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 $p^k \not\equiv 1 \pmod q$.
To do this, we first recall the following lemma.
\begin{lemma*}
Let $\Phi_p(X) = 1 + X + X^2 + \dots + X^{p-1}$.
For any integer $a$, if $q$ is a prime divisor of $\Phi_p(a)$ other than $p$,
then $a \pmod q$ has order $p$. (In particular, $q \equiv 1 \pmod p$.)
\end{lemma*}
\begin{proof}
We have $a^p-1 \equiv 0 \pmod q$, so either the order is $1$ or $p$.
If it is $1$, then $a \equiv 1 \pmod q$, so $q \mid \Phi_p(1) = p$, hence $q = p$.
\end{proof}
% Wishfully we hope the order of $p$ is $p$ and $k \nmid p$.
Now the idea is to extract a prime factor $q$
from the cyclotomic polynomial
\[ \Phi_p(p) = \frac{p^p-1}{p-1} \equiv 1+p \pmod{p^2} \]
such that $q \not\equiv 1 \pmod{p^2}$;
hence $k \not\equiv 0 \pmod p$,
and as $p \pmod q$ has order $p$ we have $p^k \not\equiv 1 \pmod q$. |
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 $BMR$ and $CNR$
have a common point lying on the side $BC$. | 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, C);
filldraw(A--B--C--cycle, opacity(0.1)+lightblue, blue);
draw(B--N, blue);
draw(C--M, blue);
filldraw(circumcircle(A, M, N), opacity(0.1)+lightgreen, heavygreen);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$H$", H, dir(H));
dot("$M$", M, dir(M));
dot("$N$", N, dir(N));
dot("$O$", O, dir(O));
dot("$R$", R, dir(R));
// dot("$T$", T, dir(T));
/* TSQ Source:
A = dir 130
B = dir 210
C = dir 330
H = A+B+C
M = foot C A B
N = foot B A C
O = midpoint B--C
R = extension A incenter A B C O midpoint M--N
T = extension H R B C
A--B--C--cycle 0.1 lightblue / blue
B--N blue
C--M blue
circumcircle A M N 0.1 lightgreen / heavygreen
R--T--B blue
*/
\end{asy}
\end{center}
In fact, since the bisector of $\angle MON$
is just the perpendicular bisector of $\ol{MN}$,
the point $R$ is actually just the arc midpoint
of $\widehat{MN}$ of $(AMN)$ as desired. |
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 $n = \deg f$ (which is even).
As $n \ge 7 \implies (8/7)^n > 2$, this means that we must have $n \le 6$.
Moreover, taking modulo $7$ we have $3^n + 5^n \equiv 6 \pmod 7$
which gives $n \equiv 2, 4 \pmod 6$.
Thus $\deg P \le 4$, which (combined with $P$ even) resolves the problem. |
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)*unitsquare);
draw(shift(2,1)*unitsquare);
draw(shift(2,2)*unitsquare);
\end{asy}
\end{center}
Which $m \times n$ rectangles can be tiled by hooks? | 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 hence by hooks).
\end{claim*}
\begin{proof}
If $3 \mid m$ and $4 \mid n$, this is clear.
Else suppose $12 \mid m$ but $3 \nmid n$, $4 \nmid n$.
Then $n \ge 7$, so it can be written in the form
$3a+4b$ for nonnegative integers $a$ and $b$, which permits a tiling.
\end{proof}
We now prove these conditions are necessary.
It is not hard to see that $m,n \notin \{1,2,5\}$ is necessary.
We thus turn our attention to divisibility conditions.
Each hook has a \emph{hole}, and if we associate each hook with
the one that fills its hole, we get a bijective pairing of hooks.
Thus the number of cells is divisible by $12$,
and the cells come grouped into two possible shapes,
which we will call \textbf{tiles} shown below,
(rotations and reflections permitted).
