Didactic to Constructive: Turning Expert Solutions into Learnable Reasoning
Paper • 2602.02405 • Published • 2
problem_id stringlengths 16 19 | problem stringlengths 69 2.04k | solution stringlengths 60 23.9k |
|---|---|---|
EGMO-2012-notes_1 | Let $ABC$ be a triangle with circumcenter $O$.
The points $D$, $E$, $F$ lie in the interiors
of the sides $BC$, $CA$, $AB$ respectively,
such that $\ol{DE} \perp \ol{CO}$ and $\ol{DF} \perp \ol{BO}$.
Let $K$ be the circumcenter of triangle $AFE$.
Prove that the lines $\ol{DK}$ and $\ol{BC}$ are perpendicular. | First, note
$\dang EDF = 180\dg - \dang BOC = 180\dg - 2A$,
so $\dang FDE = 2A$.
\begin{center}
\begin{asy}
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);
pair O = Drawing("O", origin, dir(45));... |
EGMO-2012-notes_2 | Let $n$ be a positive integer.
Find the greatest possible integer $m$, in terms of $n$,
with the following property: a table with $m$ rows and $n$ columns can be
filled with real numbers in such a manner that for any
two different rows $\left[ a_1, a_2, \dots, a_n \right]$
and $\left[ b_1, b_2, \dots, b_n \right]$ the ... | \paragraph{Answer.} The largest $m$ is $m(n) = 2^n$.
\paragraph{Construction.} Choose the table with all $2^n$ binary vectors.
\paragraph{Bound.}
We will proceed by induction on $n$, with the base case $n=1$ being immediate.
Take the first column, and assume the entries lie in some interval $I = [r, r+1]$.
Then the r... |
EGMO-2012-notes_3 | Solve over $\RR$ the functional equation
\[ f\left( yf(x + y) + f(x) \right) = 4x + 2yf(x + y). \] | The only solution is $f(x) \equiv 2x$ which obviously works.
Let $P(x,y)$ be the given condition. Then:
\begin{itemize}
\ii Note $P(x,0) \implies f(f(x)) = 4x$; in particular $f$ is bijective.
\begin{itemize}
\ii \dots This also implies $f(4x) = f(f(f(x))) = 4f(x)$.
\ii Taking $x=0$ gives $f(0) = 4f(0) \im... |
EGMO-2012-notes_4 | A set $A$ of integers is called \emph{sum-full} if $A \subseteq A + A$.
A set $A$ of integers is said to be \emph{zero-sum-free} if $0$
is the only integer that cannot be expressed as the
sum of the elements of a finite nonempty subset of $A$.
Does there exist a sum-full zero-sum-free set of integers? | The answer is YES, such a set exists, and is given by
\[ A = \left\{ 1, -2, 3, -5, 8, -13, 21, -34, 55, \dots \right\}. \]
To prove every number can be expressed as a
(possibly empty) subset of $A$, we have the following stronger claim.
\begin{claim*}
Let $N$ be an integer (possibly zero or negative).
Let $F$ deno... |
EGMO-2012-notes_5 | The numbers $p$ and $q$ are prime and satisfy
\[ \frac{p}{p+1} + \frac{q+1}{q} = \frac{2n}{n+2} \]
for some positive integer $n$.
Find all possible values of $q-p$. | We clean up the numerators as
\[ \left( 1-\frac{1}{p+1} \right) + \left( 1 + \frac{1}{q} \right) = 2 - \frac{4}{n+2} \]
which simplifies as
\[ \frac{4}{n+2} = \frac{1}{p+1} - \frac{1}{q} \]
which reduces as
\[ \frac{4}{n+2} = \frac{q-p-1}{q(p+1)} \]
At this point $n$ is a free parameter.
So, this motivates us to rewrit... |
EGMO-2012-notes_6 | There are infinitely many people registered on the social network Mugbook.
Some pairs of (different) users are registered as friends, but each person has only finitely many friends.
Every user has at least one friend.
(Friendship is symmetric; that is, if $A$ is a friend of $B$, then $B$ is a friend of $A$.)
Each perso... | First note the following statement is true by induction:
\begin{claim*}
Someone who is an $n$-best friend is also a $k$-best friend for all $1 \le k < n$.
\end{claim*}
\begin{proof}
Induction on $(k,n)$.
\end{proof}
For part (a), suppose Dan is a popular person.
Let the friends of Dan be $P_1$, $P_2$, \dots, $P_m$... |
EGMO-2012-notes_7 | Let $ABC$ be an acute-angled triangle with circumcircle $\Gamma$ and orthocenter $H$. Let $K$ be a point of $\Gamma$ on the other side of $\ol{BC}$ from $A$. Let $L$ be the reflection of $K$ across $\ol{AB}$, and let $M$ be the reflection of $K$ across $\ol{BC}$. Let $E$ be the second point of intersection of $\Gamma$ ... | \begin{center}
\begin{asy}
size(9cm);
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
draw(A--B--C--cycle);
pair Ap = -B*C/A;
pair Cp = -A*B/C;
draw(A--Ap);
draw(C--Cp);
draw(unitcircle);
pair K = dir(230);
pair M = reflect(B,C)*K;
pair blah = reflect(A,B... |
EGMO-2012-notes_8 | A word is a finite sequence of letters from some alphabet.
A word is \emph{repetitive} if it is a concatenation of at least two identical subwords
(for example, $ababab$ and $abcabc$ are repetitive, but $ababa$ and $aabb$ are not).
Prove that if a word has the property that swapping any two adjacent letters
makes the w... | A word with the property that swapping any two adjacent letters
makes the word repetitive, but for which all letters the same,
will called a \emph{unicorn}.
The problem asks to show there are no unicorns.
So assume for contradiction there exists a unicorn.
We proceed in three steps.
\paragraph{Step 1. Reduction to bi... |
EGMO-2013-notes_1 | The side $BC$ of the triangle $ABC$
is extended beyond $C$ to $D$ so that $CD = BC$.
The side $CA$ is extended beyond $A$ to $E$ so that $AE = 2CA$.
Prove that if $AD=BE$ then the triangle $ABC$ is right-angled. | Let ray $DA$ meet $\ol{BE}$ at $M$.
Consider the triangle $EBD$.
Since the point lies on median $\ol{EC}$, and $EA = 2AC$,
it follows that $A$ is the centroid of $\triangle EBD$.
\begin{center}
\begin{asy}
pair A = dir(50);
pair B = dir(180);
pair C = dir(0);
draw(A--B--C--cycle, black+1.4);
pair D = 2*C-B;
pair E = 3... |
EGMO-2013-notes_2 | Determine all integers $m$ for which the $m \times m$ square can be dissected
into five rectangles, the side lengths of which are
the integers $1$, $2$, \dots, $10$ in some order. | The answer is that this is possible if and only if $m = 11$ or $m = 13$.
\paragraph{Constructions for $m=11$ and $m=13$.}
See the figures below.
\begin{center}
\begin{asy}
size(12cm);
picture p11, p13;
pen b = black+2.5;
filldraw(p11, box((0,0), (11,11)), palegreen, b);
filldraw(p11, box((0,0), (8,5)), palecyan, b);
f... |
EGMO-2013-notes_3 | Let $n$ be a positive integer.
\begin{enumerate}
\ii[(a)] Prove that there exists a set $S$ of $6n$ positive integers
such that the least common multiple of any two is at most $32n^2$.
\ii[(b)] Show that every set $T$ of $6n$ positive integers
contains two elements with least common multiple exceeding $9n^2$.
\... | For part (a), let
\[ S = \{1, 2, \dots, 4n\} \cup \{4n+2, 4n+4, \dots, 8n\} \]
which works.
For (b),
the main idea is as follows.
Let $x_0 < x_1 < \dots < x_m$ be these integers.
Assume for contradiction the LCM's are less than $L$.
The main idea is the estimate
\[ L \ge \frac{x_i x_{i+1}}{\gcd(x_i, x_{i+1})} \ge \fra... |
EGMO-2013-notes_4 | Find all positive integers $a$ and $b$ for which there are three consecutive integers at which the polynomial $P(n) = \frac1b(n^5+a)$ takes integer values. | The answer is
\begin{itemize}
\ii Either $b = 1$; or
\ii $a \equiv \pm 1 \pmod{11}$ and $b = 11$ (and $a > 0$).
