% © Evan Chen
% Downloaded from https://web.evanchen.cc/
\documentclass[11pt]{scrartcl}
\usepackage[sexy]{evan}
\ihead{\footnotesize\textbf{\thetitle}}
\ohead{\footnotesize\href{http://web.evanchen.cc}{\ttfamily web.evanchen.cc},
updated \today}
\title{IMO 2010 Solution Notes}
\date{\today}
\begin{document}
\maketitle
\begin{abstract}
This is a compilation of solutions
for the 2010 IMO.
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
Find all functions $f \colon \RR \to \RR$
such that for all $x,y \in \RR$,
\[ f(\left\lfloor x\right\rfloor y)
= f(x)\left\lfloor f(y)\right\rfloor. \]
\item %% Problem 2
Let $I$ be the incenter of a triangle $ABC$ and let $\Gamma$ be its circumcircle.
Let line $AI$ intersect $\Gamma$ again at $D$.
Let $E$ be a point on arc $\widehat{BDC}$ and $F$ a point on side $BC$ such that
\[ \angle BAF = \angle CAE < \tfrac12 \angle BAC. \]
Finally, let $G$ be the midpoint of $\ol{IF}$.
Prove that $\ol{DG}$ and $\ol{EI}$ intersect on $\Gamma$.
\item %% Problem 3
Find all functions $g : \ZZ_{>0} \to \ZZ_{>0}$ such that
\[ \left( g(m)+n \right)\left( g(n)+m \right) \]
is always a perfect square.
\item %% Problem 4
Let $P$ be a point interior to triangle $ABC$ (with $CA \neq CB$).
The lines $AP$, $BP$ and $CP$ meet again its circumcircle $\Gamma$
at $K$, $L$, $M$, respectively.
The tangent line at $C$ to $\Gamma$ meets the line $AB$ at $S$.
Show that from $SC = SP$ follows $MK = ML$.
\item %% Problem 5
Each of the six boxes $B_1$, $B_2$, $B_3$, $B_4$, $B_5$, $B_6$
initially contains one coin.
The following two types of operations are allowed:
\begin{enumerate}
\ii Choose a non-empty box $B_j$, $1\leq j \leq 5$,
remove one coin from $B_j$ and add two coins to $B_{j+1}$;
\ii Choose a non-empty box $B_k$, $1\leq k \leq 4$,
remove one coin from $B_k$ and swap the contents
(possibly empty) of the boxes $B_{k+1}$ and $B_{k+2}$.
\end{enumerate}
Determine if there exists a finite sequence of operations of the allowed types,
such that the five boxes $B_1$, $B_2$, $B_3$, $B_4$, $B_5$ become empty,
while box $B_6$ contains exactly $2010^{2010^{2010}}$ coins.
\item %% Problem 6
Let $a_1, a_2, a_3, \ldots$ be a sequence of positive real numbers, and $s$ be a positive integer, such that
\[
a_n =
\max \{ a_k + a_{n-k} \mid 1 \leq k \leq n-1 \}
\text{ for all $n > s$}.
\]
Prove there exist positive integers $\ell \leq s$ and $N$, such that
\[
a_n =
a_{\ell} + a_{n - \ell} \text{ for all $n \ge N$}.
\]
\end{enumerate}
\pagebreak
\section{Solutions to Day 1}
\subsection{IMO 2010/1}
\textsl{Available online at \url{https://aops.com/community/p1935849}.}
\begin{mdframed}[style=mdpurplebox,frametitle={Problem statement}]
Find all functions $f \colon \RR \to \RR$
such that for all $x,y \in \RR$,
\[ f(\left\lfloor x\right\rfloor y)
= f(x)\left\lfloor f(y)\right\rfloor. \]
\end{mdframed}
The only solutions are $f(x) \equiv c$,
where $c = 0$ or $1 \le c < 2$.
It's easy to see these work.