\begin{center}
\begin{asy}
unitsize(0.5 cm);
filldraw(unitsquare, white, black);
filldraw(shift(0,1)*unitsquare, white, black);
filldraw(shift(0,2)*unitsquare, white, black);
filldraw(shift(1,2)*unitsquare, white, black);
filldraw(shift(2,1)*unitsquare, white, black);
filldraw(shift(2,2)*unitsquare, white, black);
filldraw(shift(1,0)*unitsquare, gray, black);
filldraw(shift(1,1)*unitsquare, gray, black);
filldraw(shift(2,0)*unitsquare, gray, black);
filldraw(shift(3,0)*unitsquare, gray, black);
filldraw(shift(3,1)*unitsquare, gray, black);
filldraw(shift(3,2)*unitsquare, gray, black);
\end{asy}
\qquad
\begin{asy}
unitsize(0.5 cm);
filldraw(unitsquare, white, black);
filldraw(shift(0,1)*unitsquare, white, black);
filldraw(shift(0,2)*unitsquare, white, black);
filldraw(shift(1,2)*unitsquare, white, black);
filldraw(shift(2,1)*unitsquare, white, black);
filldraw(shift(2,2)*unitsquare, white, black);
filldraw(shift(1,0)*unitsquare, gray, black);
filldraw(shift(1,1)*unitsquare, gray, black);
filldraw(shift(1,-1)*unitsquare, gray, black);
filldraw(shift(0,-1)*unitsquare, gray, black);
filldraw(shift(-1,-1)*unitsquare, gray, black);
filldraw(shift(-1,0)*unitsquare, gray, black);
\end{asy}
\end{center}
In particular, the total number of cells is divisible by $12$.
Thus the problem is reduced to proving that:
\begin{claim*}
if a $6a \times 2b$ rectangle is tiled by tiles,
then at least one of $a$ and $b$ is even.
\end{claim*}
\begin{proof}
Note that the tiles can be further classified into two types:
\begin{itemize}
\ii \textbf{First type}: These tiles have exactly four columns,
each with exactly three cells (an odd number).
Moreover, all rows have an even number of cells (either $2$ or $4$).
\ii \textbf{Second type}: vice-versa.
These tiles have exactly four rows,
each with exactly three cells (an odd number).
Moreover, all rows have an odd number of cells.
\end{itemize}
We claim that any tiling uses an even number of each type, which is enough.
By symmetry we prove an even number of first-type tiles.
Color red every fourth column of the rectangle.
The number of cells colored red is even.
The tiles of the second type cover an even number of red cells,
and the tiles of the first type cover an odd number of red cells.
Hence the number of tiles of the first type must be even.
\end{proof}
\begin{remark*}
This shows that a rectangle can be tiled by hooks
if and only if it can be tiled by $3 \times 4$ rectangles.
But there exist tilings which do not decompose into $3 \times 4$;
see e.g.\ \url{https://aops.com/community/c6h14023p99881}.
\end{remark*} |
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 $1 \le i < j < k \le n$. | 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} + \frac{t_j}{t_i} \right) \\
&\ge n + \sum_{1 \le i < j \le 3}
\left( \frac{t_i}{t_j} + \frac{t_j}{t_i} \right)
+ \sum_{\substack{1 \le i < j \le n \\ j > 3 }} 2 \\
&= n + 2\left( \binom n2-3 \right)
+ \left( \frac ab + \frac ba \right)
+ \frac{a+b}{c} + \frac{c}{b} + \frac{c}{a} \\
&\ge n + 2\left( \binom n2-3 \right) + 2
+ \frac{a+b}{c} + c \cdot \frac{4}{a+b}
\end{align*}
So, we conclude that
\[ \frac{a+b}{c} + \frac{4c}{a+b} < 5 \]
which rearranges to
\[ \left( 4c-(a+b) \right)\left( c-(a+b) \right) < 0. \]
This is enough to imply $c < a+b$.
\begin{remark*}
A variant of the same argument allows one to improve
the left-hand side to $(n+\sqrt{10}-3)^2$.
One does so by writing
\[ \text{RHS} \ge \left( \sqrt{\left( a+b+c \right)
\left( \frac1a+\frac1b+\frac1c \right)} + (n-3) \right)^2 \]
and estimating the square root as in the previous approach.
In addition, $(n+\sqrt{10}-3)^2$ is best possible,
as seen by taking $(a,b,c) = (2,1,1)$ and $t_4 = t_5 = \dots = \frac25 \sqrt{10}$.
\end{remark*} |
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 the point $P$ does not lie on line $BD$
and we assume this henceforth.
\paragraph{Barycentric solution (from teenager Evan).}
Apply barycentric coordinates to $\triangle PBD$
(nondegenerate by the non-bisect condition)
with $P = (1,0,0)$, $B = (0,1,0)$ and $D = (0,0,1)$.
Define $a = BD$, $b = DP$ and $c = PB$.