\end{itemize}
We consider only the case $b>1$.
Suppose that $P$ is a prime power dividing $b$.
Evidently $P$ is not even.
Then there exists an $n$ for which
\[ n^5 \equiv (n+1)^5 \equiv (n-1)^5 \pmod{P}.... |
EGMO-2013-notes_5 | Let $\Omega$ be the circumcircle of the triangle $ABC$.
The circle $\omega$ is tangent to the sides $AC$ and $BC$,
and it is internally tangent to the circle $\Omega$ at the point $P$.
A line parallel to $AB$ intersecting the
interior of triangle $ABC$ is tangent to $\omega$ at $Q$.
Prove that $\angle ACP = \angle QCB$... | First, let us extend $\ol{AQ}$ to meet $\ol{BC}$ at $Q_1$.
By homothety, we see that $Q_1$ is just the
contact point of the $A$-excircle with $\ol{BC}$.
\begin{center}
\begin{asy}
size(7cm);
pair A = Drawing("A", dir(100), dir(100));
pair B = Drawing("B", dir(210), dir(190));
pair C = Drawing("C", dir(330), di... |
EGMO-2013-notes_6 | Snow White and the Seven Dwarves are living in their house in the forest.
On each of 16 consecutive days, some of the dwarves worked in the diamond mine
while the remaining dwarves collected berries in the forest.
No dwarf performed both types of work on the same day.
On any two different (not necessarily consecutive) ... | Let $Q_n$ denote the vector space $\left\{ 0,1 \right\}^n$.
For a vector $v$, let $v[i]$ denote the $i$th component.
We may identify each day with a vector $v_k$,
where $v_k[i] = 0$ if dwarf $i$ worked in the diamond mine,
and $v_k[i] = 1$ otherwise.
Let $V = \left\{ v_1, v_2, \dots, v_{16} \right\}$
be the subset of $... |
EGMO-2014-notes_1 | Determine all real constants $t$ such that whenever $a$, $b$ and $c$ are the lengths of sides of a triangle, then so are $a^2+bct$, $b^2+cat$, $c^2+abt$. | \paragraph{Answer.}
We will show the answer is exactly \[ 2/3 \le t \le 2. \]
\paragraph{Proof.}
Apply the substitution $a=y+z$, $b=z+x$, $c=x+y$.
We need the inequality
\begin{align*}
a^2 + bct & < b^2 + ca t + c^2 + ab t \\
(y+z)^2 + (x+y)(x+z) t &< (x+y)^2+(x+z)^2 + (y+z)(2x+y+z) t \\
\iff [x^2-(xy+xz+yz+y^2+... |
EGMO-2014-notes_2 | Let $D$ and $E$ be points in the interiors of sides $AB$ and $AC$,
respectively, of a triangle $ABC$, such that $DB = BC = CE$.
Let the lines $CD$ and $BE$ meet at $F$.
Prove that the incenter $I$ of triangle $ABC$,
the orthocenter $H$ of triangle $DEF$ and
the midpoint $M$ of the arc $BAC$ of the circumcircle of trian... | \paragraph{First solution (Cynthia Du).}
Let $BI$ and $CI$ meet the circumcircle again at $M_B$, $M_C$.
Observe that we have the spiral congruence
\[ \triangle MDB \sim \triangle MEC \]
from $\dang MBD = \dang MBA = \dang MCA = \dang MCE$
and $BD = EC$, $BM = CM$.
That is, $M$ is the Miquel point of $BDEC$.
\begin{cen... |
EGMO-2014-notes_3 | We denote the number of positive divisors of a positive integer $m$ by $d(m)$
and the number of distinct prime divisors of $m$ by $\omega(m)$.
Let $k$ be a positive integer.
Prove that there exist infinitely many positive integers $n$
such that $\omega(n) = k$ and
$d(n)$ does not divide $d(a^2+b^2)$
for any positive in... | Let $n = 2^{p-1}t$, where $t \equiv 5 \pmod 6$, $\omega(t) = k-1$,
and $p \gg t$ is a sufficiently large prime.
Let $a+b=n$ and $a^2+b^2=c$.
We claim that $p \nmid d(c)$,
which solves the problem since $p \mid d(n)$.
First, note that $3 \nmid a^2+b^2$, since $3 \nmid n$.
Next, note that $c < 2n^2 < 5^{p-1}$ (since $p ... |
EGMO-2014-notes_4 | Determine all positive integers $n\geq 2$ for which there exist integers
$x_1$, $x_2$, \dots, $x_{n-1}$ satisfying the condition that
if $0<i<n$, $0<j<n$, $i\neq j$ and $n$ divides $2i+j$, then $x_i<x_j$. | The answer is $n = 2^k$ and $n = 3 \cdot 2^k$, for each $k \ge 0$ (excluding $n=1$).
We work with the set $S = \{1, 2, \dots, n-1\} \bmod n$
of nonzero residues modulo $n$ instead.
We define the relation $\prec$ on $S$ to mean that
$2i + j \equiv 0 \pmod n$ and $i \neq j$, for $i,j \in S$.
Then the problem satisfies t... |
EGMO-2014-notes_5 | Let $n$ be a positive integer.
We have $n$ boxes where each box contains a non-negative number of pebbles.
In each move we are allowed to take two pebbles from a box we choose,
throw away one of the pebbles and put the other pebble in another box we choose.
An initial configuration of pebbles is called \emph{solvable} ... | The point of the problem is to characterize all the solvable configurations.
We claim that it is given by the following:
\begin{claim*}
A configuration $(a_1, \dots, a_n)$ is solvable if and only if
\[ \sum_{i=1}^n \left\lceil \frac{a_i}{2} \right\rceil \ge n. \]
\end{claim*}
\begin{proof}
The proof is by induct... |
EGMO-2014-notes_6 | Solve over $\RR$ the functional equation
\[ f(y^2+2xf(y)+f(x)^2) = (y+f(x)) (x+f(y)). \]
\end{enumerate} | A key motivation throughout the problem
is that the left-hand side is asymmetric
while the right-hand side is symmetric.
Thus any time we plug in two values for $x$ and $y$
we will also plug in the opposite pair.
Once $f$ is injective this will basically kill the problem.
First, we prove the following.
\begin{lemma*}
... |
EGMO-2015-notes_1 | Let $\triangle ABC$ be an acute-angled triangle,
and let $D$ be the foot of the altitude from $C$.
The angle bisector of $\angle ABC$ intersects $CD$ at $E$ and
meets the circumcircle $\omega$ of $\triangle ADE$ again at $F$.
If $\angle ADF = 45^{\circ}$, show that $CF$ is tangent to $\omega$. | \begin{center}
\begin{asy}
pair A = dir(0);
pair B = dir(180);
pair F = dir(65);
pair D = extension(A, B, F, dir(205));
pair K = F*F;
pair C = extension(D, D+dir(90), B, K);
pair E = extension(B, F, C, D);
filldraw(unitcircle, palecyan+opacity(0.2), heavyblue);
filldraw(circumcircle(A, D, E), palered+opacity(0.2), red... |
EGMO-2015-notes_3 | Let $n, m$ be integers greater than $1$,
and let $a_1, a_2, \dots, a_m$ be positive integers
not greater than $n^m$.
Prove that there exist integers $b_1, b_2, \dots, b_m$ not greater than $n$ such that
\[ \gcd(a_1 + b_1, a_2 + b_2, \dots, a_m + b_m) < n. \] | In fact, we will prove that it's possible to choose $b_i \in \{0,1\}$!
Assume not, and all GCD's are at least $n$.
Consider the choices:
\begin{itemize}
\ii $b_1 = \dots = b_m = 0$.
\ii $b_1 = b_2 = \dots = b_{k-1} = b_{k+1} = b_m = 0$ and $b_k = 1$,
for some $2 \le k \le m$.
\end{itemize}
This generates $m$ gcd... |
EGMO-2015-notes_4 | Determine whether or not there exists an infinite sequence
$a_1$, $a_2$, \dots\ of positive integers satisfying
\[ a_{n+2} = a_{n+1} + \sqrt{a_{n+1}+a_{n}} \]
for every positive integer $n$. | In fact, we will show the following stronger result:
the largest $N$ for which one can find $(a_1, \dots, a_N)$ satisfying
(for all $1 \le n \le N-2$) is actually $N = 5$.