Plug in $x=0$ to get $f(0) = f(0) \left\lfloor f(y) \right\rfloor$,
so either
\[ 1 \le f(y) < 2 \quad \forall y
\qquad\text{or}\qquad f(0) = 0 \]
In the first situation,
plug in $y=0$ to get $f(x) \left\lfloor f(0) \right\rfloor = f(0)$,
thus $f$ is constant.
Thus assume henceforth $f(0) = 0$.
Now set $x=y=1$ to get
\[ f(1) = f(1) \left\lfloor f(1) \right\rfloor \]
so either $f(1) = 0$ or $1 \le f(1) < 2$.
We split into cases:
\begin{itemize}
\ii If $f(1) = 0$, pick $x=1$ to get $f(y) \equiv 0$.
\ii If $1 \le f(1) < 2$,
then $y=1$ gives
\[ f(\left\lfloor x \right\rfloor) = f(x) \]
from $y=1$, in particular $f(x) = 0$ for $0 \le x < 1$.
Choose $(x,y) = \left( 2, \half \right)$ to get
$f(1) = f(2) \left\lfloor f\left( \half \right) \right\rfloor = 0$.
\end{itemize}
\pagebreak
\subsection{IMO 2010/2}
\textsl{Available online at \url{https://aops.com/community/p1935927}.}
\begin{mdframed}[style=mdpurplebox,frametitle={Problem statement}]
Let $I$ be the incenter of a triangle $ABC$ and let $\Gamma$ be its circumcircle.
Let line $AI$ intersect $\Gamma$ again at $D$.
Let $E$ be a point on arc $\widehat{BDC}$ and $F$ a point on side $BC$ such that
\[ \angle BAF = \angle CAE < \tfrac12 \angle BAC. \]
Finally, let $G$ be the midpoint of $\ol{IF}$.
Prove that $\ol{DG}$ and $\ol{EI}$ intersect on $\Gamma$.
\end{mdframed}
Let $\ol{EI}$ meet $\Gamma$ again at $K$.
Then it suffices to show that $\ol{KD}$ bisects $\ol{IF}$.
Let $\ol{AF}$ meet $\Gamma$ again at $H$, so $\ol{HE} \parallel \ol{BC}$.
By Pascal theorem on \[ AHEKDD \]
we then obtain that $P = \ol{AH} \cap \ol{KD}$ lies on a line through $I$
parallel to $\ol{BC}$.
Let $I_A$ be the $A$-excenter,
and set $Q = \ol{I_AF} \cap \ol{IP}$, and $T = \ol{AIDI_A} \cap \ol{BFC}$.
Then
\[ -1 = (AI;TI_A) \overset{F} = (IQ;\infty P) \]
where $\infty$ is the point at infinity along $\ol{IPQ}$.
Thus $P$ is the midpoint of $\ol{IQ}$.
Since $D$ is the midpoint of $\ol{II_A}$ by ``Fact 5'',
it follows that $\ol{DP}$ bisects $\ol{IF}$.
