id stringlengths 12 12 | source stringclasses 6
values | problem stringlengths 103 997 | solution stringlengths 744 10.4k | rubric stringlengths 273 4.06k |
|---|---|---|---|---|
USEMO-2019-1 | https://web.evanchen.cc/exams/report-usemo-2019.pdf | Let \(ABCD\) be a cyclic quadrilateral. A circle centered at \(O\) passes through \(B\) and \(D\) and meets lines \(BA\) and \(BC\) again at points \(E\) and \(F\) (distinct from \(A, B, C\)). Let \(H\) denote the orthocenter of triangle \(DEF\). Prove that if lines \(AC, DO, EF\) are concurrent, then triangles \(ABC\)... | Define \(G\) as the intersection of \(\overline{AC}\) and \(\overline{EF}\).
extbf{Claim} â Quadrilateral \(DCGF\) is cyclic.
extit{First proof.} Because
\[ ngle DCG = ngle DCA = ngle DBA = ngle DBE = ngle DFE = ngle DFG. \]
extit{Second proof.} Follows since \(D\) is Miquel point of \(GABF\).
extbf{Cla... | Most solutions are worth 0 or 7. The following partial items are available but extit{not additive}:
- 5 points for proving if \(AC, DO, EF\) are concurrent implies \(AC \perp BD\) but not finishing.
- 4 points for solutions for which spiral similarity justification is entirely absent, but which would be complete if t... |
USEMO-2019-2 | https://web.evanchen.cc/exams/report-usemo-2019.pdf | Let \( \mathbb{Z}[x] \) denote the set of single-variable polynomials in \( x \) with integer coefficients. Find all functions \( heta : \mathbb{Z}[x] o \mathbb{Z}[x] \) (i.e. functions taking polynomials to polynomials) such that
- for any polynomials \( p, q \in \mathbb{Z}[x] \), \( heta(p + q) = heta(p) + heta... | The answer is that
\[ heta(x) = r(x) \cdot p(\pm x + c) \]
for any choice of \( c \in \mathbb{Z} \), \( r(x) \) without an integer root, with the choice of sign fixed. For the converse direction we present two approaches.
**First solution.**
Let \( r(x) = heta(1) \), which has no integer root since the constant 1... | For solutions which are not complete, the following items are available but *not additive*:
- 0 points for the correct answer.
- 1 point for proving \( heta(1) \) divides \( heta(P) \) over \( \mathbb{Q}[x] \)
- 1 point for showing \( heta(x), heta(x^2), \ldots \) all have a common integer root, or that every pair... |
USEMO-2019-3 | https://web.evanchen.cc/exams/report-usemo-2019.pdf | Consider an infinite grid \( \mathcal{G} \) of unit square cells. A \emph{chessboard polygon} is a simple polygon (i.e. not self-intersecting) whose sides lie along the gridlines of \( \mathcal{G} \).
Nikolai chooses a chessboard polygon \( F \) and challenges you to paint some cells of \( \mathcal{G} \) green, such t... | The answer is YES, the task can be made impossible.
The solution is split into three parts. First, we describe a âpolygon with holesâ \( F \). In the second part we prove that this \( F \) works. Finally, we show how to take care of the holes to obtain a true polygon.
**Part 1. Construction.** Choose large intege... | - 0 points for the correct answer
- 6 points for a solution that uses a set of grid cells which do not form a polygon, but is otherwise correct (this includes sets which are not connected)
- 7 points for a correct solution
Any partial credit is done on case-by-case basis. |
USEMO-2019-4 | https://web.evanchen.cc/exams/report-usemo-2019.pdf | Prove that for any prime \( p \), there exists a positive integer \( n \) such that
\[
1^n + 2^{n-1} + 3^{n-2} + \cdots + n^1 \equiv 2020 \pmod{p}.
\] | The idea is to pick \( n = c \cdot p \cdot (p - 1) \) for suitable integer \( c \). In what follows, everything is written modulo \( p \).
**Claim** â When \( n = c \cdot p \cdot (p - 1) \), the left-hand side is equal to
\[
c \cdot \sum_{a=0}^{p-2} \sum_{b=1}^{p-1} b^a = c \cdot \left[1^0 + 2^0 + \cdots + (p-1)^0 \... | For solutions which are not complete, the following items apply but are not additive:
- 0 points for no progress.
- 0 points for considering \( n = cp(p - 1) \) and nothing else.
- 0 points for partial steps in computing this sum, e.g. using primitive roots or the geometric series but not tying it together.
- 1 point ... |
USEMO-2019-5 | https://web.evanchen.cc/exams/report-usemo-2019.pdf | Let \( P \) be a regular polygon, and let \( \mathcal{V} \) be the set of its vertices. Each point in \( \mathcal{V} \) is colored red, white, or blue. A subset of \( \mathcal{V} \) is \emph{patriotic} if it contains an equal number of points of each color, and a side of \( P \) is \emph{dazzling} if its endpoints are ... | We prove the contrapositive: if there is no way to split \( \mathcal{V} \) into two patriotic sets, then the number of dazzling edges is odd.