This largest $N$ is obtained
for example by $(a_1,a_2,a_3,a_4,a_5) = (477,7,29,35,43)$.
\begin{remark*}
Basically, the idea is to choose $a_3$ fir... |
EGMO-2015-notes_5 | Let $m, n$ be positive integers with $m > 1$.
Anastasia partitions the integers $1, 2, \dots , 2m$ into $m$ pairs.
Boris then chooses one integer from each pair
and finds the sum of these chosen integers.
Prove that Anastasia can select the pairs
so that Boris cannot make his sum equal to $n$. | Overall the idea is to try find a few constructions which eliminate most of the cases,
then clean out the last few ones leftover.
% (It's like using reavers on zerglings:
% knock out 90\% of the lings with a couple scarabs
% and then have the zealots finish the rest.)
\begin{itemize}
\ii If $n \notin [m^2, m^2+m]$, ... |
EGMO-2015-notes_6 | Let $H$ be the orthocenter and $G$ be the centroid of
acute-angled triangle $ABC$ with $AB \neq AC$.
The line $AG$ intersects the circumcircle
of $ABC$ at $A$ and $P$.
Let $P'$ be the reflection of $P$ in the line $BC$.
Prove that $\angle CAB = 60$ if and only if $HG = GP'$.
\end{enumerate} | The following complex numbers solution was given by Stefan Tudose.
First
\[ \frac{pa(b+c)-bc(p+a)}{pa-bc}
= \frac{b+c}{2} \implies p
= -\frac{2bc-ab-ac}{bc(2a-b-c)}. \]
Then
\[ p' = b+c-bc\ol p = \frac{ab+ac-b^2-c^2}{2a-b-c}. \]
Now let $D$ be the midpoint of $\ol{HP'}$.
Then
\begin{align*}
d = \frac{h+p'}{2} & = \f... |
EGMO-2016-notes_1 | Let $n$ be an odd positive integer,
and let $x_1$, $x_2$, \dots, $x_n$ be nonnegative real numbers.
Show that
\[ \min(x_i^2+x_{i+1}^2) \le \max(2x_jx_{j+1}) \]
where $1\le i,j\le n$ and $x_{n+1}=x_1$. | It suffices to exhibit a single pair $(i,j)$ where
\[ x_i^2 + x_{i+1}^2 \le 2 x_j x_{j+1}. \]
Since $n$ is odd we can find three adjacent $x_k$'s,
call them $a$, $b$, and $c$ (with $b$ in middle) such that $a \ge b \ge c$.
Then indeed
\[ 2ab \ge b^2+c^2 \]
so we're done.
\begin{remark*}
When $n$ is even, counterexa... |
EGMO-2016-notes_2 | Let $ABCD$ be a cyclic quadrilateral,
and let diagonals $AC$ and $BD$ intersect at $X$.
Let $C_1,D_1$ and $M$ be the midpoints of segments $CX$, $DX$ and $CD$, respectively.
Lines $AD_1$ and $BC_1$ intersect at $Y$, and line $MY$ intersects diagonals $AC$ and $BD$
at different points $E$ and $F$, respectively.
Prove th... | We present two approaches.
\paragraph{First approach through isogonality lemma.}
Note $ABC_1D_1$ is cyclic.
By the ``isogonality lemma'' applied to $\triangle YC_1D_1$,
it follows that $YX$ and $YM$ are isogonal
with respect to $\triangle YC_1D_1$.
Then
\[ \angle EXY = \angle XC_1Y + \angle C_1YX
= \angle AD_1X + \ang... |
EGMO-2016-notes_3 | Let $m$ be a positive integer. Consider a $4m \times 4m$ array of square unit cells.
Two different cells are \emph{related} to each other
if they are in either the same row or in the same column.
No cell is related to itself.
Some cells are colored blue, such that every cell is
related to at least two blue cells.
Deter... | We'll prove that the answer is at least $6m$.
The construction is to take
diagonal copies of the matrix
\[ \begin{bmatrix}
& 1 & 1 & 1 \\
1 &&& \\
1 &&& \\
1 &&&
\end{bmatrix}. \]
We present two proofs this is optimal.
\paragraph{Bipartite graph (Kevin Sun).}
Consider the bipartite subgraph
\[ H \subseteq K_... |
EGMO-2016-notes_4 | Two circles $\omega_1$ and $\omega_2$, of equal radius
intersect at different points $X_1$ and $X_2$.
Consider a circle $\omega$ externally tangent to $\omega_1$ at $T_1$
and internally tangent to $\omega_2$ at point $T_2$.
Prove that lines $X_1T_1$ and $X_2T_2$ intersect at a point lying on $\omega$. | We present two solutions, one by homothety and another by inversion.
\begin{center}
\begin{asy}
/*
Converted from GeoGebra by User:Azjps using Evan's magic cleaner
https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py
*/
size(8cm);
pen zzttqq = rgb(0.6,0.2,0.);
pen dcrutc = rgb(0.86274,0.07... |
EGMO-2016-notes_6 | Let $S$ be the set of all positive integers $n$ such that $n^4$ has a divisor
in the range $n^2 + 1$, $n^2+2$, \dots, $n^2 + 2n$.
Prove that there are infinitely many elements of $S$ of
each of the forms $7m$, $7m+1$, $7m+2$, $7m+5$, $7m+6$
and no elements of $S$ of the form $7m+3$ and $7m+4$,
where $m$ is an integer.
... | Start from the implication
\[ n^2 + k \mid n^4 \iff n^2 + k \mid k^2. \]
Since $1 \le k \le 2n$, in fact the quotient $\frac{k^2}{n^2+k}$
can only take values from $1$ to $3$.
In other words, $S$ is the set of integers $n$ for which at least one equation
\begin{align*}
n^2 + k &= k^2 \\
2(n^2 + k) &= k^2 \\
3(n^2... |
EGMO-2017-notes_1 | Let $ABCD$ be a convex quadrilateral
with $\angle DAB = \angle BCD = 90\dg$ and $\angle ABC > \angle CDA$.
Let $Q$ and $R$ be points on segments $BC$ and $CD$,
respectively, such that line $QR$ intersects lines $AB$ and $AD$
at points $P$ and $S$, respectively.
It is given that $PQ=RS$.
Let the midpoint of $BD$ be $M$ ... | The condition is equivalent to $N$ being the midpoint
of both $\ol{PS}$ and $\ol{QR}$ simultaneously.
(Thus triangles $BAD$ and $BCD$ play morally dual roles.)
\begin{center}
\begin{asy}
import graph; size(12cm);
pen zzttqq = rgb(0.6,0.2,0.); pen cqcqcq = rgb(0.75294,0.75294,0.75294);
draw((-1.83576,5.20298)--(-4.,1.5... |
EGMO-2017-notes_2 | Find the smallest positive integer $k$ for which
there exists a coloring of the positive integers
$\ZZ_{>0}$ with $k$ colors and a function $f \colon \ZZ_{>0} \to \ZZ_{>0}$
with the following two properties:
\begin{enumerate}[(i)]
\ii For all positive integers $m$, $n$ of the same color,
$f(m+n)=f(m)+f(n)$.
\ii T... | Answer: $k=3$.
Construction for $k=3$:
let
\[ f(n) =
\begin{cases} n/3 & n\equiv 0 \pmod 3 \\
n & \text{else} \end{cases} \]
and color the integers modulo $3$.
Now we prove that for $k=2$ a function $f$ obeying (i) must be linear,
even if $f \colon \ZZ_{>0} \to \RR_{>0}$.
Call the colors blue/red and WLOG $f(1) =... |
EGMO-2017-notes_3 | There are $2017$ lines in the plane such that
no three of them go through the same point.
Turbo the snail sits on a point on exactly one of the lines
and starts sliding along the lines in the following fashion:
she moves on a given line until she reaches an intersection of two lines.
At the intersection, she follows he... | \textbf{Color the regions} of the plane black and white in alternating colors.
Then:
\begin{claim*}
Turbo will always move in one orientation around black regions
and the other orientation around white regions.
\end{claim*}
This completes the proof, even if Turbo may pick
which of left/right she follows each time!
