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\title{USAMO 2015 Solution Notes}
\date{\today}
\begin{document}
\maketitle
\begin{abstract}
This is a compilation of solutions
for the 2015 USAMO.
Some of the solutions are my own work,
but many are from the official solutions provided by the organizers
(for which they hold any copyrights),
and others were found by users on the Art of Problem Solving forums.
These notes will tend to be a bit more advanced and terse than the ``official''
solutions from the organizers.
In particular, if a theorem or technique is not known to beginners
but is still considered ``standard'', then I often prefer to
use this theory anyways, rather than try to work around or conceal it.
For example, in geometry problems I typically use directed angles
without further comment, rather than awkwardly work around configuration issues.
Similarly, sentences like ``let $\mathbb{R}$ denote the set of real numbers''
are typically omitted entirely.
Corrections and comments are welcome!
\end{abstract}
\tableofcontents
\newpage
\addtocounter{section}{-1}
\section{Problems}
\begin{enumerate}[\bfseries 1.]
\item %% Problem 1
Solve in integers the equation
\[ x^2+xy+y^2 = \left(\frac{x+y}{3}+1\right)^3. \]
\item %% Problem 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 $M$ denote the midpoint of chord $\ol{ST}$.
As $X$ varies on segment $\ol{PQ}$, show that $M$ moves along a circle.
\item %% Problem 3
Let $S = \left\{ 1,2,\dots,n \right\}$, where $n \ge 1$.
Each of the $2^n$ subsets of $S$ is to be colored red or blue.
(The subset itself is assigned a color and not its individual elements.)
For any set $T \subseteq S$,
we then write $f(T)$ for the number of subsets of $T$ that are blue.
Determine the number of colorings that satisfy the following condition:
for any subsets $T_1$ and $T_2$ of $S$,
\[ f(T_1)f(T_2) = f(T_1 \cup T_2)f(T_1 \cap T_2). \]
\item %% Problem 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 rectangle,
i.e.\ in positions $(i, k)$, $(i, l)$, $(j, k)$, $(j, l)$
for some $1\leq i, j, k, l\leq n$, such that $i e$.
\end{claim*}
\begin{proof}
We have $e^5 = a^4 + b^4 \le a^5 + b^5 < (ac+bd)^5 = p^5$.
\end{proof}
Thus the above equation implies $p \le \max(a-d, a+d, a^2+d^2) = a^2+d^2$.
Similarly, $p \le b^2+c^2$.
So \[ ac+bd = p \le \min \left\{ a^2+d^2, b^2+c^2 \right\} \]
or by subtraction
\[ 0 \le \min \left\{ a(a-c) + d(d-b),
b(b-d) + c(c-a) \right\}. \]
But since $a^4+b^4 = c^4+d^4$ the numbers $a-c$ and $d-b$
should have the same sign, and so this is an obvious contradiction.
\pagebreak
\subsection{USAMO 2015/6, proposed by Iurie Boreico}
\textsl{Available online at \url{https://aops.com/community/p4774023}.}
\begin{mdframed}[style=mdpurplebox,frametitle={Problem statement}]
Consider $0<\lambda<1$, 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 $\frac{n(n+1)}{2}\lambda$.
\end{mdframed}
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 \# A_n - (\# A_1 + \dots + \# A_{n-1}) \\
&= n(n \lambda - x_n) - \left[ (\lambda - x_1)
+ (2\lambda - x_2) + \dots + ((n-1)\lambda - x_{n-1}) \right] \\
&= \frac{n(n+1)}{2} \lambda - n x_n
+ (x_1 + \dots + x_{n-1}).
\end{align*}
This means that for all sufficiently large $n$, say $n \ge N_0$, we have
\[ x_{n} < \frac{x_1 + \dots + x_{n-1}}{n}
\qquad \forall n \ge N_0. \]
In particular, each $x_n$ is the less
than the average of all preceding terms.
Intuitively this means $x_n$ should become close to each other,
since they are also nonnegative.
However, we have a second condition we haven't used yet:
the ``integer'' condition implies
\[
\left\lvert x_{n+1} - x_n \right\rvert
= \left\lvert \lambda - \#\{n \in A \} \right\rvert
> \eps
\]
for some fixed $\eps > 0$,
namely $\eps = \min \left\{ \lambda, 1 - \lambda \right\}$.
Using the fact that consecutive terms differ by some fixed $\eps$,
we will derive a contradiction.
If we let $M$ be the average of $x_1$, \dots, $x_{N_0}$,
then we ought to have
\[ x_n < M \qquad \forall n > N_0. \]
Hence for $n > N_0$ we have $x_n + x_{n+1} < 2M - \eps$,
and so for large enough $n$ the average
must drop to just above $M - \half \eps$.
Thus for some large $N_1 > N_0$, we will have
\[ x_n < M - \frac13 \eps \qquad \forall n > N_1. \]
If we repeat this argument then with a large $N_2 > N_1$, we obtain
\[ x_n < M - \frac23 \eps \qquad \forall n > N_2 \]
and so on and so forth.
This is a clear contradiction.
\begin{remark*}
Note that if $A = \{2,2,3,4,5,\dots\}$ and $\lambda = 1$ then contradiction.
So the condition that $0 < \lambda < 1$ cannot be dropped,
and (by scaling) neither can the condition that $A \subseteq \ZZ$.
\end{remark*}
\begin{remark*}
[Suggested by Zhao Ting-wei]
Despite the relation
\[ x_{n} < \frac{x_1 + \dots + x_{n-1}}{n}
\qquad \forall n \ge N_0 \]
implying that $x_n$ is bounded,
it does not alone imply that $x_n$ converges,
not even to some nonzero value.
Zhao Ting-Wei showed me that one can have a sequence which is zero
``every so often'' yet where the average is nonzero.
A counterexample is given explicitly by
\[
x_n
= \begin{cases}
1000 & n = 1 \\
0 & n \text{ is a power of $10$} \\
1 + \frac 1n & \text{otherwise}
\end{cases}
\]
which does not have a limit.
For completeness, let's check this ---
let $H_n$ denote the $n$'th harmonic number, and compute
\begin{align*}
\sum_1^{n-1} x_n
&= 1000 + (n-1) + H_{n-1} - \sum_{k=1}^{\left\lfloor \log_{10} n \right\rfloor} \left( 1 + \frac{1}{10^k} \right) \\
&> n + 999 + H_{n-1} - \log_{10} n - \left( 1 + \frac{1}{10} + \dots \right) \\
&> n + 997 + H_{n-1} - \log_{10} n > n + 1
\end{align*}
so $1 + \frac 1n < \frac 1n \sum_1^{n-1} x_n$ as needed.
\end{remark*}
\pagebreak
\end{document}