\begin{center}
\begin{asy}
size(9cm);
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
pair D = dir(-90);
pair I = incenter(A, B, C);
pair I_A = 2*D-I;
pair E = dir(310);
pair K = -E+2*foot(origin, E, I);
pair H = B*C/E;
pair F = extension(B, C, A, H);
pair G = extension(D, K, I, F);
pair P = extension(A, F, D, G);
pair Q = extension(I, P, I_A, F);
draw(A--B--C--cycle, lightblue);
filldraw(unitcircle, opacity(0.1)+lightcyan, lightblue);
draw(A--I_A, lightblue);
draw(A--H--E--K--D--cycle, orange);
draw((D+0.3)--(D-0.3), orange);
draw(I--Q--I_A, red);
draw(I--F, blue);
pair T = extension(B, C, A, D);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(220));
dot("$I$", I, dir(10));
dot("$I_A$", I_A, dir(I_A));
dot("$E$", E, dir(E));
dot("$K$", K, dir(K));
dot("$H$", H, dir(H));
dot("$F$", F, dir(F));
dot("$G$", G, dir(350));
dot("$P$", P, dir(45));
dot("$Q$", Q, dir(Q));
dot("$T$", T, dir(T));
/* TSQ Source:
!size(9cm);
A = dir 110
B = dir 210
C = dir 330
D = dir -90 R220
I = incenter A B C R10
I_A = 2*D-I
E = dir 310
K = -E+2*foot origin E I
H = B*C/E
F = extension B C A H
G = extension D K I F R350
P = extension A F D G R45
Q = extension I P I_A F
A--B--C--cycle lightblue
unitcircle 0.1 lightcyan / lightblue
A--I_A lightblue
A--H--E--K--D--cycle orange
(D+0.3)--(D-0.3) orange
I--Q--I_A red
I--F blue
T = extension B C A D
*/
\end{asy}
\end{center}
\pagebreak
\subsection{IMO 2010/3, proposed by Gabriel Carroll (USA)}
\textsl{Available online at \url{https://aops.com/community/p1935854}.}
\begin{mdframed}[style=mdpurplebox,frametitle={Problem statement}]
Find all functions $g : \ZZ_{>0} \to \ZZ_{>0}$ such that
\[ \left( g(m)+n \right)\left( g(n)+m \right) \]
is always a perfect square.
\end{mdframed}
For $c \ge 0$, the function $g(n) = n+c$ works;
we prove this is the only possibility.
First, the main point of the problem is that
\begin{claim*}
We have
$g(n) \equiv g(n') \pmod p \implies n \equiv n' \pmod p$.
\end{claim*}
\begin{proof}
Pick a large integer $M$ such that
\[ \nu_p(M+g(n)), \quad \nu_p(M+g(n')) \quad
\text{are both odd}. \]
(It's not hard to see this is always possible.)
Now, since each of
\begin{align*}
\left( M + g(n) \right)&\left( n + g(M) \right) \\
\left( M + g(n') \right)&\left( n' + g(M) \right)
\end{align*}
is a square, we get $g(n) \equiv g(n') \equiv -M \pmod p$.
\end{proof}
This claim implies that
\begin{itemize}
\ii The numbers $g(n)$ and $g(n+1)$ differ by $\pm 1$ for any $n$, and
\ii The function $g$ is injective.
\end{itemize}
It follows $g$ is a linear function with slope $\pm 1$, hence done.
\pagebreak
\section{Solutions to Day 2}
\subsection{IMO 2010/4}
\textsl{Available online at \url{https://aops.com/community/p1936916}.}
\begin{mdframed}[style=mdpurplebox,frametitle={Problem statement}]
Let $P$ be a point interior to triangle $ABC$ (with $CA \neq CB$).
The lines $AP$, $BP$ and $CP$ meet again its circumcircle $\Gamma$
at $K$, $L$, $M$, respectively.
The tangent line at $C$ to $\Gamma$ meets the line $AB$ at $S$.
Show that from $SC = SP$ follows $MK = ML$.
\end{mdframed}
We present two solutions using harmonic bundles.
\paragraph{First solution (Evan Chen)}
Let $N$ be the antipode of $M$, and let $NP$ meet $\Gamma$ again at $D$.
Focus only on $CDMN$ for now (ignoring the condition).