Let \( \zeta = - rac{1}{2} + rac{\sqrt{3}}{2}i \) be a root of unity. Read the \( n \) vertices of the polygon in order starting from any point. In the complex plane, start fro... | ⢠0 points for an incorrect solution.
⢠0 points for a 1-dimensional tracker argument, to prove e.g. the problem with two colors instead of three.
⢠1 point for making a broken-line polygon (with any set of angles) and not finishing. This includes both:
â Making a polygon using 1, \( \omega \), \( \omega^2 ... |
USEMO-2019-6 | https://web.evanchen.cc/exams/report-usemo-2019.pdf | Let \( ABC \) be an acute scalene triangle with circumcenter \( O \) and altitudes \( AD, BE, CF \). Let \( X, Y, Z \) be the midpoints of \( AD, BE, CF \). Lines \( AD \) and \( YZ \) intersect at \( P \), lines \( BE \) and \( ZX \) intersect at \( Q \), and lines \( CF \) and \( XY \) intersect at \( R \).
Suppose ... | Solution 1
(Radical axis approach)
The main idea is to show that \( (DEF) \) and \( (XYZ) \) has radical axis \( A'D' \).
Let \( H \) be the orthocenter of \( riangle ABC \). We'll let \( (AH), (BH), (CH) \) denote the circles with diameters \( AH, BH, CH \).
extbf{Claim.} Points \( H, D, Y, Z \) are cyclic.
ex... | As this problem is difficult, there are not many correct solutions, so we may manually flag any unusual cases to review as a group. Hence, this rubric is very sparse on details, outlining only the common 0⺠or 7⻠cases.
The following partial items are not additive:
- 0 points for proving just the concurrence of t... |
USEMO-2020-1 | https://web.evanchen.cc/exams/report-usemo-2020.pdf | Which positive integers can be written in the form
\[
rac{\operatorname{lcm}(x, y) + \operatorname{lcm}(y, z)}{\operatorname{lcm}(x, z)}
\]
for positive integers \(x, y, z\)? | Let \(k\) be the desired value, meaning
\[
-k \operatorname{lcm}(x, z) + \operatorname{lcm}(x, y) + \operatorname{lcm}(y, z) = 0.
\]
Our claim is that the possible values are even integers.
Indeed, if \(k\) is even, it is enough to take \((x, y, z) = (1, k/2, 1)\).
For the converse direction we present a few approaches... | νâ solutions
⢠0 points for stating that even integers are the only solutions
⢠1 point for the construction for even integers
⢠1 points for considering νâ of x, y and z and resolving at least one substantial case, such as νâ(y) being maximal (additive). This point can also be awarded if done with a gen... |
USEMO-2020-2 | https://web.evanchen.cc/exams/report-usemo-2020.pdf | Calvin and Hobbes play a game. First, Hobbes picks a family \( \mathcal{F} \) of subsets of \( \{1, 2, \ldots, 2020\} \), known to both players. Then, Calvin and Hobbes take turns choosing a number from \( \{1, 2, \ldots, 2020\} \) which is not already chosen, with Calvin going first, until all numbers are taken (i.e.,... | The answer is \( 4^{1010} - 3^{1010} \). In general, if 2020 is replaced by \( 2n \), the answer is \( 4^n - 3^n \).
**Construction.** The construction is obtained as follows: pair up the numbers as \( \{1,2\}, \{3,4\}, \ldots, \{2019,2020\} \). Whenever Calvin picks a number from one pair, Hobbes elects to pick the o... | - **1 point** for a correct construction, with the correct strategy for Hobbes which results in the correct answer (but they need not correctly calculate the answer, as long as the strategy is right).
- **2 points** for claiming a statement (no proof needed) of the form âfor each integer \( k \), Calvin can ensure a... |
USEMO-2020-3 | https://web.evanchen.cc/exams/report-usemo-2020.pdf | Let \( ABC \) be an acute triangle with circumcenter \( O \) and orthocenter \( H \). Let \( \Gamma \) denote the circumcircle of triangle \( ABC \), and \( N \) the midpoint of \( \overline{OH} \). The tangents to \( \Gamma \) at \( B \) and \( C \), and the line through \( H \) perpendicular to line \( AN \), determi... | We begin by introducing several notations. The orthic triangle is denoted \( DEF \) and the tangential triangle is denoted \( T_aT_bT_c \). The reflections of \( H \) across the sides are denoted \( H_a, H_b, H_c \). We also define the crucial points \( P \) and \( Q \) as the poles of \( H_cB \) and \( H_bC \) with re... | As usual, incomplete computational approaches earn partial credits only based on the amount of synthetic progress which is made.
