problem_id stringlengths 16 19 | problem stringlengths 69 2.04k | solution stringlengths 60 23.9k |
|---|---|---|
USAMO-2008-notes_5 | Three nonnegative real numbers $r_1$, $r_2$, $r_3$ are written on a blackboard.
These numbers have the property
that there exist integers $a_1$, $a_2$, $a_3$, not all zero,
satisfying $a_1r_1 + a_2r_2 + a_3r_3 = 0$.
We are permitted to perform the following operation:
find two numbers $x$, $y$ on the blackboard with $x... | We first show we can decrease the quantity
$\left\lvert a_1 \right\rvert + \left\lvert a_2 \right\rvert + \left\lvert a_3 \right\rvert$
as long as $0 \notin \left\{ a_1,a_2,a_3 \right\}$.
Assume $a_1 > 0$ and $r_1 > r_2 > r_3$
without loss of generality and consider two cases.
\begin{itemize}
\ii Suppose $a_2 > 0$ o... |
USAMO-2008-notes_6 | At a certain mathematical conference,
every pair of mathematicians are either friends or strangers.
At mealtime, every participant eats in one of two large dining rooms.
Each mathematician insists upon eating in a room which
contains an even number of his or her friends.
Prove that the number of ways that the mathemati... | Take the obvious graph interpretation
where we are trying to $2$-color a graph.
Let $A$ be the adjacency matrix of the graph over $\FF_2$,
except the diagonal of $A$ has $\deg v \pmod 2$ instead of zero.
Then let $\vec d$ be the main diagonal.
Splittings then correspond to $A \vec v = \vec d$.
It's then immediate that ... |
USAMO-2009-notes_1 | Given circles $\omega_1$ and $\omega_2$ intersecting at points $X$ and $Y$,
let $\ell_1$ be a line through the center of $\omega_1$
intersecting $\omega_2$ at points $P$ and $Q$
and let $\ell_2$ be a line through the center of $\omega_2$
intersecting $\omega_1$ at points $R$ and $S$.
Prove that if $P$, $Q$, $R$, and $S... | Let $r_1$, $r_2$, $r_3$ denote the
circumradii of $\omega_1$, $\omega_2$, and $\omega_3$, respectively,
where $\omega_3$ is the circle through $P$, $Q$, $R$, $S$.
\begin{center}
\begin{asy}
pair O_1 = dir(220);
pair O_2 = dir(320);
pair O = dir(110);
pair T = foot(O, O_1, O_2);
pair X = midpoint(O--T);
pair Y = 2*T-X;... |
USAMO-2009-notes_2 | Let $n$ be a positive integer.
Determine the size of the largest subset of $\{ -n, -n+1, \dots, n-1, n\}$
which does not contain three elements $a$, $b$, $c$ (not necessarily distinct)
satisfying $a+b+c=0$. | The answer is $n$ with $n$ even and $n+1$ with $n$ odd;
the construction is to take all odd numbers.
To prove this is maximal, it suffices to show it for $n$ even;
we do so by induction on even $n \ge 2$ with the base case being trivial.
Letting $A$ be the subset, we consider three cases:
\begin{enumerate}[(i)]
\ii ... |
USAMO-2009-notes_3 | We define a \emph{chessboard polygon} to be a simple polygon whose sides are
situated along lines of the form $x=a$ or $y=b$, where $a$ and $b$ are integers.
These lines divide the interior into unit squares, which are shaded alternately
gray and white so that adjacent squares have different colors.
To tile a chessboar... | \paragraph{Proof of (a).} This is easier, and by induction.
Let $\mathcal P$ denote the chessboard polygon which can be tiled by dominoes.
Consider a \emph{lower-left} square $s$ of the polygon,
and WLOG is it white (other case similar).
Then we have two cases:
\begin{itemize}
\ii If there exists a domino tiling of ... |
USAMO-2009-notes_4 | For $n \ge 2$, let $a_1$, $a_2$, \dots, $a_n$ be positive real numbers such that
\[
\left( a_1 + a_2 + \dots + a_n \right)
\left( \frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_n} \right)
\le \left( n + \half \right)^2.
\]
Prove that $\max\left( a_1, \dots, a_n \right) \le 4 \min\left( a_1, \dots, a_n \right)... | Assume $a_1$ is the largest and $a_2$ is the smallest.
Let $M = a_1/a_2$.
We wish to show $M \le 4$.
In left-hand side of given, write as $a_2+a_1 + \dots + a_n$.
By Cauchy Schwarz, one obtains
\begin{align*}
\left( n+\half \right)^2 &\ge
\left( a_2 + a_1 + a_3 + \dots + a_n \right)
\left( \frac{1}{a_1} + \... |
USAMO-2009-notes_5 | Trapezoid $ABCD$, with $\ol{AB} \parallel \ol{CD}$,
is inscribed in circle $\omega$ and point $G$ lies inside triangle $BCD$.
Rays $AG$ and $BG$ meet $\omega$ again at points $P$ and $Q$, respectively.
Let the line through $G$ parallel to $\ol{AB}$ intersect
$\ol{BD}$ and $\ol{BC}$ at points $R$ and $S$, respectively.
... | Perform an inversion around $B$ with arbitrary radius,
and denote the inverse of a point $Z$ with $Z^\ast$.
\begin{center}
\begin{asy}
unitsize(2cm);
pair A = Drawing("A", dir(130), dir(130));
pair B = Drawing("B", dir(50), dir(50));
pair C = Drawing("C", dir(-20), dir(-20));
pair D = Drawing("D"... |
USAMO-2009-notes_6 | Let $s_1, s_2, s_3, \dots$ be an infinite,
nonconstant sequence of rational numbers,
meaning it is not the case that $s_1 = s_2 = s_3 = \dots$.
Suppose that $t_1, t_2, t_3, \dots$ is also an infinite,
nonconstant sequence of rational numbers
with the property that $(s_i - s_j)(t_i - t_j)$
is an integer for all $i$ and ... | First we eliminate the silly edge case:
\begin{claim*}
For some $i$ and $j$, we have
$(s_i - s_j)(t_i - t_j) \neq 0$.
\end{claim*}
\begin{proof}
Assume not.
WLOG $s_1 \neq s_2$, then $t_1 = t_2$.
Now select $i$ such that $t_i \neq t_1 = t_2$.
Then either $t_i - s_1 \neq 0$
or $t_i - s_2 \neq 0$, contradic... |
USAMO-2010-notes_1 | Let $AXYZB$ be a convex pentagon inscribed in a semicircle of diameter $AB$.
Denote by $P$, $Q$, $R$, $S$ the feet of the perpendiculars
from $Y$ onto lines $AX$, $BX$, $AZ$, $BZ$, respectively.
Prove that the acute angle formed by lines $PQ$ and $RS$ is half the size of $\angle XOZ$,
where $O$ is the midpoint of segme... | We present two possible approaches.
The first approach is just ``bare-hands'' angle chasing.
The second approach requires more insight but makes it clearer what is going on;
it shows the intersection point of lines $PQ$ and $RS$ is the foot
from the altitude from $Y$ to $AB$ using Simson lines.
The second approach also... |
USAMO-2010-notes_2 | There are $n$ students standing in a circle, one behind the other.
The students have heights $h_1<h_2<\dots <h_n$.
If a student with height $h_k$ is standing directly
behind a student with height $h_{k-2}$ or less,
the two students are permitted to switch places.
Prove that it is not possible to make more than
$\binom{... | The main claim is the following observation,
which is most motivated in the situation $j-i = 2$.
\begin{claim*}
The students with heights $h_i$ and $h_j$
switch at most $|j-i|-1$ times.
\end{claim*}
\begin{proof}
By induction on $d = |j-i|$, assuming $j > i$.
For $d = 1$ there is nothing to prove.
For $d \ge... |
USAMO-2010-notes_3 | The $2010$ positive real numbers $a_1$, $a_2$, \dots , $a_{2010}$
satisfy the inequality $a_i a_j \le i+j$
for all $1 \le i < j \le 2010$.
Determine, with proof, the largest possible value
of the product $a_1a_2\dots a_{2010}$. | The answer is $3 \times 7 \times 11 \times \dots \times 4019$,
which is clearly an upper bound
(and it's not too hard to show this is the lowest
number we may obtain by multiplying $1005$ equalities together;
this is essentially the rearrangement inequality).
The tricky part is the construction.
Intuitively we want $a_... |
USAMO-2010-notes_4 | Let $ABC$ be a triangle with $\angle A = 90\dg$.
Points $D$ and $E$ lie on sides $AC$ and $AB$, respectively,
such that $\angle ABD = \angle DBC$ and $\angle ACE = \angle ECB$.
Segments $BD$ and $CE$ meet at $I$.
Determine whether or not it is possible for segments $AB$, $AC$, $BI$, $ID$, $CI$, $IE$
to all have integer... | The answer is no.
We prove that it is not even possible that $AB$, $AC$, $CI$, $IB$ are all integers.
\begin{center}
\begin{asy}
size(5cm);
pair B = dir(140);
pair A = conj(B);
pair C = -B;
filldraw(A--B--C--cycle, opacity(0.1)+lightblue, blue);
pair I = incenter(A, B, C);
pair D = extens... |
USAMO-2010-notes_5 | Let $q = \frac{3p-5}{2}$ where $p$ is an odd prime, and let
\[ S_q = \frac{1}{2\cdot 3 \cdot 4} + \frac{1}{5\cdot 6 \cdot 7}
+ \dots + \frac{1}{q(q+1)(q+2)}. \]
Prove that if $\frac{1}{p}-2S_q = \frac{m}{n}$ for integers $m$ and $n$,
then $m - n$ is divisible by $p$. | By partial fractions, we have
\[
\frac{2}{(3k-1)(3k)(3k+1)}
= \frac{1}{3k-1} - \frac{2}{3k} + \frac{1}{3k+1}.
