problem
string | solution
string | answer
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int64 |
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The graph of $y=e^{x+1}+e^{-x}-2$ has an axis of symmetry. What is the reflection of the point $(-1,\tfrac{1}{2})$ over this axis?
$\textbf{(A) }\left(-1,-\frac{3}{2}\right)\qquad\textbf{(B) }(-1,0)\qquad\textbf{(C) }\left(-1,\frac{1}{2}\right)\qquad\textbf{(D) }\left(0,\frac{1}{2}\right)\qquad\textbf{(E) }\left(3,\frac{1}{2}\right)$
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The line of symmetry is probably of the form $x=a$ for some constant $a$. A vertical line of symmetry at $x=a$ for a function $f$ exists if and only if $f(a-b)=f(a+b)$; we substitute $a-b$ and $a+b$ into our given function and see that we must have
\[e^{a-b+1}+e^{-(a-b)}-2=e^{a+b+1}+e^{-(a+b)}-2\]
for all real $b$. Simplifying:
If $e^{a+1}-e^{-a}\neq0$, then $e^{-b}=e^b$ for all real $b$; this is clearly impossible, so let $e^{a+1}-e^{-a}=0\implies a+1=-a\implies a=-\dfrac12$. Thus, our line of symmetry is $x=-\dfrac12$, and reflecting $\left(-1,\dfrac12\right)$ over this line gives $\boxed{\textbf{(D) }\left(0,\dfrac12\right)}.$
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\left(0,\frac{1}{2}\right)
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https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_13
| 2,024 |
The numbers, in order, of each row and the numbers, in order, of each column of a $5 \times 5$ array of integers form an arithmetic progression of length $5{.}$ The numbers in positions $(5, 5), \,(2,4),\,(4,3),$ and $(3, 1)$ are $0, 48, 16,$ and $12{,}$ respectively. What number is in position $(1, 2)?$\[\begin{bmatrix} . & ? &.&.&. \\ .&.&.&48&.\\ 12&.&.&.&.\\ .&.&16&.&.\\ .&.&.&.&0\end{bmatrix}\]
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Start from the $0$. Going up, let the common difference be $a$, and going left, let the common difference be $b$. Therefore, we have\[\begin{bmatrix} . & ? &.&.&4a \\ .&.&.&48&3a\\ 12&.&.&.&2a\\ .&.&16&.&a\\ 4b&3b&2b&b&0\end{bmatrix}\]Looking at the third column, we can see that the common difference going up is $16-2b$. We fill this in:\[\begin{bmatrix} . & ? &64-6b&.&4a \\ .&.&48-4b&48&3a\\ 12&.&32-2b&.&2a\\ .&.&16&.&a\\ 4b&3b&2b&b&0\end{bmatrix}\]Looking at the second row, $48$ has two values beside it, so we can write\[48=\dfrac{48-4b+3a}{2}\rightarrow96=48-4b+3a\rightarrow48=-4b+3a,\]and we can do the same with the third row, which gives\[32-2b=\dfrac{12+2a}{2}\rightarrow32-2b=6+a\rightarrow26=a+2b.\]Now we have the system of equations\[48=-4b+3a\]\[26=a+2b,\]and solving it gives $a=20,b=3$, therefore we can now fill in the grid with actual numbers. But before doing that, note that we're only looking for a value in the first row, and because we already have two known values in that row, we can find the common difference for that row and not focus on anything else.
Focusing only on the first row yields\[\begin{bmatrix} . & ? &46&.&80\end{bmatrix}\]This means that the common difference from right to left is $\dfrac{80-46}{2}=17$. Therefore, the desired value is $46-17=\boxed{\text{(C) }29}$
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29
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https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_14
| 2,024 |
The roots of $x^3 + 2x^2 - x + 3$ are $p, q,$ and $r.$ What is the value of\[(p^2 + 4)(q^2 + 4)(r^2 + 4)?\]
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You can factor $(p^2 + 4)(q^2 + 4)(r^2 + 4)$ as $(p+2i)(p-2i)(q+2i)(q-2i)(r+2i)(r-2i)$.
For any polynomial $f(x)$, you can create a new polynomial $f(x+2)$, which will have roots that instead have the value subtracted.
Substituting $x-2$ and $x+2$ into $x$ for the first polynomial, gives you $10i-5$ and $-10i-5$ as $c$ for both equations. Multiplying $10i-5$ and $-10i-5$ together gives you $\boxed{\textbf{(D) }125}$.
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125
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https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_15
| 2,024 |
A set of $12$ tokens — $3$ red, $2$ white, $1$ blue, and $6$ black — is to be distributed at random to $3$ game players, $4$ tokens per player. The probability that some player gets all the red tokens, another gets all the white tokens, and the remaining player gets the blue token can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?
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We have $\binom{12}{4,4,4}$ ways to handle the red/white/blue balls distribution on the denominator. Now we simply $\binom{6}{1}$ $\binom{5}{2}$ $3!$ for the numerator in order to handle the black balls and distinguishable persons. The solution is therefore $\frac {6 \cdot 6 \cdot 10}{70 \cdot 45 \cdot 11} = \frac {4}{385}$ or $4+385=\boxed{\textbf{(C) }389}.$
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389
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https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_16
| 2,024 |
Integers $a$, $b$, and $c$ satisfy $ab + c = 100$, $bc + a = 87$, and $ca + b = 60$. What is $ab + bc + ca$?
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Subtracting the first two equations yields $(a-c)(b-1)=13$. Notice that both factors are integers, so $b-1$ could equal one of $13,1,-1,-13$ and $b=14,2,0,-12$. We consider each case separately: 3 For $b=0$, from the second equation, we see that $a=87$. Then $87c=60$, which is not possible as $c$ is an integer, so this case is invalid.
For $b=2$, we have $2c+a=87$ and $ca=58$, which by experimentation on the factors of $58$ has no solution, so this is also invalid.
For $b=14$, we have $14c+a=87$ and $ca=46$, which by experimentation on the factors of $46$ has no solution, so this is also invalid.
Thus, we must have $b=-12$, so $a=12c+87$ and $ca=72$. Thus $c(12c+87)=72$, so $c(4c+29)=24$. We can simply trial and error this to find that $c=-8$ so then $a=-9$. The answer is then $(-9)(-12)+(-12)(-8)+(-8)(-9)=108+96+72=\boxed{\textbf{(D) }276}$.
