.
Figure 2:
Remark(PSC): The number of the unit squares in a diamond can be evaluated alternatively with the formula
$$
2 \times(1+3+5+\ldots+47)+49=2 \times 24^{2}+49=1201
$$
## Geometry
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"problem_match": "\n## C3.",
"solution_match": "\nSolution."
}
|
9f26cf5c-1725-5fe2-83be-052e2ef811ea
| 605,178
|
The point $M$ lies on the side $A B$ of the circumscribed quadrilateral $A B C D$. The points $I_{1}, I_{2}$, and $I_{3}$ are the incenters of $\triangle M B C, \triangle M C D$, and $\triangle M D A$. Show that the points $M, I_{1}, I_{2}$, and $I_{3}$ lie on a circle.
|
Lemma. Let $I$ be the incenter of $\triangle A B C$ and let the points $P$ and $Q$ lie on the lines $A B$ and $A C$. Then the points $A, I, P$, and $Q$ lie on a circle if and only if
$$
\overline{B P}+\overline{C Q}=B C
$$
where $\overline{B P}$ equals $|B P|$ if $P$ lies in the ray $B A \rightarrow$ and $-|B P|$ if it does not, and similarly for $\overline{C Q}$.
Proof of the lemma. We shall only consider the case when $P$ and $Q$ lie in the segments $A B$ and $A C$. All other cases are treated analogously.
Suppose that $A, I, P$, and $Q$ lie on a circle. Let $D$ and $E$ be the contact points of the incircle of $\triangle A B C$ with $A B$ and $A C$. We have that $\angle P I Q=180^{\circ}-\alpha$, so $\angle D I P=\angle E I Q$ and, therefore, $\triangle D I P \simeq \triangle E I Q$. This gives us $D P=E Q$ and $B P+C Q=B D+C E=B C$, as needed.
The converse is established by following the foregoing chain of inequalities in reverse.
Let the circumcircle of $\triangle M I_{1} I_{3}$ meet the lines $A B, C M$, and $D M$ for the second time at $P, Q$, and $R$. By the lemma, $\overline{B P}+\overline{C Q}=B C$ and $\overline{D R}+\overline{A P}=D A$. Therefore, $\overline{C Q}+\overline{D R}=B C+D A-\overline{B P}-\overline{A P}=$ $B C+D A-A B$. Since $A B C D$ is circumscribed, this is equal to $C D$, and, by the lemma, the proof is complete.
Second solution. Let $\omega_{1}, \omega_{2}$, and $\omega_{3}$ be the incircles of $\triangle M B C, \triangle M C D$, and $\triangle M D A$.
The common internal tangent $t_{1}$ of $\omega_{1}$ and $\omega_{2}$ equals
$\left[\right.$ tangent from $M$ to $\left.\omega_{2}\right]-\left[\right.$ tangent from $M$ to $\left.\omega_{1}\right]=\frac{1}{2}(M C+M D-C D-M B-M C+B C)$.
Analogously, the common internal tangent $t_{2}$ of $\omega_{2}$ and $\omega_{3}$ equals
$\left[\right.$ tangent from $M$ to $\left.\omega_{2}\right]-\left[\right.$ tangent from $M$ to $\left.\omega_{3}\right]=\frac{1}{2}(M C+M D-C D-M D-M A+D A)$.
Finally, the common external tangent $t_{3}$ of $\omega_{1}$ and $\omega_{3}$ equals
$\left[\right.$ tangent from $M$ to $\left.\omega_{1}\right]-\left[\right.$ tangent from $M$ to $\left.\omega_{3}\right]=\frac{1}{2}(M B+M C-B C+M D+M A-D A)$.
Since $A B C D$ is circumscribed, we have $A B+C D=B C+D A$, and, therefore, $t_{1}+t_{2}=t_{3}$. It follows from this that $\omega_{1}, \omega_{2}$, and $\omega_{3}$ have a common tangent $s$ (which separates $\omega_{2}$ from $\omega_{1}$ and $\omega_{3}$ ).
Let $\triangle M K L$ be the triangle formed by the lines $M C, M D$, and $s$. Then, since $I_{1} I_{2}$ and $I_{2} I_{3}$ are external angle bisectors in it, we have $\angle I_{1} I_{2} I_{3}=90^{\circ}-\frac{1}{2} \angle K M L=180^{\circ}-\angle I_{1} M I_{3}$ and, therefore, $M I_{1} I_{2} I_{3}$ is cyclic.
$\operatorname{Remark}(\mathbf{P S C}):$ In the first solution, the angle $\alpha$ refers to the angle $\angle B A C$.
$\operatorname{Remark}(\mathbf{P S C})$ : Referring to solution 1 , the inequality $t_{3} \leq t_{1}+t_{2}$ holds, the equality is possible if and only if the circles $\omega_{1}, \omega_{2}$, and $\omega_{3}$ have a common tangent.