\]
Thus
\begin{align*}
2S_q
&= \left( \frac12 - \frac23 + \frac14 \right)
+ \left( \frac 15 - \frac26 + \frac17 \right)
+ \dots
+ \left( \frac1q - \frac2{q+1} + \frac{1}{q+2} \right) \\
&= \left... |
USAMO-2011-notes_1 | Let $a$, $b$, $c$ be positive real numbers
such that $a^2+b^2+c^2+(a+b+c)^2 \le 4$. Prove that
\[ \frac{ab+1}{(a+b)^2} + \frac{bc+1}{(b+c)^2} + \frac{ca+1}{(c+a)^2} \ge 3. \] | The condition becomes $2 \ge a^2+b^2+c^2 + ab+bc+ca$.
Therefore,
\begin{align*}
\sum_{\text{cyc}} \frac{2ab+2}{(a+b)^2}
&\ge \sum_{\text{cyc}} \frac{2ab+(a^2+b^2+c^2+ab+bc+ca)}{(a+b)^2} \\
&= \sum_{\text{cyc}} \frac{(a+b)^2+(c+a)(c+b)}{(a+b)^2} \\
&= 3 + \sum_{\text{cyc}} \frac{(c+a)(c+b)}{(a+b)^2} \\
&\ge 3 ... |
USAMO-2011-notes_2 | An integer is assigned to each vertex of a regular pentagon
so that the sum of the five integers is $2011$.
A turn of a solitaire game consists of subtracting an integer $m$
(not necessarily positive) from each of the integers at two neighboring vertices
and adding $2m$ to the opposite vertex, which is not adjacent
to ... | Call the vertices $0$, $1$, $2$, $3$, $4$ in order.
First, notice that the quantity
\[ S \coloneq N_1 + 2N_2 + 3N_3 + 4N_4 \pmod 5 \]
is invariant, where $N_i$ is the amount at vertex $i$.
This immediately implies that at most one vertex can win,
since in a winning situation all $N_i$ are $0$ except for one, which is $... |
USAMO-2011-notes_3 | In hexagon $ABCDEF$, which is nonconvex but not self-intersecting,
no pair of opposite sides are parallel.
The internal angles satisfy
$\angle A=3\angle D$, $\angle C=3\angle F$, and $\angle E=3\angle B$.
Furthermore $AB=DE$, $BC=EF$, and $CD=FA$.
Prove that diagonals $\ol{AD}$, $\ol{BE}$, and $\ol{CF}$
are concurrent. | We present the official solution.
We say a hexagon is \emph{satisfying}
if it obeys the six conditions;
note that intuitively we expect three degrees of freedom
for satisfying hexagons.
Main idea:
\begin{claim*}
In a satisfying hexagon,
$B$, $D$, $F$ are reflections of $A$, $C$, $E$ across
the sides of $\triangle ACE$... |
USAMO-2011-notes_4 | Consider the assertion that for each positive integer $n\geq2$,
the remainder upon dividing $2^{2^n}$ by $2^n-1$ is a power of $4$.
Either prove the assertion or find (with proof) a counterexample. | We claim $n = 25$ is a counterexample.
Since $2^{25} \equiv 2^0 \pmod{2^{25}-1}$, we have
\[ 2^{2^{25}} \equiv 2^{2^{25} \bmod{25}}
\equiv 2^7 \bmod{2^{25}-1} \]
and the right-hand side is actually the remainder,
since $0 < 2^7 < 2^{25}$.
But $2^7$ is not a power of $4$.
\begin{remark*}
Really, the problem is just... |
USAMO-2011-notes_5 | Let $P$ be a point inside convex quadrilateral $ABCD$.
Points $Q_1$ and $Q_2$ are located within $ABCD$ such that
\begin{align*}
\angle Q_1BC=\angle ABP, & \qquad \angle Q_1CB=\angle DCP, \\
\angle Q_2AD=\angle BAP, & \qquad \angle Q_2DA=\angle CDP.
\end{align*}
Prove that $\ol{Q_1Q_2} \parallel \ol{AB}$
if and only if... | If $\ol{AB} \parallel \ol{CD}$ there is nothing to prove.
Otherwise let $X = \ol{AB} \cap \ol{CD}$.
Then the $Q_1$ and $Q_2$ are the isogonal conjugates of $P$
with respect to triangles $XBC$ and $XAD$.
Thus $X$, $Q_1$, $Q_2$ are collinear,
on the isogonal of $\ol{XP}$ with respect to
$\angle DXA = \angle CXB$. |
USAMO-2011-notes_6 | Let $A$ be a set with $|A|=225$, meaning that $A$ has $225$ elements.
Suppose further that there are eleven subsets $A_1, \dots, A_{11}$ of $A$
such that $|A_i|=45$ for $1\leq i\leq11$ and $|A_i\cap A_j|=9$
for $1\leq i<j\leq11$.
Prove that $|A_1\cup A_2\cup\dots\cup A_{11}|\geq 165$,
and give an example for which equa... | Ignore the $225$ --- it is irrelevant.
Denote the elements of $A_1 \cup \dots \cup A_{11}$
by $a_1$, \dots, $a_n$,
and suppose that $a_i$ appears $x_i$ times among $A_i$
for each $1 \le i \le n$ (so $1 \le x_i \le 11$).
Then we have
\[ \sum_{i=1}^{11} x_i = \sum_{i=1}^{11} |A_i| = 45 \cdot 11 \]
and
\[ \sum_{i=1}^{11}... |
USAMO-2012-notes_1 | Find all integers $n \ge 3$ such that
among any $n$ positive real numbers
$a_1$, $a_2$, \dots, $a_n$ with
\[ \max(a_1,a_2,\dots,a_n)
\le n \cdot \min(a_1,a_2,\dots,a_n), \]
there exist three that are the side lengths
of an acute triangle. | The answer is all $n \ge 13$.
Define $(F_n)$ as the sequence of Fibonacci numbers,
by $F_1 = F_2 = 1$ and $F_{n+1} = F_n + F_{n-1}$.
We will find that Fibonacci numbers show up naturally
when we work through the main proof,
so we will isolate the following calculation now
to make the subsequent solution easier to read... |
USAMO-2012-notes_2 | A circle is divided into congruent arcs by $432$ points.
The points are colored in four colors such that
some $108$ points are colored red,
some $108$ points are colored green,
some $108$ points are colored blue,
and the remaining $108$ points are colored yellow.
Prove that one can choose three points of each color in ... | First, consider the $431$ possible non-identity rotations
of the red points, and count overlaps with green points.
If we select a rotation randomly,
then each red point lies over a green point
with probability $\frac{108}{431}$;
hence the expected number of red-green incidences is
\[ \frac{108}{431} \cdot 108 > 27 \]
a... |
USAMO-2012-notes_3 | Determine which integers $n > 1$ have the property
that there exists an infinite sequence $a_1$, $a_2$, $a_3$, \dots
of nonzero integers such that the equality
\[ a_k + 2a_{2k} + \dots + na_{nk} = 0 \]
holds for every positive integer $k$. | Answer: all $n > 2$.
For $n = 2$, we have $a_k + 2a_{2k} = 0$,
which is clearly not possible,
since it implies $a_{2^k} = \frac{a_1}{2^{k-1}}$
for all $k \ge 1$.
For $n \ge 3$ we will construct a \emph{completely multiplicative} sequence
(meaning $a_{ij} = a_i a_j$ for all $i$ and $j$).
Thus $(a_i)$ is determined by ... |
USAMO-2012-notes_4 | Find all functions $f \colon \NN \to \NN$ such that
$f(n!) = f(n)!$ for all positive integers $n$ and such that
$m-n$ divides $f(m) - f(n)$ for all distinct positive integers $m$, $n$. | Answer: $f \equiv 1$, $f \equiv 2$, and $f$ the identity.
As these obviously work, we prove these are the only ones.
By putting $n=1$ and $n=2$ we give $f(1), f(2) \in \{1,2\}$.
Also, we will use the condition
\[ m!-n! \text{ divides } f(m)! - f(n)!. \]
We consider four cases on $f(1)$ and $f(2)$,
and dispense with th... |
USAMO-2012-notes_5 | Let $P$ be a point in the plane of $\triangle ABC$,
and $\gamma$ a line through $P$.
Let $A'$, $B'$, $C'$ be the points where the
reflections of lines $PA$, $PB$, $PC$ with respect to $\gamma$
intersect lines $BC$, $CA$, $AB$ respectively.
Prove that $A'$, $B'$, $C'$ are collinear. | We present three solutions.
\paragraph{First solution (complex numbers).}
Let $p=0$ and set $\gamma$ as the real line.
Then $A'$ is the intersection of $bc$ and $p\ol a$.