No points are awarded for just drawing a diagram or simple observations.
Follow the notation in the typeset official solution. The following rubric items are totally additive:
(a) **1 poin... |
USEMO-2020-4 | https://web.evanchen.cc/exams/report-usemo-2020.pdf | A function \( f \) from the set of positive real numbers to itself satisfies
\[
f(x + f(y) + xy) = xf(y) + f(x + y)
\]
for all positive real numbers \( x \) and \( y \). Prove that \( f(x) = x \) for all positive real numbers \( x \). | Solution 1
We first begin with the following observation.
**Claim** â We must have \( f(y) \ge y \) for all \( y > 0 \).
*Proof.* Otherwise, choose \( 0 < x < 1 \) satisfying that \( f(y) = (1 - x) \cdot y \). Then plugging this \( P(x, y) \) gives \( xf(y) = 0 \), contradiction. \(\square\)
Now, we make the subs... | General remarks
- Unlike most functional equations, this one doesnât really have that many steps of âpartial progress.â As such, the steps below are mainly intermediate claims, not specific equations.
- This problem does not ask the contestant to find all functions \( f \) that satisfy the given property. As a ... |
USEMO-2020-5 | https://web.evanchen.cc/exams/report-usemo-2020.pdf | The sides of a convex 200-gon \( A_1A_2 \ldots A_{200} \) are colored red and blue in an alternating fashion. Suppose the extensions of the red sides determine a regular 100-gon, as do the extensions of the blue sides.
Prove that the 50 diagonals \( A_1A_{101},\ A_3A_{103},\ \ldots,\ A_{99}A_{199} \) are concurrent. | Let \( B_1 \ldots B_{100} \) and \( R_1 \ldots R_{100} \) be the regular 100-gons (oriented counterclockwise), and define \( X_i = A_{2i+1} = \overline{B_iB_{i+1}} \cap \overline{R_iR_{i+1}} \) for all \( i \), where all indices are taken modulo 100.
We wish to show that \( X_1X_{51}, \ldots, X_{50}X_{100} \) are concu... | Most solutions are worth 0 or 7.
- 0 points for no progress, special cases, etc.
- 5â6 points for any tiny slip which the contestant could have easily repaired
- 7 points for a correct solution
For solutions which are not complete, the following items are additive:
- 1 point for considering the spiral similarity t... |
USEMO-2020-6 | https://web.evanchen.cc/exams/report-usemo-2020.pdf | Prove that for every odd integer \( n > 1 \), there exist integers \( a, b > 0 \) such that, if we let \[ Q(x) = (x + a)^2 + b, \] then the following conditions hold:
- we have \( \gcd(a,n) = \gcd(b,n) = 1 \);
- the number \( Q(0) \) is divisible by \( n \); and
- the numbers \( Q(1), Q(2), Q(3), \ldots \) each have a... | Let \( p_1 < p_2 < \cdots < p_m \) denote the odd primes dividing \( n \) and call these primes small. The construction is based on the following idea:
**Claim** â For each \( i = 1, \ldots, m \), we can choose a prime \( q_i \equiv 1 \pmod{4} \) such that
\[
\left( \frac{p_j}{q_i} \right) = \begin{cases}
-1 & \text... | In general, not much partial credit is expected for this problem.
The heart of the problem can be thought of as studying the equation
\[
X^2 + b = p_1^{e_1} \cdots p_k^{e_k}
\]
where \( p_1, \ldots, p_k \) are a fixed set of primes, and showing that the equation cannot hold for all sufficiently large \( X \).
- **No ... |
USEMO-2021-1 | https://web.evanchen.cc/exams/report-usemo-2021.pdf | Let \( n \) be a positive integer and consider an \( n imes n \) grid of real numbers. Determine the greatest possible number of cells \( c \) in the grid such that the entry in \( c \) is both strictly greater than the average of \( c \)'s column and strictly less than the average of \( c \)'s row. | The answer is \((n-1)^2\). An example is given by the following construction, shown for \(n = 5\), which generalizes readily. Here, the lower-left \((n-1) imes (n-1)\) square gives a bound.
\[
egin{bmatrix}
-1 & -1 & -1 & -1 & 0 \
0 & 0 & 0 & 0 & 1 \
0 & 0 & 0 & 0 & 1 \
0 & 0 & 0 & 0 & 1 \
0 & 0 & 0 & 0 & 1
\end{bma... | In all approaches, extbf{1 point} is awarded for a correct construction, and extbf{6 points} for the proof that \((n - 1)^2\) is best possible; hence \(1 + 6 = 7\). No points are awarded for answer alone.