\be... |
EGMO-2017-notes_4 | Let $n\geq1$ be an integer and let $t_1<t_2<\dots<t_n$ be positive integers.
In a group of $t_n+1$ people, some games of chess are played.
Two people can play each other at most once.
Prove that it is possible for the following two conditions to hold at the same time:
\begin{enumerate}[(i)]
\ii The number of games pl... | Phrased in graph theory,
the problem asks to produce a simple graph $G$ on $t_n+1$ vertices
such that all degrees are in the set $\{t_1, \dots, t_n\}$
and each degree appears at least once.
We proceed by induction on $n$.
If $n = 1$, take a clique on $t_1+1$ vertices.
For $n = 2$, take a clique on $t_1$ vertices and
a... |
EGMO-2017-notes_5 | An $n$-tuple $(a_1,a_2,\dots,a_n)$ of positive integers is \emph{expensive} if
\[ (a_1+a_2)(a_2+a_3)\dots(a_{n-1}+a_n)(a_n+a_1) = 2^{2k-1} \]
for some positive integer $k$.
\begin{enumerate}
\ii[(a)] Find all integers $n \ge 2$ for which there exists an expensive $n$-tuple.
\ii[(b)] Prove that each odd integer $m \... | For part (a), the answer is odd $n$,
with construction given by setting $a_1 = \dots = a_n = 1$.
We show by induction that even $n$ all fail,
with $n=2$ being plain.
Construct a regular $n$-gon whose vertices are labeled $a_i$
and edges are labeled with $a_i + a_j$.
\begin{center}
\begin{asy}
size(5cm);
int[] x = ... |
EGMO-2017-notes_6 | Let $ABC$ be an acute scalene triangle.
The reflections of the centroid $G$ and the circumcenter $O$ of $ABC$ in its sides $BC$, $CA$, $AB$
are denoted by $G_1$, $G_2$, $G_3$ and $O_1$, $O_2$, $O_3$, respectively.
Show that the circumcircles of triangles $G_1G_2C$, $G_1G_3B$, $G_2G_3A$,
$O_1O_2C$, $O_1O_3B$, $O_2O_3A$ ... | Here is an approach with complex numbers.
Let $P$ be an arbitrary point.
Let $P_B$ and $P_C$ be the reflections of $P$ across $AB$ and $AC$
and let $Q_B$ and $Q_C$ be the second intersections of
lines $AP_B$ and $AP_C$ with the circumcircle.
Then we will compute the intersection of
$(A P_B P_C)$ and $(A Q_B Q_C) \equiv... |
EGMO-2018-notes_1 | Let $ABC$ be a triangle with $CA=CB$ and $\angle{ACB}=120^{\circ}$,
and let $M$ be the midpoint of $AB$.
Let $P$ be a variable point of the circumcircle of $ABC$,
and let $Q$ be the point on the segment $CP$ such that $QP = 2QC$.
It is given that the line through $P$ and perpendicular to $AB$
intersects the line $MQ$ a... | Since $\vec N = 3\vec Q - 2\vec M$,
it suffices to show $Q$ moves on a fixed circle.
That fixed circle is the image of $(CAB)$ under a homothety at $C$ with ratio $1/3$,
so we are done.
\begin{center}
\begin{asy}
pair C = dir(90);
pair A = dir(150);
pair B = dir(30);
pair M = midpoint(A--B);
pair P = dir(-40);
pair Q =... |
EGMO-2018-notes_2 | Consider the set
\[ A = \left\{ 1 + \frac 1k : k = 1, 2, 3, \dots \right\}. \]
For every integer $x \ge 2$, let $f(x)$ denote
the minimum integer such that $x$ can be written as the product
of $f(x)$ elements of $A$ (not necessarily distinct).
Prove that there are infinitely many pairs of integers $x \ge 2$
and $y \ge ... | One of many constructions: let $n = 2^e+1$
for $e \equiv 5 \pmod{10}$
and let $x = 11$, $y = n/11$ be our two integers.
We prove two lemmas:
\begin{claim*}
For any $m \ge 2$ we have
$f(m) \ge \left\lceil \log_2 m \right\rceil$.
\end{claim*}
\begin{proof}
This is obvious.
\end{proof}
It follows that $f(n) = e+1$... |
EGMO-2018-notes_3 | The $n$ contestants of EGMO are named $C_1$, $C_2$, \dots, $C_n$.
After the competition,
they queue in front of the restaurant according to the following rules.
\begin{itemize}
\ii The Jury chooses the initial order of the contestants in the queue.
\ii Every minute, the Jury chooses an integer $i$ with $1 \leq i \l... | The maximum money is
$1 + 3 + 7 + \dots + (2^{n-1}-1) = 2^n-n-1$,
which is finite.
Call the 1-euro process a jump,
and let $x_i$ denote the number of times that $C_i$ jumps.
Note that:
\begin{itemize}
\ii Whenever $C_i$ jumps it must jump over some $C_j$ with $j > i$.
\ii Contestant $C_i$ can jump over a $C_j$ wit... |
EGMO-2018-notes_4 | Let $n \ge 3$ be an integer.
Several non-overlapping dominoes are placed on an $n \times n$ board.
The \emph{value} of a row or column is the number of dominoes
that cover at least one cell of that row or column.
A domino configuration is called \emph{balanced} if there exists
some $k \ge 1$ such that every row and col... | The answer is $2n/3$ when $n \equiv 0 \pmod 3$, and $2n$ otherwise.
\paragraph{Proof this is best possible.}
To prove these are best possible, assume there are $d$ dominoes.
\begin{claim*}
In any balanced configuration, we always have $k \cdot 2n = 3d$.
\end{claim*}
\begin{proof}
Consider counting the number of or... |
EGMO-2018-notes_5 | Let $\Gamma $ be the circumcircle of triangle $ABC$.
A circle $\Omega$ is tangent to the line segment $AB$
and is tangent to $\Gamma$ at a point
lying on the same side of the line $AB$ as $C$.
The angle bisector of $\angle BCA$ intersects $\Omega$
at two different points $P$ and $Q$. Prove that $\angle ABP = \angle QBC... | If we let $M$ denote the midpoint of arc $\widehat{AB}$
then the inversion at $M$ with radius $MA = MB$ fixes $\Omega$,
so it swaps $P$ and $Q$, thus
\[ \dang MPB = \dang QBM. \]
\begin{center}
\begin{asy}
pair A = dir(210);
pair B = dir(330);
pair C = dir(110);
pair T = dir(20);
pair M = dir(270);
pair S = extension(T... |
EGMO-2018-notes_6 | Fix a real number $0 < t < \half$.
\begin{enumerate}
\ii [(a)]
Prove that there exists a positive integer $n$
such that for every set $S$ of $n$ positive integers,
the following holds:
there exist distinct $x, y \in S$ and \emph{nonnegative} integer $m \ge 0$
such that $|x-my| \le ty$.
\ii [(b)]
Determi... | \paragraph{Solution to (a).}
Assume not. Let $S = \left\{ s_1 < \dots < s_n \right\}$.
Consider
\[ 1 > \frac{s_1}{s_2} > \frac{s_1}{s_3} > \dots > \frac{s_1}{s_n} > t. \]
Note that two of the fractions above are within a factor of
$t^{1/(n-1)}$ of each other; taking $n$ large enough so that
$t^{1/(n-1)} \ge 1-t$ gives ... |
EGMO-2019-notes_1 | Find all triples $(a, b, c)$ of real numbers such that $ab + bc + ca = 1$ and
\[ a^2b + c = b^2c + a = c^2a + b. \] | Answer: $(\pm 1/\sqrt3, \pm 1/\sqrt3, \pm1/\sqrt3)$
where the signs correspond,
and $(\pm1, \pm1, 0)$ and permutations where the signs correspond.
These work and we prove that is all.
We begin by eliminating the condition
via homogenization:
the first equality now reads
\begin{align*}
a^2b+c\left[ ab+bc+ca \right] &... |
EGMO-2019-notes_2 | Let $n$ be a positive integer.
Dominoes are placed on a $2n \times 2n$ board
in such a way that every cell of the board is
(orthogonally) adjacent to exactly one cell covered by a domino.
For each $n$, determine the largest number of dominoes
that can be placed in this way. | The answer is $\binom{n+1}{2}$ and a construction is shown below.