So, we get
\[ a' = \frac{\ol a(\ol b c - b \ol c)}{(\ol b - \ol c)\ol a-(b-c) a}. \]
\begin{center}
\begin{asy}
size(4cm);
pair A = dir(1... |
USAMO-2012-notes_6 | For integer $n \ge 2$,
let $x_1$, $x_2$, \dots, $x_n$ be real numbers satisfying
\[ x_1 + x_2 + \dots + x_n = 0
\quad\text{and}\quad x_1^2 + x_2^2 + \dots + x_n^2 = 1. \]
For each subset $A\subseteq\{1, 2, \dots, n\}$, define $S_A=\sum_{i\in A} x_i$.
(If $A$ is the empty set, then $S_A=0$.)
Prove that for any positi... | Let $\eps_i$ be a coin flip of $0$ or $1$.
Then we have
\begin{align*} \mathbb E[S_A^2]
&= \mathbb E\left[ \left( \sum \eps_i x_i \right)^2 \right]
= \sum_i \mathbb E[\eps_i^2] x_i^2
+ \sum_{i < j} \mathbb E[\eps_i\eps_j] 2x_i x_j \\
&= \half \sum x_i^2 + \half \sum x_i x_j
= \half + \half \sum_{i<j} x_i x_j
... |
USAMO-2013-notes_1 | In triangle $ABC$,
points $P$, $Q$, $R$ lie on sides $BC$, $CA$, $AB$, respectively.
Let $\omega_A$, $\omega_B$, $\omega_C$ denote the
circumcircles of triangles $AQR$, $BRP$, $CPQ$, respectively.
Given the fact that segment $AP$ intersects
$\omega_A$, $\omega_B$, $\omega_C$ again at $X$, $Y$, $Z$ respectively,
prove t... | Let $M$ be the concurrence point of $\omega_A$, $\omega_B$, $\omega_C$ (by Miquel's theorem).
\begin{center}
\begin{asy}
size(8cm);
defaultpen(fontsize(9pt));
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
pair P = 0.4*B+0.6*C;
pair Q = 0.4*C+0.6*A;
pair R = 0.7*A+0.3*B;
... |
USAMO-2013-notes_2 | For a positive integer $n\geq 3$ plot $n$ equally spaced points around a circle.
Label one of them $A$, and place a marker at $A$.
One may move the marker forward in a clockwise direction
to either the next point or the point after that.
Hence there are a total of $2n$ distinct moves available; two from each point.
Let... | We present two similar approaches.
\paragraph{First solution.}
Imagine the counter is moving along
the set $S = \{0, 1, \dots, 2n\}$ instead,
starting at $0$ and ending at $2n$, in jumps of length $1$ and $2$.
We can then record the sequence of moves as a matrix of the form
\[
\begin{bmatrix}
p_0 & p_1 & p_2 & \... |
USAMO-2013-notes_3 | Let $n$ be a positive integer.
There are $\tfrac{n(n+1)}{2}$ tokens,
each with a black side and a white side,
arranged into an equilateral triangle,
with the biggest row containing $n$ tokens.
Initially, each token has the white side up.
An operation is to choose a line parallel to the sides of the triangle,
and flip a... | The answer is
\[
\max_C f(C) =
\begin{cases}
6k & n = 4k \\
6k+1 & n = 4k+1 \\
6k+2 & n = 4k+2 \\
6k+3 & n = 4k+3.
\end{cases}
\]
The main point of the problem is actually to determine
all linear dependencies among the $3n$ possible moves
(since the moves commute and applying a move twice
is the s... |
USAMO-2013-notes_4 | Find all real numbers $x,y,z \ge 1$ satisfying
\[ \min \left( \sqrt{x+xyz}, \sqrt{y+xyz}, \sqrt{z+xyz} \right)
= \sqrt{x-1} + \sqrt{y-1} + \sqrt{z-1}. \] | Set $x = 1+a$, $y = 1+b$, $z = 1+c$
which eliminates the $x,y,z \ge 1$ condition.
Assume without loss of generality that $a \leq b \leq c$.
Then the given equation rewrites as
\[ \sqrt{(1+a)\left( 1+(1+b)(1+c) \right)} = \sqrt a + \sqrt b + \sqrt c. \]
In fact, we are going to prove the left-hand side always exceeds t... |
USAMO-2013-notes_5 | Let $m$ and $n$ be positive integers.
Prove that there exists a positive integer $c$
such that $cm$ and $cn$ have the same nonzero decimal digits. | One-line spoiler: $142857$.
More verbosely,
the idea is to look at the decimal representation of $1/D$, $m/D$, $n/D$
for a suitable denominator $D$, which have a ``cyclic shift'' property
in which the digits of $n/D$ are the digits of $m/D$ shifted by $3$.
\begin{remark*}
[An example to follow along]
Here is an e... |
USAMO-2014-notes_1 | Let $a$, $b$, $c$, $d$ be real numbers such that $b-d \ge 5$ and all zeros
$x_1$, $x_2$, $x_3$, and $x_4$ of the polynomial $P(x)=x^4+ax^3+bx^2+cx+d$ are real.
Find the smallest value the product $(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1)$ can take. | The answer is $\boxed{16}$.
This can be achieved by taking $x_1 = x_2 = x_3 = x_4 = 1$,
whence the product is $2^4 = 16$, and $b-d = 5$.
We now show the quantity is always at least $16$.
We prove:
\begin{claim*}
We always have $(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1) = (b-d-1)^2 + (a-c)^2$.
\end{claim*}
\begin{proof}
... |
USAMO-2014-notes_2 | Find all $f \colon \ZZ \to \ZZ$ such that
\[ xf\left( 2f(y)-x \right) + y^2f\left( 2x-f(y) \right)
= \frac{f(x)^2}{x} + f\left( yf(y) \right) \]
for all $x,y \in \ZZ$ such that $x \neq 0$. | The answer is $f(x) \equiv 0$ and $f(x) \equiv x^2$. Check that these work.
Now let's prove these are the only solutions.
Put $y=0$ to obtain
\[ x f\left( 2f(0)-x \right) = \frac{f(x)^2}{x} + f(0). \]
The nicest part of the problem is the following step:
\begin{claim*}
We have $f(0)=0$.
\end{claim*}
\begin{proof}
... |
USAMO-2014-notes_3 | Prove that there exists an infinite set of points
\[ \dots, \; P_{-3}, \; P_{-2},\; P_{-1},\; P_0,\; P_1,\; P_2,\; P_3,\; \dots \]
in the plane with the following property:
For any three distinct integers $a$, $b$, and $c$,
points $P_a$, $P_b$, and $P_c$ are collinear if and only if $a+b+c=2014$. | The construction
\[ P_n = \left( n-\frac{2014}{3},
\left( n-\frac{2014}{3} \right)^3 \right) \]
works fine, and follows from the following claim:
\begin{claim*}
If $x$, $y$, $z$ are distinct real numbers
then the points $(x,x^3)$, $(y,y^3)$, $(z,z^3)$
are collinear if and only if $x+y+z=0$.
\end{claim*}
\begin{... |
USAMO-2014-notes_4 | Let $k$ be a positive integer.
Two players $A$ and $B$ play a game on an infinite grid of regular hexagons.
Initially all the grid cells are empty.
Then the players alternately take turns with $A$ moving first.
In her move, $A$ may choose two adjacent hexagons in the grid
which are empty and place a counter in both of ... | The answer is $k = 6$.
\paragraph{Proof that $A$ cannot win if $k=6$.}
We give a strategy for $B$ to prevent $A$'s victory.
Shade in every third cell, as shown in the figure below.
Then $A$ can never cover two shaded cells simultaneously on her turn.
Now suppose $B$ always removes a counter on a shaded cell
(and other... |
USAMO-2014-notes_5 | Let $ABC$ be a triangle with orthocenter $H$ and
let $P$ be the second intersection of the circumcircle of
triangle $AHC$ with the internal bisector of $\angle BAC$.
Let $X$ be the circumcenter of triangle $APB$
and let $Y$ be the orthocenter of triangle $APC$.
Prove that the length of segment $XY$ is equal to
the circ... | \begin{center}
\begin{asy}
size(8cm);
pair A = Drawing("A", dir(110), dir(110));
pair B = Drawing("B", dir(210), dir(210));
pair C = Drawing("C", dir(330), dir(330));
draw(A--B--C--cycle);
draw(unitcircle);
pair Q = dir(30); // 330 + 1/2 * (330-210) (mod 360)
Drawing("Q", Q, Q);
pair P = A+C-A*C*conj(... |
USAMO-2014-notes_6 | Prove that there is a constant $c>0$ with the following property:
If $a$, $b$, $n$ are positive integers such that $\gcd(a+i, b+j)>1$
for all $i, j \in \{0, 1, \dots, n\}$, then
\[ \min\{a, b\}> (cn)^{n/2}. \]
\end{enumerate} | Let $N = n+1$ and assume $N$ is (very) large.
We construct an $N \times N$ with cells $(i,j)$
where $0 \le i, j \le n$ and in each cell
place a prime $p$ dividing $\gcd (a+i, b+j)$.
The central claim is at least $50\%$ of the primes
in this table exceed $0.001n^2$.
We count the maximum number of squares they could occ... |
USAMO-2015-notes_1 | Solve in integers the equation
\[ x^2+xy+y^2 = \left(\frac{x+y}{3}+1\right)^3. \] | We do the trick of setting $a=x+y$ and $b=x-y$.
This rewrites the equation as
\[ \frac14\left( (a+b)^2+(a+b)(a-b)+(a-b)^2 \right) = \left( \frac a3 + 1 \right)^3 \]
where $a,b \in \ZZ$ have the same parity.