The 6 points can be decomposed according to the following approaches. All the items, including the deductions, a... |
USEMO-2021-2 | https://web.evanchen.cc/exams/report-usemo-2021.pdf | Find all integers \( n \geq 1 \) such that \( 2^n - 1 \) has exactly \( n \) positive integer divisors. | The valid \( n \) are \( 1, 2, 4, 6, 8, 16, 32 \). They can be verified to work through inspection, using the well known fact that the Fermat prime \( F_i = 2^{2^i} + 1 \) is indeed prime for \( i = 0, 1, \ldots, 4 \) (but not prime when \( i = 5 \)).
We turn to the proof that these are the only valid values of \( n \... | In this rubric, none of the items are additive: neither the positive items nor the deductions. Hence an incomplete solution receives the largest positive item, while a complete solution receives 7 minus the largest deduction. Deductions do not apply to solutions scoring at most 2 points.
Common items for both solution... |
USEMO-2021-3 | https://web.evanchen.cc/exams/report-usemo-2021.pdf | Let \( A_1C_2B_1A_2C_1B_2 \) be an equilateral hexagon. Let \( O_1 \) and \( H_1 \) denote the circumcenter and orthocenter of \( riangle A_1B_1C_1 \), and let \( O_2 \) and \( H_2 \) denote the circumcenter and orthocenter of \( riangle A_2B_2C_2 \). Suppose that \( O_1
e O_2 \) and \( H_1
e H_2 \). Prove that the... | Let \( riangle X_1Y_1Z_1 \) and \( riangle X_2Y_2Z_2 \) be the medial triangles of \( riangle A_1B_1C_1 \) and \( riangle A_2B_2C_2 \). The first simple observation is as follows.
\begin{quote}
\textbf{Claim} â \( Y_1, Z_1, Y_2, Z_2 \) are concyclic.
\end{quote}
\textit{Proof.} The distance from each of \( Y_1,... | Because the number of solutions with any substantial progress was low, there was no official rubric written for this problem. Instead, the graders first identified all papers that were plausible candidates for partial marks, if any. Then they discussed each individual paper case by case. |
USEMO-2021-4 | https://web.evanchen.cc/exams/report-usemo-2021.pdf | Let \( ABC \) be a triangle with circumcircle \( \omega \), and let \( X \) be the reflection of \( A \) in \( B \). Line \( CX \) meets \( \omega \) again at \( D \). Lines \( BD \) and \( AC \) meet at \( E \), and lines \( AD \) and \( BC \) meet at \( F \). Let \( M \) and \( N \) denote the midpoints of \( AB \) a... | The answer is no, they never intersect.
**Solution 1 (Classical solution, by author)**
Let \( P \) denote the midpoint of \( \overline{AD} \), which
- lies on \( \overline{BN} \), since \( BN \parallel CX \); and
- lies on \( (AMN) \), since itâs homothetic to \( (ABC) \) through \( A \) with factor \( \tfrac{1}{2} ... | The following things might happen:
(a) It is stated/conjectured that answer is NO.
(b) It is mentioned that \( \overline{EF} \) is the polar of \( X \) wrt \( \omega \).
(c) Point \( K = \overline{EF} \cap \overline{AB} \) is introduced.
(d) It is mentioned that \( rac{AK}{KB} = 2 \).
(e) Center \( O \) of \( \om... |
USEMO-2021-5 | https://web.evanchen.cc/exams/report-usemo-2021.pdf | Given a polynomial \( p(x) \) with real coefficients, we denote by \( S(p) \) the sum of the squares of its coefficients. For example, \( S(20x + 21) = 20^2 + 21^2 = 841 \).
Prove that if \( f(x), g(x), \) and \( h(x) \) are polynomials with real coefficients satisfying the identity \( f(x) \cdot g(x) = h(x)^2 \), the... | Solution 1
Claim â Let \( p \) be a polynomial with real coefficients, and \( n > \deg p \) an integer. Then
\[
S(p) = \frac{1}{n} \sum_{k=0}^{n-1} \left|p\left(e^{2\pi i k/n}\right)\right|^2.
\]
Proof. Note that
\[
\left|p\left(e^{2\pi i k/n}\right)\right|^2 = p\left(e^{2\pi i k/n}\right) \cdot p\left(e^{-2\pi i k... | Roots of unity solution
The following items apply but are not additive:
- 0 points for proving the inequality for special cases. For example when \( f \) is a quadratic polynomial, cubic polynomial etc.
- 0 points for expressing the polynomials as \( f(x) = p(x)^2r(x) \), \( g(x) = q(x)^2r(x) \) and \( h(x) = p(x)q(x... |
USEMO-2021-6 | https://web.evanchen.cc/exams/report-usemo-2021.pdf | A \emph{bagel} is a loop of \( 2a + 2b + 4 \) unit squares which can be obtained by cutting a concentric \( a imes b \) hole out of an \( (a+2) imes (b+2) \) rectangle, for some positive integers \( a \) and \( b \). (The side of length \( a \) of the hole is parallel to the side of length \( a+2 \) of the rectangle.... | The answer is \( lpha = 3/2 \).