Remark(PSC): This problem was selected as G1 relative to the other Geometry problems proposed. PSC thinks the difficulty level of this problem is medium.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
The point $M$ lies on the side $A B$ of the circumscribed quadrilateral $A B C D$. The points $I_{1}, I_{2}$, and $I_{3}$ are the incenters of $\triangle M B C, \triangle M C D$, and $\triangle M D A$. Show that the points $M, I_{1}, I_{2}$, and $I_{3}$ lie on a circle.
|
Lemma. Let $I$ be the incenter of $\triangle A B C$ and let the points $P$ and $Q$ lie on the lines $A B$ and $A C$. Then the points $A, I, P$, and $Q$ lie on a circle if and only if
$$
\overline{B P}+\overline{C Q}=B C
$$
where $\overline{B P}$ equals $|B P|$ if $P$ lies in the ray $B A \rightarrow$ and $-|B P|$ if it does not, and similarly for $\overline{C Q}$.
Proof of the lemma. We shall only consider the case when $P$ and $Q$ lie in the segments $A B$ and $A C$. All other cases are treated analogously.
Suppose that $A, I, P$, and $Q$ lie on a circle. Let $D$ and $E$ be the contact points of the incircle of $\triangle A B C$ with $A B$ and $A C$. We have that $\angle P I Q=180^{\circ}-\alpha$, so $\angle D I P=\angle E I Q$ and, therefore, $\triangle D I P \simeq \triangle E I Q$. This gives us $D P=E Q$ and $B P+C Q=B D+C E=B C$, as needed.
The converse is established by following the foregoing chain of inequalities in reverse.
Let the circumcircle of $\triangle M I_{1} I_{3}$ meet the lines $A B, C M$, and $D M$ for the second time at $P, Q$, and $R$. By the lemma, $\overline{B P}+\overline{C Q}=B C$ and $\overline{D R}+\overline{A P}=D A$. Therefore, $\overline{C Q}+\overline{D R}=B C+D A-\overline{B P}-\overline{A P}=$ $B C+D A-A B$. Since $A B C D$ is circumscribed, this is equal to $C D$, and, by the lemma, the proof is complete.
Second solution. Let $\omega_{1}, \omega_{2}$, and $\omega_{3}$ be the incircles of $\triangle M B C, \triangle M C D$, and $\triangle M D A$.
The common internal tangent $t_{1}$ of $\omega_{1}$ and $\omega_{2}$ equals
$\left[\right.$ tangent from $M$ to $\left.\omega_{2}\right]-\left[\right.$ tangent from $M$ to $\left.\omega_{1}\right]=\frac{1}{2}(M C+M D-C D-M B-M C+B C)$.
Analogously, the common internal tangent $t_{2}$ of $\omega_{2}$ and $\omega_{3}$ equals
$\left[\right.$ tangent from $M$ to $\left.\omega_{2}\right]-\left[\right.$ tangent from $M$ to $\left.\omega_{3}\right]=\frac{1}{2}(M C+M D-C D-M D-M A+D A)$.
Finally, the common external tangent $t_{3}$ of $\omega_{1}$ and $\omega_{3}$ equals
$\left[\right.$ tangent from $M$ to $\left.\omega_{1}\right]-\left[\right.$ tangent from $M$ to $\left.\omega_{3}\right]=\frac{1}{2}(M B+M C-B C+M D+M A-D A)$.
Since $A B C D$ is circumscribed, we have $A B+C D=B C+D A$, and, therefore, $t_{1}+t_{2}=t_{3}$. It follows from this that $\omega_{1}, \omega_{2}$, and $\omega_{3}$ have a common tangent $s$ (which separates $\omega_{2}$ from $\omega_{1}$ and $\omega_{3}$ ).
Let $\triangle M K L$ be the triangle formed by the lines $M C, M D$, and $s$. Then, since $I_{1} I_{2}$ and $I_{2} I_{3}$ are external angle bisectors in it, we have $\angle I_{1} I_{2} I_{3}=90^{\circ}-\frac{1}{2} \angle K M L=180^{\circ}-\angle I_{1} M I_{3}$ and, therefore, $M I_{1} I_{2} I_{3}$ is cyclic.
$\operatorname{Remark}(\mathbf{P S C}):$ In the first solution, the angle $\alpha$ refers to the angle $\angle B A C$.
$\operatorname{Remark}(\mathbf{P S C})$ : Referring to solution 1 , the inequality $t_{3} \leq t_{1}+t_{2}$ holds, the equality is possible if and only if the circles $\omega_{1}, \omega_{2}$, and $\omega_{3}$ have a common tangent.
Remark(PSC): This problem was selected as G1 relative to the other Geometry problems proposed. PSC thinks the difficulty level of this problem is medium.
|
{
"resource_path": "Balkan_Shortlist/segmented/en-2016_bmo_shortlist.jsonl",
"problem_match": "\n## G1.",
"solution_match": "\nSolution."
}
|
83353965-80b6-5f53-853d-2abbb5188686
| 605,190
|
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