problem_id stringlengths 16 19 | problem stringlengths 69 2.04k | solution stringlengths 60 23.9k |
|---|---|---|
IMO-2003-notes_6 | Let $p$ be a prime number.
Prove that there exists a prime number $q$ such that for every integer $n$,
the number $n^p-p$ is not divisible by $q$.
\end{enumerate} | By orders, we must have $q=pk+1$ for this to be possible
(since if $q \not \equiv 1 \pmod p$, then $n^p$ can be any residue modulo $q$).
Since $p \equiv n^p \pmod q \implies p^k \equiv 1 \pmod q$,
it suffices to prevent the latter situation from happening.
So we need a prime $q \equiv 1 \pmod p$ such that $p^k \not\eq... |
IMO-2004-notes_1 | Let $ABC$ be an acute-angled triangle with $AB\neq AC$.
The circle with diameter $BC$ intersects the sides $AB$ and $AC$
at $M$ and $N$ respectively.
Denote by $O$ the midpoint of the side $BC$.
The bisectors of the angles $\angle BAC$ and $\angle MON$ intersect at $R$.
Prove that the circumcircles of the triangles $BM... | By Miquel's theorem it's enough to show $AMRN$ is cyclic.
\begin{center}
\begin{asy}
pair A = dir(130);
pair B = dir(210);
pair C = dir(330);
pair H = A+B+C;
pair M = foot(C, A, B);
pair N = foot(B, A, C);
pair O = midpoint(B--C);
pair R = extension(A, incenter(A, B, C), O, midpoint(M--N));
pair T = extension(H, R, B, ... |
IMO-2004-notes_2 | Find all polynomials $P$ with real coefficients such that
for all reals $a$, $b$, $c$ such that $ab+bc+ca = 0$, we have
\[ P(a-b) + P(b-c) + P(c-a) = 2P(a+b+c). \] | The answer is \[ P(x) = \alpha x^4 + \beta x^2 \]
which can be checked to work, for any real numbers $\alpha$ and $\beta$.
It is easy to obtain that $P$ is even and $P(0) = 0$.
The trick is now to choose $(a,b,c) = (6x,3x,-2x)$
and then compare the leading coefficients to get
\[ 3^n + 5^n + 8^n = 2 \cdot 7^n \]
for $n... |
IMO-2004-notes_3 | Define a ``hook'' to be a figure made up of six unit squares
as shown below in the picture,
or any of the figures obtained by applying rotations
and reflections to this figure.
\begin{center}
\begin{asy}
unitsize(0.5 cm);
draw(unitsquare);
draw(shift(0,1)*unitsquare);
draw(shift(0,2)*unitsquare);
draw(shift(1,2)*unitsq... | The answer is that one requires:
\begin{itemize}
\ii $\{1,2,5\} \notin \{m,n\}$,
\ii $3 \mid m$ or $3 \mid n$,
\ii $4 \mid m$ or $4 \mid n$.
\end{itemize}
First, we check all of these work, in fact we claim:
\begin{claim*}
Any rectangle satisfying these conditions
can be tiled by $3 \times 4$ rectangles (and... |
IMO-2004-notes_4 | Let $n \ge 3$ be an integer
and $t_1$, $t_2$, \dots, $t_n$ positive real numbers such that
\[ n^2+1 > \left(t_1 + t_2 + \dots + t_n\right)
\left( \frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n} \right). \]
Show that $t_i$, $t_j$, $t_k$ are the sides of a triangle
for all $i$, $j$, $k$ with $1 \le i < j < k \le ... | Let $a = t_1$, $b = t_2$, $c = t_3$.
Expand:
\begin{align*}
n^2+1 &> \left(t_1 + t_2 + \dots + t_n\right)
\left( \frac{1}{t_1} + \dots + \frac{1}{t_n} \right) \\
&= n + \sum_{1 \le i < j \le n}
\left( \frac{t_i}{t_j} + \frac{t_j}{t_i} \right) \\
&= n + \sum_{1 \le i < j \le n}
\left( \frac{t_i}{t_j} +... |
IMO-2004-notes_5 | In a convex quadrilateral $ABCD$,
the diagonal $BD$ bisects neither the angle $ABC$ nor the angle $CDA$.
The point $P$ lies inside $ABCD$ and satisfies
\[\angle PBC=\angle DBA \quad\text{and}\quad \angle PDC=\angle BDA. \]
Prove that $ABCD$ is a cyclic quadrilateral if and only if $AP=CP$. | We show two solutions.
We note that the first hypothesis cannot be dropped, because $ABCD$ being a kite
with $BA=BC$ and $DA=DC$ with center $P$ is a counterexample
(where $AP=CP$ but $ABCD$ is not cyclic).
However, the condition that $BD$ bisects neither angle $\angle B$ nor $\angle D$
is equivalent to requiring that ... |
IMO-2004-notes_6 | We call a positive integer \emph{alternating} if every two consecutive digits
in its decimal representation are of different parity.
Find all positive integers $n$ which have an alternating multiple.
\end{enumerate} | If $20 \mid n$, then clearly $n$ has no alternating
multiple since the last two digits are both even.
We will show the other values of $n$ all work.
The construction is just rush-down do-it.
The meat of the solution is the two following steps.
\begin{claim*}
[Tail construction]
For every even integer $w \ge 2$,
... |
IMO-2005-notes_1 | Six points are chosen on the sides of an equilateral triangle $ABC$:
$A_1$, $A_2$ on $BC$, $B_1$, $B_2$ on $CA$ and $C_1$, $C_2$ on $AB$,
such that they are the vertices of a
convex hexagon $A_1A_2B_1B_2C_1C_2$ with equal side lengths.
Prove that the lines $A_1B_2$, $B_1C_2$ and $C_1A_2$ are concurrent. | The six sides of the hexagon, when oriented, comprise
six vectors with vanishing sum.
However note that \[ \overrightarrow{A_1A_2}
+ \overrightarrow{B_1B_2}
+ \overrightarrow{C_1C_2} = 0. \]
Thus
\[ \overrightarrow{A_2B_1} + \overrightarrow{B_2C_1} +
\overrightarrow{C_2A_1} = 0 \]
and since three unit vectors with vani... |
IMO-2005-notes_2 | Let $a_1$, $a_2$, \dots\ be a sequence of integers
with infinitely many positive and negative terms.
Suppose that for every positive integer $n$
the numbers $a_1$, $a_2$, \dots, $a_n$
leave $n$ different remainders upon division by $n$.
Prove that every integer occurs exactly once in the sequence. | Obviously every integer appears at most once
(otherwise take $n$ much larger).
So we will prove every integer appears at least once.
\begin{claim*}
For any $i < j$ we have $\left\lvert a_i-a_j \right\rvert < j$.
\end{claim*}
\begin{proof}
Otherwise, let $n = \left\lvert a_i-a_j \right\rvert \neq 0$.
Then $i,j \i... |
IMO-2005-notes_3 | Let $x,y,z > 0$ satisfy $xyz\geq 1$. Prove that
\[ \frac { x^5-x^2 }{x^5+y^2+z^2}
+ \frac {y^5-y^2}{x^2+y^5+z^2}
+ \frac {z^5-z^2}{x^2+y^2+z^5} \geq 0. \] | Negating both sides and adding $3$ eliminates the minus signs:
\[ \sum_{\text{cyc}} \frac{1}{x^5+y^2+z^2}
\le \frac{3}{x^2+y^2+z^2}. \]
Thus we only need to consider the case $xyz = 1$.
Direct expansion and Muirhead works now!
As advertised, once we show it suffices to analyze if $xyz=1$
the inequality becomes more ... |
IMO-2005-notes_4 | Determine all positive integers relatively
prime to all the terms of the infinite sequence
\[ a_n = 2^n+3^n+6^n-1, \quad n \ge 1. \] | The answer is $1$ only (which works).
It suffices to show there are no primes.
For the primes $p=2$ and $p=3$, take $a_2=48$.
For any prime $p \ge 5$ notice that
\begin{align*}
a_{p-2} &= 2^{p-2} + 3^{p-2} + 6^{p-2} - 1 \\
&\equiv \frac 12 + \frac 13 + \frac 16 - 1 \pmod p \\
&\equiv 0 \pmod p
\end{align*}
so no... |
IMO-2005-notes_5 | Let $ABCD$ be a fixed convex quadrilateral
with $BC=DA$ and $\ol{BC} \nparallel \ol{DA}$.
Let two variable points $E$ and $F$ lie on the
sides $BC$ and $DA$, respectively, and satisfy $BE=DF$.
The lines $AC$ and $BD$ meet at $P$,
the lines $BD$ and $EF$ meet at $Q$,
the lines $EF$ and $AC$ meet at $R$.