In what follows, â\( Y \) is about \( X \)â means that \( Y = [1+o(1)]X \). Equivalently, \( \lim_{n o \infty} Y/X = 1 \).
1. Intuitively, both of these say that \( X \) and \( Y \) become closer and closer together as \( n \) grows. This is fine for the problem since only suffic... | No points are awarded for the answer \( lpha = 3/2 \) alone. However, the following items are possible:
- A complete construction is worth extbf{2 points}. A student can earn extbf{1 point} of this item for stating the answer \( lpha = 3/2 \) roughly describing the idea of the construction (that is, to find a âr... |
USEMO-2022-1 | https://web.evanchen.cc/exams/report-usemo-2022.pdf | A \emph{stick} is defined as a \(1 imes k\) or \(k imes 1\) rectangle for any integer \(k \geq 1\). We wish to partition the cells of a \(2022 imes 2022\) chessboard into \(m\) non-overlapping sticks, such that any two of these \(m\) sticks share at most one unit of perimeter. Determine the smallest \(m\) for which ... | In general, with 2022 replaced by \(n\), we will prove the answer is
\[
m = egin{cases}
1 & ext{if } n = 1 \
3 & ext{if } n = 2 \
rac{1}{2}(n^2 - 2n + 7) & ext{if } n \geq 3 ext{ and } n ext{ is odd} \
rac{1}{2}(n^2 - 2n + 8) & ext{if } n \geq 3 ext{ and } n ext{ is even}
\end{cases}
\]
with the following co... | For solutions which are not complete, the following items are available and are additive:
(i) \(+1\) point is awarded for stating the correct answer which is 2,042,224 (or the answer for general \(n\)), even with no justification.
(ii) \(+1\) point is awarded for giving a working construction.
(iii) \(+1\) point for... |
USEMO-2022-2 | https://web.evanchen.cc/exams/report-usemo-2022.pdf | A function \( \psi : \mathbb{Z} o \mathbb{Z} \) is said to be \emph{zero-requiem} if for any positive integer \( n \) and any integers \( a_1, \ldots, a_n \) (not necessarily distinct), the sums \( a_1 + a_2 + \cdots + a_n \) and \( \psi(a_1) + \psi(a_2) + \cdots + \psi(a_n) \) are not both zero.
Let \( f \) and \( g... | Solution 1 (rephrased from several contestants).
Assume for contradiction that neither \( f \circ f \) nor \( g \circ g \) is zero-requiem, meaning that there exist \( c_1, \ldots, c_m \) and \( w_1, \ldots, w_n \) such that
\[
0 = \sum_{i=1}^m c_i = \sum_{i=1}^m f(c_i) + \sum_{i=1}^m g(c_i)
\]
\[
0 = \sum_{i=1}^n w_i ... | (i) 4 points: Proving the first lemma in the first solution
(ii) 1 point: Conjecturing the first lemma in the first solution
(iii) 0 points: Neither \( \sum_{i=1}^k f(a_i) \) nor \( \sum_{i=1}^k g(a_i) \) is zero.
(iv) 1 point: For showing that for all \( (a_n)_n \) such \( a_1 + \cdots + a_n = 0 \), \( f(a_1) + \cd... |
USEMO-2022-3 | https://web.evanchen.cc/exams/report-usemo-2022.pdf | Point \( P \) lies in the interior of a triangle \( ABC \). Lines \( AP, BP, \) and \( CP \) meet the opposite sides of triangle \( ABC \) at points \( A', B', \) and \( C' \), respectively. Let \( P_A \) be the midpoint of the segment joining the incenters of triangles \( BPC' \) and \( CPB' \), and define points \( P... | Solution 1 (authorâs).
We will need a couple of lemmas.
**Lemma 3.3.1 (âSparrow lemmaâ)**
Let \( I \) be the incenter of triangle \( ABC \). Point \( U \) lies on ray \( \overrightarrow{AB} \) and point \( V \) lies on ray \( \overrightarrow{CA} \) beyond \( A \) so that \( AU - AV = AB + AC - BC \). Then poin... | As usual, incomplete computational approaches earn partial credits only based on the amount of synthetic progress which is made. No points are awarded for just drawing a diagram or simple observations.
