% © Evan Chen
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\ihead{\footnotesize\textbf{\thetitle}}
\ohead{\footnotesize\href{http://web.evanchen.cc}{\ttfamily web.evanchen.cc},
updated \today}
\title{IMO 2011 Solution Notes}
\date{\today}
\begin{document}
\maketitle
\begin{abstract}
This is a compilation of solutions
for the 2011 IMO.
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 any set $A = \{a_1, a_2, a_3, a_4\}$ of four distinct
positive integers, we denote the sum $a_1+a_2+a_3+a_4$ by $s_A$.
Let $n_A$ denote the number of pairs $(i,j)$ with $1 \le i < j \le 4$
for which $a_i + a_j$ divides $s_A$.
Find all sets $A$ of four distinct positive integers which achieve
the largest possible value of $n_A$.
\item %% Problem 2
Let $\mathcal{S}$ be a finite set of at least two points in the plane.
Assume that no three points of $\mathcal S$ are collinear.
A \emph{windmill} is a process that starts with a
line $\ell$ going through a single point $P \in \mathcal S$.
The line rotates clockwise about the \emph{pivot} $P$ until the first time
that the line meets some other point belonging to $\mathcal S$.
This point, $Q$, takes over as the new pivot,
and the line now rotates clockwise about $Q$,
until it next meets a point of $\mathcal S$.
This process continues indefinitely.
Show that we can choose a point $P$ in $\mathcal S$ and
a line $\ell$ going through $P$ such that the resulting windmill
uses each point of $\mathcal S$ as a pivot infinitely many times.
\item %% Problem 3
Let $f \colon \RR \to \RR$ be a real-valued function
defined on the set of real numbers that satisfies
\[ f(x+y) \leq yf(x) + f(f(x))\]
for all real numbers $x$ and $y$.
Prove that $f(x) = 0$ for all $x \leq 0$.
\item %% Problem 4
Let $n > 0$ be an integer.
We are given a balance and $n$ weights of weight $2^0$, $2^1$, \dots, $2^{n-1}$.
We are to place each of the $n$ weights on the balance, one after another,
in such a way that the right pan is never heavier than the left pan.
At each step we choose one of the weights
that has not yet been placed on the balance,
and place it on either the left pan or the right pan,
until all of the weights have been placed.
Determine the number of ways in which this can be done.
\item %% Problem 5
Let $f \colon \ZZ \to \ZZ_{>0}$ be a function such that
$f(m-n) \mid f(m) - f(n)$ for $m,n \in \ZZ$.
Prove that if $m,n \in \ZZ$ satisfy $f(m) \le f(n)$
then $f(m) \mid f(n)$.
\item %% Problem 6
Let $ABC$ be an acute triangle with circumcircle $\Gamma$.
Let $\ell$ be a tangent line to $\Gamma$, and let $\ell_a$, $\ell_b$, $\ell_c$ be the lines obtained
by reflecting $\ell$ in the lines $BC$, $CA$, and $AB$, respectively.
Show that the circumcircle of the triangle determined by the lines $\ell_a$, $\ell_b$, and $\ell_c$
is tangent to the circle $\Gamma$.
\end{enumerate}
\pagebreak
\section{Solutions to Day 1}
\subsection{IMO 2011/1, proposed by Fernando Campos (MEX)}
\textsl{Available online at \url{https://aops.com/community/p2363530}.}
\begin{mdframed}[style=mdpurplebox,frametitle={Problem statement}]
Given any set $A = \{a_1, a_2, a_3, a_4\}$ of four distinct
positive integers, we denote the sum $a_1+a_2+a_3+a_4$ by $s_A$.
Let $n_A$ denote the number of pairs $(i,j)$ with $1 \le i < j \le 4$
for which $a_i + a_j$ divides $s_A$.
Find all sets $A$ of four distinct positive integers which achieve
the largest possible value of $n_A$.