This becomes
\[ 3a^2+b^2 = 4\left( \frac a3 + 1 \right)^3 \]
which is enough to imply $3 \mid a$, so let $a = 3c$... |
USAMO-2015-notes_2 | Quadrilateral $APBQ$ is inscribed in circle $\omega$ with
$\angle P = \angle Q = 90^{\circ}$ and $AP = AQ < BP$.
Let $X$ be a variable point on segment $\ol{PQ}$.
Line $AX$ meets $\omega$ again at $S$ (other than $A$).
Point $T$ lies on arc $AQB$ of $\omega$ such that $\ol{XT}$
is perpendicular to $\ol{AX}$.
Let $M$ de... | We present three solutions,
one by complex numbers, two more synthetic.
(A fourth solution using median formulas is also possible.)
Most solutions will prove that the center of the fixed circle
is the midpoint of $\ol{AO}$
(with $O$ the center of $\omega$);
this can be recovered empirically by letting
\begin{itemize}
... |
USAMO-2015-notes_4 | Steve is piling $m\geq 1$ indistinguishable stones
on the squares of an $n\times n$ grid.
Each square can have an arbitrarily high pile of stones.
After he finished piling his stones in some manner,
he can then perform \emph{stone moves}, defined as follows.
Consider any four grid squares, which are corners of a rectan... | The answer is $\binom{m+n-1}{n-1}^2$.
The main observation is that the ordered sequence of column counts
(i.e.\ the number of stones in the first, second, etc.\ column)
is invariant under stone moves, as does the analogous sequence of row counts.
\paragraph{Definitions.}
Call these numbers $(c_1, c_2, \dots, c_n)$
and... |
USAMO-2015-notes_5 | Let $a$, $b$, $c$, $d$, $e$ be distinct positive integers
such that $a^4+b^4=c^4+d^4=e^5$.
Show that $ac+bd$ is a composite number. | Assume to the contrary that $p = ac+bd$, so that
\begin{align*}
ac &\equiv -bd \pmod p \\
\implies a^4c^4 &\equiv b^4d^4 \pmod p \\
\implies a^4 (e^5 - d^4) &\equiv (e^5 - a^4) d^4 \pmod p \\
\implies a^4 e^5 &\equiv d^4 e^5 \pmod p \\
\implies e^5(a^4-d^4) &\equiv 0 \pmod p
\end{align*}
and hence \[ p \mid e... |
USAMO-2015-notes_6 | Fix $0 < \lambda < 1$, and let $A$ be a multiset of positive integers.
Let $A_n = \{a\in A: a\leq n\}$.
Assume that for every $n \in \NN$, the multiset $A_n$ contains at most $n\lambda$ numbers.
Show that there are infinitely many $n\in\NN$ for
which the sum of the elements in $A_n$ is at most $\frac{n(n+1)}{2}\lambda$... | For brevity, $\#S$ denotes $|S|$.
Let $x_n = n\lambda - \#A_n \ge 0$.
We now proceed by contradiction by assuming
the conclusion fails for $n$ large enough; that is,
\begin{align*}
\frac{n(n+1)}{2}\lambda
&< \sum_{a \in A_n} a \\
&= 1(\#A_1-\#A_0)
+ 2(\#A_2 - \#A_1)
+ \dots + n(\#A_n - \#A_{n-1}) \\
&= n \#... |
USAMO-2016-notes_1 | Let $X_1$, $X_2$, \dots, $X_{100}$ be a sequence of
mutually distinct nonempty subsets of a set $S$.
Any two sets $X_i$ and $X_{i+1}$ are disjoint and their union is not the whole set $S$,
that is, $X_i \cap X_{i+1} = \emptyset$ and $X_i \cup X_{i+1} \neq S$,
for all $i \in \{1, \dots, 99\}$.
Find the smallest possible... | Solution with Danielle Wang: the answer is that $|S| \ge 8$.
\paragraph{Proof that $|S| \ge 8$ is necessary.}
Since we must have $2^{|S|} \geq 100$, we must have $|S| \geq 7$.
To see that $|S| = 8$ is the minimum possible size,
consider a chain on the set $S = \{1, 2, \dots, 7\}$
satisfying $X_i \cap X_{i+1} = \empty... |
USAMO-2016-notes_2 | Prove that for any positive integer $k$,
\[ (k^2)!\cdot\displaystyle\prod_{j=0}^{k-1}\frac{j!}{(j+k)!} \]
is an integer. | We show the exponent of any given prime $p$ is nonnegative in the expression.
Recall that the exponent of $p$ in $n!$ is equal to
$\sum_{i \ge 1} \left\lfloor n/p^i \right\rfloor$.
In light of this, it suffices to show
that for any prime power $q$, we have
\[
\left\lfloor \frac{k^2}{q} \right\rfloor
+ \sum_{j=0}^{k-1} ... |
USAMO-2016-notes_3 | Let $ABC$ be an acute triangle
and let $I_B$, $I_C$, and $O$ denote its
$B$-excenter, $C$-excenter, and circumcenter, respectively.
Points $E$ and $Y$ are selected on $\ol{AC}$ such that
$\angle ABY = \angle CBY$ and $\ol{BE} \perp \ol{AC}$.
Similarly, points $F$ and $Z$ are selected on $\ol{AB}$ such that
$\angle ACZ ... | We present two solutions.
\paragraph{First solution.}
Let $I_A$ denote the $A$-excenter and $I$ the incenter.
Then let $D$ denote the foot of the altitude from $A$.
Suppose the $A$-excircle is tangent to $\ol{BC}$, $\ol{AB}$, $\ol{AC}$
at $A_1$, $B_1$, $C_1$ and let
$A_2$, $B_2$, $C_2$ denote the reflections of $I_A$ ... |
USAMO-2016-notes_4 | Find all functions $f \colon \RR \to \RR$ such that
for all real numbers $x$ and $y$,
\[ (f(x)+xy) \cdot f(x-3y) + (f(y)+xy) \cdot f(3x-y) = (f(x+y))^2. \] | We claim that the only two functions satisfying
the requirements are $f(x) \equiv 0$ and $f(x) \equiv x^2$.
These work.
First, taking $x=y=0$ in the given yields $f(0) = 0$,
and then taking $x=0$ gives $f(y)f(-y) = f(y)^2$.
So also $f(-y)^2 = f(y)f(-y)$, from which we conclude $f$ is even.
Then taking $x = -y$ gives
\... |
USAMO-2016-notes_5 | An equilateral pentagon $AMNPQ$ is inscribed in triangle $ABC$
such that $M \in \ol{AB}$, $Q \in \ol {AC}$, and $N,P \in \ol{BC}$.
Let $S$ be the intersection of $\ol{MN}$ and $\ol{PQ}$.
Denote by $\ell$ the angle bisector of $\angle MSQ$.
Prove that $\ol{OI}$ is parallel to $\ell$,
where $O$ is the circumcenter of tr... | \paragraph{First solution (complex).}
In fact, we only need $AM = AQ = NP$ and $MN = QP$.
We use complex numbers with $ABC$ the unit circle,
assuming WLOG that $A$, $B$, $C$ are labeled counterclockwise.
Let $x$, $y$, $z$ be the complex numbers corresponding to the arc midpoints
of $BC$, $CA$, $AB$, respectively; thus... |
USAMO-2016-notes_6 | Integers $n$ and $k$ are given, with $n \ge k \ge 2$.
You play the following game against an evil wizard.
The wizard has $2n$ cards; for each $i=1,\dots,n$, there are two cards labeled $i$.
Initially, the wizard places all cards face down in a row, in unknown order.
You may repeatedly make moves of the following form: ... | The game is winnable if and only if $k < n$.
First, suppose $2 \le k < n$. Query the cards in positions $\left\{ 1, \dots, k \right\}$,
then $\left\{ 2, \dots, k+1 \right\}$, and so on, up to $\left\{ 2n-k+1, 2n \right\}$.
Indeed, by taking the difference of the $i$th and $(i+1)$st query,
we can deduce the value of th... |
USAMO-2017-notes_1 | Prove that there exist infinitely many pairs of
relatively prime positive integers $a,b > 1$
for which $a+b$ divides $a^b+b^a$. | One construction: let $d \equiv 1 \pmod 4$, $d > 1$.
Let $x = \frac{d^d+2^d}{d+2}$. Then set
\[ a = \frac{x+d}{2}, \qquad
b = \frac{x-d}{2}. \]
To see this works, first check that $b$ is odd and $a$ is even.
Let $d = a-b$ be odd.
Then:
\begin{align*}
a+b \mid a^b+b^a &\iff
(-b)^b + b^a \equiv 0 \pmod{a+b} \\
&\... |
USAMO-2017-notes_2 | Let $m_1$, $m_2$, \dots, $m_n$ be a collection of $n$ positive integers,
not necessarily distinct.
For any sequence of integers $A = (a_1, \dots, a_n)$
and any permutation $w = w_1, \dots, w_n$ of $m_1, \dots, m_n$,
define an $A$-inversion of $w$ to be a pair of
entries $w_i, w_j$ with $i < j$ for which
one of the foll... | The following solution was posted by Michael Ren,
and I think it is the most natural one
(since it captures all the combinatorial ideas
using a $q$-generating function that is easier to think about,
and thus makes the problem essentially a long computation).
Denote by $M$ our multiset of $n$ positive integers.