**Solution 1 Rubric**
(i) 2 points are awarded for proving the points \( P, I_C, U_A, V_A \) are concyclic (and the ... |
USEMO-2022-4 | https://web.evanchen.cc/exams/report-usemo-2022.pdf | Let \(ABCD\) be a cyclic quadrilateral whose opposite sides are not parallel. Suppose points \(P, Q, R, S\) lie in the interiors of segments \(AB, BC, CD, DA\), respectively, such that
\[
\angle PDA = \angle PCB, \quad \angle QAB = \angle QDC, \quad \angle RBC = \angle RAD, \quad \text{and} \quad \angle SCD = \angle SB... | Solution 1 (authorâs).
Let \(U = \overline{AD} \cap \overline{BC}\) and \(V = \overline{AB} \cap \overline{CD}\).
**Claim.** We have \(US = UQ\) and \(VP = VR\).
*Proof.* We have
\[
\angle BSA = \angle BAS + \angle SBA = \angle BCD + \angle DCS = \angle BCS
\]
hence
\[
US^2 = UB \cdot UC.
\]
Similarly, \(UQ^2 = UA ... | §3.4b Marking scheme
As usual, incomplete computational approaches earn partial credits only based on the amount of synthetic progress which is made. No points are awarded for just drawing a diagram or simple observations.
There are two major paths a solution can follow:
⢠One which introduces and works with \(\ov... |
USEMO-2022-5 | https://web.evanchen.cc/exams/report-usemo-2022.pdf | Let \( au(n) \) denote the number of positive integer divisors of a positive integer \( n \) (for example, \( au(2022) = 8 \)). Given a polynomial \( P(X) \) with integer coefficients, we define a sequence \( a_1, a_2, \ldots \) of nonnegative integers by setting
\[
a_n = egin{cases}
\gcd(P(n), au(P(n))) & ext{if ... | We claim the answer is no, such \( P \) does not exist.
Clearly we may assume \( P \) is nonconstant with positive leading coefficient. Fix \( P \) and fix constants \( n_0, c > 0 \) such that \( c = P(n_0) > 0 \). We are going to prove that infinitely many terms of the sequence are at most \( c \).
We start with the... | None of the following items are additive.
(i) 0 points for claiming the answer is no.
(ii) 0 points for solving the problem for linear polynomials.
(iii) 1 point for claiming infinitely many terms of the sequence have value \( c \) for some constant \( c \) in terms of \( P \) whose \(
u_p \)'s are âwell-behavedâ... |
USEMO-2022-6 | https://web.evanchen.cc/exams/report-usemo-2022.pdf | Find all positive integers \( k \) for which there exists a \emph{nonlinear} function \( f : \mathbb{Z} o \mathbb{Z} \) which satisfies
\[
f(a) + f(b) + f(c) = rac{f(a-b) + f(b-c) + f(c-a)}{k}
\]
for any integers \( a, b, c \) with \( a + b + c = 0 \). | The complete set of solutions is given by
- For \( k = 1 \), \( f(x) \equiv C_1 x + C_2(x mod 2) + C_3 \).
- For \( k = 3 \), \( f(x) \equiv C_1 x + C_2 x^2 \).
- For \( k = 9 \), \( f(x) \equiv C_1 x + C_2 x^4 \).
- For all other \( k \), only \( f(x) \equiv C_1 x \).
Here \( C_1, C_2, C_3 \) are arbitrary integers.... | In this rubric, a student earns up to 2 points for giving valid constructions, and up to 5 points for proving those solutions are the only ones. These points for the construction are additive with those for the proof.
The construction has three steps:
- Proving \( k = 3 \) works (writing something like: just expand an... |
USEMO-2023-1 | https://web.evanchen.cc/exams/report-usemo-2023.pdf | A positive integer \( n \) is called \emph{beautiful} if, for every integer \( 4 \leq b \leq 10000 \), the base-\( b \) representation of \( n \) contains the consecutive digits 2, 0, 2, 3 (in this order, from left to right). Determine whether the set of all beautiful integers is finite. | We show there are infinitely many beautiful integers. Here are three different approaches.
**Solution 1** (One constructive approach):
We will construct an increasing sequence of positive integers
\[
N_4 < N_5 < N_6 < \cdots
\]
such that for every \( k = 4, 5, \ldots \), the number \( N_k \) contains 2023\(_b\) in eve... | Marking scheme for inductions and constructions (common approach)
The following remark from the solution packet is key to understanding the rubric:
The essential difficulty in this problem arises from the fact that different bases may share overlapping prime factors, so the Chinese Remainder Theorem does not immediat... |
USEMO-2023-2 | https://web.evanchen.cc/exams/report-usemo-2023.pdf | Each point in the plane is labeled with a real number. Show that there exist two distinct points \( P \) and \( Q \) whose labels differ by less than the distance from \( P \) to \( Q \). | Let \( f : \mathbb{R}^2 o \mathbb{R} \) be the labeling, and suppose for contradiction the difference in labels for any points \( P, Q \in \mathbb{R}^2 \) is at least their distance.
egin{quote}
extbf{Claim} â Let \( I \) be a closed interval of length 1. For any \( arepsilon > 0 \), the pre-image
\[
f^{-1}(I) :... | For all solutions, the following are not awarded marks:
⢠Proving the statement for all continuous labellings.