\end{mdframed}
There are two curves of solutions,
namely $\{x,5x,7x,11x\}$ and $\{x,11x,19x,29x\}$,
for any positive integer $x$,
achieving $n_A = 4$ (easy to check).
We'll show that $n_A \le 4$
and equality holds only in one of the curves.
Let $A = \{a < b < c < d\}$.
\begin{claim*}
We have $n_A \le 4$ with equality iff
\[ a+b \mid c+d, \quad a+c \mid b+d, \quad a+d = b+c. \]
\end{claim*}
\begin{proof}
Note $a+b \mid s_A \iff a+b \mid c+d$ etc.
Now $c+d \nmid a+b$ and $b+d \nmid a+c$ for size reasons,
so we already have $n_A \le 4$;
moreover $a+d \mid b+c$ and $b+c \mid a+d$
if and only if $a+d = b+c$.
\end{proof}
We now show the equality curve is the one above.
\[ a+c \mid b+d \iff a+c \mid -a+2b+c \iff a+c \mid 2(b-a). \]
Since $a+c > |b-a|$, so we must have $a+c=2(b-a)$.
So we now have
\begin{align*}
c &= 2b-3a \\
d &= b+c-a = 3b+c-4a.
\end{align*}
The last condition is
\[ a+b \mid c+d = 5b-7a \iff a+b \mid 12a. \]
Now, let $x = \gcd(a,b)$.
The expressions for $c$ and $d$ above imply that
$x \mid c,d$ so we may scale down so that $x = 1$.
Then $\gcd(a+b,a) = \gcd(a,b) = 1$ and so $a+b \mid 12$.
We have $c > b$, so $3a < b$.
The only pairs $(a,b)$ with $3a < 2b$, $\gcd(a,b) = 1$
and $a+b \mid 12$
are $(a,b) \in \left\{ (1,5), (1,11) \right\}$
which give the solutions earlier.
\pagebreak
\subsection{IMO 2011/2, proposed by Geoff Smith (UNK)}
\textsl{Available online at \url{https://aops.com/community/p2363537}.}
\begin{mdframed}[style=mdpurplebox,frametitle={Problem statement}]
Let $\mathcal{S}$ be a finite set of at least two points in the plane.
Assume that no three points of $\mathcal S$ are collinear.
A \emph{windmill} is a process that starts with a
line $\ell$ going through a single point $P \in \mathcal S$.
The line rotates clockwise about the \emph{pivot} $P$ until the first time
that the line meets some other point belonging to $\mathcal S$.
This point, $Q$, takes over as the new pivot,
and the line now rotates clockwise about $Q$,
until it next meets a point of $\mathcal S$.
This process continues indefinitely.
Show that we can choose a point $P$ in $\mathcal S$ and
a line $\ell$ going through $P$ such that the resulting windmill
uses each point of $\mathcal S$ as a pivot infinitely many times.
\end{mdframed}
Orient $\ell$ in some direction,
and color the plane such that its left half is red
and right half is blue.
The critical observation is that:
\begin{claim*}
The number of points on the red side of $\ell$ does not change,
nor does the number of points on the blue side
(except at a moment when $\ell$ contains two points).
\end{claim*}
Thus, if $|\mathcal S| = n+1$,
it suffices to pick the initial configuration
so that there are $\left\lfloor n/2 \right\rfloor$
red and $\left\lceil n/2 \right\rceil$ blue points.
Then when the line $\ell$ does a full $180\dg$ rotation,
the red and blue sides ``switch'',
so the windmill has passed through every point.
(See official shortlist for verbose write-up;
this is deliberately short to make a point.)
\pagebreak
\subsection{IMO 2011/3, proposed by Igor Voronovich, BLR}
\textsl{Available online at \url{https://aops.com/community/p2363539}.}
\begin{mdframed}[style=mdpurplebox,frametitle={Problem statement}]
Let $f \colon \RR \to \RR$ be a real-valued function
defined on the set of real numbers that satisfies
\[ f(x+y) \leq yf(x) + f(f(x))\]
for all real numbers $x$ and $y$.