Prove that the c... | Let $M$ be the Miquel point of complete quadrilateral $ADBC$;
in other words, let $M$ be the second intersection point
of the circumcircles of $\triangle APD$ and $\triangle BPC$.
(A good diagram should betray this secret;
all the points are given in the picture.)
This makes lots of sense since we know $E$ and $F$
will... |
IMO-2005-notes_6 | In a mathematical competition $6$ problems were posed to the contestants.
Each pair of problems was solved by more than $\frac{2}{5}$ of the contestants.
Nobody solved all 6 problems.
Show that there were at least $2$ contestants
who each solved exactly $5$ problems.
\end{enumerate} | Assume not and at most one contestant solved five problems.
By adding in solves,
we can assume WLOG that one contestant solved problems one through five,
and every other contestant solved four of the six problems.
We split the remaining contestants based on whether they solved P6.
Let $a_i$ denote the number of contes... |
IMO-2006-notes_1 | Let $ABC$ be a triangle with incenter $I$.
A point $P$ in the interior of the triangle satisfies
\[ \angle PBA + \angle PCA = \angle PBC + \angle PCB. \]
Show that $AP \ge AI$ and that equality holds if and only if $P=I$. | The condition rewrites as
\[
\angle PBC + \angle PCB
= (\angle B - \angle PBC)
+ (\angle C - \angle PCB)
\implies
\angle PBC + \angle PCB = \frac{\angle B + \angle C}{2}
\]
which means that
\[ \angle BPC = 180\dg - \frac{\angle B + \angle C}{2}
= 90\dg + \frac{\angle A}{2}
= \angle BIC.
\]
Since $P$ and $... |
IMO-2006-notes_2 | Let $P$ be a regular $2006$-gon.
A diagonal is called \emph{good} if its endpoints
divide the boundary of $P$ into two parts,
each composed of an odd number of sides of $P$.
The sides of $P$ are also called \emph{good}.
Suppose $P$ has been dissected into triangles by $2003$ diagonals,
no two of which have a common poi... | Call a triangle with the desired property \emph{special}.
We prove the maximum number of special triangles is $1003$,
achieved by paring up the sides of the polygon.
We present two solutions for the upper bound.
Both of them rely first on two geometric notes:
\begin{itemize}
\ii In a special triangle, the good sides... |
IMO-2006-notes_3 | Determine the least real number $M$ such that the inequality
\[ \left\lvert ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2) \right\rvert
\leq M\left( a^2+b^2+c^2 \right)^2 \]
holds for all real numbers $a$, $b$ and $c$. | It's the same as
\[ \left\lvert (a-b)(b-c)(c-a)(a+b+c) \right\rvert
\le M \left( a^2+b^2+c^2 \right)^2. \]
Let $x=a-b$, $y=b-c$, $z=c-a$, $s=a+b+c$.
Then we want to have
\[
\left\lvert xyzs \right\rvert
\le \frac{M}{9} (x^2+y^2+z^2+s^2)^2.
\]
Here $x+y+z=0$.
Now if $x$ and $y$ have the same sign,
we can replace th... |
IMO-2006-notes_4 | Determine all pairs $(x,y)$ of integers such that
\[ 1 + 2^x + 2^{2x+1} = y^2. \] | Answers: $(0, \pm 2)$, $(4, \pm 23)$, which work.
Assume $x \ge 4$.
\[ 2^x \left( 1 + 2^{x+1} \right)
= 2^x + 2^{2x+1} = y^2 - 1 = (y-1)(y+1). \]
So either:
\begin{itemize}
\ii $y = 2^{x-1} m + 1$ for some odd $m$, so
\[ 1 + 2^{x+1} = m\left( 2^{x-2}m+1 \right)
\implies 2^x = \frac{4(1-m)}{m^2-8}. \]
\ii $... |
IMO-2006-notes_5 | Let $P(x)$ be a polynomial of degree $n > 1$
with integer coefficients and let $k$ be a positive integer.
Consider the polynomial
\[ Q(x) = P(P(\dots P(P(x)) \dots )) \] where $P$ occurs $k$ times.
Prove that there are at most $n$ integers $t$ such that $Q(t) = t$. | First, we prove that:
\begin{claim*}
[Putnam 2000 et al]
If a number is periodic under $P$
then in fact it's fixed by $P \circ P$.
\end{claim*}
\begin{proof}
Let $x_1$, $x_2$, \dots, $x_n$ be a minimal orbit.
Then
\[ x_i - x_{i+1} \mid P(x_i) - P(x_{i+1})
= x_{i+1} - x_{i+2} \]
and so on cyclically.
... |
IMO-2006-notes_6 | Assign to each side $b$ of a convex polygon $P$
the maximum area of a triangle that has $b$ as a side and is contained in $P$.
Show that the sum of the areas assigned to the sides of $P$ is at least twice the area of $P$.
\end{enumerate} | We say a polygon in \emph{almost convex}
if all its angles are at most $180\dg$.
Note that given any convex or almost convex polygon,
we can take any side $b$ and add another vertex on it, and the sum of the labels doesn't change
(since the label of a side is the length of the side times the distance of the farthest p... |
IMO-2007-notes_1 | Real numbers $a_1$, $a_2$, \dots, $a_n$ are fixed.
For each $1 \le i \le n$ we let
$d_i = \max\{a_j : 1 \le j \le i\} - \min\{a_j : i \le j \le n\}$
and let $d = \max \{d_i : 1 \le i \le n\}$.
\begin{enumerate}[(a)]
\ii Prove that for any real numbers $x_1 \le \dots \le x_n$ we have
\[
\max \left\{ \left\lvert... | Note that we can dispense of $d_i$ immediately
by realizing that the definition of $d$ just says
\[ d = \max_{1 \le i \le j \le n} \left( a_i - a_j \right). \]
If $a_1 \le \dots \le a_n$ are already nondecreasing
then $d = 0$ and there is nothing to prove
(for the equality case, just let $x_i = a_i$),
so we will no lo... |
IMO-2007-notes_2 | Consider five points $A$, $B$, $C$, $D$ and $E$
such that $ABCD$ is a parallelogram and $BCED$ is a cyclic quadrilateral.
Let $\ell$ be a line passing through $A$.
Suppose that $\ell$ intersects the interior of the segment $DC$ at $F$
and intersects line $BC$ at $G$.
Suppose also that $EF = EG = EC$.
Prove that $\ell$ ... | Let $M$, $N$, $P$ denote the midpoints of $\ol{CF}$, $\ol{CG}$, $\ol{AC}$
(noting $P$ is also the midpoint of $\ol{BD}$).
By a homothety at $C$ with ratio $\half$,
we find $\ol{MNP}$ is the image of line $\ell \equiv \ol{AGF}$.
\begin{center}
\begin{asy}
pair C = dir(200);
pair D = dir(340);
pair E = dir(240);
pair K... |
IMO-2007-notes_3 | In a mathematical competition some competitors are (mutual) friends.
Call a group of competitors a \emph{clique} if each two of them are friends.
Given that the largest size of a clique is even,
prove that the competitors can be arranged into two rooms
such that the largest size of a clique contained in one room
is the... | Take the obvious graph interpretation $G$.
We paint red any vertices in one of the maximal cliques $K$,
which we assume has $2r$ vertices, and paint the remaining vertices green.
We let $\alpha(\bullet)$ denote the clique number.
Initially, let the two rooms $A = K$, $B = G-K$.
\begin{claim*}
We can move at most $r$... |
IMO-2007-notes_4 | In triangle $ABC$ the bisector of $\angle BCA$
meets the circumcircle again at $R$,
the perpendicular bisector of $\ol{BC}$ at $P$,
and the perpendicular bisector of $\ol{AC}$ at $Q$.
The midpoint of $\ol{BC}$ is $K$ and the midpoint of $\ol{AC}$ is $L$.
Prove that the triangles $RPK$ and $RQL$ have the same area. | We first begin by proving the following claim.
\begin{claim*}
We have $CQ = PR$ (equivalently, $CP = QR$).
\end{claim*}
\begin{proof}
Let $O = \ol{LQ} \cap \ol{KP}$ be the circumcenter.
Then
\[ \dang OPQ = \dang KPC = 90\dg - \dang PCK
= 90\dg - \dang LCQ = \dang \dang CQL = \dang PQO. \]
Thus $OP = OQ$.
... |
IMO-2007-notes_5 | Let $a$ and $b$ be positive integers.
Show that if $4ab - 1$ divides $(4a^{2} - 1)^{2}$, then $a = b$. | As usual,
\[ 4ab-1 \mid (4a^2-1)^2 \iff 4ab-1 \mid (4ab \cdot a-b)^2
\iff 4ab-1 \mid (a-b)^2. \]
Then we use a typical Vieta jumping argument.