For each domino, its \emph{aura} consists of all the cells
which are adjacent to a cell of the domino.
There are up to eight squares in each aura,
but some auras could be cut off by the boundary of the board,
which means that there could be as few as fi... |
EGMO-2019-notes_3 | Let $ABC$ be a triangle such that $\angle CAB > \angle ABC$, and let $I$ be its incenter.
Let $D$ be the point on segment $BC$ such that $\angle CAD = \angle ABC$.
Let $\omega$ be the circle tangent to $AC$ at $A$ and passing through $I$.
Let $X$ be the second point of intersection of $\omega$ and the circumcircle of $... | Here is a cross-ratio/trig solution:
rare for me to do one of these,
but this problem called out to me that way.
As usual, let $\alpha = \angle BAC$, etc.
Let $T$ be the foot of the bisector of $\angle BAD$ on $\ol{BD}$,
so that
\[ \frac{TB}{TC} = \frac{AB \sin \angle BAT}{AC \sin \angle CAT}
= \frac{AB \sin \frac{\... |
EGMO-2019-notes_4 | Let $ABC$ be a triangle with incenter $I$.
The circle through $B$ tangent to $AI$ at $I$
meets side $AB$ again at $P$.
The circle through $C$ tangent to $AI$ at $I$
meets side $AC$ again at $Q$.
Prove that $PQ$ is tangent to the incircle of $ABC$. | Let $E$ and $F$ be the tangency points of the incircle on $AC$ and $AB$.
By angle chasing,
\[ \angle PIF = \angle AIF - \angle AIP
= \left( 90\dg - \half \angle A \right) - \half \angle B = \half \angle C. \]
Similarly, $\angle EIQ = \half \angle B$.
\begin{center}
\begin{asy}
pair A = dir(110);
pair B = dir(210);
p... |
EGMO-2019-notes_5 | Let $n\ge 2$ be an integer,
and let $a_1, a_2, \dots , a_n$ be positive integers.
Show that there exist positive integers $b_1, b_2, \dots, b_n$
satisfying the following three conditions:
\begin{enumerate}[(a)]
\ii $a_i\le b_i$ for $i=1, 2, \dots, n$;
\ii the remainders of $b_1$, $b_2$, \dots, $b_n$
on division b... | Note that if $a_i > n$,
we can replace $a_i$ with $a_i-n$ and $b_i$ with $b_i-n$,
and nothing changes.
So we may as well assume $a_i \in \{1, \dots, n\}$ for each $i$.
We choose our $b_i$'s in the following way.
Draw an $n \times n$ grid and in the $i$th
column fill in the bottom $a_i-1$ cells red.
We can select $b_i$... |
EGMO-2019-notes_6 | On a circle, Alina draws $2019$ chords, the endpoints of which are all different.
A point is considered marked if it is either
\begin{enumerate}[(i)]
\ii one of the $4038$ endpoints of a chord; or
\ii an intersection point of at least two chords.
\end{enumerate}
Of the $4038$ points meeting criterion (i),
Alina lab... | Only the labels mod $3$ matter at all.
Assume for contradiction no blue labels are divisible by $3$.
Let $e_{ij}$ denote the number of segments
joining $i \pmod 3$ to $j \pmod 3$.
By double-counting (noting that points in (ii) are
counted an even number of times
but points in (i) are counted once)
we derive that
\begi... |
EGMO-2020-notes_1 | The positive integers $a_0$, $a_1$, $a_2$, \dots, $a_{3030}$ satisfy
\[ 2a_{n + 2} = a_{n + 1} + 4a_n \text{ for } n = 0, 1, 2, \dots, 3028. \]
Prove that at least one of the numbers
$a_0$, $a_1$, $a_2$, \dots, $a_{3030}$ is divisible by $2^{2020}$. | The idea is this:
\begin{itemize}
\ii All terms $a_0$, \dots, $a_{3030}$ are integers
(divisible by $2^0=1$);
\ii Hence all terms $a_1$, \dots, $a_{3029}$ are divisible by $2$,
\ii Hence all terms $a_1$, \dots, $a_{3028}$ are divisible by $4$,
\ii Hence all terms $a_2$, \dots, $a_{3027}$ are divisible by $8$,... |
EGMO-2020-notes_2 | Find all lists $(x_1, x_2, \dots, x_{2020})$ of non-negative real numbers
such that the following three conditions are all satisfied:
\begin{itemize}
\ii $x_1 \le x_2 \le \dots \le x_{2020}$;
\ii $x_{2020} \le x_1 + 1$;
\ii there is a permutation $(y_1, y_2, \dots, y_{2020})$
of $(x_1, x_2, \dots, x_{2020})$ such that... | The main point of the problem is to prove an inequality.
\begin{claim*}
If $a$ and $b$ are real numbers with $|a-b| \le 1$,
then \[ (a+1)^2(b+1)^2 \geq 4(a^3+b^3). \]
Moreover, equality holds only when $\{a,b\}=\{1,0\}$ or $\{a,b\}=\{2,1\}$.
\end{claim*}
\begin{proof}
Write
\begin{align*}
a^3+b^3 &= (a+b)... |
EGMO-2020-notes_3 | Let $ABCDEF$ be a convex hexagon such that $\angle A = \angle C = \angle E$ and $\angle B = \angle D = \angle F$
and the (interior) angle bisectors of $\angle A$, $\angle C$, and $\angle E$ are concurrent.
Prove that the (interior) angle bisectors of $\angle B$, $\angle D$, and $\angle F$ must also be concurrent. | In general, if hexagon $ABCDEF$ has $\angle A = \angle C = \angle E$ and $\angle B = \angle D = \angle F$,
then its sides can be extended to form two
equilateral triangles $PQR$ and $XYZ$, as shown.
\begin{center}
\begin{asy}
size(5cm);
pair P = dir(90);
pair Q = dir(210);
pair R = dir(330);
pair O = (0.1,0.2);
pair ... |
EGMO-2020-notes_4 | A permutation of the integers $1, 2, \dots, m$ is called \emph{fresh} if there exists
no positive integer $k < m$ such that the first $k$ numbers
in the permutation are $1, 2, \dots, k$ in some order.
Let $f_m$ be the number of fresh permutations of the integers $1, 2, \dots, m$.
Prove that $f_n \ge n \cdot f_{n - 1}$ ... | For every fresh permutation on $(1, 2, \dots, n-1)$
we generate $n$ fresh permutations on $(1, 2, \dots, n)$ in the following way:
\begin{itemize}
\ii Insert $n$ in the $k$th position for $k=1, 2, \dots, n-1$;
\ii Replace $n-1$ with $n$ and append $n-1$ to the end.
\end{itemize}
For example, $3142$ would generate $... |
EGMO-2020-notes_5 | Triangle $ABC$ has circumcircle $\Gamma$ and obeys $\angle BCA > 90^{\circ}$.
There is a point $P$ in the interior of the line segment $AB$
such that $PB = PC$ and the length of $PA$ equals the radius of $\Gamma$.
The perpendicular bisector of $\ol{PB}$ intersects $\Gamma$ at the points $D$ and $E$.
Prove $P$ is the in... | Let $O$ be the center of $\Gamma$ and $M$
the arc midpoint of $\ol{DE}$.
\begin{claim*}
Quadrilateral $APMO$ is a rhombus.
\end{claim*}
\begin{proof}
Since $PA = MO$ and both are perpendicular to $\ol{DE}$,
it follows $APMO$ is a parallelogram.
In fact though, because $AO = MO$, we get the rhombus.
\end{proof}
... |
EGMO-2020-notes_6 | Find all integers $m > 1$ for which the sequence $(a_n)_{n\ge1}$
defined recursively by
\[ a_{n+2} = m(a_{n+1} + a_n) - a_{n-1} \]
with initial conditions $a_1 = a_2 = 1$ and $a_3 = 4$
contains only perfect squares.
\end{enumerate} | The answer is $m=2$ and $m=10$.
The verification they work is left as an exercise.
Now compute the first several terms:
\begin{align*}
a_1 &= 1 \\
a_2 &= 1 \\
a_3 &= 4 \\
a_4 &= 5m-1 \\
a_5 &= 5m^2+3m-1 \\
a_6 &= 5m^3+8m^2-2m-4.