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276
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https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_17
| 2,024 |
Cyclic quadrilateral $ABCD$ has lengths $BC=CD=3$ and $DA=5$ with $\angle CDA=120^\circ$. What is the length of the shorter diagonal of $ABCD$?
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First, $\angle CBA=60 ^\circ$ by properties of cyclic quadrilaterals.
Let $AC=u$. Apply the Law of Cosines on $\triangle ACD$:\[u^2=3^2+5^2-2(3)(5)\cos120^\circ\]\[u=7\]
Let $AB=v$. Apply the Law of Cosines on $\triangle ABC$:\[7^2=3^2+v^2-2(3)(v)\cos60^\circ\]\[v=\frac{3\pm13}{2}\]\[v=8\]
By Ptolemy’s Theorem,\[AB \cdot CD+AD \cdot BC=AC \cdot BD\]\[8 \cdot 3+5 \cdot 3=7BD\]\[BD=\frac{39}{7}\]Since $\frac{39}{7}<7$, The answer is $\boxed{\textbf{(D) }\frac{39}{7}}$.
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\frac{39}{7}
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https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_19
| 2,024 |
Points $P$ and $Q$ are chosen uniformly and independently at random on sides $\overline {AB}$ and $\overline{AC},$ respectively, of equilateral triangle $\triangle ABC.$ Which of the following intervals contains the probability that the area of $\triangle APQ$ is less than half the area of $\triangle ABC?$
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Let $\overline{AP}=x$ and $\overline{AQ}=y$. Applying the sine formula for a triangle's area, we see that\[[\Delta APQ]=\dfrac12\cdot x\cdot y\sin(\angle PAQ)=\dfrac{xy}2\sin(60^\circ)=\dfrac{\sqrt3}4xy.\]
Without loss of generality, we let $AB=BC=CA=1$, and thus $[\Delta ABC]=\dfrac{\sqrt3}4$; we therefore require $\dfrac{\sqrt3}4xy\le\dfrac12\cdot\dfrac{\sqrt3}4\implies xy\le\dfrac12$ for $0\le x,y\le1$. (Note: You can skip most of this by using triangle area ratios ~A_MatheMagician.)
A quick rough sketch of $y=\dfrac1{2x}$ on the square given by $x,y\in[0,1]$ reveals that the curve intersects the boundaries at $(0.5,1)$ and $(1,0.5)$, and it is actually quite (very) obvious that the area bounded by the inequality $xy\le0.5$ and the aforementioned unit square is more than $\dfrac34$ but less than $\dfrac78$ (cf. the diagram below). Thus, our answer is $\boxed{\textbf{(D) }\left(\dfrac34,\dfrac78\right]}$.
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\left(\dfrac34,\dfrac78\right]
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https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_20
| 2,024 |
Suppose that $a_1 = 2$ and the sequence $(a_n)$ satisfies the recurrence relation\[\frac{a_n -1}{n-1}=\frac{a_{n-1}+1}{(n-1)+1}\]for all $n \ge 2.$ What is the greatest integer less than or equal to\[\sum^{100}_{n=1} a_n^2?\]
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Multiply both sides of the recurrence to find that $n(a_n-1)=(n-1)(a_{n-1}+1)=(n-1)(a_{n-1}-1)+(n-1)(2)$.
Let $b_n=n(a_n-1)$. Then the previous relation becomes\[b_n=b_{n-1}+2(n-1)\]
We can rewrite this relation for values of $n$ until $1$ and use telescoping to derive an explicit formula:
\[b_n=b_{n-1}+2(n-1)\]\[b_{n-1}=b_{n-2}+2(n-2)\]\[b_{n-2}=b_{n-3}+2(n-3)\]\[\cdot\]\[\cdot\]\[\cdot\]\[b_2=b_1+2(1)\]
Summing the equations yields:
\[b_n+b_{n-1}+\cdots+b_2=b_{n-1}+b_{n-2}+\cdots+b_1+2((n-1)+(n-2)+\cdots+1)\]\[b_n-b_1=2\cdot\frac{n(n-1)}{2}\]\[b_n-1=n(n-1)\]\[b_n=n(n-1)+1\]
Now we can substitute $a_n$ back into our equation:
\[n(a_n-1)=n(n-1)+1\]\[a_n-1=n-1+\frac{1}{n}\]\[a_n=n+\frac{1}{n}\]\[a_n^2=n^2+\frac{1}{n^2}+2\]
Thus the sum becomes
\[\sum^{100}_{n=1} a_n^2= \sum^{100}_{n=1} n^2+ \sum^{100}_{n=1} \frac{1}{n^2}+ \sum^{100}_{n=1} 2\]
We know that $\sum^{100}_{n=1} n^2=\frac{100\cdot101\cdot201}{6}=338350$, and we also know that $\sum^{100}_{n=1} 2=200$, so the requested sum is equivalent to $\sum^{100}_{n=1} \frac{1}{n^2}+338550$. All that remains is to calculate $\sum^{100}_{n=1} \frac{1}{n^2}$, and we know that this value lies between $1$ and $2$ (see the note below for a proof). Thus,
\[\sum^{100}_{n=1} a_n^2= \sum^{100}_{n=1} \frac{1}{n^2}+338550<2+338550=338552\]\[\sum^{100}_{n=1} a_n^2= \sum^{100}_{n=1} \frac{1}{n^2}+338550>1+338550=338551\]
so
\[\sum^{100}_{n=1} a_n^2\in(338551,338552)\]
and thus the answer is $\boxed{\textbf{(B) }338551}$.