Define \[ k = \frac{(a-b)^2}{4ab-1}. \]
Note that $k = 0 \iff a = b$.
So we will prove that $k > 0$ leads to a contradiction.
Indeed, suppose $(a, b)$ is a minimal solution w... |
IMO-2007-notes_6 | Let $n$ be a positive integer.
Consider
\[ S = \left\{ (x,y,z) \mid
x,y,z \in \{ 0, 1, \dots, n\}, \;
x+y+z > 0 \right\} \]
as a set of $(n+1)^3-1$ points in the three-dimensional space.
Determine the smallest possible number of planes,
the union of which contains $S$ but does not include $(0,0,0)$.
\end{enumerate... | The answer is $3n$.
Here are two examples of constructions with $3n$ planes:
\begin{itemize}
\ii $x+y+z=i$ for $i=1,\dots,3n$.
\ii $x=i$, $y=i$, $z=i$ for $i=1,\dots,n$.
\end{itemize}
Suppose for contradiction we have $N < 3n$ planes.
Let them be $a_i x + b_i y + c_i z + 1 = 0$, for $i = 1, \dots, N$.
Define the po... |
IMO-2008-notes_1 | Let $H$ be the orthocenter of an acute-angled triangle $ABC$.
The circle $\Gamma_{A}$ centered at the midpoint of $\ol{BC}$ and passing
through $H$ intersects the sideline $BC$ at points $A_1$ and $A_2$.
Similarly, define the points $B_1$, $B_2$, $C_1$, and $C_2$.
Prove that six points $A_1$, $A_2$, $B_1$, $B_2$, $C_1... | We show two solutions.
\paragraph{First solution using power of a point.}
Let $D$, $E$, $F$ be the centers of $\Gamma_A$, $\Gamma_B$, $\Gamma_C$
(in other words, the midpoints of the sides).
We first show that $B_1$, $B_2$, $C_1$, $C_2$ are concyclic.
It suffices to prove that $A$
lies on the radical axis of the circ... |
IMO-2008-notes_2 | Let $x$, $y$, $z$ be real numbers with $xyz = 1$, all different from $1$.
Prove that
\[ \frac{x^2}{(x-1)^2} + \frac{y^2}{(y-1)^2} + \frac{z^2}{(z-1)^2} \ge 1 \]
and show that equality holds for infinitely many choices
of rational numbers $x$, $y$, $z$. | Let $x=a/b$, $y=b/c$, $z=c/a$, so we want to show
\[ \left(\frac{a}{a-b}\right)^2+\left(\frac{b}{b-c}\right)^2
+\left(\frac{c}{c-a}\right)^2\ge 1.\]
A boring computation shows this is equivalent to
\[ \frac{(a^2b+b^2c+c^2a-3abc)^2}{(a-b)^2(b-c)^2(c-a)^2} \ge 0 \]
which proves the inequality
(and it is unsurprising we... |
IMO-2008-notes_3 | Prove that there are infinitely many positive integers $n$
such that $n^2+1$ has a prime factor greater than $2n + \sqrt{2n}$. | The idea is to pick the prime $p$ first!
Select any large prime $p \ge 2013$,
and let $h = \left\lceil \sqrt p \right\rceil$.
We will try to find an $n$ such that
\[ n \le \frac 12 (p-h) \quad \text{and} \quad p \mid n^2+1. \]
This implies $p \ge 2n+\sqrt{p}$
which is enough to ensure $p \ge 2n + \sqrt{2n}$.
Assume $... |
IMO-2008-notes_4 | Find all functions $f$ from the positive reals to the positive reals such that
\[ \frac{f(w)^2 + f(x)^2}{f(y^2)+f(z^2)} = \frac{w^2+x^2}{y^2+z^2} \]
for all positive real numbers $w$, $x$, $y$, $z$ satisfying $wx=yz$. | The answers are $f(x) \equiv x$ and $f(x) \equiv 1/x$.
These work, so we show they are the only ones.
First, setting $(t,t,t,t)$ gives $f(t^2) = f(t)^2$.
In particular, $f(1) = 1$.
Next, setting $(t, 1, \sqrt t, \sqrt t)$ gives
\[ \frac{f(t)^2 + 1}{2f(t)} = \frac{t^2 + 1}{2t} \]
which as a quadratic implies $f(t) \in ... |
IMO-2008-notes_5 | Let $n$ and $k$ be positive integers with $k \geq n$ and $k - n$ an even number.
There are $2n$ lamps labelled $1$, $2$, \dots, $2n$ each of which can be either on or off.
Initially all the lamps are off.
We consider sequences of steps: at each step one of the lamps is switched
(from on to off or from off to on).
Let $... | The answer is $2^{k-n}$.
Consider the following map $\Psi$ from $N$-sequences to $M$-sequences:
\begin{itemize}
\ii change every instance of $n+1$ to $1$;
\ii change every instance of $n+2$ to $2$;
\ii[$\vdots$]
\ii change every instance of $2n$ to $n$.
\end{itemize}
(For example, suppose $k=9$, $n=3$;
then $1... |
IMO-2008-notes_6 | Let $ABCD$ be a convex quadrilateral with $BA \neq BC$.
Denote the incircles of triangles $ABC$ and $ADC$
by $\omega_1$ and $\omega_2$ respectively.
Suppose that there exists a circle $\omega$ tangent
to ray $BA$ beyond $A$ and to the ray $BC$ beyond $C$,
which is also tangent to the lines $AD$ and $CD$.
Prove that the... | By the external version of Pitot theorem, the existence
of $\omega$ implies that
\[ BA + AD = CB + CD. \]
Let $\ol{PQ}$ and $\ol{ST}$ be diameters of $\omega_1$ and $\omega_2$
with $P, T \in \ol{AC}$.
Then the length relation on $ABCD$ implies that $P$ and $T$
are reflections about the midpoint of $\ol{AC}$.
Now orien... |
IMO-2009-notes_1 | Let $n, k \ge 2$ be positive integers and let $a_1$, $a_2$, $a_3$, \dots, $a_k$
be distinct integers in the set $\left\{ 1,2,\dots,n \right\}$
such that $n$ divides $a_i(a_{i+1} - 1)$ for $i = 1,2,\dots,k-1$.
Prove that $n$ does not divide $a_k(a_1 - 1)$. | We proceed indirectly and assume that
\[ a_i (a_{i+1}-1) \equiv 0 \pmod n \]
for $i = 1, \dots, k$ (indices taken modulo $k$).
We claim that this implies all the $a_i$ are equal modulo $n$.
Let $q = p^e$ be any prime power dividing $n$.
Then, $a_1 (a_2 - 1) \equiv 0 \pmod q$, so $p$ divides either $a_1$ or $a_2-1$.
\b... |
IMO-2009-notes_2 | Let $ABC$ be a triangle with circumcenter $O$.
The points $P$ and $Q$ are interior points of the sides $CA$ and $AB$ respectively.
Let $K$, $L$, $M$ be the midpoints of $\ol{BP}$, $\ol{CQ}$, $\ol{PQ}$.
Suppose that $\ol{PQ}$ is tangent to the circumcircle of $\triangle KLM$.
Prove that $OP = OQ$. | By power of a point, we have $-AQ \cdot QB = OQ^2 - R^2$
and $-AP \cdot PC = OP^2 - R^2$.
Therefore, it suffices to show $AQ \cdot QB = AP \cdot PC$.
\begin{center}
\begin{asy}
pair A = dir(70);
pair B = dir(210);
pair C = dir(330);
pair P = 0.4*A+0.6*C;
pair Q = 0.25*B+0.75*A;
filldraw(A--B--C--cycle, opacity(0.2)+... |
IMO-2009-notes_3 | Suppose that $s_1,s_2,s_3, \dotsc$ is a strictly increasing sequence of
positive integers such that the sub-sequences
$s_{s_1}$, $s_{s_2}$, $s_{s_3}$, \dots
and $s_{s_1 + 1}$, $s_{s_2 + 1}$, $s_{s_3 + 1}$, \dots
are both arithmetic progressions.
Prove that the sequence $s_1$, $s_2$, $s_3$, \dots\ is itself an arithmeti... | We present two solutions.
\paragraph{First solution (Alex Zhai).}
Let $s(n) \coloneq s_n$ and write
\begin{align*}
s(s(n)) &= Dn + A \\
s(s(n)+1) &= D'n + B.
\end{align*}
In light of the bounds $s(s(n)) \le s(s(n)+1) \le s(s(n+1))$
we right away recover $D = D'$ and $A \le B$.
Let $d_n = s(n+1)-s(n)$.
