% © Evan Chen
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\ihead{\footnotesize\textbf{\thetitle}}
\ohead{\footnotesize\href{http://web.evanchen.cc}{\ttfamily web.evanchen.cc},
updated \today}
\title{JMO 2015 Solution Notes}
\date{\today}
\begin{document}
\maketitle
\begin{abstract}
This is a compilation of solutions
for the 2015 JMO.
The ideas of the solution are a mix of my own work,
the solutions provided by the competition organizers,
and solutions found by the community.
However, all the writing is maintained by me.
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
Given a sequence of real numbers,
a move consists of choosing two terms
and replacing each with their arithmetic mean.
Show that there exists a sequence of $2015$ distinct real numbers
such that after one initial move
is applied to the sequence --- no matter what move --- there
is always a way to continue with a finite sequence of moves
so as to obtain in the end a constant sequence.
\item %% Problem 2
Solve in integers the equation
\[ x^2+xy+y^2 = \left(\frac{x+y}{3}+1\right)^3. \]
\item %% Problem 3
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 4
Find all functions $f \colon \QQ \to \QQ$ such that
\[ f(x)+f(t)=f(y)+f(z) \]
for all rational numbers $x 0$
\begin{align*}
f(a) + f(a+3d) &= f(a+d) + f(a+2d) \\
f(a-d) + f(a+2d) &= f(a) + f(a+d) \\
\intertext{which imply}
f(a-d) + f(a+3d) &= 2f(a+d).
\end{align*}
Thus we conclude that for arbitrary $x$ and $y$ we have
\[ f(x) + f(y) = 2f\left( \frac{x+y}{2} \right) \]
thus $f$ satisfies Jensen functional equation over $\QQ$, so linear.
The solution can be made to avoid appealing to Jensen's functional equation;
here is a presentation of such a solution based on the official ones.
Let $d > 0$ be a positive integer, and let $n$ be an integer.
Consider the two equations
\begin{align*}
f\left( \frac{2n-1}{2d} \right) + f\left( \frac{2n+2}{2d} \right)
&= f\left( \frac{2n}{2d} \right) + f\left( \frac{2n+1}{2d} \right) \\
f\left( \frac{2n-2}{2d} \right) + f\left( \frac{2n+1}{2d} \right)
&= f\left( \frac{2n-1}{2d} \right) + f\left( \frac{2n}{2d} \right)
\end{align*}
Summing them and simplifying implies that
\[ f\left( \frac{n-1}{d} \right) + f\left( \frac{n+1}{d} \right)
= 2 f \left( \frac nd \right) \]
or equivalently
\[ f\left( \frac nd \right) - f\left( \frac{n-1}{d} \right)
= f\left( \frac{n+1}{d} \right) - f\left( \frac nd \right). \]
This implies that on the set of rational numbers
with denominator dividing $d$, the function $f$ is linear.
In particular, we should have
$f\left( \frac nd \right) = f(0) + \frac nd f(1)$
since $\frac nd$, $0$, $1$ have denominators dividing $d$.
This is the same as saying $f(q) = f(0) + q (f(1)-f(0))$
for any $q \in \QQ$, which is what we wanted to prove.
\pagebreak
\subsection{JMO 2015/5, proposed by Sungyoon Kim}
\textsl{Available online at \url{https://aops.com/community/p4774099}.}
\begin{mdframed}[style=mdpurplebox,frametitle={Problem statement}]
Let $ABCD$ be a cyclic quadrilateral.
Prove that there exists a point $X$ on segment $\ol{BD}$
such that $\angle BAC=\angle XAD$ and $\angle BCA=\angle XCD$
if and only if there exists a point $Y$ on segment $\ol{AC}$
such that $\angle CBD=\angle YBA$ and $\angle CDB=\angle YDA$.
\end{mdframed}
Both conditions are equivalent to $ABCD$ being harmonic.
Here is a complex solution. Extend $U$ and $V$ and shown. Thus $u = bd/a$ and $v = bd/c$.
\begin{center}
\begin{asy}
size(4.5cm);
pair A = Drawing("A", dir(110), dir(110));
pair B = Drawing("B", dir(160), dir(160));
pair D = Drawing("D", dir( 20), dir( 20));
pair C = Drawing("C", dir(240), dir(240));
draw(unitcircle);
draw(A--B--C--D--cycle);
draw(B--D);
pair U = -conj(A);
pair V = -conj(C);
Drawing("U", U, U);
Drawing("V", V, V);
draw(C--U);
draw(A--V);
Drawing("X", extension(C,U,A,V), dir(0));
draw(A--U, dashed);
draw(C--V, dashed);
draw(A--C);
\end{asy}
\end{center}
Note $\ol{AV} \cap \ol{CU}$ lies on the
perpendicular bisector of $\ol{BD}$ unconditionally.
Then $X$ exists as described if and only if
the midpoint of $\ol{BD}$ lies on $\ol{AV}$.
In complex numbers this is $a + v = m + av \ol m$, or
\[
a + \frac{bd}{c} = \frac{b+d}{2} + \frac{abd}{c} \cdot \frac{b+d}{2bd}
\iff 2(ac+bd) = (b+d)(a+c)
\]
which is symmetric.
\pagebreak
\subsection{JMO 2015/6, proposed by Maria Monks Gillespie}
\textsl{Available online at \url{https://aops.com/community/p4774079}.}
\begin{mdframed}[style=mdpurplebox,frametitle={Problem statement}]
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