Prove that $f(x) = 0$ for all $x \leq 0$.
\end{mdframed}
We begin by rewriting the given as
\[ f(z) \le (z-x)f(x) + f(f(x)) \quad
\forall x,z \in \RR \qquad (\heartsuit) \]
(which is better anyways since control over inputs to $f$
is more valuable).
We start by eliminating the double $f$:
let $z = f(w)$ to get
\[ f(f(w)) \le (f(w)-x)f(x) + f(f(x)) \]
and then use the \emph{symmetry trick} to write
\[ f(f(x)) \le (f(x)-w)f(w) + f(f(w)) \]
so that when we sum we get
\[ wf(w) + xf(x) \le 2f(x)f(w). \]
%(We remark now that if we set $w = 0$,
%we find that $xf(x) \le 2f(x)f(0)$ for all $x$;
%since we know $f(0)$ should be zero, this is pretty good.)
Next we use \emph{cancellation trick}:
set $w = 2f(x)$ in the above to get
\[ xf(x) \le 0 \quad \forall x \in \RR. \qquad (\spadesuit) \]
\begin{claim*}
For every $p \in \RR$, we have $f(p) \le 0$.
\end{claim*}
\begin{proof}
Assume $f(p) > 0$ for some $p \in \RR$.
Then for any negative number $z$,
\[ 0 \overset{(\spadesuit)}{\le} f(z)
\overset{(\heartsuit)}{\le} (z-p)f(p) + f(f(p)). \]
which is false if we let $z \to -\infty$.
\end{proof}
Together with $(\spadesuit)$ we derive $f(x) = 0$ for $x < 0$.
Finally, letting $x$ and $z$ be any negative numbers
in $(\heartsuit)$, we get $f(0) \ge 0$, so $f(0) = 0$ too.
\begin{remark*}
As another corollary of the claim, $f(f(x)) = 0$ for all $x$.
\end{remark*}
\begin{remark*}
A nontrivial example of a working $f$
is to take
\[ f(x) = \begin{cases}
-\exp(\exp(\exp(x))) & x > 0 \\
0 & x \le 0.
\end{cases} \]
or some other negative function growing rapidly in absolute value
for $x > 0$.
\end{remark*}
\pagebreak
\section{Solutions to Day 2}
\subsection{IMO 2011/4, proposed by Morteza Saghafian (IRN)}
\textsl{Available online at \url{https://aops.com/community/p2365036}.}
\begin{mdframed}[style=mdpurplebox,frametitle={Problem statement}]
Let $n > 0$ be an integer.
We are given a balance and $n$ weights of weight $2^0$, $2^1$, \dots, $2^{n-1}$.
We are to place each of the $n$ weights on the balance, one after another,
in such a way that the right pan is never heavier than the left pan.
At each step we choose one of the weights
that has not yet been placed on the balance,
and place it on either the left pan or the right pan,
until all of the weights have been placed.
Determine the number of ways in which this can be done.
\end{mdframed}
The answer is $a_n = (2n-1)!!$.
We refer to what we're counting as a \emph{valid $n$-sequence}:
an order of which weights to place,
and whether to place them on the left or right pan.
We use induction, with $n=1$ being obvious.
Now consider the weight $2^0 = 1$.
\begin{itemize}
\ii If we delete it from any valid $n$-sequence,
we get a valid $(n-1)$-sequence with all weights doubled.
\ii Given a valid $(n-1)$-sequence with all weights doubled,
we may insert $2^0 = 1$ it into $2n-1$ ways.
Indeed, we may insert it anywhere, and designate it either left or right,
except we may not designate right if we choose to insert
$2^0 = 1$ at the very beginning.
\end{itemize}
Consequently, we have that
\[ a_n = (2n-1) \cdot a_{n-1}. \]
Since $a_1 = 1$, the conclusion follows.
\begin{remark*}
[Gabriel Levin]
An alternate approach can be done by considering the heaviest weight $2^{n-1}$
instead of the lightest weight $2^0=1$.