Note that $... |
IMO-2009-notes_4 | Let $ABC$ be a triangle with $AB = AC$.
The angle bisectors of $\angle CAB$ and $\angle ABC$
meet the sides $BC$ and $CA$ at $D$ and $E$, respectively.
Let $K$ be the incenter of triangle $ADC$.
Suppose that $\angle BEK = 45^\circ$.
Find all possible values of $\angle CAB$. | Here is the solution presented in my book \emph{EGMO}.
Let $I$ be the incenter of $ABC$,
and set $\angle DAC = 2x$ (so that $0\dg < x < 45\dg$).
From $\angle AIE = \angle DIC$, it is easy to compute
\[
\angle KIE = 90\dg - 2x, \;
\angle ECI = 45\dg -x, \;
\angle IEK = 45\dg, \;
\angle KEC = 3x. \]
Having chase... |
IMO-2009-notes_5 | Find all functions $f \colon \ZZ_{>0} \to \ZZ_{>0}$
such that for positive integers $a$ and $b$, the numbers
\[ a, \qquad f(b), \qquad f(b+f(a)-1) \]
are the sides of a non-degenerate triangle. | The only function is the identity function (which works).
We prove it is the only one.
Let $P(a,b)$ denote the given statement.
\begin{claim*}
We have $f(1) = 1$, and $f(f(n)) = n$.
(In particular $f$ is a bijection.)
\end{claim*}
\begin{proof}
Note that \[ P(1,b) \implies f(b) = f(b+f(1)-1). \]
Otherwise, th... |
IMO-2009-notes_6 | Let $a_1$, $a_2$, \dots, $a_n$ be distinct positive integers and
let $M$ be a set of $n-1$ positive integers not containing $s = a_1 + \dots + a_n$.
A grasshopper is to jump along the real axis, starting at the point $0$ and
making $n$ jumps to the right with lengths $a_1$, $a_2$, \dots, $a_n$ in some order.
Prove that... | The proof is by induction on $n$.
Assume $a_1 < \dots < a_n$ and call each element of $M$ a \emph{mine}.
Let $x = s - a_n$.
We consider four cases, based on whether $x$ has a mine
and whether there is a mine past $x$.
\begin{itemize}
\ii If $x$ has no mine, and there is a mine past $x$,
then at most $n-2$ mines in ... |
IMO-2010-notes_1 | Find all functions $f \colon \RR \to \RR$ such that for all $x,y \in \RR$,
\[ f(\left\lfloor x\right\rfloor y) = f(x)\left\lfloor f(y)\right\rfloor. \] | 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) ... |
IMO-2010-notes_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... | 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 ... |
IMO-2010-notes_3 | Find all functions $g \colon \ZZ_{>0} \to \ZZ_{>0}$ such that
\[ \left( g(m)+n \right)\left( g(n)+m \right) \]
is always a perfect square. | 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')... |
IMO-2010-notes_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 $MK = ML$. | 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 $... |
IMO-2010-notes_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... | 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 $(5,1... |
IMO-2010-notes_6 | Let $a_1, a_2, a_3, \dots$ 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 ... | 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 t... |
IMO-2011-notes_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 ... | 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 ... |
IMO-2011-notes_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... | 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 p... |
IMO-2011-notes_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 $x \leq 0$. | 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... |
IMO-2011-notes_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 $f(m) \mid f(n)$. | Let $P(m,n)$ denote the given assertion.
First, we claim $f$ is even.
This is straight calculation:
\begin{itemize}
\ii $P(x,0) \implies f(x) \mid f(x)-f(0)
\implies f(x) \mid M \coloneq f(0)$.
\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)... |
IMO-2011-notes_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... | 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... |
IMO-2012-notes_1 | Let $ABC$ be a triangle and $J$ the center of the $A$-excircle.
This excircle is tangent to the side $BC$ at $M$,
and to the lines $AB$ and $AC$ at $K$ and $L$, respectively.
The lines $LM$ and $BJ$ meet at $F$, and the lines $KM$ and $CJ$ meet at $G$.
Let $S$ be the point of intersection of the lines $AF$ and $BC$,
an... | We employ barycentric coordinates with reference $\triangle ABC$.
As usual $a = BC$, $b = CA$, $c = AB$, $s = \half(a+b+c)$.
It's obvious that $K = ( -(s-c): s : 0)$, $M = ( 0 : s-b : s-c)$.
Also, $J = (-a : b : c)$.
We then obtain
\[ G = \left( -a: b : \frac{-as + (s-c)b}{s-b} \right). \]
It follows that
\[ T = \left... |
IMO-2012-notes_2 | Let $a_2$, $a_3$, \dots, $a_n$ be positive reals with product $1$,
where $n \ge 3$.
Show that
\[ (1+a_2)^2 (1+a_3)^3 \dots (1+a_n)^n > n^n. \] | Try the dumbest thing possible: by AM-GM,
\begin{align*}
(1 + a_2)^2 &\ge 2^2 a_2 \\
(1 + a_3)^3 = \left( \half + \half + a_3 \right)^3 &\ge \frac{3^3}{2^2} a_3 \\
(1 + a_4)^4 = \left( \frac13 + \frac13 + \frac13 + a_4 \right)^4
&\ge \frac{4^4}{3^3} a_4 \\
&\vdotswithin=
\end{align*}
and so on.
Multiplying ... |
IMO-2012-notes_3 | The liar's guessing game is a game played between two players $A$ and $B$.
The rules of the game depend on two fixed positive integers $k$ and $n$
which are known to both players.
At the start of the game $A$
chooses integers $x$ and $N$ with $1 \le x \le N$.
Player $A$ keeps $x$ secret, and truthfully tells $N$ to pl... | Call the players Alice and Bob.
\textbf{Part (a)}: We prove the following.
\begin{claim*}
If $N \ge 2^k+1$, then in $2k+1$ questions,
Bob can rule out some number in $\{1, \dots, 2^k+1\}$
form being equal to $x$.
\end{claim*}
\begin{proof}
First, Bob asks the question $S_0 = \{ 2^k+1 \}$
until Alice answers ``yes'... |
IMO-2012-notes_4 | Find all functions $f \colon \ZZ \to \ZZ$ such that,
for all integers $a$, $b$, $c$ that satisfy $a+b+c=0$,
the following equality holds:
\[ f(a)^2+f(b)^2+f(c)^2 = 2f(a)f(b)+2f(b)f(c)+2f(c)f(a). \] | Answer: for arbitrary $k \in \ZZ$, we have
\begin{enumerate}[(i)]
\ii $f(x) = kx^2$,
\ii $f(x) = 0$ for even $x$, and $f(x) = k$ for odd $x$, and
\ii $f(x) = 0$ for $x \equiv 0 \pmod 4$,
$f(x) = k$ for odd $x$, and $f(x) = 4k$ for $x \equiv 2 \pmod 4$.
\end{enumerate}
These can be painfully seen to work.
(It's ... |
IMO-2012-notes_5 | Let $ABC$ be a triangle with $\angle BCA = 90\dg$,
and let $D$ be the foot of the altitude from $C$.
Let $X$ be a point in the interior of the segment $CD$.
Let $K$ be the point on the segment $AX$ such that $BK = BC$.
Similarly, let $L$ be the point on the segment $BX$ such that $AL = AC$.
Let $M = \ol{AL} \cap \ol{BK... | Let $\omega_A$ and $\omega_B$ be the circles through $C$
centered at $A$ and $B$;
extend rays $AK$ and $BL$ to hit $\omega_B$ and $\omega_A$ again at $K^\ast$, $L^\ast$.
By radical center $X$,
we have $KLK^{\ast}L^{\ast}$ is cyclic,
say with circumcircle $\omega$.
\begin{center}
\begin{asy}
size(10cm);
pair ... |
IMO-2012-notes_6 | Find all positive integers $n$
for which there exist non-negative integers $a_1, a_2, \dots, a_n$
such that
\[ \frac{1}{2^{a_1}} + \frac{1}{2^{a_2}} + \dots + \frac{1}{2^{a_n}}
= \frac{1}{3^{a_1}} + \frac{2}{3^{a_2}} + \dots + \frac{n}{3^{a_n}}
= 1. \]
\end{enumerate} | The answer is $n \equiv 1, 2 \pmod 4$.
To see these are necessary,
note that taking the latter equation modulo $2$ gives
\[ 1 = \frac{1}{3^{a_1}} + \frac{2}{3^{a_2}} + \dots + \frac{n}{3^{a_n}}
\equiv 1 + 2 + .. + n \pmod 2. \]
Now we prove these are sufficient.
The following nice construction was posted on AOPS
by th... |
IMO-2013-notes_1 | Let $k$ and $n$ be positive integers.