⢠Assuming there exists a labelling \( f : \mathbb{R}^2 o \mathbb{R} \) such that \( |f(x) - f(y)| \ge |x - y| \) for the sake of contradiction.
⢠Proving \( f \) is injective.
For solutions similar... |
USEMO-2023-3 | https://web.evanchen.cc/exams/report-usemo-2023.pdf | Canmoo is trying to do constructions, but doesnât have a ruler or compass. Instead, Canmoo has a device that, given four distinct points \( A, B, C, P \) in the plane, will mark the isogonal conjugate of \( P \) with respect to triangle \( ABC \), if it exists. Show that if two points are marked on the plane, then Ca... | We assume Canmoo can mark points in arbitrarily general position.
We first prove two claims showing that reflection around a point is possible. We will only use the second claim in what proceeds (so with the second claim proven, we can forget about the first one.)
extbf{Claim â} Given any three points \( X, B, C \... | For the most part, solutions could be read on a case-by-case basis, because not many solutions have nontrivial progress. We standardize the following extbf{non-additive} benchmarks:
- extbf{1 point} for showing \( X \) can be reflected over \( YZ \).
- extbf{0 points} for showing \( X \) can be reflected over a poi... |
USEMO-2023-4 | https://web.evanchen.cc/exams/report-usemo-2023.pdf | Let \( ABC \) be an acute triangle with orthocenter \( H \). Points \( A_1, B_1, C_1 \) are chosen in the interiors of sides \( BC, CA, AB \), respectively, such that \( riangle A_1B_1C_1 \) has orthocenter \( H \). Define \( A_2 = \overrightarrow{AH} \cap B_1C_1 \), \( B_2 = \overrightarrow{BH} \cap C_1A_1 \), and \(... | Solution 1: Power of a point solution, by Nikolai Beluhov.
In this solution, all lengths are signed. Let \( riangle DEF \) be the orthic triangle of \( riangle ABC \), and \( riangle D_1E_1F_1 \) be the orthic triangle of \( riangle A_1B_1C_1 \). We define two common quantities, through power of a point:
\[ k := ... | General rules:
- As usual, incomplete computational approaches earn partial credits only based on the amount of synthetic progress which is made.
- No points are awarded for just drawing a diagram or simple observations.
- No points are deducted for configuration issues (such as not using directed angles) and minor ty... |
USEMO-2023-5 | https://web.evanchen.cc/exams/report-usemo-2023.pdf | Let \( n \geq 2 \) be an integer. A cube of size \( n \times n \times n \) is dissected into \( n^3 \) unit cubes. A nonzero real number is written at the center of each unit cube so that the sum of the \( n^2 \) numbers in each slab of size \( 1 \times n \times n \), \( n \times 1 \times n \), or \( n \times n \times ... | We show this can never happen.
Suppose, for the sake of contradiction, that such a plane \( lpha \) did exist. Let \( Oxyz \) be a Cartesian coordinate system whose origin \( O \) lies in \( lpha \) and whose axes are parallel to the edges of our cube. Let the equation of \( lpha \) in this coordinate system be \( ... | In what follows, we let \( f(x, y, z) \) be the number written at \( (x, y, z) \) with \( 1 \leq x, y, z \leq n \). Also let the required plane be \( ax + by + cz - d = 0 \).
**Partial items for 0+ solutions**
The following partial items apply for incomplete solutions and are additive.
- **0 points** for claiming the... |
USEMO-2023-6 | https://web.evanchen.cc/exams/report-usemo-2023.pdf | Let \( n \geq 2 \) be a fixed integer.
(a) Determine the largest positive integer \( m \) (in terms of \( n \)) such that there exist complex numbers \( r_1, \ldots, r_n \), not all zero, for which
\[
\prod_{k=1}^n (r_k + 1) = \prod_{k=1}^n (r_k^2 + 1) = \cdots = \prod_{k=1}^n (r_k^m + 1) = 1.
\]
(b) For this value o... | For part (a) the answer is \( m = 2^n - 2 \); for part (b) the answer is \( 2^n \).
**Construction for (a).** For \( m = 2^n - 2 \), fix \( \omega := \exp\left(\frac{2\pi i}{2^n - 1}\right) \) and set
\[
r_j = \omega^{2^j}, \quad j = 1, 2, \ldots, n, \quad m = 2^n - 2.
\]
We can expand to see that the
\[
\prod_{k=1}^n... | Part (a) will be graded out of 5 points and part (b) will be graded out of 2 points. The scores of the two parts will be added for the final score out of 7.