Then $C$ and $D$ are feet of altitudes in $\triangle MNP$;
it is well-known that the circumcircle of $\triangle CDP$
is orthogonal to $\Gamma$
(passing through the orthocenter of $\triangle MPN$).
\begin{center}
\begin{asy}
pair N = dir(100);
pair M = -N;
pair L = dir(150);
pair K = N*N/L;
pair C = dir(122);
pair D = conj(C);
pair P = extension(D, N, M, C);
pair A = -K+2*foot(origin, K, P);
pair B = -L+2*foot(origin, L, P);
pair S = circumcenter(C, P, D);
filldraw(unitcircle, opacity(0.1)+lightgreen, heavygreen);
draw(C--S--D, heavygreen);
draw(A--C--B--S, lightblue);
draw(A--D--B, lightblue);
filldraw(L--N--K--M--cycle, opacity(0.1)+lightred, lightred);
draw(M--N, lightred);
draw(K--L, lightred);
draw(A--K, dashed+orange);
draw(B--L, dashed+orange);
draw(C--M, dashed+orange);
draw(D--N, dashed+orange);
draw(S--P, heavygreen);
draw(arc(S,abs(S-D),-60,60), heavygreen);
dot("$N$", N, dir(N));
dot("$M$", M, dir(M));
dot("$L$", L, dir(170));
dot("$K$", K, dir(K));
dot("$C$", C, dir(80));
dot("$D$", D, dir(280));
dot("$P$", P, dir(P));
dot("$A$", A, dir(220));
dot("$B$", B, dir(B));
dot("$S$", S, dir(S));
/* TSQ Source:
N = dir 100
M = -N
L = dir 150 R170
K = N*N/L
C = dir 122 R80
D = conj(C) R280
P = extension D N M C
A = -K+2*foot origin K P R220
B = -L+2*foot origin L P
S = circumcenter C P D
unitcircle 0.1 lightgreen / heavygreen
C--S--D heavygreen
A--C--B--S lightblue
A--D--B lightblue
L--N--K--M--cycle 0.1 lightred / lightred
M--N lightred
K--L lightred
A--K dashed orange
B--L dashed orange
C--M dashed orange
D--N dashed orange
S--P heavygreen
!draw(arc(S,abs(S-D),-60,60), heavygreen);
*/
\end{asy}
\end{center}
Now, we are given that point $S$ is such that $\ol{SC}$
is tangent to $\Gamma$, and $SC = SP$.
It follows that $S$ is the circumcenter of $\triangle CDP$,
and hence $\ol{SC}$ and $\ol{SD}$ are tangents to $\Gamma$.
Then $-1 = (AB;CD) \overset{P}{=} (KL;MN)$.
Since $\ol{MN}$ is a diameter, this implies $MK = ML$.
\begin{remark*}
I think it's more natural to come up with
this solution in reverse.
Namely, suppose we define the points the other way:
let $\ol{SD}$ be the other tangent, so $(AB;CD) = -1$.
Then project through $P$ to get $(KL;MN) = -1$,
where $N$ is the second intersection of $\ol{DP}$.
However, if $ML = MK$ then $KMLN$ must be a kite.
Thus one can recover the solution in reverse.
\end{remark*}
\paragraph{Second solution (Sebastian Jeon)}
We have \[ SP^2 = SC^2 = SA \cdot SB
\implies
\dang SPA = \dang PBA = \dang LBA = \dang LKA = \dang LKP \]
(the latter half is Reim's theorem).
Therefore $\ol{SP}$ and $\ol{LK}$ are \emph{parallel}.
Now, let $\ol{SP}$ meet $\Gamma$ again at $X$ and $Y$,
and let $Q$ be the antipode of $P$ on $(S)$.
Then
\[ SP^2 = SQ^2 = SX \cdot SY
\implies (PQ;XY) = -1 \implies \angle QCP = 90\dg \]
that $\ol{CP}$ bisects $\angle XCY$.
Since $\ol{XY} \parallel \ol{KL}$,
it follows $\ol{CP}$ bisects to $\angle LCK$ too.
\pagebreak
\subsection{IMO 2010/5, proposed by Netherlands}
\textsl{Available online at \url{https://aops.com/community/p1936917}.}
\begin{mdframed}[style=mdpurplebox,frametitle={Problem statement}]
Each of the six boxes $B_1$, $B_2$, $B_3$, $B_4$, $B_5$, $B_6$
initially contains one coin.