Prove that there exist positive integers $m_1$, \dots, $m_k$
such that
\[ 1 + \frac{2^k-1}{n} = \left( 1 + \frac{1}{m_1} \right) \left( 1 + \frac{1}{m_2} \right)
\dots \left( 1 + \frac{1}{m_k} \right). \] | By induction on $k \ge 1$.
When $k = 1$ there is nothing to prove.
For the inductive step, if $n$ is even, write
\[
\frac{n + (2^k-1)}{n}
= \left( 1 + \frac{1}{n + (2^k-2)} \right) \cdot \frac{\frac n2 + (2^{k-1}-1)}{\frac n2}
\]
and use inductive hypothesis on the second term.
On the other hand if $n$ is odd then... |
IMO-2013-notes_3 | Let the excircle of triangle $ABC$ opposite
the vertex $A$ be tangent to the side $BC$ at the point $A_1$.
Define the points $B_1$ on $CA$ and $C_1$ on $AB$ analogously,
using the excircles opposite $B$ and $C$, respectively.
Suppose that the circumcenter of triangle $A_1B_1C_1$ lies
on the circumcircle of triangle $AB... | We ignore for now the given condition
and prove the following important lemma.
\begin{lemma*}
Let $(AB_1C_1)$ meet $(ABC)$ again at $X$.
From $BC_1 = B_1C$ follows $XC_1 = XB_1$,
and $X$ is the midpoint of major arc $\widehat{BC}$.
\end{lemma*}
\begin{proof}
This follows from the fact that we have
a spiral s... |
IMO-2013-notes_4 | Let $ABC$ be an acute triangle with orthocenter $H$,
and let $W$ be a point on the side $\ol{BC}$, between $B$ and $C$.
The points $M$ and $N$ are the feet of the altitudes
drawn from $B$ and $C$, respectively.
Suppose $\omega_1$ is the circumcircle of triangle $BWN$
and $X$ is a point such that $\ol{WX}$ is a diameter... | We present two solutions, an elementary one
and then an advanced one by moving points.
\paragraph{First solution, classical.}
Let $P$ be the second intersection of $\omega_1$ and $\omega_2$;
this is the Miquel point, so $P$ also lies on the circumcircle of $AMN$,
which is the circle with diameter $\ol{AH}$.
\begin{ce... |
IMO-2013-notes_5 | Suppose a function $f \colon \QQ_{>0} \to \RR$ satisfies:
\begin{enumerate}
\ii [(i)] If $x,y \in \QQ_{>0}$, then $f(x)f(y) \ge f(xy)$.
\ii [(ii)] If $x,y \in \QQ_{>0}$, then $f(x+y) \ge f(x) + f(y)$.
\ii [(iii)] There exists a rational number $a > 1$ with $f(a) = a$.
\end{enumerate}
Prove that $f(x) = x$ for all... | First, we dispense of negative situations by proving:
\begin{claim*}
For any integer $n > 0$, we have $f(n) \ge n$.
\end{claim*}
\begin{proof}
Note by induction on (ii) we have $f(nx) \ge n f(x)$.
Taking $(x,y) = (a,1)$ in (i) gives $f(1) \ge 1$,
and hence $f(n) \ge n$.
\end{proof}
\begin{claim*}
The $f$ tak... |
IMO-2013-notes_6 | Let $n \ge 3$ be an integer, and consider a circle with $n + 1$ equally spaced points marked on it.
Consider all labellings of these points with the numbers
$0, 1, \dots , n$ such that each label is used exactly once;
two such labellings are considered to be the same if
one can be obtained from the other by a rotation ... | First, here are half of the beautiful labellings up to reflection for $n = 6$,
just for concreteness.
\begin{center}
\begin{asy}
size(9cm);
pair g(int n) { return dir(90 + 60*n); }
picture ring(int a, int b, int c, int d, int e, real r) {
picture pic = new picture;
draw(pic, unitcircle);
draw(pic, dir(90)--dir... |
IMO-2014-notes_1 | Let $a_0 < a_1 < a_2 < \dotsb$ be an infinite sequence of positive integers.
Prove that there exists a unique integer $n\geq 1$ such that
\[ a_n < \frac{a_0+a_1+a_2+\dotsb+a_n}{n} \le a_{n+1}. \] | Fedor Petrov presents the following nice solution.
Let us define the sequence
\[ b_n = \left( a_n - a_{n-1} \right) + \dots + \left( a_n - a_1 \right). \]
Since $(a_n)_n$ is increasing, we get $(b_n)_n$ is strictly increasing,
and moreover $b_1 = 0$.
The problem requires an $n$ such that
\[ b_n < a_0 \le b_{n+1} \]
whi... |
IMO-2014-notes_3 | Convex quadrilateral $ABCD$ has $\angle ABC = \angle CDA = 90\dg$.
Point $H$ is the foot of the perpendicular from $A$ to $\ol{BD}$.
Points $S$ and $T$ lie on sides $AB$ and $AD$,
respectively, such that $H$ lies inside triangle $SCT$ and
\[ \angle CHS - \angle CSB = 90^{\circ},
\quad \angle THC - \angle DTC = 90^{\c... | \paragraph{First solution (mine).}
First we rewrite the angle condition in a suitable way.
\begin{claim*}
We have $\angle ATH = \angle TCH + 90\dg$.
Thus the circumcenter of $\triangle CTH$ lies on $\ol{AD}$.
Similarly the circumcenter of $\triangle CSH$ lies on $\ol{AB}$.
\end{claim*}
\begin{proof}
\begin{alig... |
IMO-2014-notes_4 | Let $P$ and $Q$ be on segment $BC$ of an acute triangle $ABC$
such that $\angle PAB=\angle BCA$ and $\angle CAQ=\angle ABC$.
Let $M$ and $N$ be points on $\ol{AP}$ and $\ol{AQ}$,
respectively, such that $P$ is the midpoint of $\ol{AM}$
and $Q$ is the midpoint of $\ol{AN}$.
Prove that $\ol{BM}$ and $\ol{CN}$ meet on the... | We give three solutions.
\paragraph{First solution by harmonic bundles.}
Let $\ol{BM}$ intersect the circumcircle again at $X$.
\begin{center}
\begin{asy}
pair A = dir(70);
pair B = dir(190);
pair C = dir(350);
filldraw(unitcircle, opacity(0.1)+palecyan, lightblue);
filldraw(A--B--C--cycle, opacity(0.1)+heavycyan, bl... |
IMO-2014-notes_5 | For every positive integer $n$,
the Bank of Cape Town issues coins of denomination $\frac 1n$.
Given a finite collection of such coins (of not necessarily different denominations)
with total value at most $99 + \frac12$, prove that it is possible to split
this collection into $100$ or fewer groups, such that each group... | We'll prove the result
for at most $k - \frac{k}{2k+1}$ with $k$ groups.
First, perform the following optimizations.
\begin{itemize}
\ii If any coin of size $\frac{1}{2m}$ appears twice,
then replace it with a single coin of size $\frac{1}{m}$.
\ii If any coin of size $\frac{1}{2m+1}$ appears $2m+1$ times,
grou... |
IMO-2014-notes_6 | A set of lines in the plane is in \emph{general position}
if no two are parallel and no three pass through the same point.
A set of lines in general position cuts the plane into regions,
some of which have finite area; we call these its \emph{finite regions}.
Prove that for all sufficiently large $n$,
in any set of $n$... | Suppose we have colored $k$ of the lines blue, and that
it is not possible to color any additional lines.
That means any of the $n-k$ non-blue lines
is the side of some finite region with
an otherwise entirely blue perimeter.
For each such line $\ell$, select one such region,
and take the next counterclockwise vertex;
... |
IMO-2015-notes_1 | We say that a finite set $\mathcal{S}$ of points in the plane
is \emph{balanced} if,
for any two different points $A$ and $B$ in $\mathcal{S}$,
there is a point $C$ in $\mathcal{S}$ such that $AC=BC$.
We say that $\mathcal{S}$ is \emph{center-free} if for
any three different points $A$, $B$ and $C$ in $\mathcal{S}$,
th... | For part (a), take a circle centered at a point $O$,
and add $n-1$ additional points by adding pairs of points
separated by an arc of $60^{\circ}$ or similar triples.
An example for $n = 6$ is shown below.
\begin{center}
\begin{asy}
size(4cm);
draw(unitcircle);
dotfactor *= 1.5;
pair O = (0,0);
dot("$O$", O, ... |
IMO-2015-notes_2 | Find all positive integers $a$, $b$, $c$ such that
each of $ab-c$, $bc-a$, $ca-b$ is a power of $2$
(possibly including $2^0=1$). | Here is the solution of \textbf{Telv Cohl},
which is the shortest solution I am aware of.