Since $A$ and $C$ are isogonal conjugates with respect to $\triangle PBD$,
we set \[ A = (au : bv : cw) \qquad\text{and}\qquad
C = \left( \frac au : \frac bv : \frac cw \right). \]
For brevity define $M = au + bv + cw$ and $N = avw + bwu + cuv$.
We now compute each condition.
\begin{claim*}
Quadrilateral $ABCD$ is cyclic if and only if $N^2 = u^2 M^2$.
\end{claim*}
\begin{proof}
W know a circle through $B$ and $D$
is a locus of points
with \[ \frac{a^2yz+b^2zx+c^2xy}{x(x+y+z)} \]
is equal to some constant.
Therefore quadrilateral $ABCD$ is cyclic if and only if
$\frac{abc \cdot N}{au \cdot M}$
is equal to $\frac{abc \cdot uvw \cdot M}{avw \cdot N}$
which rearranges to $N^2 = u^2M^2$.
\end{proof}
\begin{claim*}
We have $PA = PC$ if and only if $N^2 = u^2 M^2$.
\end{claim*}
\begin{proof}
We have the displacement vector
$\overrightarrow{PA}
= \tfrac{1}{M} \left( bv+cw , -bv ,-cw \right)$.
Therefore,
\begin{align*}
M^2 \cdot \left\lvert PA \right\rvert^2
&= -a^2(bv)(cw) + b^2(cw)(bv+cw) + c^2(bv)(bv+cw) \\
&= bc(-a^2vw + (bw+cv)(bv+cw)).
\end{align*}
In a similar way
(by replacing $u$, $v$, $w$ with their inverses)
we have
\begin{align*}
\left( \frac{N}{uvw} \right)^2 \cdot \left\lvert PC \right\rvert^2
&= (vw)^{-2} \cdot bc(-a^2vw + (bv+cw)(bw+cv)) \\
\iff N^2 \cdot \left\lvert PC \right\rvert^2
& = u^2 bc(-a^2vw + (bw+cv)(bv+cw))
\end{align*}
These are equal if and only if $N^2 = u^2M^2$, as desired.
\end{proof}
\paragraph{Angle chasing solution (Evan Chen and Petko Lazarov).}
The solution consists of two parts.
The first part is that by angle chasing, we will prove that
\[ \dang PAC = \dang ACP \iff \dang BAC + \dang CBD + \dang DCA + \dang ADB = 0.
\qquad (\spadesuit). \]
A careless reader would be forgiven for thinking that $(\spadesuit)$
implies the problem or at least one direction,
but it turns out the situation is more subtle.
The second part analyzes the angle conditions more carefully
and provides a complete proof.
\subparagraph{Proof of the equivalence $(\spadesuit)$ by angle chasing.}
We start with the following unconditional claim, valid for any quadrilateral.
\begin{claim*}
[Isogonal conjugation]
Let $ABCD$ and $P$ be as in the problem statement.
Then $\dang APB + \dang CPD = 180\dg$.
\end{claim*}
\begin{proof}
The angles in the statement imply that $A$ and $C$
are isogonal conjugates with respect to $\triangle PBC$.
Thus, lines $PA$ and $PC$ are isogonal with respect to $\angle BPC$, as needed.
\end{proof}
\begin{center}
\begin{asy}
import geometry;
size(9cm);
pair A = (-0.50099,1.69422);
pair B = (-1.8,-0.6);
pair C = (0.09182,-1.79600);
pair D = (1.55802,-0.60384);
pair P = (-1.04359,-0.19363);
draw(circumcircle(A,B,C), gray+dashed);
filldraw(A--B--C--D--cycle, opacity(0.2)+yellow, black);
draw(B--D);
draw(A--C);
draw(C--P--A, brown);
draw(B--P--D, gray);
markangle(radius=15, n=2, P, A, C, red, StickIntervalMarker(1,1, red));
markangle(radius=15, n=2, A, C, P, red, StickIntervalMarker(1,1, red));
markangle(radius=10, n=3, A, P, B, deepgreen, StickIntervalMarker(1,2, deepgreen));
markangle(radius=10, n=3, C, P, D, deepgreen, StickIntervalMarker(1,1, deepgreen));
markangle(radius=12, n=1, B, A, C, blue);
markangle(radius=12, n=1, C, B, D, blue);
markangle(radius=12, n=1, D, C, A, blue);
markangle(radius=12, n=1, A, D, B, blue);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$P$", P, dir(45));
\end{asy}
\end{center}
Next we rewrite the two angles $\dang APB$ and $\dang CPD$ in the claim
(colored green with three rings)
so that their only dependence on $P$
is through the angles $\dang PAC$ and $\dang CAP$ (colored red with two rings),
as follows:
\begin{align*}
-\dang APB &= \dang PBA + \dang BAP = \dang PBA + (\dang BAC - \dang PAC) \\
&= \dang CBD + \dang BAC - \dang PAC \\
-\dang CPD &= \dang PDC + \dang DCP = \dang PDC + (\dang DCA + \dang ACP) \\
&= \dang ADB + \dang DCA + \dang ACP.