This gives a more complicated recursion:
\[ a_n = \sum_{k=0}^{n-1} \binom{n-1}{k} 2^{n-k-1} (n-k-1)! a_k \]
by summing over the index $k$ corresponding to the number of weights
placed before the heaviest weight $2^{n-1}$ is placed.
This recursion can then be rewritten as
\begin{align*}
a_n &= \sum_{k=0}^{n-1} \binom{n-1}{k} 2^{n-k-1} (n-k-1)! a_k \\
&= (n-1)! \sum_{k=0}^{n-1} \frac{2^{n-k-1}a_k}{k!} \\
&= 2(n-1)! \sum_{k=0}^{n-1} \frac{2^{(n-1)-k-1}a_k}{k!} \\
&= 2(n-1)(n-2)! (\sum_{k=0}^{(n-1)-1} \frac{2^{(n-1)-k-1}a_k}{k!}) + 2(n-1)!
\frac{\frac12 a_{n-1}}{(n-1)!}
&= 2(n-1) a_{n-1} + a_{n-1}
&= (2n-1)a_{n-1}
\end{align*}
and since $a_1 = 1$, we have $a_n = (2n-1)!!$ by induction on $n$.
\end{remark*}
\pagebreak
\subsection{IMO 2011/5, proposed by Mahyar Sefidgaran (IRN)}
\textsl{Available online at \url{https://aops.com/community/p2365041}.}
\begin{mdframed}[style=mdpurplebox,frametitle={Problem statement}]
Let $f \colon \ZZ \to \ZZ_{>0}$ be a function such that
$f(m-n) \mid f(m) - f(n)$ for $m,n \in \ZZ$.
Prove that if $m,n \in \ZZ$ satisfy $f(m) \le f(n)$
then $f(m) \mid f(n)$.
\end{mdframed}
Let $P(m,n)$ denote the given assertion.
First, we claim $f$ is even.
This is straight calculation:
\begin{itemize}
\ii $P(x,0) \implies f(x) \mid f(x)-f(0)
\implies f(x) \mid M \coloneqq f(0)$.
\ii $P(0,x) \implies f(-x) \mid M-f(x) \implies
f(-x) \mid f(x)$.
Analogously, $f(x) \mid f(-x)$.
So $f(x) = f(-x)$ and $f$ is even.
\end{itemize}
\begin{claim*}
Let $x$, $y$, $z$ be integers with $x+y+z=0$.
Then among $f(x)$, $f(y)$, $f(z)$,
two of them are equal and divide the third.
\end{claim*}
\begin{proof}
Let $a = f(\pm x)$, $b = f(\pm y)$, $c = f(\pm z)$
be positive integers.
Note that
\begin{align*}
a &\mid b-c \\
b &\mid c-a \\
c &\mid a-b
\end{align*}
from $P(y,-z)$ and similarly.
WLOG $c = \max(a,b,c)$; then $c > |a-b|$
so $a=b$. Thus $a=b \mid c$ from the first two.
\end{proof}
This implies the problem,
by taking $x$ and $y$ in the previous claim
to be the integers $m$ and $n$.
\begin{remark*}
At \url{https://aops.com/community/c6h418981p2381909},
Davi Medeiros gives the following characterization
of functions $f$ satisfying the hypothesis.
Pick $f(0)$, $k$ positive integers,
a chain $d_1 \mid d_2 \mid \dots \mid d_k$ of divisors of $f(0)$,
and positive integers $a_1,a_2,\dots,a_{k-1}$,
greater than $1$ (if $k=1$, $a_i$ doesn't exist, for every $i$).