We will prove the only solutions are $(2,2,2)$, $(2,2,3)$,
$(2,6,11)$ and $(3,5,7)$ and permutations.
WLOG assume $a \ge b \ge c > 1$, so $ab-c \ge ca-b \ge bc-a$.
We consider the following cases:
\begin{itemize}
\ii If $a$ is ... |
IMO-2015-notes_3 | Let $ABC$ be an acute triangle with $AB > AC$.
Let $\Gamma$ be its circumcircle, $H$ its orthocenter, and $F$ the foot of the altitude from $A$.
Let $M$ be the midpoint of $\ol{BC}$.
Let $Q$ be the point on $\Gamma$ such that $\angle HQA = 90\dg$
and let $K$ be the point on $\Gamma$ such that $\angle HKQ = 90\dg$.
Assu... | Let $L$ be on the nine-point circle with $\angle HML = 90^{\circ}$.
The negative inversion at $H$ swapping $\Gamma$ and nine-point circle maps
\[ A \longleftrightarrow F, \quad
Q \longleftrightarrow M, \quad
K \longleftrightarrow L. \]
In the inverted statement, we want line $ML$ to be tangent to $(AQL)$.
\begin{c... |
IMO-2015-notes_4 | Triangle $ABC$ has circumcircle $\Omega$ and circumcenter $O$.
A circle $\Gamma$ with center $A$
intersects the segment $BC$ at points $D$ and $E$,
such that $B$, $D$, $E$, and $C$ are all different
and lie on line $BC$ in this order.
Let $F$ and $G$ be the points of intersection of $\Gamma$ and $\Omega$,
such that $A$... | Since $\ol{AO} \perp \ol{FG}$ for obvious reasons,
we will only need to show that $XF = XG$,
or that $\dang KFG = \dang LGF$.
Let line $FG$ meet $(BDF)$ and $(CGE)$
again at $F_2$ and $G_2$.
\begin{center}
\begin{asy}
size(10cm);
pair A = dir(105);
pair B = dir(200);
pair C = dir(340);
real r = 1.337;
pair F = IP(CR(... |
IMO-2015-notes_5 | Solve the functional equation
\[ f(x+f(x+y)) + f(xy) = x + f(x+y) + yf(x) \]
for $f \colon \RR \to \RR$. | The answers are $f(x) \equiv x$ and $f(x) \equiv 2-x$.
Obviously, both of them work.
Let $P(x,y)$ be the given assertion.
We also will let $S = \{t \mid f(t) = t\}$
be the set of fixed points of $f$.
\begin{itemize}
\ii From $P(0,0)$ we get $f(f(0)) = 0$.
\ii From $P(0,f(0))$ we get $2f(0) = f(0)^2$
and hence $f... |
IMO-2015-notes_6 | The sequence $a_1,a_2,\dots$ of integers satisfies the conditions:
\begin{enumerate}[(i)]
\ii $1\le a_j\le2015$ for all $j\ge1$,
\ii $k+a_k\neq \ell+a_\ell$ for all $1\le k<\ell$.
\end{enumerate}
Prove that there exist two positive integers $b$ and $N$ for which
\[ \left\lvert\sum_{j=m+1}^n (a_j-b) \right\rvert \le... | We give two equivalent solutions with different presentations,
one with ``arrows'' and the other by ``juggling''.
\paragraph{First solution (arrows).}
Consider the map
\[ f \colon k \mapsto k + a_k. \]
This map is injective, so if we draw all arrows of the form $k \mapsto f(k)$
we get a partition of $\NN$ into one or ... |
IMO-2016-notes_1 | In convex pentagon $ABCDE$ with $\angle B > 90\dg$,
let $F$ be a point on $\ol{AC}$ such that $\angle FBC = 90\dg$.
It is given that $FA=FB$, $DA=DC$, $EA=ED$,
and rays $\ol{AC}$ and $\ol{AD}$ trisect $\angle BAE$.
Let $M$ be the midpoint of $\ol{CF}$.
Let $X$ be the point such that $AMXE$ is a parallelogram.
Show that... | Here is a ``long'' solution which I think
shows where the ``power'' in the configuration comes from
(it should be possible to come up with shorter solutions
by cutting more directly to the desired conclusion).
Throughout the proof, we let
\[ \theta = \angle FAB = \angle FBA = \angle DAC = \angle DCA
= \angle EAD = \a... |
IMO-2016-notes_2 | Find all integers $n$ for which each cell of $n \times n$ table
can be filled with one of the letters $I$, $M$ and $O$ in such a way that:
\begin{itemize}
\item In each row and column, one third of the entries are $I$,
one third are $M$ and one third are $O$; and
\item in any diagonal, if the number of entries on the ... | The answer is $n$ divisible by $9$.
First we construct $n=9$ and by extension every multiple of $9$.
\[
\begin{array}{|ccc|ccc|ccc|} \hline
I & I & I & M & M & M & O & O & O \\
M & M & M & O & O & O & I & I & I \\
O & O & O & I & I & I & M & M & M \\\hline
I & I & I & M & M & M & O & O & O \\
M & M & M & O ... |
IMO-2016-notes_3 | Let $P=A_1A_2\dots A_k$ be a convex polygon in the plane.
The vertices $A_1$, $A_2$, \dots, $A_k$ have integral coordinates
and lie on a circle. Let $S$ be the area of $P$.
An odd positive integer $n$ is given such that
the squares of the side lengths of $P$ are integers divisible by $n$.
Prove that $2S$ is an integer ... | Solution by Jeck Lim:
We will prove the result just for $n = p^e$
where $p$ is an odd prime and $e \ge 1$.
The case $k=3$ is resolved by Heron's formula directly:
we have $S = \frac14\sqrt{2(a^2b^2 + b^2c^2 + c^2a^2) - a^4-b^4-c^4}$,
so if $p^e \mid \gcd(a^2,b^2,c^2)$ then $p^{2e} \mid S^2$.
Now we show we can pick a ... |
IMO-2016-notes_4 | A set of positive integers is called \emph{fragrant}
if it contains at least two elements and each of its elements
has a prime factor in common with at least one of the other elements.
Let $P(n)=n^2+n+1$.
What is the smallest possible positive integer value of $b$ such that
there exists a non-negative integer $a$ for w... | The answer is $b = 6$.
First, we prove $b \ge 6$ must hold.
It is not hard to prove the following divisibilities by Euclid:
\begin{align*}
\gcd(P(n), P(n+1)) &\mid 1 \\
\gcd(P(n), P(n+2)) &\mid 7 \\
\gcd(P(n), P(n+3)) &\mid 3 \\
\gcd(P(n), P(n+4)) &\mid 19.
\end{align*}
Now assume for contradiction $b \le 5$.
Then... |
IMO-2016-notes_5 | The equation
\[ (x-1)(x-2)\dots(x-2016)=(x-1)(x-2)\dots (x-2016) \]
is written on the board, with $2016$ linear factors on each side.
What is the least possible value of $k$ for which it is possible to
erase exactly $k$ of these $4032$ linear factors so that at least
one factor remains on each side and the resulting eq... | The answer is $2016$.
Obviously this is necessary in order to delete duplicated factors.
We now prove it suffices to deleted $2 \pmod 4$ and $3 \pmod 4$
guys from the left-hand side, and $0 \pmod 4$,
$1 \pmod 4$ from the right-hand side.
Consider the $1008$ inequalities
\begin{align*}
(x-1)(x-4) &< (x-2)(x-3) \\
(... |
IMO-2016-notes_6 | There are $n\ge 2$ line segments in the plane such that
every two segments cross and no three segments meet at a point.
Geoff has to choose an endpoint of each segment and place a frog
on it facing the other endpoint. Then he will clap his hands $n-1$ times.
Every time he claps, each frog will immediately jump forward
... | The following solution was communicated to me by Yang Liu.
Imagine taking a larger circle $\omega$ encasing
all $\binom{n}{2}$ intersection points.
Denote by $P_1$, $P_2$, \dots, $P_{2n}$ the order of the points on $\omega$
in clockwise order; we imagine placing the frogs on $P_i$ instead.
Observe that, in order for e... |
IMO-2017-notes_1 | For each integer $a_0 > 1$, define the sequence $a_0$, $a_1$, $a_2$,
\dots, by
\[
a_{n+1} =
\begin{cases}
\sqrt{a_n} & \text{if $\sqrt{a_n}$ is an integer,} \\
a_n + 3 & \text{otherwise}
\end{cases}
\]
for each $n \ge 0$.
Determine all values of $a_0$ for which there is a number $A$
such that $a_n = A$ fo... | The answer is $a_0 \equiv 0 \pmod 3$ only.