The following two types of operations are allowed:
\begin{enumerate}
\ii Choose a non-empty box $B_j$, $1\leq j \leq 5$,
remove one coin from $B_j$ and add two coins to $B_{j+1}$;
\ii Choose a non-empty box $B_k$, $1\leq k \leq 4$,
remove one coin from $B_k$ and swap the contents
(possibly empty) of the boxes $B_{k+1}$ and $B_{k+2}$.
\end{enumerate}
Determine if there exists a finite sequence of operations of the allowed types,
such that the five boxes $B_1$, $B_2$, $B_3$, $B_4$, $B_5$ become empty,
while box $B_6$ contains exactly $2010^{2010^{2010}}$ coins.
\end{mdframed}
First,
\begin{align*}
(1,1,1,1,1,1) &\to (0,3,1,0,3,1) \to (0,0,7,0,0,7) \\
&\to (0,0,6,2,0,7) \to (0,0,6,1,2,7) \to (0,0,6,1,0,11) \\
&\to (0,0,6,0,11,0) \to (0,0,5,11,0,0).
\end{align*}
and henceforth we ignore boxes $B_1$ and $B_2$,
looking at just the last four boxes;
so we write the current position as $(5,11,0,0)$.
We prove a lemma:
\begin{claim*}
Let $k \ge 0$ and $n > 0$.
From $(k,n,0,0)$ we may reach $(k-1,2^n,0,0)$.
\end{claim*}
\begin{proof}
Working with only the last three boxes for now,
\begin{align*}
(n,0,0) &\to (n-1, 2, 0) \to (n-1, 0, 4) \\
&\to (n-2, 4, 0) \to (n-2, 0, 8) \\
&\to (n-3, 8, 0) \to (n-3, 0, 16) \\
&\to \dots \to (1, 2^{n-1}, 0) \to (1, 0, 2^n) \to (0, 2^n, 0).
\end{align*}
Finally we have $(k,n,0,0) \to (k,0,2^n,0) \to (k-1,2^n, 0,0)$.
\end{proof}
Now from $(5,11,0,0)$ we go as follows:
\begin{align*} (5,11,0,0) &\to (4, 2^{11}, 0, 0)
\to \left(3, 2^{2^{11}}, 0, 0\right)
\to \left(2, 2^{2^{2^{11}}}, 0, 0\right) \\
&\to \left( 1, 2^{2^{2^{2^{11}}}}, 0, 0\right)
\to \left(0, 2^{2^{2^{2^{2^{11}}}}}, 0, 0\right).
\end{align*}
Let $A = 2^{2^{2^{2^{2^{11}}}}} > 2010^{2010^{2010}} = B$.
Then by using move 2 repeatedly on the fourth box
(i.e., throwing away several coins by swapping the empty $B_5$ and $B_6$),
we go from $(0,A,0,0)$ to $(0,B/4,0,0)$.
From there we reach $(0,0,0,B)$.
\pagebreak
\subsection{IMO 2010/6}
\textsl{Available online at \url{https://aops.com/community/p1936918}.}
\begin{mdframed}[style=mdpurplebox,frametitle={Problem statement}]
Let $a_1, a_2, a_3, \ldots$ be a sequence of positive real numbers, and $s$ be a positive integer, such that
\[
a_n =
\max \{ a_k + a_{n-k} \mid 1 \leq k \leq n-1 \}
\text{ for all $n > s$}.
\]
Prove there exist positive integers $\ell \leq s$ and $N$, such that
\[
a_n =
a_{\ell} + a_{n - \ell} \text{ for all $n \ge N$}.
\]
\end{mdframed}
Let \[ w_1 = \frac{a_1}{1}, \quad w_2 = \frac{a_2}{2},
\quad \dots, \quad w_s = \frac{a_s}{s}. \]
(The choice of the letter $w$ is for ``weight''.)