We'll define $f$ as follows:
\begin{itemize}
\ii $f(n)=d_1$, for every integer $n$ that is not divisible by $a_1$;
\ii $f(a_1n)=d_2$, for every integer $n$ that is not divisible by $a_2$;
\ii $f(a_1a_2n)=d_3$, for every integer $n$ that is not divisible by $a_3$;
\ii $f(a_1a_2a_3n)=d_4$, for every integer $n$ that is not divisible by $a_4$;
\ii \dots
\ii $f(a_1a_2 \dots a_{k-1}n)=d_k$, for every integer $n$;
\end{itemize}
\end{remark*}
\pagebreak
\subsection{IMO 2011/6, proposed by Japan}
\textsl{Available online at \url{https://aops.com/community/p2365045}.}
\begin{mdframed}[style=mdpurplebox,frametitle={Problem statement}]
Let $ABC$ be an acute triangle with circumcircle $\Gamma$.
Let $\ell$ be a tangent line to $\Gamma$, and let $\ell_a$, $\ell_b$, $\ell_c$ be the lines obtained
by reflecting $\ell$ in the lines $BC$, $CA$, and $AB$, respectively.
Show that the circumcircle of the triangle determined by the lines $\ell_a$, $\ell_b$, and $\ell_c$
is tangent to the circle $\Gamma$.
\end{mdframed}
This is a hard problem with many beautiful solutions.
The following solution is not very beautiful but not too hard to find during an olympiad,
as the only major insight it requires is the construction of $A_2$, $B_2$, and $C_2$.
\begin{center}
\begin{asy}
size(11cm);
pair A = dir(110);
dot("$A$", A, dir(A));
pair B = dir(195);
dot("$B$", B, dir(160));
pair C = dir(325);
dot("$C$", C, 1.4*dir(30));
pair P = dir(270);
dot("$P$", P, dir(P));
draw(unitcircle);
draw(A--B--C--cycle);
pair U = P+(2,0);
pair V = 2*P-U;
pair X_1 = reflect(B,C)*P;
pair Y_1 = reflect(C,A)*P;
pair Z_1 = reflect(A,B)*P;
pair X_2 = extension(B, C, U, V);
dot(X_2);
pair Y_2 = extension(C, A, U, V);
dot(Y_2);
pair Z_2 = extension(A, B, U, V);
dot(Z_2);
draw(B--Z_2, dotted);
draw(C--Y_2, dotted);
draw(C--X_2, dotted);
draw(X_2--Z_2);
pair A_1 = extension(Y_1, Y_2, Z_1, Z_2);
dot("$A_1$", A_1, dir(A_1));
pair B_1 = extension(Z_1, Z_2, X_1, X_2);
dot("$B_1$", B_1, dir(B_1));
pair C_1 = extension(X_1, X_2, Y_1, Y_2);
dot("$C_1$", C_1, dir(50));
draw(A_1--B_1--C_1--cycle, black+1);
draw(C_1--X_2, dotted);
pair O_1 = circumcenter(A_1, B_1, C_1);
draw(arc(O_1, abs(O_1-A_1), -80, 140));
pair A_2 = A*A/P;
dot("$A_2$", A_2, dir(-20));
pair B_2 = B*B/P;
dot("$B_2$", B_2, dir(130));
pair C_2 = C*C/P;
dot("$C_2$", C_2, dir(C_2));
draw(A_2--B_2--C_2--cycle, black+1);
pair T = extension(A_1, A_2, B_1, B_2);
dot("$T$", T, dir(T));
draw(T--A_1, dashed);
draw(T--B_1, dashed);
draw(T--C_1, dashed);
/*
A = dir 110
B = dir 195
C = dir 325
P = dir 270
unitcircle
A--B--C--cycle blue
U := P+(2,0)
V := 2*P-U
X = reflect(B,C)*P
Y = reflect(C,A)*P
Z = reflect(A,B)*P
Y--Z heavygreen
X1 := extension B C U V
Y1 := extension C A U V
Z1 := extension A B U V
Line Y1 Z1
A_1 = extension Y Y1 Z Z1
B_1 = extension Z Z1 X X1
C_1 = extension X X1 Y Y1
A_1--B_1--C_1--cycle red
circumcircle A_1 B_1 C_1
circumcircle B Z X dotted
circumcircle C X Y dotted
circumcircle A Y Z dotted
M = extension A_1 A*A/P B_1 B*B/P
*/
\end{asy}
\end{center}
We apply complex numbers with $\omega$ the unit circle and $p=1$. Let $A_1 = \ell_B \cap \ell_C$, and let $a_2 = a^2$ (in other words, $A_2$ is the reflection of $P$ across the diameter of $\omega$ through $A$). Define the points $B_1$, $C_1$, $B_2$, $C_2$ similarly.