\paragraph{First solution.}
We first compute the minimal term of any sequence, periodic or not.
\begin{lemma*}
Let $c$ be the smallest term in $a_n$.
Then either $c \equiv 2 \pmod 3$ or $c = 3$.
\end{lemma*}
\begin{proof}
Clearly $c \neq 1, 4$.
Assume $c \not\equiv 2... |
IMO-2017-notes_2 | Solve over $\RR$ the functional equation
\[ f\left( f(x)f(y) \right) + f(x+y) = f(xy). \] | The only solutions are $f(x) = 0$, $f(x) = x-1$ and $f(x)=1-x$,
which clearly work.
Note that
\begin{itemize}
\ii If $f$ is a solution, so is $-f$.
\ii Moreover, if $f(0)=0$ then setting $y=0$ gives $f\equiv0$.
So henceforth we assume $f(0)>0$.
\end{itemize}
\begin{claim*}
We have $f(z) = 0 \iff z =1$.
Also... |
IMO-2017-notes_3 | A hunter and an invisible rabbit play a game in the plane.
The rabbit and hunter start at points $A_0 = B_0$.
In the $n$th round of the game ($n \ge 1$), three things occur in order:
\begin{enumerate}[(i)]
\ii The rabbit moves invisibly from $A_{n-1}$ to a point $A_n$
such that $A_{n-1} A_n = 1$.
\ii The hunter h... | No, the hunter cannot.
We will show how to increase the distance in the following way:
\begin{claim*}
Suppose the rabbit is at a distance $d \ge 1$ from the hunter
at some point in time.
Then it can increase its distance to at least
$\sqrt{d^2+\half}$ in $4d$ steps
regardless of what the hunter already knows... |
IMO-2017-notes_4 | Let $R$ and $S$ be different points on a circle $\Omega$
such that $\ol{RS}$ is not a diameter.
Let $\ell$ be the tangent line to $\Omega$ at $R$.
Point $T$ is such that $S$ is the midpoint of $\ol{RT}$.
Point $J$ is chosen on minor arc $RS$ of $\Omega$ so that
the circumcircle $\Gamma$ of triangle $JST$ intersects $\e... | \paragraph{First solution (elementary).}
First, note
\[ \dang RKA = \dang RKJ = \dang RSJ = \dang TSJ = \dang TAJ = \dang TAK \]
so $\ol{RK} \parallel \ol{AT}$.
Now,
\begin{itemize}
\ii $\ol{RA}$ is tangent at $R$ iff $\triangle KRS \sim \triangle RTA$ (oppositely),
because both equate to $-\dang RKS = \dang SKR = ... |
IMO-2017-notes_5 | Fix $N \ge 1$. A collection of $N(N+1)$ soccer players of distinct
heights stand in a row.
Sir Alex Song wishes to remove $N(N-1)$ players from this row
to obtain a new row of $2N$ players in which the following $N$
conditions hold: no one stands between the two tallest players,
no one stands between the third and four... | Some opening remarks:
\textbf{location and height are symmetric to each other},
if one thinks about this problem as permutation pattern avoidance.
So while officially there are multiple solutions,
they are basically isomorphic to one another,
and I am not aware of any solution otherwise.
\begin{center}
\begin{asy}
siz... |
IMO-2017-notes_6 | An \emph{irreducible lattice point} is an ordered pair
of integers $(x,y)$ satisfying $\gcd(x,y) = 1$.
Prove that if $S$ is a finite set of irreducible lattice points
then there exists a nonconstant
\emph{homogeneous} polynomial $f(x,y)$ with integer coefficients
such that $f(x,y)=1$ for each $(x,y) \in S$.
\end{enume... | We present two solutions.
\paragraph{First solution (Dan Carmon, Israel).}
We prove the result by induction on $|S|$,
with the base case being Bezout's Lemma ($n=1$).
For the inductive step, suppose we want to add a given pair
$(a_{m+1},b_{m+1})$ to $\left\{ (a_1, \dots, a_m), (b_1, \dots, b_m) \right\}$.
\begin{claim... |
IMO-2018-notes_1 | Let $\Gamma$ be the circumcircle of acute triangle $ABC$.
Points $D$ and $E$ lie on segments $AB$ and $AC$,
respectively, such that $AD = AE$.
The perpendicular bisectors of $\ol{BD}$ and $\ol{CE}$
intersect the minor arcs $AB$ and $AC$ of $\Gamma$
at points $F$ and $G$, respectively.
Prove that the lines $DE$ and $FG$... | We present a synthetic solution from the IMO shortlist
as well as a complex numbers approach.
We also outline a trig solution (the one I found at IMO),
and a fourth solution from Derek Liu.
\paragraph{Synthetic solution (from Shortlist).}
Construct parallelograms $AXFD$ and $AEGY$,
noting that $X$ and $Y$ lie on $\Ga... |
IMO-2018-notes_2 | Find all integers $n \geq 3$ for which
there exist real numbers $a_1, a_2, \dots, a_n$ satisfying
\[ a_i a_{i+1} +1 = a_{i+2} \]
for $i=1,2, \dots, n$, where indices are taken modulo $n$. | The answer is $3 \mid n$,
achieved by $(-1,-1,2,-1,-1,2,\dots)$.
We present two solutions.
\paragraph{First solution by inequalities.}
We compute $a_i a_{i+1} a_{i+2}$ in two ways:
\begin{align*}
a_i a_{i+1} a_{i+2} &= [a_{i+2}-1]a_{i+2} = a_{i+2}^2 - a_{i+2} \\
&= a_i [a_{i+3}-1] = a_i a_{i+3} - a_i.
\end{align*}... |
IMO-2018-notes_3 | An \emph{anti-Pascal triangle} is an equilateral triangular array
of numbers such that, except for the numbers in the bottom row,
each number is the absolute value of the difference
of the two numbers immediately below it.
For example, the following array is an anti-Pascal triangle
with four rows which contains every i... | The answer is no, there is no anti-Pascal triangle
with the required properties.
Let $n = 2018$ and $N = 1+2+\dots+n$.
For every number $d$ not in the bottom row,
draw an arrow from $d$ to the larger of the two numbers below it
(i.e.\ if $d=a-b$, draw $d \to a$).
This creates an \emph{oriented forest} (which looks lik... |
IMO-2018-notes_4 | A \emph{site} is any point $(x,y)$ in the plane
for which $x,y \in \{1, \dots, 20\}$.
Initially, each of the $400$ sites is unoccupied.
Amy and Ben take turns placing stones on unoccupied sites,
with Amy going first;
Amy has the additional restriction that no two of her stones
may be at a distance equal to $\sqrt5$.
Th... | The answer is $K = 100$.
First, we show Amy can always place at least $100$ stones.
Indeed, treat the problem as a grid with checkerboard coloring.
Then Amy can choose to always play on one of the $200$ black squares.
In this way, she can guarantee half the black squares,
i.e.\ she can get $\half \cdot 200 = 100$ ston... |
IMO-2018-notes_5 | Let $a_1$, $a_2$, \dots\ be an infinite sequence of positive integers,
and $N$ a positive integer.
Suppose that for all integers $n \ge N$, the expression
\[ \frac{a_1}{a_2} + \frac{a_2}{a_3} + \dots
+ \frac{a_{n-1}}{a_n} + \frac{a_n}{a_1} \]
is an integer.
Prove that $(a_n)$ is eventually constant. | The condition implies that the difference
\[ S(n) = \frac{a_{n+1} - a_n}{a_1} + \frac{a_n}{a_{n+1}} \]
is an integer for all $n > N$.
We proceed by $p$-adic valuation only henceforth;
fix a prime $p$.
Then analyzing the $\nu_p$, we immediately get that for $n > N$:
\begin{itemize}
\ii If $\nu_p(a_n) < \nu_p(a_{n+1})$, ... |
IMO-2018-notes_6 | A convex quadrilateral $ABCD$ satisfies $AB \cdot CD = BC \cdot DA$.
Point $X$ lies inside $ABCD$ so that
\[ \angle XAB=\angle XCD \quad \text{ and } \quad \angle XBC=\angle XDA. \]
Prove that $\angle BXA + \angle DXC=180\dg$.
\end{enumerate} | We present two solutions by inversion.
The first is the official one.
The second is a solution via inversion, completed by USA5 Michael Ren.
\paragraph{Official solution by inversion.}
In what follows a convex quadrilateral is called
\emph{quasi-harmonic} if $AB \cdot CD = BC \cdot DA$.
\begin{claim*}
A quasi-harmo... |
IMO-2019-notes_1 | Solve over $\ZZ$ the functional equation
$f(2a) + 2f(b) = f(f(a+b))$. | Notice that $f(x) \equiv 0$ or $f(x) \equiv 2x+k$ work
and are clearly the only linear solutions.