\end{align*}
We will now prove:
\begin{claim*}
If $a_4 \cdot a_6$ is a squar... |
EGMO-2021-notes_1 | The number $2021$ is fantabulous.
For any positive integer $m$,
if any element of the set $\{m, 2m+1, 3m\}$ is fantabulous,
then all the elements are fantabulous.
Does it follow that the number $2021^{2021}$ is fantabulous? | Write $a \iff b$ to mean $a$ fantabulous iff $b$ fantabulous.
Notice that for any integer $n$, we have
\[ 2n \iff 4n+1 \iff 12n+3 \iff 6n+1 \iff 3n \iff n. \]
Together with the given $2n+1 \iff n$,
it follows that if any integer is fantabulous then all of them are. |
EGMO-2021-notes_2 | Find all functions $f \colon \QQ \to \QQ$
such that the equation
\[f(xf(x)+y) = f(y) + x^2\]
holds for all rational numbers $x$ and $y$. | The answers are $f(x) \equiv +x$
and $f(x) \equiv -x$ which work.
To show they are the only ones,
we follow an approach similar to \texttt{SnowPanda}.
\begin{claim*}
If $f(z) = 0$ then $z = 0$.
In other words, $f$ has at most one root, at $0$.
\end{claim*}
\begin{proof}
Take $P(z,0)$.
\end{proof}
\begin{claim*... |
EGMO-2021-notes_3 | Let $ABC$ be a triangle with an obtuse angle at $A$.
Let $E$ and $F$ be the intersections of the external bisector of angle $A$
with the altitudes of $ABC$ through $B$ and $C$ respectively.
Let $M$ and $N$ be the points on the segments $EC$ and $FB$ respectively
such that $\angle EMA = \angle BCA$ and $\angle ANF = \an... | Call the altitudes $\ol{BZ}$ and $\ol{CY}$,
and $H$ the orthocenter.
Let $W$ be the midpoint of $\ol{BC}$.
Then according to
\textbf{\href{https://aops.com/community/c6h89098p519896}{IMO Shortlist 2005 G5}},
the line $AW$ is concurrent
with $(HYZ)$, $(HEF)$, $(HBC)$ at a point $Q$.
\begin{center}
\begin{asy}
import gr... |
EGMO-2021-notes_4 | Let $ABC$ be a triangle with incenter $I$
and let $D$ be an arbitrary point on the side $BC$.
Let the line through $D$ perpendicular to $BI$ intersect $CI$ at $E$.
Let the line through $D$ perpendicular to $CI$ intersect $BI$ at $F$.
Prove that the reflection of $A$ across the line $EF$ lies on the line $BC$. | We begin as follows:
\begin{claim*}
$AEIF$ is cyclic.
\end{claim*}
\begin{proof}
Let $X = \ol{CA} \cap \ol{DF}$.
Then
\begin{align*}
\dang FXA &= \dang DXC = \dang CDX = 90\dg - \half C
= \dang AIB = \dang FIA \\
\dang EXF &= \dang EXD = \dang XDE = \dang FDE = \dang EIF
\end{align*}
which impli... |
EGMO-2021-notes_5 | A plane has a special point $O$ called the origin.
Let $P$ be a set of $2021$ points in the plane such that
\begin{itemize}
\ii no three points in $P$ lie on a line and
\ii no two points in $P$ lie on a line through the origin.
\end{itemize}
A triangle with vertices in $P$ is \emph{fat}
if $O$ is strictly inside the tr... | For every pair of points $P$ and $Q$,
draw a directed arrow $P \to Q$ if $\angle POQ$ is labeled clockwise.
This gives a tournament on $2021$ vertices,
and a triangle is fat if its three edges form a directed cycle.
Consequently, by
\href{https://aops.com/community/c6h130647p740398}{\textbf{Canada 2006/4}},
there are ... |
EGMO-2021-notes_6 | Does there exist a nonnegative integer $a$ for which the equation
\[\left\lfloor\frac{m}{1}\right\rfloor
+ \left\lfloor\frac{m}{2}\right\rfloor
+ \left\lfloor\frac{m}{3}\right\rfloor
+ \dotsb + \left\lfloor\frac{m}{m}\right\rfloor = n^2 + a \]
has more than one million different solutions
$(m, n)$ where $m$ and $... | The answer is yes.
In fact, we prove the following general version:
\begin{claim*}
Consider any function $f \colon \NN \to \NN$
satisfying $f(m) = o(m^2)$.
Then for some $a$, the equation
\[ f(m) = n^2 + a \]
has at least 1000000 solutions $(m,n)$.
\end{claim*}
\begin{proof}
We consider triples $(m,n,a)$ satisfying t... |
IMO-1997-notes_1 | In the plane there is an infinite chessboard.
For any pair of positive integers $m$ and $n$,
consider a right-angled triangle with vertices at lattice points
and whose legs, of lengths $m$ and $n$, lie along edges of the squares.
Let $S_1$ be the total area of the black part of the triangle
and $S_2$ be the total area ... | In general, we say the \emph{discrepancy} of a region in the plane
equals its black area minus its white area.
We allow negative discrepancies,
so discrepancy is additive and $f(m,n)$ equals the absolute value
of the discrepancy of a right triangle with legs $m$ and $n$.
For (a), the answers are $0$ and $1/2$ respecti... |
IMO-1997-notes_2 | Let $ABC$ be a triangle with $\angle A < \min(\angle B, \angle C)$.
The points $B$ and $C$ divide the circumcircle of the triangle into two arcs.
Let $U$ be an interior point of the arc between $B$ and $C$ which does not contain $A$.
The perpendicular bisectors of $ AB$ and $ AC$ meet the line $AU$ at $V$ and $W$, resp... | Let $\ol{BTV}$ meet the circle again at $U_1$,
so that $AU_1 UB$ is an isosceles trapezoid.
Define $U_2$ similarly.
\begin{center}
\begin{asy}
pair A = dir(110);
pair B = dir(230);
pair C = dir(310);
pair U = dir(250);
pair U_1 = A*B/U;
pair U_2 = A*C/U;
pair T = extension(B, U_1, C, U_2);
filldraw(unitcircle, opacity... |
IMO-1997-notes_3 | Let $x_1$, $x_2$, \dots, $x_n$ be real numbers satisfying the conditions:
\begin{align*}
|x_1 + x_2 + \dots + x_n| &= 1 \\
|x_i| &\le \frac{n+1}{2} \qquad \text{for } i= 1,2, \dots, n
\end{align*}
Show that there exists a permutation $y_1$, $y_2$, \dots, $y_n$
of $x_1$, $x_2$, \dots, $x_n$ such that
\[ | y_1 + 2 y_... | WLOG $\sum x_i = 1$ (by negating $x_i$) and $x_1 \le x_2 \le \dots \le x_n$.
Notice that
\begin{itemize}
\ii The largest possible value of the sum in question is
\[ A = x_1 + 2x_2 + 3x_3 + \dots + nx_n. \]
while the smallest value is
\[ B = nx_1 + (n-1)x_2 + \dots + x_n. \]
\ii Meanwhile, the \emph{average} value acros... |
IMO-1997-notes_4 | An $n \times n$ matrix whose entries come
from the set $S = \{1, 2, \dots , 2n - 1\}$
is called a \emph{silver} matrix if,
for each $i = 1, 2, \dots , n$,
the $i$-th row and the $i$-th column together
contain all elements of $S$. Show that:
\begin{enumerate}[(a)]
\ii there is no silver matrix for $n = 1997$;
\ii silver... | \paragraph{Solution to (a).}
Define a \emph{cross} to be the union of the $i$th row and $i$th column.
Every cell of the matrix not on the diagonal is contained in exactly two crosses,
while each cell on the diagonal is contained in one cross.
On the other hand, if a silver matrix existed for $n=1997$,
then each elemen... |
IMO-1997-notes_5 | Find all pairs $(a,b)$ of positive integers satisfying
\[ a^{b^2} = b^a. \] | The answer is $(1,1)$, $(16,2)$ and $(27,3)$.
We assume $a,b > 1$ for convenience.
Let $T$ denote the set of non perfect powers other than $1$.
\begin{claim*}
Every integer greater than $1$
is uniquely of the form $t^n$ for some $t \in T$, $n \in \NN$.