We claim the right choice of $\ell$
is the one maximizing $w_\ell$.
Our plan is to view each $a_n$ as a linear combination
of the weights $w_1, \dots, w_s$ and track their coefficients.
To this end, let's define an \emph{$n$-type}
to be a vector $T = \left< t_1, \dots, t_s\right>$
of nonnegative integers such that
\begin{itemize}
\ii $n = t_1 + \dots + t_s$; and
\ii $t_i$ is divisible by $i$ for every $i$.
\end{itemize}
We then define its \emph{valuation} as $v(T) = \sum_{i=1}^s w_i t_i$.
Now we define a $n$-type to be \emph{valid}
according to the following recursive rule.
For $1 \le n \le s$ the only valid $n$-types are
\begin{align*}
T_1 &= \left< 1, 0, 0, \dots, 0 \right> \\
T_2 &= \left< 0, 2, 0, \dots, 0 \right> \\
T_3 &= \left< 0, 0, 3, \dots, 0 \right> \\
&\vdotswithin= \\
T_s &= \left< 0, 0, 0, \dots, s \right>
\end{align*}
for $n = 1, \dots, s$, respectively.
Then for any $n > s$, an $n$-type is valid
if it can be written as the sum of a valid $k$-type
and a valid $(n-k)$-type, componentwise.
These represent the linear combinations possible in the recursion;
in other words the recursion in the problem is phrased as
\[ a_n = \max_{T \text{ is a valid $n$-type}} v(T). \]
In fact, we have the following description of valid $n$-types:
\begin{claim*}
Assume $n > s$.
Then an $n$-type $\left< t_1, \dots, t_s \right>$ is valid
if and only if either
\begin{itemize}
\ii there exist indices $i < j$ with $i+j > s$,
$t_i \ge i$ and $t_j \ge j$; or
\ii there exists an index $i > s/2$
with $t_i \ge 2i$.
\end{itemize}
\end{claim*}
\begin{proof}
Immediate by forwards induction on $n > s$
that all $n$-types have this property.
The reverse direction is by downwards induction on $n$.
Indeed if $\sum_i \frac{t_i}{i} > 2$,
then we may subtract off on of $\{T_1, \dots, T_s\}$
while preserving the condition;
and the case $\sum_i \frac{t_i}{i} = 2$
is essentially by definition.
\end{proof}
\begin{remark*}
The claim is a bit confusingly stated in its two cases;
really the latter case should be thought of as the situation
$i=j$ but requiring that $t_i/i$ is counted with multiplicity.
\end{remark*}
Now, for each $n > s$ we pick a valid $n$-type $T_n$
with $a_n = v(T_n)$;
if there are ties, we pick one for which the $\ell$th
entry is as large as possible.
\begin{claim*}
For any $n > s$ and index $i \neq \ell$,
the $i$th entry of $T_n$
is at most $2s + \ell i$.
\end{claim*}
\begin{proof}
If not, we can go back $i\ell$ steps to get
a valid $(n-i\ell)$-type $T$
achieved by decreasing the $i$th entry of $T_n$ by $i \ell$.
But then we can add $\ell$ to the $\ell$th entry $i$
times to get another $n$-type $T'$ which obviously
has valuation at least as large,
but with larger $\ell$th entry.
\end{proof}
Now since all other entries in $T_n$ are bounded,
eventually the sequence $(T_n)_{n > s}$
just consists of repeatedly
adding $1$ to the $\ell$th entry, as required.
\begin{remark*}
One big step is to consider $w_k = a_k / k$.
You can get this using wishful thinking
or by examining small cases.
(In addition this normalization makes it easier
to see why the largest $w$ plays an important role,
since then in the definition of type,
the $n$-types all have a sum of $n$.
Unfortunately, it makes the characterization
of valid $n$-types somewhat clumsier too.)
\end{remark*}
\pagebreak
\end{document}