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338,551
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https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_21
| 2,024 |
What is the value of\[\tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {3\pi}{16} + \tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {5\pi}{16}+\tan^2 \frac {3\pi}{16} \cdot \tan^2 \frac {7\pi}{16}+\tan^2 \frac {5\pi}{16} \cdot \tan^2 \frac {7\pi}{16}?\]
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First, notice that
\[\tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {3\pi}{16} + \tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {5\pi}{16}+\tan^2 \frac {3\pi}{16} \cdot \tan^2 \frac {7\pi}{16}+\tan^2 \frac {5\pi}{16} \cdot \tan^2 \frac {7\pi}{16}\]
\[=(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2 \frac{5\pi}{16})\]
Here, we make use of the fact that
\[\tan^2 x+\tan^2 (\frac{\pi}{2}-x)\]\[=(\tan x+\tan (\frac{\pi}{2}-x))^2-2\]\[=\left(\frac{\sin x}{\cos x}+\frac{\sin (\frac{\pi}{2}-x)}{\cos (\frac{\pi}{2}-x)}\right)^2-2\]\[=\left(\frac{\sin x \cos (\frac{\pi}{2}-x)+\sin (\frac{\pi}{2}-x) \cos x}{\cos x \cos (\frac{\pi}{2}-x)}\right)^2-2\]\[=\left(\frac{\sin \frac{\pi}{2}}{\cos x \cos (\frac{\pi}{2}-x)}\right)^2-2\]\[=\left(\frac{1}{\cos x \sin x}\right)^2-2\]\[=\left(\frac{2}{\sin 2x}\right)^2-2\]\[=\frac{4}{\sin^2 2x}-2\]
Hence,
\[(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2 \frac{4\pi}{16})\]\[=\left(\frac{4}{\sin^2 \frac{\pi}{8}}-2\right)\left(\frac{4}{\sin^2 \frac{3\pi}{8}}-2\right)\]
Note that
\[\sin^2 \frac{\pi}{8}=\frac{1-\cos \frac{\pi}{4}}{2}=\frac{2-\sqrt{2}}{4}\]
\[\sin^2 \frac{3\pi}{8}=\frac{1-\cos \frac{3\pi}{4}}{2}=\frac{2+\sqrt{2}}{4}\]
Hence,
\[\left(\frac{4}{\sin^2 \frac{\pi}{8}}-2\right)\left(\frac{4}{\sin^2 \frac{3\pi}{8}}-2\right)\]
\[=\left(\frac{16}{2-\sqrt{2}}-2\right)\left(\frac{16}{2+\sqrt{2}}-2\right)\]
\[=(14+8\sqrt{2})(14-8\sqrt{2})\]
\[=68\]
Therefore, the answer is $\fbox{\textbf{(B) } 68}$.
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68
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https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_23
| 2,024 |
A $\textit{disphenoid}$ is a tetrahedron whose triangular faces are congruent to one another. What is the least total surface area of a disphenoid whose faces are scalene triangles with integer side lengths?
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Notice that any scalene $\textit{acute}$ triangle can be the faces of a $\textit{disphenoid}$. (See proof in Solution 2.)
As a result, we simply have to find the smallest area a scalene acute triangle with integer side lengths can take on. This occurs with a $4,5,6$ triangle (notice that if you decrease the value of any of the sides the resulting triangle will either be isosceles, degenerate, or non-acute). For this triangle, the semiperimeter is $\frac{15}{2}$, so by Heron’s Formula:
The surface area is simply four times the area of one of the triangles, or $\boxed{\textbf{(D) }15\sqrt{7}}$.
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15\sqrt{7}
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https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_24
| 2,024 |
A graph is $\textit{symmetric}$ about a line if the graph remains unchanged after reflection in that line. For how many quadruples of integers $(a,b,c,d)$, where $|a|,|b|,|c|,|d|\le5$ and $c$ and $d$ are not both $0$, is the graph of\[y=\frac{ax+b}{cx+d}\]symmetric about the line $y=x$?
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Symmetric about the line $y=x$ implies that the inverse function $y^{-1}=y$. Then we split the question into several cases to find the final answer.
Case 1: $c=0$
Then $y=\frac{a}{d}x+\frac{b}{d}$ and $y^{-1}=\frac{d}{a}x-\frac{b}{a}$. Giving us $\frac{a}{d}=\frac{d}{a}$ and $\frac{b}{d}=-\frac{b}{a}$
Therefore, we obtain 2 subcases: $b\neq 0, a+d=0$ and $b=0, a^2=d^2$
Case 2: $c\neq 0$
Then $y^{-1}=\frac{b-dx}{cx-a}=\frac{(cx-a)(-\frac{d}{c})+b-\frac{ad}{c}}{cx-a}=-\frac{d}{c}+\frac{b-\frac{ad}{c}}{cx-a}$
And $y=\frac{(cx+d)(\frac{a}{c})+b-\frac{ad}{c}}{cx+d}=\frac{a}{c}+\frac{b-\frac{ad}{c}}{cx+d}$
So $\frac{a}{c}=-\frac{d}{c}$, or $a=-d$ ($c\neq 0$), and substitute that into $\frac{b-\frac{ad}{c}}{cx-a}=\frac{b-\frac{ad}{c}}{cx+d}$ gives us:
$bc-ad\neq 0$ (Otherwise $y=\frac{a}{c}$, $y^{-1}=-\frac{d}{c}=\frac{a}{c}$, and is not symmetric about $y=x$)
Therefore we get three cases:
Case 1.1: $c= 0, b\neq 0, d\neq 0, a+d=0$
We have 10 choice of $b$, 10 choice of $d$ and each choice of $d$ has one corresponding choice of $a$. In total $10\times 10=100$ ways.
Case 1.2: $c= 0, b = 0, d\neq 0, a^2=d^2$
We have 10 choice for $d$ ($d\neq 0$), each choice of $d$ has 2 corresponding choice of $a$, thus $10\times 2=20$ ways.
Case 2: $c\neq 0, bc-ad\neq 0, a=-d$
$a=0$: $10\times 10=100$ ways.
$a=\pm 1$: $(11\times 10-2)\times 2=216$ ways.
$a=\pm 2$: $(11\times 10-2)\times 2=216$ ways.
$a=\pm 3$: $(11\times 10-2)\times 2=216$ ways.
$a=\pm 4$: $(11\times 10-6)\times 2=208$ ways.
$a=\pm 5$: $(11\times 10-2)\times 2=216$ ways.
In total $100+208+216\times 4= 1172$ ways.
So the answer is $100+20+1172= \boxed{\textbf{(B) }1292}$
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1292
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https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_25
| 2,024 |
In a long line of people arranged left to right, the 1013th person from the left is also the 1010th person from the right. How many people are in the line?
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If the person is the 1013th from the left, that means there is 1012 people to their left. If the person is the 1010th from the right, that means there is 1009 people to their right. Therefore, there are $1012 + 1 + 1009 = \boxed{\textbf{(B) } 2022}$ people in line.