**Scoring for (a)**
For incomplete solutions, the following items are available but **not additive**:
- 1 point for the correct answer of \( m = 2^n - 2 \) in p... |
USEMO-2024-1 | https://web.evanchen.cc/exams/report-usemo-2024.pdf | There are 1001 stacks of coins \( S_1, S_2, \ldots, S_{1001} \). Initially, stack \( S_k \) has \( k \) coins for each \( k = 1, 2, \ldots, 1001 \). In an operation, one selects an ordered pair \( (i, j) \) of indices \( i \) and \( j \) satisfying \( 1 \le i < j \le 1001 \) subject to two conditions:
- The stacks \( ... | The answer is \( 500 \cdot 501 = 250500 \). Our solution is split into two parts.
### Construction
Firstly, we will give a valid construction. We start by performing operations \( (1001, 1000), (1001, 999), \ldots, (1001, 1) \), in order. By induction, at each step \( (1001, j) \), \( S_{1001} \) will have \( j + 1 \... | The solution is split into two parts: the lower bound (construction), worth **4 points**, and the upper bound, worth **3 points**. These parts are completely additive.
In general, minor errors will be worth a deduction, but please message the channel when you find any that are not included in the rubric so that we can... |
USEMO-2024-2 | https://web.evanchen.cc/exams/report-usemo-2024.pdf | Let \( k \) be a fixed positive integer. For each integer \( 1 \le i \le 4 \), let \( x_i \) and \( y_i \) be positive integers such that their least common multiple is \( k \). Suppose that the four points \( (x_1, y_1), (x_2, y_2), (x_3, y_3), (x_4, y_4) \) are the vertices of a non-degenerate rectangle in the Cartes... | It suffices to prove that \( v_p(x_1x_2x_3x_4) \) is even for each prime \( p \mid k \). Since the four points form a rectangle, we have
\[
egin{aligned}
x_1 + x_3 &= x_2 + x_4 ag{3.1} \
y_1 + y_3 &= y_2 + y_4 ag{3.2} \
(x_1 - x_3)^2 + (y_1 - y_3)^2 &= (x_2 - x_4)^2 + (y_2 - y_4)^2 ag{3.3} \
x_2x_4 - x_1x_3 &= y_1... | Recall the four equations
\[
egin{aligned}
x_1 + x_3 &= x_2 + x_4 \
y_1 + y_3 &= y_2 + y_4 \
(x_1 - x_3)^2 + (y_1 - y_3)^2 &= (x_2 - x_4)^2 + (y_2 - y_4)^2 \
x_2x_4 - x_1x_3 &= y_1y_3 - y_2y_4
\end{aligned}
\]
which were numbered (3.1), (3.2), (3.3), and (3.4) in the provided solution.
The following are awarded mar... |
USEMO-2024-3 | https://web.evanchen.cc/exams/report-usemo-2024.pdf | Let ABC be an acute triangle with incenter I. Two distinct points P and Q are chosen on the circumcircle of ABC such that
\[ngle API = ngle AQI = 45^\circ.\]
Lines PQ and BC meet at S. Let H denote the foot of the altitude from A to BC. Prove that \(ngle AHI = ngle ISH.\) | We give three solutions.
### Solution 1 (via Tebault circles from the author)
Construct the Tebault circles \(\omega_1\) and \(\omega_2\) which are tangent to (ABC), side BC, and cevian AH.
The key claim is that P and Q coincide with the tangency points of the circles we just drew:
> **Claim** â Points P and Q are... | For all solutions, the following are not awarded marks:
- Rephrasing the angle condition in terms of PI \cap (ABC) and QI \cap (ABC).
- Swapping \(ngle AHI\) with some other angles, even if they are used in the official solutions.
For solutions not using Tebault circles, the following items are not additive:
- **2 ... |
USEMO-2024-4 | https://web.evanchen.cc/exams/report-usemo-2024.pdf | Find all sequences \( a_1, a_2, \ldots \) of nonnegative integers such that for all positive integers \( n \), the polynomial
\[
1 + x^{a_1} + x^{a_2} + \cdots + x^{a_n}
\]
has at least one integer root. (Here \( x^0 = 1. \)) | The only answer is \( a_1 = 1 \) and \( a_2 = a_3 = \cdots = 0 \).
Itâs clear that this works because for each \( n \), the requested integer root is \( x = -n \). We now prove this is the only solution.
In general, let
\[ F_n(x) := 1 + x^{a_1} + \cdots + x^{a_n}. \]
> **Claim** â Let \( p \) be any prime. Then
... | For all solutions, the following are not awarded marks:
- Getting the correct answer with no explanation.
- Showing that all roots are negative.
For correct solutions:
- **7 points** for a complete solution that shows \( a_1 = 1 \) and \( a_{i > 1} = 0 \) is the only possible solution.
Solutions that are not comple... |
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