We now prove all solutions are linear.
Let $P(a,b)$ be the assertion.
\begin{claim*}
For each $x \in \ZZ$ we have $f(2x) = 2f(x) - f(0)$.
\end{claim*}
\begin{proof}
Compare $P(0,x)$ and $P(x,0)$.
\end{proof}
Now, $P(a... |
IMO-2019-notes_2 | In triangle $ABC$ point $A_1$ lies on side $BC$
and point $B_1$ lies on side $AC$.
Let $P$ and $Q$ be points on segments $AA_1$ and $BB_1$,
respectively, such that $\ol{PQ} \parallel \ol{AB}$.
Point $P_1$ is chosen on ray $PB_1$ beyond $B_1$
such that $\angle PP_1C = \angle BAC$.
Point $Q_1$ is chosen on ray $QA_1$ bey... | We present two solutions.
\paragraph{First solution by bary (Evan Chen).}
Let $PB_1$ and $QA_1$ meet line $AB$ at $X$ and $Y$.
Since $\ol{XY} \parallel \ol{PQ}$ it is equivalent
to show $P_1XYQ_1$ is cyclic (Reim's theorem).
Note the angle condition implies $P_1CXA$ and $Q_1CYB$ are cyclic.
Letting $T = \ol{PX} \cap... |
IMO-2019-notes_3 | A social network has $2019$ users, some pairs of which are friends (friendship is symmetric).
If $A$, $B$, $C$ are three users such that $AB$ are friends and $AC$ are friends but $BC$ is not,
then the administrator may perform the following operation:
change the friendships such that $BC$ are friends, but $AB$ and $AC$... | We take the obvious graph formulation
and call the move a \emph{toggle}.
\begin{claim*}
Let $G$ be a connected graph.
Then one can toggle $G$ without disconnecting the graph,
unless $G$ is a clique, a cycle, or a tree.
\end{claim*}
\begin{proof}
Assume $G$ is connected and not a tree, so it has a cycle.
Take... |
IMO-2019-notes_4 | Solve over positive integers the equation
\[ k! = \prod_{i=0}^{n-1} (2^n-2^i) = (2^n-1)(2^n-2)(2^n-4) \dots (2^n-2^{n-1}). \] | The answer is $(n,k) =(1,1)$ and $(n,k) = (2,3)$ which work.
Let $A = \prod_i (2^n-2^k)$, and assume $A = k!$ for some $k \ge 3$.
Recall by exponent lifting that
\[ \nu_3(2^t-1) = \begin{cases}
0 & t \text{ odd} \\
1 + \nu_3(t/2) & t \text{ even}.
\end{cases} \]
Thus, we can compute
\begin{align*}
k > \nu_... |
IMO-2019-notes_5 | Let $n$ be a positive integer.
Harry has $n$ coins lined up on his desk, which can show either heads or tails.
He does the following operation: if there are $k$ coins which show heads and $k > 0$,
then he flips the $k$th coin over; otherwise he stops the process.
(For example, the process starting with $THT$ would be
$... | The answer is \[ E_n = \half (1 + \dots + n) = \frac14 n(n+1) \]
which is finite.
We'll represent the operation by a
directed graph $G_n$ on vertices $\{0,1\}^n$
(each string points to its successor)
with $1$ corresponding to heads and $0$ corresponding to tails.
For $b \in \{0,1\}$ we let $\ol b = 1-b$,
and denote bi... |
IMO-2019-notes_6 | Let $ABC$ be a triangle with incenter $I$ and incircle $\omega$.
Let $D$, $E$, $F$ denote the tangency points of $\omega$ with $\ol{BC}$, $\ol{CA}$, $\ol{AB}$.
The line through $D$ perpendicular to $\ol{EF}$ meets $\omega$ again at $R$ (other than $D$),
and line $AR$ meets $\omega$ again at $P$ (other than $R$).
Suppos... | We present five solutions.
\paragraph{First solution by complex numbers (Evan Chen, with Yang Liu).}
We use complex numbers with $D=x$, $E=y$, $F=z$.
\begin{center}
\begin{asy}
pair I = origin;
pair D = dir(110);
pair E = dir(210);
pair F = dir(330);
draw(D--E--F--cycle, blue);
filldraw(unitcircle, opacity(0.1)+lightc... |
IMO-2020-notes_1 | Consider the convex quadrilateral $ABCD$.
The point $P$ is in the interior of $ABCD$.
The following ratio equalities hold:
\[\angle PAD:\angle PBA:\angle DPA
= 1:2:3
= \angle CBP:\angle BAP:\angle BPC.\]
Prove that the following three lines meet in a point:
the internal bisectors of angles $\angle ADP$ and $\angle ... | Let $O$ denote the circumcenter of $\triangle PAB$.
We claim it is the desired concurrency point.
\begin{center}
\begin{asy}
pair O = origin;
pair P = dir(118);
pair A = dir(190);
pair B = dir(350);
pair C = extension(P/dir(64), B, P, B*dir(192));
pair D = extension(P*dir(36), A, P, A/dir(108));
filldraw(unitcircle, o... |
IMO-2020-notes_2 | Let $a \ge b \ge c \ge d > 0$ be real numbers satisfying $a+b+c+d=1$.
Prove that
\[ (a+2b+3c+4d) a^a b^b c^c d^d < 1. \] | By weighted AM-GM we have
\[ a^a b^b c^c d^d \le \sum_{\text{cyc}} \frac{a}{a+b+c+d} \cdot a
= a^2+b^2+c^2+d^2. \]
So, it is enough to prove that
\[ (a^2+b^2+c^2+d^2)(a+2b+3c+4d) \le 1 = (a+b+c+d)^3. \]
Expand both sides to get
\[
\begin{array}{cccc}
+a^3 &+ b^2a &+ c^2a & +d^2a \\
+2a^2b &+ 2b^3 &+ 2bc^2 ... |
IMO-2020-notes_3 | There are $4n$ pebbles of weights $1, 2, 3, \dots, 4n$.
Each pebble is coloured in one of $n$ colours
and there are four pebbles of each colour.
Show that we can arrange the pebbles into two piles
so the total weights of both piles are the same,
and each pile contains two pebbles of each colour. | The first key idea is the deep fact that
\[ 1+4n = 2+(4n-1) = 3+(4n-2) = \dotsb. \]
So, place all four pebbles of the same colour in a box (hence $n$ boxes).
For each $k=1,2,\dots,2n$
we tape a piece of string between pebble $k$ and $4n+1-k$.
To solve the problem, it suffices to paint each string
either blue or green ... |
IMO-2020-notes_4 | There is an integer $n > 1$.
There are $n^2$ stations on a slope of a mountain, all at different altitudes.
Each of two cable car companies, $A$ and $B$, operates $k$ cable cars;
each cable car provides a transfer from one of the stations
to a higher one (with no intermediate stops).
The $k$ cable cars of $A$ have $k$ ... | Answer: $k = n^2 - n + 1$.
When $k = n^2-n$,
the construction for $n=4$ is shown below which generalizes readily.
(We draw $A$ in red and $B$ in blue.)
\begin{center}
\begin{asy}
pair P(int i, int j) { return (i+j/10, j+i/10); }
dotfactor *= 2;
for (int i=0; i<4; ++i) {
for (int j=0; j<4; ++j) {
dot( P(i,j... |
IMO-2020-notes_5 | A deck of $n > 1$ cards is given.
A positive integer is written on each card.
The deck has the property that the arithmetic mean of the
numbers on each pair of cards is also the
geometric mean of the numbers on some collection of one or more cards.
For which $n$ does it follow that the numbers on the cards are all equa... | The assertion is true for all $n$.
\bigskip
\textbf{Setup (boilerplate).}
Suppose that $a_1$, \dots, $a_n$ satisfy the required properties
but are not all equal.
Let $d = \gcd(a_1, \dots, a_n) > 1$
then replace $a_1$, \dots, $a_n$ by
$\frac{a_1}{d}$, \dots, $\frac{a_n}{d}$.
Hence without loss of generality we may ass... |
IMO-2020-notes_6 | Consider an integer $n > 1$, and a set $\mathcal S$ of $n$ points
in the plane such that the distance between any two different points
in $\mathcal S$ is at least $1$.
Prove there is a line $\ell$ separating $\mathcal S$
such that the distance from any point of $\mathcal S$ to $\ell$
is at least $\Omega(n^{-1/3})$.
(A... | We present the official solution given by the Problem Selection Committee.
Let's suppose that among all projections
of points in $\mathcal S$ onto some line $m$,
the maximum possible distance between two consecutive projections is $\delta$.
We will prove that $\delta \ge \Omega(n^{-1/3})$,
solving the problem.
We mak... |
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