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2022
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https://artofproblemsolving.com/wiki/index.php/2024_AMC_12B_Problems/Problem_1
| 2,024 |
What is $10! - 7! \cdot 6!$
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$10! = 10 \cdot 9 \cdot 8 \cdot 7! = 720 \cdot 7!$
$6! \cdot 7! = 720 \cdot 7!$
Therefore, the equation is equal to $720 \cdot 7! - 720 \cdot 7! = \boxed{\textbf{(B) }0}$
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0
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https://artofproblemsolving.com/wiki/index.php/2024_AMC_12B_Problems/Problem_2
| 2,024 |
For how many integer values of $x$ is $|2x| \leq 7 \pi$
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$\pi = 3.14159\dots$ is slightly less than $\dfrac{22}{7} = 3.\overline{142857}$. So $7\pi \approx 21.9$ The inequality expands to be $-21.9 \le 2x \le 21.9$. We find that $x$ can take the integer values between $-10$ and $10$ inclusive. There are $\boxed{\text{E. }21}$ such values.
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21
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https://artofproblemsolving.com/wiki/index.php/2024_AMC_12B_Problems/Problem_3
| 2,024 |
Balls numbered 1, 2, 3, ... are deposited in 5 bins, labeled A, B, C, D, and E, using the following procedure. Ball 1 is deposited in bin A, and balls 2 and 3 are deposited in bin B. The next 3 balls are deposited in bin C, the next 4 in bin D, and so on, cycling back to bin A after balls are deposited in bin E. (For example, balls numbered 22, 23, ..., 28 are deposited in bin B at step 7 of this process.) In which bin is ball 2024 deposited?
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Consider the triangular array of numbers:\[1\]\[2, 3\]\[4, 5, 6\]\[7, 8, 9, 10\]\[11, 12, 13, 14, 15\]\[\vdots\].
The numbers in a row congruent to $1 \bmod{5}$ will be in bucket A. Similarly, the numbers in a row congruent to $2, 3, 4, 0 \bmod{5}$ will be in buckets B, C, D, and E respectively. Note that the $n^\text{th}$ row ends with the $n^\text{th}$ triangle number, $\frac{n(n+1)}{2}$.
We must find values of $n$ that make $\frac{n(n+1)}{2}$ close to $2024$.\[\frac{n(n+1)}{2} \approx 2024\]\[n(n+1) \approx 4048\]\[n^2 \approx 4048\]\[n \approx 63\]
Trying $n = 63$ we finde that $\frac{n(n+1)}{2} = 2016$. Since $2016$ will be the last ball in row $63$, ball $2024$ will be in row $64$. Since $64 \equiv 4 \bmod{5}$, ball $2024$ will be placed in bucket $\boxed{\text{D. } D}$
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D
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https://artofproblemsolving.com/wiki/index.php/2024_AMC_12B_Problems/Problem_4
| 2,024 |
In the following expression, Melanie changed some of the plus signs to minus signs:\[1+3+5+7+...+97+99\]When the new expression was evaluated, it was negative. What is the least number of plus signs that Melanie could have changed to minus signs?
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Recall that the sum of the first $n$ odd numbers is $n^2$. Thus\[1 + 3 + 5 + 7+ \dots + 97 + 99 = 50^2 = 2500.\]
If we want to minimize the number of sign flips to make the number negative, we must flip the signs corresponding to the values with largest absolute value. This will result in the inequality\[1 + 3 + 5 +\dots + (2n - 3) + (2n - 1) - (2n + 1) - (2n + 3)-\dots - 97 - 99 < 0.\]
The positive section of the sum will contribute $n^2$, and the negative section will contribute $-(2500-n^2) = (n^2 - 2500)$. The inequality simplifies to\[n^2 + (n^2 - 2500) < 0\]\[2n^2 < 2500\]\[n^2 < 1250\]The greatest positive value of $n$ satisfying the inequality is $n = 35$, corresponding to $35$ positive numbers, and $\boxed{\text{B. } 15}$ negatives.
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15
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https://artofproblemsolving.com/wiki/index.php/2024_AMC_12B_Problems/Problem_5
| 2,024 |
The national debt of the United States is on track to reach $5\times10^{13}$ dollars by $2033$. How many digits does this number of dollars have when written as a numeral in base 5? (The approximation of $\log_{10} 5$ as $0.7$ is sufficient for this problem)
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Generally, number of digits of a number $n$ in base $b$:\[\text{Number of digits} = \lfloor \log_b n \rfloor + 1\]In this question, it is $\lfloor \log_{5} 5\times 10^{13}\rfloor+1$. Note that\[\log_{5} 5\times 10^{13}=1+\frac{13}{\log_{10} 5}\]\[\approx 1+\frac{13}{0.7}\]\[\approx 19.5\]
Hence, our answer is $\fbox{\textbf{(B) } 20}$
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20
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https://artofproblemsolving.com/wiki/index.php/2024_AMC_12B_Problems/Problem_6
| 2,024 |
What value of $x$ satisfies\[\frac{\log_2x \cdot \log_3x}{\log_2x+\log_3x}=2?\]
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We have
\begin{align*}
\log_2x\cdot\log_3x&=2(\log_2x+\log_3x) \\
1&=\frac{2(\log_2x+\log_3x)}{\log_2x\cdot\log_3x} \\
1&=2(\frac{1}{\log_3x}+\frac{1}{\log_2x}) \\
1&=2(\log_x3+\log_x2) \\
\log_x6&=\frac{1}{2} \\
x^{\frac{1}{2}}&=6 \\
x&=36
\end{align*}
so $\boxed{\textbf{(C) }36}$
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36
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https://artofproblemsolving.com/wiki/index.php/2024_AMC_12B_Problems/Problem_8
| 2,024 |
A dartboard is the region B in the coordinate plane consisting of points $(x, y)$ such that $|x| + |y| \le 8$. A target T is the region where $(x^2 + y^2 - 25)^2 \le 49$. A dart is thrown and lands at a random point in B. The probability that the dart lands in T can be expressed as $\frac{m}{n} \cdot \pi$, where $m$ and $n$ are relatively prime positive integers. What is $m + n$?