Define ... |
USAMO-2017-notes_3 | Let $ABC$ be a scalene triangle with circumcircle $\Omega$ and incenter $I$.
Ray $AI$ meets $\ol{BC}$ at $D$ and $\Omega$ again at $M$;
the circle with diameter $\ol{DM}$ cuts $\Omega$ again at $K$.
Lines $MK$ and $BC$ meet at $S$, and $N$ is the midpoint of $\ol{IS}$.
The circumcircles of $\triangle KID$ and $\triangl... | Let $W$ be the midpoint of $\ol{BC}$,
let $X$ be the point on $\Omega$ opposite $M$.
Observe that $\ol{KD}$ passes through $X$,
and thus lines $BC$, $MK$, $XA$ concur at
the orthocenter of $\triangle DMX$, which we call $S$.
Denote by $I_A$ the $A$-excenter of $ABC$.
Next, let $E$ be the foot of the altitude from $I$ ... |
USAMO-2017-notes_4 | Let $P_1$, $P_2$, \dots, $P_{2n}$ be $2n$ distinct points on the
unit circle $x^2+y^2=1$, other than $(1,0)$.
Each point is colored either red or blue,
with exactly $n$ red points and $n$ blue points.
Let $R_1$, $R_2$, \dots, $R_n$ be any ordering of the red points.
Let $B_1$ be the nearest blue point to $R_1$ travelin... | We present two solutions, one based on
swapping and one based on an invariant.
\paragraph{First ``local'' solution by swapping two points.}
Let $1 \le i < n$ be any index and consider the two red points
$R_i$ and $R_{i+1}$.
There are two blue points $B_i$ and $B_{i+1}$ associated with them.
\begin{claim*}
If we swa... |
USAMO-2017-notes_5 | Find all real numbers $c > 0$ such that there exists a labeling
of the lattice points in $\ZZ^2$ with positive integers for which:
\begin{itemize}
\ii only finitely many distinct labels occur, and
\ii for each label $i$, the distance between any two points
labeled $i$ is at least $c^i$.
\end{itemize} | The answer is $c < \sqrt2$. Here is a solution with Calvin Deng.
The construction for any $c < \sqrt 2$ can be done as follows.
Checkerboard color the lattice points and label the black ones with $1$.
The white points then form a copy of $\ZZ^2$ again
scaled up by $\sqrt 2$ so we can repeat the procedure with $2$
on h... |
USAMO-2017-notes_6 | Find the minimum possible value of
\[ \frac{a}{b^3+4} + \frac{b}{c^3+4} + \frac{c}{d^3+4} + \frac{d}{a^3+4} \]
given that $a$, $b$, $c$, $d$ are nonnegative real numbers
such that $a+b+c+d=4$.
\end{enumerate} | The minimum $\frac23$ is achieved
at $(a,b,c,d) = (2,2,0,0)$ and cyclic permutations.
The problem is an application of the tangent line trick:
we observe the miraculous identity
\[ \frac{1}{b^3+4} \ge \frac14 - \frac{b}{12} \]
since $12-(3-b)(b^3+4) = b(b+1)(b-2)^2 \ge 0$.
Moreover,
\[ ab+bc+cd+da = (a+c)(b+d)
\le \... |
USAMO-2018-notes_1 | Let $a$, $b$, $c$ be positive real numbers such that $a+b+c = 4\sqrt[3]{abc}$.
Prove that
\[ 2(ab+bc+ca) + 4 \min (a^2, b^2, c^2) \ge a^2 + b^2 + c^2. \] | WLOG let $c = \min(a,b,c) = 1$ by scaling.
The given inequality becomes equivalent to
\[ 4ab + 2a + 2b + 3 \ge (a+b)^2 \qquad \forall a+b = 4(ab)^{1/3}-1. \]
Now, let $t = (ab)^{1/3}$ and eliminate $a+b$ using the condition, to get
\[ 4t^3 + 2(4t-1) + 3 \ge (4t-1)^2
\iff 0 \le 4t^3 - 16t^2 + 16t = 4t(t-2)^2 \]
which ... |
USAMO-2018-notes_2 | Find all functions $f \colon (0,\infty) \to (0,\infty)$ such that
\[
f\left( x+\frac 1y \right)
+ f\left( y+\frac 1z \right)
+ f\left( z+\frac 1x \right)
= 1
\]
for all $x,y,z > 0$ with $xyz = 1$. | The main part of the problem is to show all solutions are linear.
As always, let $x=b/c$, $y=c/a$, $z=a/b$
(classical inequality trick).
Then the problem becomes
\[ \sum_{\text{cyc}} f\left( \frac{b+c}{a} \right) = 1. \]
Let $f(t) = g(\frac{1}{t+1})$,
equivalently $g(s) = f(1/s-1)$.
Thus $g \colon (0,1) \to (0,1)$
whic... |
USAMO-2018-notes_3 | Let $n \ge 2$ be an integer, and let $\{a_1, \dots, a_m\}$ denote
the $m = \varphi(n)$ integers less than $n$ and relatively prime to $n$.
Assume that every prime divisor of $m$ also divides $n$.
Prove that $m$ divides $a_1^k + \dots + a_m^k$ for every positive
integer $k$. | For brevity, given any $n$, we let $A(n) = \left\{ 1 \le x \le n, \gcd(x,n) = 1 \right\}$
(thus $|A(n)| = \varphi(n)$).
Also, let $S(n,k) = \sum_{a \in A(n)} a^k$.
We will prove the stronger statement (which eliminates the hypothesis on $n$).
\begin{claim*}
Let $n \ge 2$ be arbitrary (and $k \ge 0$).
If $p \mid n$... |
USAMO-2018-notes_4 | Let $p$ be a prime, and let $a_1$, \dots, $a_p$ be integers.
Show that there exists an integer $k$ such that the numbers
\[ a_1 + k, \; a_2 + 2k, \; \dots, \; a_p + pk \]
produce at least $\half p$ distinct remainders upon division by $p$. | For each $k = 0, \dots, p-1$ let $G_k$ be the graph
on $\{1, \dots, p\}$ where we join $\{i,j\}$ if and only if
\[ a_i + ik \equiv a_j + jk \pmod p
\iff k \equiv - \frac{a_i - a_j}{i-j} \pmod p. \]
So we want a graph $G_k$ with at least $\half p$ connected components.
However, each $\{i,j\}$ appears in exactly one g... |
USAMO-2018-notes_5 | Let $ABCD$ be a convex cyclic quadrilateral with
$E = \ol{AC} \cap \ol{BD}$, $F = \ol{AB} \cap \ol{CD}$,
$G = \ol{DA} \cap \ol{BC}$.
The circumcircle of $\triangle ABE$
intersects line $CB$ at $B$ and $P$,
and the circumcircle of $\triangle ADE$
intersects line $CD$ at $D$ and $Q$.
Assume $C$, $B$, $P$, $G$
and $C$, $Q... | We present three general routes.
(The second route, using the fact that $\ol{AC}$
is an angle bisector, has many possible variations.)
\paragraph{First solution (Miquel points).}
This is indeed a Miquel point problem,
but the main idea is to focus on the self-intersecting
cyclic quadrilateral $PBQD$ as the key player,... |
USAMO-2018-notes_6 | Let $a_n$ be the number of permutations $(x_1, \dots, x_n)$ of $(1, \dots, n)$
such that the ratios $x_k / k$ are all distinct.
Prove that $a_n$ is odd for all $n \ge 1$.
\end{enumerate} | This is the official solution; the proof has two main insights.
The first idea:
\begin{lemma*}
If a permutation $x$ works, so does the inverse permutation.
\end{lemma*}
Thus it suffices to consider permutations $x$
in which all cycles have length at most $2$.
Of course, there can be at most one fixed point (since t... |
USAMO-2019-notes_1 | A function $f \colon \NN \to \NN$ satisfies
\[ \underbrace{f(f(\dots f}_{f(n)\text{ times}} (n)\dots)) %chktex 9
= \frac{n^2}{f(f(n))} \]
for all positive integers $n$.
What are all possible values of $f(1000)$? | Actually, we classify all such functions: $f$ can be any function which fixes odd integers
and acts as an involution on the even integers.
In particular, $f(1000)$ may be any even integer.
It's easy to check that these all work, so now we check they are the only solutions.
\begin{claim*}
$f$ is injective.
\end{clai... |
USAMO-2019-notes_2 | Let $ABCD$ be a cyclic quadrilateral satisfying $AD^2 + BC^2 = AB^2$.
The diagonals of $ABCD$ intersect at $E$.
Let $P$ be a point on side $\ol{AB}$ satisfying $\angle APD = \angle BPC$.
Show that line $PE$ bisects $\ol{CD}$. | Here are three solutions.
The first two are similar although the first one makes use of symmedians.
The last solution by inversion is more advanced.
\paragraph{First solution using symmedians.}
We define point $P$ to obey
\[ \frac{AP}{BP} = \frac{AD^2}{BC^2} = \frac{AE^2}{BE^2} \]
so that $\ol{PE}$ is the $E$-symmedia... |
USAMO-2019-notes_3 | Let $K$ be the set of positive integers not containing the decimal digit $7$.
Determine all polynomials $f(x)$ with nonnegative coefficients
such that $f(x) \in K$ for all $x \in K$. | The answer is only the obvious ones:
$f(x) = 10^e x$, $f(x) = k$,
and $f(x) = 10^e x + k$, for any choice of $k \in K$
and $e > \log_{10} k$ (with $e \ge 0$).