\end{claim*}
\begin{proof}
Clear.
\end{proof}
Let $a = s^m$, $b... |
IMO-1997-notes_6 | For each positive integer $n$,
let $f(n)$ denote the number of ways of representing $n$
as a sum of powers of 2 with nonnegative integer exponents.
Representations which differ only in the ordering
of their summands are considered to be the same.
For instance, $f(4) = 4$,
because the number $4$ can be represented in th... | It's clear that $f$ is non-decreasing.
By sorting by the number of $1$'s we used,
we have the equation
\[ f(N) =
f\left( \left\lfloor \frac N2 \right\rfloor \right)
+ f\left( \left\lfloor \frac N2 \right\rfloor -1 \right)
+ f\left( \left\lfloor \frac N2 \right\rfloor -2 \right)
+ \dots
+ f(1) + f(0). \quad ... |
IMO-1998-notes_1 | A convex quadrilateral $ABCD$ has perpendicular diagonals.
The perpendicular bisectors of the sides $AB$ and $CD$ meet
at a unique point $P$ inside $ABCD$.
Prove that the quadrilateral $ABCD$ is cyclic
if and only if triangles $ABP$ and $CDP$ have equal areas. | If $ABCD$ is cyclic, then $P$ is the circumcenter,
and $\angle APB + \angle PCD = 180\dg$.
The hard part is the converse.
\begin{center}
\begin{asy}
import graph; size(10cm);
pen zzttqq = rgb(0.6,0.2,0.); pen aqaqaq = rgb(0.62745,0.62745,0.62745); pen cqcqcq = rgb(0.75294,0.75294,0.75294);
draw((2.28154,-9.32038)--(8.... |
IMO-1998-notes_2 | In a competition, there are $a$ contestants
and $b$ judges, where $b \ge 3$ is an odd integer.
Each judge rates each contestant as either ``pass'' or ``fail''.
Suppose $k$ is a number such that for any two judges,
their ratings coincide for at most $k$ contestants.
Prove that
\[ \frac ka \ge \frac{b-1}{2b}. \] | This is a ``routine'' problem with global ideas.
We count pairs of coinciding ratings,
i.e.\ the number $N$ of tuples
\[(\{J_1, J_2\}, C) \]
of two distinct judges and a contestant
for which the judges gave the same rating.
On the one hand, if we count by the judges,
we have \[ N \le \binom b2 k \]
by he problem condi... |
IMO-1998-notes_4 | Determine all pairs $(x,y)$ of positive integers
such that $x^{2}y+x+y$ is divisible by $xy^{2}+y+7$. | The answer is $(7k^2,7k)$ for all $k \ge 1$,
as well as $(11,1)$ and $(49,1)$.
We are given $xy^2+y+7 \mid x^2y+x+y$.
Multiplying the right-hand side by $y$ gives
\[ xy^2+y+7 \mid x^2y^2+xy+y^2 \]
Then subtracting $x$ times the left-hand side gives
\[ xy^2+y+7 \mid y^2-7x. \]
We consider cases based on the sign of $y^... |
IMO-1998-notes_5 | Let $I$ be the incenter of triangle $ABC$.
Let the incircle of $ABC$ touch
the sides $BC$, $CA$, and $AB$ at $K$, $L$, and $M$, respectively.
The line through $B$ parallel to $MK$ meets the lines
$LM$ and $LK$ at $R$ and $S$, respectively.
Prove that angle $RIS$ is acute. | Observe that $\triangle MKL$ is acute with circumcenter $I$.
We now present two proofs.
\paragraph{First simple proof (grobber).}
The problem is equivalent to showing $BI^2 > BR \cdot BS$.
But from
\[ \triangle BRK \sim \triangle MKL \sim \triangle BLS \]
we conclude
\[ BR = t \cdot \frac{MK}{ML},
\qquad BS = t \cdo... |
IMO-1998-notes_6 | Classify all functions $f \colon \NN \to \NN$
satisfying the identity
\[ f(n^2 f(m)) = m f(n)^2. \]
\end{enumerate} | Let $\mathcal P$ be the set of primes,
and let $g \colon \mathcal P \to \mathcal P$ be any involution on them.
Extend $g$ to a completely multiplicative function on $\NN$.
Then $f(n) = d g(n)$ is a solution for any $d \in \NN$
which is fixed by $g$.
It's straightforward to check these all work,
since $g \colon \NN \to... |
IMO-1999-notes_1 | A set $S$ of points from the space will be called
completely symmetric if it has at least three elements
and fulfills the condition that for every two distinct points $A$ and $B$ from $S$,
the perpendicular bisector plane of the segment $AB$ is a plane of symmetry for $S$.
Prove that if a completely symmetric set is fi... | Let $G$ be the centroid of $S$.
\begin{claim*}
All points of $S$ lie on a sphere $\Gamma$ centered at $G$.
\end{claim*}
\begin{proof}
Each perpendicular bisector plane passes through $G$.
So if $A,B \in S$ it follows $GA = GB$.
\end{proof}
\begin{claim*}
Consider any plane passing through three or more points... |
IMO-1999-notes_2 | Find the least constant $C$ such that for any integer $n > 1$ the inequality
\[\sum_{1 \le i < j \le n} x_i x_j (x_i^2 + x_j^2)
\le C \left( \sum_{1 \le i \le n} x_i \right)^4\]
holds for all real numbers $x_1, \dots, x_n \ge 0$.
Determine the cases of equality. | Answer: $C = \frac 18$, with equality when two $x_i$ are equal
and the remaining $x_i$ are equal to zero.
We present two proofs of the bound.
\paragraph{First solution by smoothing.}
Fix $\sum x_i = 1$.
The sum on the left-hand side can be interpreted as
$\sum_{i=1}^n x_i^3 \sum_{j \neq i} x_j = \sum_{i=1}^n x_i^3(1-... |
IMO-1999-notes_3 | Let $n$ be an even positive integer.
Find the minimal number of cells on the $n \times n$ board
that must be marked so that any cell
(marked or not marked) has a marked neighboring cell. | For every marked cell, consider the marked cell adjacent
to it; in this way we have a \emph{domino} of two cells.
For each domino, its \emph{aura} consists of all the cells
which are adjacent to a cell of the domino.
There are up to eight squares in each aura,
but some auras could be cut off by the boundary of the boar... |
IMO-1999-notes_4 | Find all pairs of positive integers $(x,p)$
such that $p$ is a prime and $x^{p-1}$ is a divisor of $(p-1)^{x}+1$. | If $p = 2$ then $x \in \{1,2\}$,
and if $p = 3$ then $x \in \{1,3\}$, since this is IMO 1990/3.
Also, $x = 1$ gives a solution for any prime $p$.
We show that there are no other solutions.
Assume $x > 1$ and let $q$ be smallest prime divisor of $x$.
We have $q > 2$ since $(p-1)^x+1$ is odd.
Then
\[ (p-1)^x \equiv -1 \... |
IMO-1999-notes_5 | Two circles $\Omega_{1}$ and $\Omega_{2}$ touch internally the circle
$\Omega$ in $M$ and $N$ and the center of $\Omega_{2}$ is on $\Omega_{1}$.
The common chord of the circles $\Omega_{1}$ and $\Omega_{2}$
intersects $\Omega$ in $A$ and $B$
Lines $MA$ and $MB$ intersects $\Omega_{1}$ in $C$ and $D$.
Prove that $\Omega... | Let $P$ and $Q$ be the centers of $\Omega_1$ and $\Omega_2$.
Let line $MQ$ meet $\Omega_1$ again at $W$,
the homothetic image of $Q$ under $\Omega_1 \to \Omega$.
Meanwhile, let $T$ be the intersection of segment $PQ$
with $\Omega_2$, and let $L$ be its homothetic image on $\Omega$.
Since $\ol{PTQ} \perp \ol{AB}$, it ... |
IMO-1999-notes_6 | Find all the functions $f \colon \RR \to \RR$ such that
\[f(x-f(y))=f(f(y))+xf(y)+f(x)-1\]
for all $x,y \in \RR$.
\end{enumerate} | The answer is $f(x) = -\half x^2+1$
which obviously works.