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Inequalities of the form $|x|+|y| \le 8$ are well-known and correspond to a square in space with centre at origin and vertices at $(8, 0)$, $(-8, 0)$, $(0, 8)$, $(0, -8)$. The diagonal length of this square is clearly $16$, so it has an area of\[\frac{1}{2} \cdot 16 \cdot 16 = 128\]Now,\[(x^2 + y^2 - 25)^2 \le 49\]Converting to polar form,\[r^2 - 25 \le 7 \implies r \le \sqrt{32},\]and\[r^2 - 25 \ge -7\implies r\ge \sqrt{18}.\]
The intersection of these inequalities is the circular region $T$ for which every circle in $T$ has a radius between $\sqrt{18}$ and $\sqrt{32}$, inclusive. The area of such a region is thus $\pi(32-18)=14\pi.$ The requested probability is therefore $\frac{14\pi}{128} = \frac{7\pi}{64},$ yielding $(m,n)=(7,64).$ We have $m+n=7+64=\boxed{\textbf{(B)}\ 71}.$
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71
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https://artofproblemsolving.com/wiki/index.php/2024_AMC_12B_Problems/Problem_9
| 2,024 |
A list of $9$ real numbers consists of $1$, $2.2$, $3.2$, $5.2$, $6.2$, and $7$, as well as $x$, $y$ , and $z$ with $x$ $\le$ $y$ $\le$ $z$. The range of the list is $7$, and the mean and the median are both positive integers. How many ordered triples ($x$, $y$, $z$) are possible?
|
The sum of the six existing numbers is $24.8$. For the mean to be an integer, the sum of the remaining three numbers $x,y,z$ and the remaining six must be divisible by $9$; thus $x+y+z$ must equal either $2.2,11.2,20.2$. However, one of $x,y,z$ must be the median since the middle four numbers in the original set are all non-integral. As a result, we can eliminate $2.2$ since the sum would be too small to allow for one of $x,y,z$ to be the median (notice that negative numbers are not permitted since the range would be greater than $7$). This leaves two cases.
For the $x+y+z=20.2$ case, notice that all three must be less than or equal to $8$ due to the range restriction. However, also notice that $y$ and $z$ must be on the higher end of the range, meaning that the new median would have to fall between the new middle numbers $5.2$ and $6.2$. Thus $x=6$, and we can manipulate the numbers to make the range $7$ by making $z=8$ and thus $y=6.2$, providing one case in $(6,6.2,8)$.
For the $x+y+z=11.2$ case, we note that the average of the three would fall near the median, signaling that one of the numbers would be the median, one would fall lesser than the median, and one would reach greater than the median. Thus we let the median be $y$. Since adding a number to each side of the set would not change the median, we know that the median must fall between the two middle numbers $3.2,5.2$. Then we find two triples $(0.1,4,7.1)$ and $(0,5,6.2)$, resulting in $\boxed{\textbf{(C) }3}$.
|
3
|
https://artofproblemsolving.com/wiki/index.php/2024_AMC_12B_Problems/Problem_10
| 2,024 |
Let $x_n = \sin^2(n^{\circ})$. What is the mean of $x_1,x_2,x_3,\dots,x_{90}$?
|
Add up $x_1$ with $x_{89}$, $x_2$ with $x_{88}$, and $x_i$ with $x_{90-i}$. Notice\[x_i+x_{90-i}=\sin^2(i^{\circ})+\sin^2((90-i)^{\circ})=\sin^2(i^{\circ})+\cos^2(i^{\circ})=1\]by the Pythagorean identity. Since we can pair up $1$ with $89$ and keep going until $44$ with $46$, we get\[x_1+x_2+\dots+x_{90}=44+x_{45}+x_{90}=44+\left(\frac{\sqrt{2}}{2}\right)^2+1^2=\frac{91}{2}\]Hence the mean is $\boxed{\textbf{(E) }\frac{91}{180}}$
|
\frac{91}{180}
|
https://artofproblemsolving.com/wiki/index.php/2024_AMC_12B_Problems/Problem_11
| 2,024 |
Suppose $z$ is a complex number with positive imaginary part, with real part greater than $1$, and with $|z| = 2$. In the complex plane, the four points $0$, $z$, $z^{2}$, and $z^{3}$ are the vertices of a quadrilateral with area $15$. What is the imaginary part of $z$?
|
By making a rough estimate of where $z$, $z^2$, and $z^3$ are on the complex plane, we can draw a pretty accurate diagram (like above.)
Here, points $Z_1$, $Z_2$, and $Z_3$ lie at the coordinates of $z$, $z^2$, and $z^3$ respectively, and $O$ is the origin.
We're given $|z|=2$, so $|z^2|=|z|^2=4$ and $|z^3|=|z|^3 = 8$. This gives us $OZ_1=2$, $OZ_2=4$, and $OZ_3=8$.
Additionally, we know that $\angle{Z_1OZ_2}\cong\angle{Z_2OZ_3}$ (since every power of $z$ rotates around the origin by the same angle.) We set these angles equal to $\theta$.
We have that
Since this is equal to $15$, we have $20\sin\theta=15$, so $\sin\theta=\frac{3}{4}$.
Thus, $\text{Im}(z)=|z|\sin(\theta)=2(\frac{3}{4})=\boxed{\textbf{(D) }\frac{3}{2}}$.
|
\frac{3}{2}
|
https://artofproblemsolving.com/wiki/index.php/2024_AMC_12B_Problems/Problem_12
| 2,024 |
There are real numbers $x,y,h$ and $k$ that satisfy the system of equations\[x^2 + y^2 - 6x - 8y = h\]\[x^2 + y^2 - 10x + 4y = k\]What is the minimum possible value of $h+k$?
|
Adding up the first and second equation, we get:\begin{align*} h + k &= 2x^2 + 2y^2 - 16x - 4y \\ &= 2(x^2 - 8x) + 2(y^2 - 2y) \\ &= 2(x^2 - 8x) + 2(y^2 - 2y) \\ &= 2(x^2 - 8x + 16) - (2)(16) + 2(y^2 - 2y + 1) - (2)(1) \\ &= 2(x - 4)^2 + 2(y - 1)^2 - 34 \end{align*}All squared values must be greater than or equal to $0$. As we are aiming for the minimum value, we set the two squared terms to be $0$.
This leads to $\min(h + k) = 0 + 0 - 34 = \boxed{\textbf{(C)} -34}$
|
-34
|
https://artofproblemsolving.com/wiki/index.php/2024_AMC_12B_Problems/Problem_13
| 2,024 |
How many different remainders can result when the $100$th power of an integer is divided by $125$?
|
First note that the Euler's totient function of $125$ is $100$. We can set up two cases, which depend on whether a number is relatively prime to $125$.
If $n$ is relatively prime to $125$, then $n^{100} \equiv 1 \pmod{125}$ because of Euler's Totient Theorem.