Now assume $f$ satisfies $f(K) \subseteq K$;
such polynomials will be called \emph{stable}.
We first prove the following claim
which reduces the problem to the ... |
USAMO-2019-notes_4 | Let $n$ be a nonnegative integer.
Determine the number of ways to choose sets
$S_{ij} \subseteq \{1, 2, \dots, 2n\}$,
for all $0 \le i \le n$ and $0 \le j \le n$
(not necessarily distinct), such that
\begin{itemize}
\ii $|S_{ij}| = i+j$, and
\ii $S_{ij} \subseteq S_{kl}$ if $0 \le i \le k \le n$
and $0 \le j \le ... | The answer is $(2n)! \cdot 2^{n^2}$.
First, we note that
$\varnothing = S_{00} \subsetneq S_{01} \subsetneq \dots \subsetneq S_{nn}
= \left\{ 1, \dots, 2n \right\}$
and thus multiplying by $(2n)!$
we may as well assume $S_{0i} = \left\{ 1, \dots, i \right\}$
and $S_{in} = \left\{ 1, \dots, n+i \right\}$.
We illustrate ... |
USAMO-2019-notes_6 | Find all polynomials $P$ with real coefficients such that
\[ \frac{P(x)}{yz} + \frac{P(y)}{zx} + \frac{P(z)}{xy} = P(x-y) + P(y-z) + P(z-x) \]
for all nonzero real numbers $x$, $y$, $z$ obeying $2xyz = x+y+z$.
\end{enumerate} | The given can be rewritten as saying that
\begin{align*}
Q(x,y,z) &\coloneq xP(x) + yP(y) + zP(z) \\ &- xyz \left( P(x-y) + P(y-z) + P(z-x) \right)
\end{align*}
is a polynomial vanishing whenever $xyz \neq 0$ and $2xyz = x+y+z$, for real numbers $x$, $y$, $z$.
\begin{claim*}
This means $Q(x,y,z)$ vanishes also for ... |
USAMO-2020-notes_1 | Let $ABC$ be a fixed acute triangle
inscribed in a circle $\omega$ with center $O$.
A variable point $X$ is chosen on minor arc $AB$ of $\omega$,
and segments $CX$ and $AB$ meet at $D$.
Denote by $O_1$ and $O_2$ the circumcenters of
triangles $ADX$ and $BDX$, respectively.
Determine all points $X$ for which
the area of... | We prove $[OO_1O_2] \ge \frac14 [ABC]$,
with equality if and only if $\ol{CX} \perp \ol{AB}$.
\paragraph{First approach (Bobby Shen).}
We use two simultaneous inequalities:
\begin{itemize}
\ii Let $M$ and $N$ be the midpoints of $CX$ and $DX$.
Then $MN$ equals the length of the $O$-altitude of $\triangle OO_1O_2$,... |
USAMO-2020-notes_2 | An empty $2020 \times 2020 \times 2020$ cube is given,
and a $2020 \times 2020$ grid of square unit cells is drawn on each of its six faces.
A \emph{beam} is a $1 \times 1 \times 2020$ rectangular prism.
Several beams are placed inside the cube subject to the following conditions:
\begin{itemize}
\item The two $1 \time... | \paragraph{Answer.} $3030$ beams.
\paragraph{Construction.}
We first give a construction with $3n/2$ beams for any $n \times n \times n$ box,
where $n$ is an even integer.
Shown below is the construction for $n=6$, which generalizes.
(The left figure shows the cube in 3d;
the right figure shows a direct view of the th... |
USAMO-2020-notes_3 | Let $p$ be an odd prime.
An integer $x$ is called a \emph{quadratic non-residue}
if $p$ does not divide $x-t^2$ for any integer $t$.
Denote by $A$ the set of all integers $a$
such that $1 \le a < p$,
and both $a$ and $4-a$ are quadratic non-residues.
Calculate the remainder
when the product of the elements of $A$ is d... | The answer is that $\prod_{a \in A} a \equiv 2 \pmod p$
regardless of the value of $p$.
In the following solution,
we work in $\FF_p$ always
and abbreviate ``quadratic residue'' and ``non-quadratic residue''
to ``QR'' and ``non-QR'', respectively.
We define
\begin{align*}
A &= \left\{ a \in \FF_p \mid a, 4-a \text{ ... |
USAMO-2020-notes_4 | Suppose that $(a_1, b_1)$, $(a_2, b_2)$, \dots, $(a_{100}, b_{100})$
are distinct ordered pairs of nonnegative integers.
Let $N$ denote the number of pairs of integers $(i,j)$ satisfying
$1 \le i < j \le 100$ and $\left\lvert a_ib_j - a_jb_i \right\rvert = 1$.
Determine the largest possible value of $N$
over all possib... | The answer is $197$.
In general, if $100$ is replaced by $n \ge 2$ the answer is $2n-3$.
The idea is that if we let $P_i = (a_i, b_i)$ be a point
in the coordinate plane, and let $O = (0,0)$
then we wish to maximize the number of triangles
$\triangle O P_i P_j$ which have area $1/2$.
Call such a triangle \emph{good}.
... |
USAMO-2020-notes_5 | A finite set $S$ of points in the coordinate plane
is called \emph{overdetermined} if $|S| \ge 2$
and there exists a nonzero polynomial $P(t)$,
with real coefficients and of degree at most $|S|-2$,
satisfying $P(x)=y$ for every point $(x,y) \in S$.
For each integer $n \ge 2$,
find the largest integer $k$ (in terms of ... | We claim the answer is $k = 2^{n-1}-n$.
We denote the $n$ points by $A$.
Throughout the solution we will repeatedly use the following fact:
\begin{lemma*}
If $S$ is a finite set of points in the plane
there is at most one polynomial with real coefficients
and of degree at most $|S|-1$
whose graph passes throug... |
USAMO-2020-notes_6 | Let $n \geq 2$ be an integer.
Let $x_1 \ge x_2 \ge \dots \ge x_n$ and $y_1 \ge y_2 \ge \dots \ge y_n$
be $2n$ real numbers such that
\begin{align*}
0 &= x_1 + x_2 + \dots + x_n = y_1 + y_2 + \dots + y_n, \\
\text{and}\quad
1 &= x_1^2 + x_2^2 + \dots + x_n^2 = y_1^2 + y_2^2 + \dots + y_n^2.
\end{align*}
Prove that... | We present two approaches.
In both approaches, it's helpful motivation that for even $n$, equality occurs at
\begin{align*}
(x_i) &= \Big(
\underbrace{\frac{1}{\sqrt n}, \dots, \frac{1}{\sqrt n}}_{n/2},
\underbrace{-\frac{1}{\sqrt n}, \dots, -\frac{1}{\sqrt n}}_{n/2} \Big) \\
(y_i) &= \Big( \frac{n-1}{\sqrt... |
USAMO-2021-notes_1 | Rectangles $BCC_1B_2$, $CAA_1C_2$, and $ABB_1A_2$ are erected
outside an acute triangle $ABC$. Suppose that
\[ \angle BC_1C + \angle CA_1A + \angle AB_1B = 180^\circ. \]
Prove that lines $B_1C_2$, $C_1A_2$, and $A_1B_2$ are concurrent. | The angle condition implies the circumcircles of the three
rectangles concur at a single point $P$.
\begin{center}
\begin{asy}
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
pair P = 0.2*dir(190);
filldraw(A--B--C--cycle, opacity(0.2)+lightcyan, blue);
pair X = circumcenter(P, B, C);
pair Y = circumcenter(P... |
USAMO-2021-notes_2 | The Planar National Park is a undirected 3-regular planar graph
(i.e.\ all vertices have degree $3$).
A visitor walks through the park as follows:
she begins at a vertex and starts walking along an edge.
When she reaches the other endpoint, she turns left.
On the next vertex she turns right, and so on,
alternating left... | The answer is $3$.
We consider the trajectory of the visitor as an ordered sequence of \emph{turns}.
A turn is defined by specifying a vertex, the incoming edge, and the outgoing edge.
Hence there are six possible turns for each vertex.
\begin{claim*}
Given one turn in the sequence, one can reconstruct the entire s... |
USAMO-2021-notes_4 | A finite set $S$ of positive integers has the property that,
for each $s\in S$, and each positive integer divisor $d$ of $s$,
there exists a unique element $t\in S$ satisfying $\gcd(s,t) = d$.
(The elements $s$ and $t$ could be equal.)
Given this information, find all possible values for the
number of elements of $S$. | The answer is that $|S|$ must be a power of $2$ (including $1$),
or $|S| = 0$ (a trivial case we do not discuss further).
\paragraph{Construction.}
For any nonnegative integer $k$, a construction for $|S| = 2^k$ is given by
\[ S = \left\{
(p_1 \text{ or } q_1)
\times
(p_2 \text{ or } q_2)
\times
\d... |
USAMO-2021-notes_5 | Let $n \ge 4$ be an integer.
Find all positive real solutions to the following
system of $2n$ equations:
\begin{align*}
a_1 &= \frac{1}{a_{2n}} + \frac{1}{a_{2}}, & a_2 &= a_1 + a_3, \\[1ex]
a_3 &= \frac{1}{a_{2}} + \frac{1}{a_{4}}, & a_4 &= a_3 + a_5, \\[1ex]
a_5 &= \frac{1}{a_{4}} + \frac{1}{a_{6}}, & a_6 &= a_... | The answer is that the only solution is
$(1,2,1,2,\dots,1,2)$ which works.