We claim that $\ol{A_1A_2}$, $\ol{B_1B_2}$, $\ol{C_1C_2}$ concur at a point on $\Gamma$.
We begin by finding $A_1$. If we reflect the points $1+i$ and $1-i$ over $\ol{AB}$, then we get two points $Z_1$, $Z_2$ with
\begin{align*}
z_1 &= a+b-ab(1-i) = a+b-ab+abi \\
z_2 &= a+b-ab(1+i) = a+b-ab-abi. \\
\intertext{Therefore,}
z_1 - z_2 &= 2abi \\
\ol{z_1}z_2 - \ol{z_2}{z_1}
&= -2i \left( a+b+\frac1a+\frac1b-2 \right).
\end{align*}
Now $\ell_C$ is the line $\ol{Z_1Z_2}$,
so with the analogous equation $\ell_B$ we obtain:
\begin{align*}
a_1 &= \frac{ -2i\left( a+b+\frac1a+\frac1b-2 \right)\left( 2ac i \right) +
2i\left( a+c+\frac1a+\frac1c-2 \right)(2abi) }
{ \left( -\frac{2}{ab}i \right)
\left( 2ac i \right) - \left( -\frac{2}{ac}i \right) \left( 2abi \right)} \\
&= \frac{\left[ c-b \right]a^2 + \left[ \frac cb - \frac bc - 2c + 2b \right]a + (c-b) }{\frac cb - \frac bc} \\
&= a + \frac{(c-b)\left[ a^2-2a+1 \right]}{(c-b)(c+b)/bc} \\
&= a + \frac{bc}{b+c} (a-1)^2.
\end{align*}
Then the second intersection of $\ol{A_1A_2}$ with $\omega$ is given by
\begin{align*}
\frac{a_1-a_2}{1-a_2\ol{a_1}}
&= \frac{a+\frac{bc}{b+c}(a-1)^2-a^2}{1-a-a^2 \cdot \frac{(1-1/a)^2}{b+c}} \\
&= \frac{a + \frac{bc}{b+c}(1-a)}{1 - \frac{1}{b+c}(1-a)} \\
&= \frac{ab+bc+ca - abc}{a+b+c-1}.
\end{align*}
Thus, the claim is proved.
Finally, it suffices to show $\ol{A_1B_1} \parallel \ol{A_2B_2}$.
One can also do this with complex numbers;
it amounts to showing $a^2-b^2$, $a-b$, $i$
(corresponding to $\ol{A_2 B_2}$, $\ol{A_1 B_1}$, $\ol{PP}$)
have their arguments an arithmetic progression, equivalently
\[ \frac{(a-b)^2}{i(a^2-b^2)} \in \RR
\iff
\frac{(a-b)^2}{i(a^2-b^2)}
= \frac{\left( \frac 1a-\frac1b \right)^2}
{\frac1i\left(\frac{1}{a^2}-\frac{1}{b^2}\right)}
\]
which is obvious.
\begin{remark*}
One can use directed angle chasing for this last part too.
Let $\ol{BC}$ meet $\ell$ at $K$ and $\ol{B_2C_2}$ meet $\ell$ at $L$.
Evidently
\begin{align*}
-\dang B_2LP &= \dang LPB_2 + \dang PB_2L \\
&= 2 \dang KPB + \dang PB_2C_2 \\
&= 2 \dang KPB + 2\dang PBC \\
&= -2\dang PKB \\
&= \dang PKB_1 \\
\end{align*}
as required.
\end{remark*}
\pagebreak
\end{document}