If $n$ is not relatively prime to $125$, it must be have a factor of $5$. Express $n$ as $5m$, where $m$ is some integer. Then $n^{100} \equiv (5m)^{100} \equiv 5^{100}\cdot m^{100} \equiv 125 \cdot 5^{97} \cdot m^{100} \equiv 0 \pmod{125}$.
Therefore, $n^{100}$ can only be congruent to $0$ or $1 \pmod{125}$. Our answer is $\boxed{2}$.
|
2
|
https://artofproblemsolving.com/wiki/index.php/2024_AMC_12B_Problems/Problem_14
| 2,024 |
A triangle in the coordinate plane has vertices $A(\log_21,\log_22)$, $B(\log_23,\log_24)$, and $C(\log_27,\log_28)$. What is the area of $\triangle ABC$?
|
We rewrite: $A(0,1)$ $B(\log _{2} 3, 2)$ $C(\log _{2} 7, 3)$.
From here we setup Shoelace Theorem and obtain: $\frac{1}{2}(2(\log _{2} 3) - log _{2} 7)$.
Following log properties and simplifying gives $\boxed{\textbf{(B) }\log_2 \frac{3}{\sqrt{7}}}$.
|
\log_2 \frac{3}{\sqrt{7}}}
|
https://artofproblemsolving.com/wiki/index.php/2024_AMC_12B_Problems/Problem_15
| 2,024 |
A group of $16$ people will be partitioned into $4$ indistinguishable $4$-person committees. Each committee will have one chairperson and one secretary. The number of different ways to make these assignments can be written as $3^{r}M$, where $r$ and $M$ are positive integers and $M$ is not divisible by $3$. What is $r$?
|
There are ${16 \choose 4}$ ways to choose the first committee, ${12 \choose 4}$ ways to choose the second, ${8 \choose 4}$ for the third, and ${4 \choose 4}=1$ for the fourth. Since the committees are indistinguishable, we need to divide the product by $4!$. Thus the $16$ people can be grouped in\[\frac{1}{4!}{16 \choose 4}{12 \choose 4}{8 \choose 4}{4 \choose 4}=\frac{16!}{(4!)^5}\]ways.
In each committee, there are $4 \cdot 3=12$ ways to choose the chairperson and secretary, so $12^4$ ways for all $4$ committees. Therefore, there are\[\frac{16!}{(4!)^5}12^4\]total possibilities.
Since $16!$ contains $6$ factors of $3$, $(4!)^5$ contains $5$, and $12^4$ contains $4$, $r=6-5+4=\boxed{\textbf{(A) }5}$.
|
5
|
https://artofproblemsolving.com/wiki/index.php/2024_AMC_12B_Problems/Problem_16
| 2,024 |
Integers $a$ and $b$ are randomly chosen without replacement from the set of integers with absolute value not exceeding $10$. What is the probability that the polynomial $x^3 + ax^2 + bx + 6$ has $3$ distinct integer roots?
|
Since $-10 \le a,b \le 10$, there are 21 integers to choose from, and $P(21,2) = 21 \times 20 = 420$ equally likely ordered pairs $(a,b)$.
Applying Vieta's formulas,
$x_1 \cdot x_2 \cdot x_3 = -6$
$x_1 + x_2+ x_3 = -a$
$x_1 \cdot x_2 + x_1 \cdot x_3 + x_3 \cdot x_2 = b$
Cases:
(1) $(x_1,x_2,x_3) = (-1,1,6) , b = -1, a=-6$ valid
(2) $(x_1,x_2,x_3) = ( 1,2,-3) , b = -7, a=0$ valid
(3) $(x_1,x_2,x_3) = (1,-2,3) , b = -5, a=-2$ valid
(4) $(x_1,x_2,x_3) = (-1,2,3) , b = 1, a=-4$ valid
(5) $(x_1,x_2,x_3) = (-1,-2,-3) , b = 11$ invalid
the total event space is $21 \cdot (21- 1)$ (choice of select a times choice of selecting b given no-replacement)
hence, our answer is $\frac{4}{21 \cdot 20} = \boxed{\textbf{(C) }\frac{1}{105}}$
|
\frac{1}{105}
|
https://artofproblemsolving.com/wiki/index.php/2024_AMC_12B_Problems/Problem_17
| 2,024 |
The Fibonacci numbers are defined by $F_1 = 1, F_2 = 1,$ and $F_n = F_{n-1} + F_{n-2}$ for $n \geq 3.$ What is\[{\frac{F_2}{F_1}} + {\frac{F_4}{F_2}} + {\frac{F_6}{F_3}} + ... + {\frac{F_{20}}{F_{10}}}?\]
|
The first $20$ terms are $F_n = 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765$
So the answer is $1 + 3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 = \boxed{(B) 319}$.
|
319
|
https://artofproblemsolving.com/wiki/index.php/2024_AMC_12B_Problems/Problem_18
| 2,024 |
Suppose $A$, $B$, and $C$ are points in the plane with $AB=40$ and $AC=42$, and let $x$ be the length of the line segment from $A$ to the midpoint of $\overline{BC}$. Define a function $f$ by letting $f(x)$ be the area of $\triangle ABC$. Then the domain of $f$ is an open interval $(p,q)$, and the maximum value $r$ of $f(x)$ occurs at $x=s$. What is $p+q+r+s$?
|
Let the midpoint of $BC$ be $M$, and let the length $BM = CM = a$. We know there are limits to the value of $x$, and these limits can probably be found through the triangle inequality. But the triangle inequality relates the third side length $BC$ to $AC$ and $AB$, and doesn't contain any information about the median. Therefore we're going to have to write the side $BC$ in terms of $x$ and then use the triangle inequality to find bounds on $x$.
We use Stewart's theorem to relate $BC$ to the median $AM$: $man + dad = bmb + cnc$. In this case $m = \frac{a}2$, $n=\frac{a}2$, $a = m+n$, $d = x$, $b = 42$, $c = 40$.
Therefore we get the equation $2a^3 + 2ax^2 = a \cdot 42^2 + a \cdot 40^2$
$2a^2 + 2x^2 = 42^2 + 40^2$.
Notice that since $20-21-29$ is a pythagorean triple, this means $2a^2 + 2x^2 = 58^2$.
\[\implies a^2 = \frac{58^2}{2}-x^2\]\[\implies a = \sqrt{\frac{58^2}{2}-x^2}\]
By triangle inequality, $2a+40>42 \implies a>1$ and $40+42>2a \implies a<41$
Let's tackle the first inequality:\[\sqrt{\frac{58^2}{2}-x^2}>1 \implies x^2 < \frac{58^2}{2}-1\]
\[\implies x^2 < \frac{40^2+42^2}{2}-1 \implies x^2<41^2\]
Here we use the property that $\frac{x^2+(x+2)^2}{2}-1 = (x+1)^2$.