We will prove $a_{2k}$ is a constant sequence,
at which point the result is obvious.
\paragraph{First approach (Andrew Gu).}
Apparently, with indices modulo $2n$, we should have
\[ a_{2k} = \frac{1}{a_{2k-2}}
+ \frac{2}{a_{2k}} + \frac{1}{a_... |
USAMO-2021-notes_6 | Let $ABCDEF$ be a convex hexagon satisfying
$\ol{AB} \parallel \ol{DE}$,
$\ol{BC} \parallel \ol{EF}$,
$\ol{CD} \parallel \ol{FA}$, and
\[ AB \cdot DE = BC \cdot EF = CD \cdot FA. \]
Let $X$, $Y$, and $Z$ be the midpoints
of $\ol{AD}$, $\ol{BE}$, and $\ol{CF}$.
Prove that
the circumcenter of $\triangle ACE$,
the circum... | We present two solutions.
\paragraph{Parallelogram solution found by contestants.}
Note that the following figure is intentionally
\emph{not} drawn to scale, to aid legibility.
We construct parallelograms $ABCE'$, etc as shown.
Note that this gives two congruent triangles $A'C'E'$ and $B'D'F'$.
(Assuming that triangle... |
USAMO-2022-notes_1 | Let $a$ and $b$ be positive integers.
Every cell of an $(a+b+1)\times (a+b+1)$ grid is colored either amber or bronze
such that there are at least $a^2+ab-b$ amber cells
and at least $b^2+ab-a$ bronze cells.
Prove that it is possible to choose $a$ amber cells and $b$ bronze cells
such that no two of the $a+b$ chosen ce... | \begin{claim*}
There exists a transversal $T_a$ with at least $a$ amber cells.
Analogously, there exists a transversal $T_b$ with at least $b$ bronze cells.
\end{claim*}
\begin{proof}
If one picks a random transversal, the expected value
of the number of amber cells is at least
\[ \frac{a^2+ab-b}{a+b+1} = (a-... |
USAMO-2022-notes_2 | Let $b\geq 2$ and $w\geq 2$ be fixed integers, and $n=b+w$.
Given are $2b$ identical black rods and $2w$ identical white rods,
each of side length $1$.
We assemble a regular $2n$-gon using these rods
so that parallel sides are the same color.
Then, a convex $2b$-gon $B$ is formed by translating the black rods,
and a c... | We are going to prove that one may swap a black rod with an adjacent white rod
(as well as the rods parallel to them)
without affecting the difference in the areas of $B-W$.
Let $\vec u$ and $\vec v$ denote the originally black and white vectors
that were adjacent on the $2n$-gon and are now going to be swapped.
Let $\... |
USAMO-2022-notes_3 | Solve over positive real numbers the functional equation
\[ f(x) = f(f(f(x)) + y) + f(xf(y)) f(x+y). \] | The answer is $f(x) \equiv c/x$ for any $c > 0$.
This works, so we'll prove this is the only solution.
The following is based on the solution posted by \texttt{pad} on AoPS.
In what follows, $f^n$ as usual denotes $f$ iterated $n$ times,
and $P(x,y)$ is the given statement.
Also, we introduce the notation $Q$ for the ... |
USAMO-2022-notes_4 | Find all pairs of primes $(p, q)$ for which $p-q$ and $pq-q$ are both perfect squares. | The answer is $(3,2)$ only.
This obviously works so we focus on showing it is the only one.
\paragraph{Approach using difference of squares (from author).}
Set
\begin{align*}
a^2 &= p-q \\
b^2 &= pq-q.
\end{align*}
Note that $0 < a < p$, and $0 < b < p$ (because $q \le p$).
Now subtracting gives
\[ \underbrace{(b-... |
USAMO-2022-notes_5 | A function $f \colon \RR \to \RR$ is
\emph{essentially increasing}
if $f(s) \leq f(t)$ holds
whenever $s\leq t$ are real numbers such that $f(s)\neq 0$ and $f(t)\neq 0$.
Find the smallest integer $k$ such that
for any $2022$ real numbers $x_1$, $x_2$, \dots, $x_{2022}$,
there exist $k$ essentially increasing functions... | The answer is $11$ and, more generally, if $2022$ is replaced by $N$
then the answer is $\left\lfloor \log_2 N \right\rfloor + 1$.
\paragraph{Bound.}
Suppose for contradiction that $2^k-1 > N$
and choose $x_n = -n$ for each $n = 1, \dots, N$.
Now for each index $1 \le n \le N$, define
\[ S(n) = \left\{
\text{indic... |
USAMO-2022-notes_6 | There are $2022$ users on a social network called Mathbook,
and some of them are Mathbook-friends.
(On Mathbook, friendship is always mutual and permanent.)
Starting now, Mathbook will only allow a new friendship to be formed
between two users if they have at least two friends in common.
What is the minimum number of ... | With $2022$ replaced by $n$, the answer is
$\left\lceil \frac 32 n \right\rceil - 2$.
\paragraph{Terminology.}
Standard graph theory terms:
starting from a graph $G$ on $n$ vertices,
we're allowed to take any $C_4$ in the graph and complete it to a $K_4$.
The problem asks the minimum number of edges needed
so that th... |
USAMO-2023-notes_1 | In an acute triangle $ABC$, let $M$ be the midpoint of $\ol{BC}$.
Let $P$ be the foot of the perpendicular from $C$ to $AM$.
Suppose that the circumcircle of triangle $ABP$
intersects line $BC$ at two distinct points $B$ and $Q$.
Let $N$ be the midpoint of $\ol{AQ}$.
Prove that $NB = NC$. | We show several different approaches.
In all solutions, let $D$ denote the foot of the altitude from $A$.
\begin{center}
\begin{asy}
size(10cm);
pair A = dir(115);
pair B = dir(210);
pair C = dir(330);
pair D = foot(A, B, C);
pair M = midpoint(B--C);
pair P = foot(C, A, M);
pair Q = 2*M-D;
filldraw(A--B--C--cycle, o... |
USAMO-2023-notes_2 | Solve over the positive real numbers the functional equation
\[ f(xy+f(x)) = xf(y) + 2. \] | The answer is $f(x) \equiv x+1$,
which is easily verified to be the only linear solution.
We show conversely that $f$ is linear. Let $P(x,y)$ be the assertion.
\begin{claim*}
$f$ is weakly increasing.
\end{claim*}
\begin{proof}
Assume for contradiction $a>b$ but $f(a)<f(b)$.
Choose $y$ such that $ay+f(a)=by+f(b... |
USAMO-2023-notes_3 | Consider an $n$-by-$n$ board of unit squares for some odd positive integer $n$.
We say that a collection $C$ of identical dominoes is a
maximal grid-aligned configuration on the board if $C$ consists of $(n^2-1)/2$
dominoes where each domino covers exactly two neighboring squares
and the dominoes don't overlap: $C$ the... | The answer is that
\[ k(C) \in \left\{ 1, 2, \dots, \left( \frac{n-1}{2} \right)^2 \right\}
\cup \left\{ \left( \frac{n+1}{2} \right)^2 \right\}. \]
Index the squares by coordinates $(x,y) \in \{1,2,\dots,n\}^2$.
We say a square is \emph{special} if it is empty or
it has the same parity in both coordinates as the em... |
USAMO-2023-notes_4 | Positive integers $a$ and $N$ are fixed,
and $N$ positive integers are written on a blackboard.
Alice and Bob play the following game.
On Alice's turn, she must replace some integer $n$ on the board with $n+a$,
and on Bob's turn he must replace some even integer $n$ on the board with $n/2$.
Alice goes first and they al... | For $N=1$, there is nothing to prove since each player has only one option each turn.
We address $N \ge 2$ only henceforth.
Let $S$ denote the numbers on the board.
\begin{claim*}
When $N \ge 2$, if $\nu_2(x) < \nu_2(a)$ for all $x \in S$,
the game must terminate no matter what either player does.
\end{claim*}
\be... |
USAMO-2023-notes_5 | Let $n\geq3$ be an integer. We say that an arrangement of the numbers
$1$, $2$, \dots, $n^2$ in an $n \times n$ table is \emph{row-valid}
if the numbers in each row can be permuted to form an arithmetic progression,
and \emph{column-valid} if the numbers in each column
can be permuted to form an arithmetic progression.... | Answer: $n$ prime only.
\paragraph{Proof for $n$ prime.}
Suppose $n = p$.
In an arithmetic progression with $p$ terms, it's easy to see that either
every term has a different residue modulo $p$ (if the common difference
is not a multiple of $p$), or all of the residues coincide
(when the common difference is a multipl... |
USAMO-2023-notes_6 | Let $ABC$ be a triangle with incenter $I$
and excenters $I_a$, $I_b$, $I_c$ opposite $A$, $B$, and $C$, respectively.