For the other direction, first note that
\[ P(f(y),y) \implies 2f(f(y)) + f(y)^2 - 1 = f(0). \]
We introduce the notation $c = \frac{f(0)-1}{2}$,
and $S = \opname{img} f$.
Then the above assertion says
\[ f(s) = -\half s^2 + (c + 1). \]
Thus, the given functio... |
IMO-2000-notes_1 | Two circles $G_1$ and $G_2$ intersect at two points $M$ and $N$.
Let $AB$ be the line tangent to these circles at $A$ and $B$,
respectively, so that $M$ lies closer to $AB$ than $N$.
Let $CD$ be the line parallel to $AB$
and passing through the point $M$,
with $C$ on $G_1$ and $D$ on $G_2$.
Lines $AC$ and $BD$ meet at ... | First, we have $\dang EAB = \dang ACM = \dang BAM$
and similarly $\dang EBA = \dang BDM = \dang ABM$.
Consequently, $\ol{AB}$ bisects $\angle EAM$ and $\angle EBM$,
and hence $\triangle EAB \cong \triangle MAB$.
\begin{center}
\begin{asy}
pair M = dir(120);
pair U = dir(170);
pair V = dir(10);
pair N = 0.3*V+0.7*U;
p... |
IMO-2000-notes_2 | Let $a$, $b$, $c$ be positive real numbers with $abc = 1$.
Show that
\[
\left( a - 1 + \frac 1b \right)
\left( b - 1 + \frac 1c \right)
\left( c - 1 + \frac 1a \right)
\le 1.
\] | Let $a = x/y$, $b = y/z$, $c = z/x$ for $x,y,z > 0$.
Then the inequality rewrites as
\[ (-x+y+z)(x-y+z)(x+y-z) \le xyz \]
which when expanded is equivalent to Schur's inequality.
Alternatively, if one wants to avoid appealing to Schur,
then the following argument works:
\begin{itemize}
\ii At most one term on the lef... |
IMO-2000-notes_3 | Let $n \ge 2$ be a positive integer
and $\lambda$ a positive real number.
Initially there are $n$ fleas on a horizontal line,
not all at the same point.
We define a move as choosing two fleas at some points $A$ and $B$,
with $A$ to the left of $B$,
and letting the flea from $A$ jump over the flea from $B$ to the point ... | The answer is $\lambda \ge \frac{1}{n-1}$.
We change the problem by replacing the fleas
with \textbf{bowling balls} $B_1$, $B_2$, \dots, $B_n$ in that order.
Bowling balls aren't exactly great at jumping,
so each move can now be described as follows:
\begin{itemize}
\ii Select two indices $i < j$.
Then ball $B_i$ move... |
IMO-2000-notes_4 | A magician has one hundred cards numbered $1$ to $100$.
He puts them into three boxes, a red one, a white one and a blue one,
so that each box contains at least one card.
A member of the audience draws two cards from two different boxes
and announces the sum of numbers on those cards.
Given this information,
the magici... | There are $2 \cdot 3! = 12$ ways, which amount to:
\begin{itemize}
\ii Partitioning the cards modulo $3$, or
\ii Placing $1$ alone in a box,
$100$ alone in a second box,
and all remaining cards in the third box.
\end{itemize}
These are easily checked to work so we prove they are the only ones.
\paragraph{First... |
IMO-2000-notes_5 | Does there exist a positive integer $n$
such that $n$ has exactly 2000 distinct prime divisors
and $n$ divides $2^n + 1$? | Answer: Yes.
We say that $n$ is \emph{Korean} if $n \mid 2^n+1$.
First, observe that $n=9$ is Korean.
Now, the problem is solved upon the following claim:
\begin{claim*}
If $n > 3$ is Korean,
there exists a prime $p$ not dividing $n$
such that $np$ is Korean too.
\end{claim*}
\begin{proof}
I claim that one can... |
IMO-2000-notes_6 | Let $\ol{AH_1}$, $\ol{BH_2}$, and $\ol{CH_3}$ be the
altitudes of an acute triangle $ABC$.
The incircle $\omega$ of triangle $ABC$ touches the sides
$BC$, $CA$ and $AB$ at $T_1$, $T_2$ and $T_3$, respectively.
Consider the reflections of the lines $H_1H_2$, $H_2H_3$, and
$H_3H_1$ with respect to the lines $T_1T_2$, $T_... | We use complex numbers with $\omega$ the unit circle.
Let $T_1 = a$, $T_2 = b$, $T_3 = c$.
The main content of the problem is to show that
the triangle in question has vertices
$ab/c$, $bc/a$, $ca/b$
(which is evident from a good diagram).
Since $A = \frac{2bc}{b+c}$, we have
\[ H_1 = \half \left( \frac{2bc}{b+c} + a ... |
IMO-2001-notes_1 | Let $ABC$ be an acute-angled triangle with $O$ as its circumcenter.
Let $P$ on line $BC$ be the foot of the altitude from $A$.
Assume that $\angle BCA \ge \angle ABC + 30\dg$.
Prove that $\angle CAB + \angle COP < 90\dg$. | The conclusion rewrites as
\begin{align*}
\angle COP &< 90\dg - \angle A = \angle OCP \\
\iff PC &< PO \\
\iff PC^2 &< PO^2 \\
\iff PC^2 &< R^2 - PB \cdot PC \\
\iff PC \cdot BC &< R^2 \\
\iff ab \cos C &< R^2 \\
\iff \sin A \sin B \cos C & < \frac14.
\end{align*}
Now
\[ \cos C \sin B
= \half \left( \s... |
IMO-2001-notes_2 | Let $a$, $b$, $c$ be positive reals. Prove that
\[ \frac{a}{\sqrt{a^2+8bc}} + \frac{b}{\sqrt{b^2+8ca}} + \frac{c}{\sqrt{c^2+8ab}} \ge 1. \] | By Holder, we have
\[
\left( \sum_{\text{cyc}} \frac{a}{\sqrt{a^2+8bc}} \right)^2
\left( \sum_{\text{cyc}} a(a^2+8bc) \right)
\ge (a+b+c)^3.
\]
So it suffices to show $(a+b+c)^3 \ge a^3+b^3+c^3+24abc$ which is clear by expanding. |
IMO-2001-notes_3 | Twenty-one girls and twenty-one boys took part in a mathematical competition.
It turned out that each contestant solved at most six problems,
and for each pair of a girl and a boy,
there was at least one problem that was solved by both the girl and the boy.
Show that there is a problem that was solved by at least three... | We will show the contrapositive.
That is, assume that
\begin{itemize}
\ii For each pair of a girl and a boy,
there was at least one problem that was
solved by both the girl and the boy.
\ii Assume every problem is either solved
mostly by girls (at most two boys)
or mostly by boys (at most two girls).
\end{i... |
IMO-2001-notes_4 | Let $n > 1$ be an odd integer and let $c_1$, $c_2$, \dots, $c_n$ be integers.
For each permutation $a = (a_1, a_2, \dots, a_n)$
of $\{1,2,\dots,n\}$, define $S(a) = \sum_{i=1}^n c_i a_i$.
Prove that there exist two permutations $a \neq b$
of $\{1,2,\dots,n\}$ such that $n!$ is a divisor of $S(a)-S(b)$. | Assume for contradiction that all the $S(a)$ are distinct modulo $n!$.
Then summing across all permutations gives
\begin{align*}
1 + 2 + \dots + n!
&\equiv \sum_a S(a) \\
&= \sum_a \sum_{i=1}^n c_i a_i \\
&= \sum_{i=1}^n c_i \sum_a a_i \\
&= \sum_{i=1}^n c_i \cdot \left( (n-1)! \cdot (1+\dots+n) \right) \\
... |
End of preview. Expand in Data Studio
e1-proof
e1-proof contains mathematics olympiad problems and proofs. The proofs were sourced from Evan Chen's website (https://web.evanchen.cc/problems.html) and are provided here for the use of the community with Evan's permission.
This dataset was collected in this work: Didactic to Constructive: Turning Expert Solutions into Learnable Reasoning. If you use this dataset, please cite this paper:
@article{mendes2026didactic,
title={Didactic to Constructive: Turning Expert Solutions into Learnable Reasoning},
author={Mendes, Ethan and Park, Jungsoo and Ritter, Alan},
journal={arXiv preprint arXiv:2602.02405},
year={2026}
}
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