Therefore in this case, $x<41$.
For the second inequality,\[\sqrt{\frac{58^2}{2}-x^2} < 41 \implies x^2 > \frac{58^2}{2}-41^2\]
\[\implies x^2 > \frac{58^2}{2}-1 + 1 - 41^2\]
\[\implies x^2 > 41^2 + 1 - 41^2 \implies x^2 > 1 \implies x > 1\]
Therefore we have $1<x<41$, so the domain of $f(x)$ is $(1,41)$.
The area of this triangle is $\frac{1}{2} 42 \cdot 40 \cdot \sin(\theta)$. The maximum value of the area occurs when the triangle is right, i.e. $\theta = 90^{\circ}$. Then the area is $\frac{1}{2} \cdot 40 \cdot 42 = 840$. The length of the median of a right triangle is half the length of it's hypotenuse, which squared is $40^2+42^2 = 58^2$. Thus the length of $x$ is $29$.
Our final answer is $1+41+840+29 = \boxed{\textbf{911 } (C)}$
|
911
|
https://artofproblemsolving.com/wiki/index.php/2024_AMC_12B_Problems/Problem_20
| 2,024 |
The measures of the smallest angles of three different right triangles sum to $90^\circ$. All three triangles have side lengths that are primitive Pythagorean triples. Two of them are $3-4-5$ and $5-12-13$. What is the perimeter of the third triangle?
|
Let $\alpha$ and $\beta$ be the smallest angles of the $3-4-5$ and $5-12-13$ triangles respectively. We have\[\tan(\alpha)=\frac{3}{4} \text{ and } \tan(\beta)=\frac{5}{12}\]Then\[\tan(\alpha+\beta)=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}\cdot\frac{5}{12}}=\frac{56}{33}\]Let $\theta$ be the smallest angle of the third triangle. Consider\[\tan{90^\circ}=\tan((\alpha+\beta)+\theta)=\frac{\frac{56}{33}+\tan{\theta}}{1-\frac{56}{33}\cdot\tan{\theta}}\]In order for this to be undefined, we need\[1-\frac{56}{33}\cdot\tan{\theta}=0\]so\[\tan{\theta}=\frac{33}{56}\]Hence the base side lengths of the third triangle are $33$ and $56$. By the Pythagorean Theorem, the hypotenuse of the third triangle is $65$, so the perimeter is $33+56+65=\boxed{\textbf{(C) }154}$.
|
154
|
https://artofproblemsolving.com/wiki/index.php/2024_AMC_12B_Problems/Problem_21
| 2,024 |
Let $\triangle{ABC}$ be a triangle with integer side lengths and the property that $\angle{B} = 2\angle{A}$. What is the least possible perimeter of such a triangle?
|
Let $AB=c$, $BC=a$, $AC=b$. According to the law of sines,\[\frac{b}{a}=\frac{\sin \angle B}{\sin \angle A}\]\[=2\cos \angle A\]
According to the law of cosines,\[\cos \angle A=\frac{b^2+c^2-a^2}{2bc}\]
Hence,\[\frac{b}{a}=\frac{b^2+c^2-a^2}{bc}\]
This simplifies to $b^2=a(a+c)$. We want to find the positive integer solution $(a, b, c)$ to this equation such that $a, b, c$ forms a triangle, and $a+b+c$ is minimized. We proceed by casework on the value of $b$. Remember that $a<a+c$.
Case $1$: $b=1$
Clearly, this case yields no valid solutions.
Case $2$: $b=2$
For this case, we must have $a=1$ and $c=3$. However, $(1, 2, 3)$ does not form a triangle. Hence this case yields no valid solutions.
Case $3$: $b=3$
For this case, we must have $a=1$ and $c=9$. However, $(1, 3, 9)$ does not form a triangle. Hence this case yields no valid solutions.
Case $4$: $b=4$
For this case, $a=1$ and $c=15$, or $a=2$ and $c=6$. As one can check, this case also yields no valid solutions
Case $5$: $b=5$
For this case, we must have $a=1$ and $c=24$. There are no valid solutions
Case $6$: $b=6$
For this case, $a=2$ and $c=16$, or $a=4$ and $c=5$, or $a=3$ and $c=9$. The only valid solution for this case is $(4, 6, 5)$, which yields a perimeter of $15$.
When $b\ge 7$, it is easy to see that $a+c>7$. Hence $a+b+c>14$, which means $a+b+c\ge15$. Therefore, the answer is $\fbox{\textbf{(C) }15}$
|
15
|
https://artofproblemsolving.com/wiki/index.php/2024_AMC_12B_Problems/Problem_22
| 2,024 |
A right pyramid has regular octagon $ABCDEFGH$ with side length $1$ as its base and apex $V.$ Segments $\overline{AV}$ and $\overline{DV}$ are perpendicular. What is the square of the height of the pyramid?
|
To find the height of the pyramid, we need the length from the center of the octagon (denote as $I$) to its vertices and the length of AV.
From symmetry, we know that $\overline{AV} = \overline{DV}$, therefore $\triangle{AVD}$ is a 45-45-90 triangle. Denote $\overline{AV}$ as $x$ so that $\overline{AD} = x\sqrt{2}$. Doing some geometry on the isosceles trapezoid $ABCD$ (we know this from the fact that it is a regular octagon) reveals that $\overline{AD}=1+2(\sqrt{2}/2)=1+\sqrt{2}$ and $\overline{AV}=(\overline{AD})/\sqrt{2}=(\sqrt{2}+2)/2$.
To find the length $\overline{IA}$, we cut the octagon into 8 triangles, each with a smallest angle of 45 degrees. Using the law of cosines on $\triangle{AIB}$ we find that ${\overline{IA}}^2=(2+\sqrt{2})/2$.