Given an arbitrary point $D$ on the circumcircle of $\triangle ABC$
that does not lie on any of the lines $II_a$, $I_bI_c$, or $BC$,
suppose the circumcircles of $\triangle DII_a$ and $\triangle DI_bI_c... | Here are two approaches.
\begin{center}
\begin{asy}
size(9cm);
pair A = dir(115);
pair B = dir(210);
pair C = dir(330);
pair D = dir(240);
pair E = extension(B, C, A, B*C/D);
pair I = incenter(A, B, C);
pair M = dir(270);
pair I_a = 2*M-I;
pair I_b = extension(B, I, C, I_a);
pair I_c = extension(C, I, B, I_a);
pair F ... |
USAMO-2024-notes_1 | Find all integers $n \geq 3$ such that the following property holds:
if we list the divisors of $n!$ in increasing order
as $1 = d_1 < d_2 < \dots < d_k = n!$, then we have
\[ d_2 - d_1 \leq d_3 - d_2 \leq \dots \leq d_k - d_{k-1}. \] | The answer is $n \in \{3,4\}$. These can be checked by listing all the divisors:
\begin{itemize}
\ii For $n=3$ we have $(1,2,3,6)$.
\ii For $n=4$ we have $(1,2,3,4,6,8,12,24)$.
\end{itemize}
We prove these are the only ones.
The numbers $5 \le n \le 12$ all fail because:
\begin{itemize}
\ii For $n = 5$ we have ... |
USAMO-2024-notes_2 | Let $S_1$, $S_2$, \dots, $S_{100}$ be finite sets of integers whose intersection is not empty.
For each non-empty $T \subseteq \{S_1, S_2, \ldots, S_{100}\}$,
the size of the intersection of the sets in $T$ is a multiple of $|T|$.
What is the smallest possible number of elements which are in at least $50$ sets? | The answer is $50 \binom{100}{50}$.
\paragraph{Rephrasing (cosmetic translation only, nothing happens yet).}
We encode with binary strings $v \in \FF_2^{100}$ of length $100$.
Write $v \subseteq w$ if $w$ has $1$'s in every component $v$ does,
and let $|v|$ denote the number of $1$'s in $v$.
Then for each $v$, we let... |
USAMO-2024-notes_3 | Let $(m,n)$ be positive integers with $n \ge 3$ and draw a regular $n$-gon.
We wish to triangulate this $n$-gon into $n-2$ triangles,
each colored one of $m$ colors, so that each color has an equal sum of areas.
For which $(m,n)$ is such a triangulation and coloring possible? | The answer is if and only if $m$ is a proper divisor of $n$.
Throughout this solution, we let $\omega = \exp\left( 2 \pi i / n \right)$
and let the regular $n$-gon have vertices $1$, $\omega$, \dots, $\omega^{n-1}$.
We cache the following frequent calculation:
\begin{lemma*}
The triangle with vertices $\omega^k$, $\... |
USAMO-2024-notes_4 | Let $m$ and $n$ be positive integers.
A circular necklace contains $mn$ beads, each either red or blue.
It turned out that no matter how the necklace was cut into $m$ blocks of $n$
consecutive beads, each block had a distinct number of red beads.
Determine, with proof, all possible values of the ordered pair $(m, n)$. | The answer is $m \leq n+1$ only.
\paragraph{Proof the task requires $m \le n+1$.}
Each of the $m$ blocks has a red bead count between $0$ and $n$,
each of which appears at most once, so $m \le n+1$ is needed.
\paragraph{Construction when $m=n+1$.}
For concreteness, here is the construction for $n=4$, which obviously ... |
USAMO-2024-notes_5 | Point $D$ is selected inside acute triangle $ABC$
so that $\angle DAC = \angle ACB$ and $\angle BDC = 90^\circ + \angle BAC$.
Point $E$ is chosen on ray $BD$ so that $AE = EC$.
Let $M$ be the midpoint of $BC$.
Show that line $AB$ is tangent to the circumcircle of triangle $BEM$. | This problem has several approaches and we showcase a collection of them.
\paragraph{The author's original solution.}
Complete isosceles trapezoid $ABQC$ (so $D \in \ol{AQ}$).
Reflect $B$ across $E$ to point $F$.
\begin{center}
\begin{asy}
size(7cm);
pair A = dir(185);
pair C = dir(355);
pair B = dir(115);
pair Q = A*... |
USAMO-2024-notes_6 | Let $n > 2$ be an integer and let $\ell \in \{1, 2,\dots, n\}$.
A collection $A_1$, \dots, $A_k$ of (not necessarily distinct) subsets of
$\{1, 2,\dots, n\}$ is called $\ell$-large if $|A_i| \ge \ell$ for all $1 \le i \le k$.
Find, in terms of $n$ and $\ell$, the largest real number $c$ such that the inequality
\[ \sum... | The answer turns out to be
\[ c = \frac{n+\ell^2-2\ell}{n(n-1)}. \]
The following solution using indicator functions was
suggested by the language reasoning system developed by Thang Luong
and the team at Google DeepMind (\url{https://dpmd.ai/imo-silver}).
\paragraph{Rewriting the sum.}
Abbreviate $[n] \coloneq \{1, \... |
USAMO-2025-notes_1 | Fix positive integers $k$ and $d$.
Prove that for all sufficiently large odd positive integers $n$,
the digits of the base-$2n$ representation of $n^k$ are all greater than $d$. | The problem actually doesn't have much to do with digits:
the idea is to pick any length $\ell \le k$,
and look at the rightmost $\ell$ digits of $n^k$;
that is, the remainder upon division by $(2n)^\ell$.
We compute it exactly:
\begin{claim*}
Let $n \ge 1$ be an odd integer, and $k \ge \ell \ge 1$ integers.
Then
... |
USAMO-2025-notes_2 | Let $n > k \ge 1$ be integers.
Let $P(x) \in \RR[x]$ be a polynomial of degree $n$
with no repeated roots and $P(0) \neq 0$.
Suppose that for any real numbers $a_0$, \dots, $a_k$
such that the polynomial $a_k x^k + \dots + a_1 x + a_0$ divides $P(x)$,
the product $a_0 a_1 \dots a_k$ is zero.
Prove that $P(x)$ has a non... | By considering any $k+1$ of the roots of $P$, we may as well assume WLOG that $n = k+1$.
Suppose that $P(x) = (x+r_1) \dots (x+r_n) \in \RR[x]$ has $P(0) \neq 0$.
Then the problem hypothesis is that each of the $n$ polynomials
(of degree $n-1$) given by
\begin{align*}
P_1(x) &= (x+r_2)(x+r_3)(x+r_4) \dots (x+r_n) \\
... |
USAMO-2025-notes_3 | Alice the architect and Bob the builder play a game.
First, Alice chooses two points $P$ and $Q$ in the plane
and a subset $\mathcal{S}$ of the plane, which are announced to Bob.
Next, Bob marks infinitely many points in the plane,
designating each a city.
He may not place two cities within distance at most one unit of... | The answer is that Alice wins.
Let's define a \emph{Bob-set} $V$ to be a set of points in the plane with no three collinear
and with all distances at least $1$.
The point of the problem is to prove the following fact.
\begin{claim*}
Given a Bob-set $V \subseteq \RR^2$, consider the \emph{Bob-graph}
with vertex set ... |
USAMO-2025-notes_4 | Let $H$ be the orthocenter of an acute triangle $ABC$,
let $F$ be the foot of the altitude from $C$ to $AB$,
and let $P$ be the reflection of $H$ across $BC$.
Suppose that the circumcircle of triangle $AFP$ intersects line $BC$
at two distinct points $X$ and $Y$.
Prove that $CX = CY$. | Let $Q$ be the antipode of $B$.
\begin{claim*}
$AHQC$ is a parallelogram, and $APCQ$ is an isosceles trapezoid.
\end{claim*}
\begin{proof}
As $\ol{AH} \perp \ol{BC} \perp \ol{CQ}$ and $\ol{CF} \perp \ol{AB} \perp \ol{AQ}$.
\end{proof}
\begin{center}
\begin{asy}
pair A = dir(110);
pair B = dir(210);
pair C = dir(33... |
USAMO-2025-notes_5 | Find all positive integers $k$ such that: for every positive integer $n$, the sum
\[ \binom n0^k + \binom n1^k + \dots + \binom nn^k \]
is divisible by $n+1$. | The answer is all even $k$.
Let's abbreviate $S(n) \coloneq \binom n0^k + \dots + \binom nn^k$ for the sum in the problem.
\paragraph{Proof that even $k$ is necessary.}
Choose $n=2$. We need $3 \mid S(2) = 2+2^k$,
which requires $k$ to be even.
\begin{remark*}
It's actually not much more difficult to just use $n =... |
USAMO-2025-notes_6 | Let $m$ and $n$ be positive integers with $m\geq n$.
There are $m$ cupcakes of different flavors arranged around a circle
and $n$ people who like cupcakes.
Each person assigns a nonnegative real number score to each cupcake,
depending on how much they like the cupcake.
Suppose that for each person $P$, it is possible t... | Arbitrarily pick any one person --- call her Pip --- and her $n$ arcs.
The initial idea is to try to apply Hall's marriage lemma to match
the $n$ people with Pip's arcs (such that each such person is happy with their matched arc).
To that end, construct the obvious bipartite graph $\mathfrak{G}$
between the people and ... |
sols-TST-IMO-2014_1 | Let $ABC$ be an acute triangle, and let $X$ be a variable interior point
on the minor arc $BC$ of its circumcircle.
Let $P$ and $Q$ be the feet of the perpendiculars from $X$ to lines $CA$ and $CB$, respectively.
Let $R$ be the intersection of line $PQ$ and the perpendicular from $B$ to $AC$.
Let $\ell$ be the line thr... | The fixed point is the orthocenter,
since $\ell$ is a Simson line.
See Lemma 4.4 of \emph{Euclidean Geometry in Math Olympiads}. |
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