Finally, using the pythagorean theorem, we can find that ${\overline{IV}}^2={\overline{AV}}^2-{\overline{IA}}^2= {((\sqrt{2}+2)/2)}^2 - (2+\sqrt{2})/2 = \boxed{(1+\sqrt{2})/2}$ which is answer choice $\boxed{B}$.
|
\frac{1+\sqrt{2}}{2}
|
https://artofproblemsolving.com/wiki/index.php/2024_AMC_12B_Problems/Problem_23
| 2,024 |
What is the number of ordered triples $(a,b,c)$ of positive integers, with $a\le b\le c\le 9$, such that there exists a (non-degenerate) triangle $\triangle ABC$ with an integer inradius for which $a$, $b$, and $c$ are the lengths of the altitudes from $A$ to $\overline{BC}$, $B$ to $\overline{AC}$, and $C$ to $\overline{AB}$, respectively? (Recall that the inradius of a triangle is the radius of the largest possible circle that can be inscribed in the triangle.)
|
First we derive the relationship between the inradius of a triangle $r$, and its three altitudes $a, b, c$. Using an area argument, we can get the following well known result\[\left(\frac{AB+BC+AC}{2}\right)r=A\]where $AB, BC, AC$ are the side lengths of $\triangle ABC$, and $A$ is the triangle's area. Substituting $A=\frac{1}{2}\cdot AB\cdot c$ into the above we get\[\frac{r}{c}=\frac{AB}{AB+BC+AC}\]Similarly, we can get\[\frac{r}{b}=\frac{AC}{AB+BC+AC}\]\[\frac{r}{a}=\frac{BC}{AB+BC+AC}\]Hence,
\begin{align}\label{e1}
\frac{1}{r}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}
\end{align}
Note that there exists a unique, non-degenerate triangle with altitudes $a, b, c$ if and only if $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are the side lengths of a non-degenerate triangle, i.e., $\frac{1}{b}+\frac{1}{c}>\frac{1}{a}$.
With this in mind, it remains to find all positive integer solutions $(r, a, b, c)$ to the above such that $\frac{1}{b}+\frac{1}{c}>\frac{1}{a}$, and $a\le b\le c\le 9$. We do this by doing casework on the value of $r$.
Since $r$ is a positive integer, $r\ge 1$. Since $a\le b\le c\le 9$, $\frac{1}{r}\ge \frac{1}{3}$, so $r\le3$. The only possible values for $r$ are 1, 2, and 3.
Case 1: $r=1$
For this case, we can't have $a\ge 4$, since $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ would be too small. When $a=3$, we must have $b=c=3$. When $a\le2$, we would have $\frac{1}{b}+\frac{1}{c}\le\frac{1}{a}$, which doesn't work. Hence this case only yields one valid solution $(1, 3, 3, 3)$
Case 2: $r=2$
For this case, we can't have $a\ge 7$, for the same reason as in Case 1. When $a=6$, we must have $b=c=6$. When $a=5$, we must have $b=5, c=10$ or $b=10, c=5$. Regardless, $10$ appears, so it is not a valid solution. When $a\le4$, $\frac{1}{b}+\frac{1}{c}\le\frac{1}{a}$. Hence, this case also only yields one valid solution $(2, 6, 6, 6)$
Case 3: $r=3$
The only possible solution is $(3, 9, 9, 9)$, and clearly it is a valid solution.
Hence the only valid solutions are $(1, 3, 3, 3), (2, 6, 6, 6), (3, 9, 9, 9)$, and our answer is $\fbox{\textbf{(B) }3}$
|
3
|
https://artofproblemsolving.com/wiki/index.php/2024_AMC_12B_Problems/Problem_24
| 2,024 |
Pablo will decorate each of $6$ identical white balls with either a striped or a dotted pattern, using either red or blue paint. He will decide on the color and pattern for each ball by flipping a fair coin for each of the $12$ decisions he must make. After the paint driesbarf, he will place the $6$ balls in an urn. Frida will randomly select one ball from the urn and note its color and pattern. The events "the ball Frida selects is red" and "the ball Frida selects is striped" may or may not be independent, depending on the outcome of Pablo's coin flips. The probability that these two events are independent can be written as $\frac mn,$ where $m$ and $n$ are relatively prime positive integers. What is $m?$ (Recall that two events $A$ and $B$ are independent if $P(A \text{ and }B) = P(A) \cdot P(B).$)
|
Let $a$ be the number of balls that are both striped and red, $x$ be the number of balls that are striped but blue, and $y$ be the number of balls that are red but dotted. Then there must be $6-a-x-y$ balls that are dotted and blue.
Let $A$ be the event, "the ball Frida selects is red", and $B$ be the event, "the ball Frida selects is striped". $A$ and $B$ are independent if and only if $\frac{a}{6}=\frac{a+x}{6}\cdot\frac{a+y}{6}$, i.e.,\[6a=(a+x)(a+y)\]
Before we continue, I'd like to clarify that the sample space $S$, under which we are computing probability that $A$ and $B$ are independent, consists of $4^6$ total outcomes, each equally likely. Notice that once given $a, x, y$, there are ${6\choose a}{6-a \choose x}{6-a-x \choose y}=\frac{6!}{a!x!y!(6-a-x-y)!}$ corresponding outcomes.
Now, it remains to find all nonnegative integer solutions $(a, x, y)$ to the above equation such that $a+x+y<6$, add up all the corresponding outcomes, and then divide by $4^6$. We can find the solutions relatively easily by doing casework on $a=0, 1, 2, 3, 4, 5, 6$. Long story short, one finds the solutions to be\[(0, 0, 0), (0, 0, 1)\cdots (0, 0, 6)\]\[(0, 1, 0), (0, 2, 0)\cdots (0, 6, 0)\]\[(1, 5, 0), (1, 0, 5), (1, 2, 1), (1, 1, 2)\]\[(2, 4, 0), (2, 0, 4) (2, 2, 1), (2, 1, 2)\]\[(3, 3, 0), (3, 0, 3)\]\[(4, 2, 0), (4, 0, 2)\]\[(5, 1, 0), (5, 0, 1)\]\[(6, 0, 0)\]
There are little tricks we can use to make the process of adding up the corresponding outcomes faster. First, we notice that the first, second, and last row of solutions contributes $2^7=128$ outcomes. Then we can take advantage of the symmetry between $x, y$ to add up the rest of the outcomes. In the end we get\[128+2(6+180+15+180+20+15+6)=972\]
Hence, the probability is $\frac{972}{4^6}=\frac{243}{2^{10}}$. Therefore, the answer is $\fbox{\textbf{(A) } 243}$
|
243
|
https://artofproblemsolving.com/wiki/index.php/2024_AMC_12B_Problems/Problem_25
| 2,024 |
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