\end{align*}
Since the claim says $\dang APB + \dang CPD = 0$, summing
lets us finally rewrite $\dang PAC - \dang APC$ in terms of only $A$, $B$, $C$, $D$:
\begin{align*}
0 &= (\dang ACP - \dang PAC) + \dang ADB + \dang DCA + \dang CBD + \dang BAC \\
\implies \dang PAC - \dang ACP &= \dang ADB + \dang DCA + \dang CBD + \dang BAC.
\end{align*}
These four latter angles are colored blue with one ring in the figure.
This proves $(\spadesuit)$.
\subparagraph{Quasi-harmonic quadrilaterals.}
To interpret the condition $(\spadesuit)$, we define a new term:
a quadrilateral $ABCD$ is \emph{quasi-harmonic} if $AB \cdot CD = BC \cdot DA$.
(See IMO 2018/6 for another problem involving quasi-harmonic quadrilaterals.)
The following two lemmas show why this condition is relevant:
\begin{lemma*}
The condition
\[ \dang BAC + \dang CBD + \dang DCA + \dang ADB = 0 \]
is equivalent to $ABCD$ being \emph{either} cyclic or quasi-harmonic or both.
\end{lemma*}
\begin{proof}
A clean approach with complex numbers
(posted at \url{https://aops.com/community/p15327734}) goes as follows:
define complex numbers $\mu = (a-b)(c-d) \neq 0$ and $\nu = (b-c)(d-a) \neq 0$.
Then \[ \mu-\nu = (a-c)(b-d) \neq 0. \]
Now the angle condition is equivalent to
\begin{align*}
\RR &\ni \frac{a-b}{d-b} \cdot \frac{b-c}{a-c} \cdot \frac{c-d}{b-d} \cdot \frac{d-a}{c-a} \\
&= \frac{\mu \nu}{(\mu - \nu)^2} \\
&= \frac{1}{\frac{\mu}{\nu} + \frac{\nu}{\mu} - 2} \\
\iff \RR &\ni \frac{\mu}{\nu} + \frac{\nu}{\mu}.
\end{align*}
In general, though $z+1/z$ is real exactly when either $|z|=1$ or $z \in \RR$.
Note $\mu/\nu \in \RR$ is equivalent to $ABCD$ cyclic
and $|\mu| = |\nu|$ is equivalent to $ABCD$ being quasiharmonic; the proof follows.
This can be also proved by inversion at $A$, though we don't write the details here.
(See also \url{https://problems.ru/view_problem_details_new.php?id=116602}).
\end{proof}
\begin{lemma*}
Quadrilateral $ABCD$ is quasi-harmonic if and only if $P$ lies on line $AC$.
\end{lemma*}
\begin{proof}
Let $X = \ol{BD} \cap \ol{AC}$.
If the isogonal of line $BD$ with respect to $\angle B$ meets line $AC$ at $Y$, then
$\frac{AX}{CX} \frac{AY}{CY} = \left( \frac{BA}{BC} \right)^2$.
Similarly if the isogonal to $\angle D$ meets line $AC$ at $Y'$, then
$\frac{AX}{CX} \frac{AY'}{CY'} = \left( \frac{DA}{DC} \right)^2$.
Hence $ABCD$ is quasi-harmonic if and only if $Y = Y'$ (that is, $Y=Y'=P$).
\end{proof}
\subparagraph{Wrap-up.}
We now show that $PA = PC$ if and only if $ABCD$ is cyclic
by cases on whether $P$ lies on $\ol{AC}$.
\begin{itemize}
\ii If $ABCD$ is \emph{not} quasi-harmonic,
then $(\spadesuit)$ implies the problem statement immediately.
Indeed, $\dang PAC = \dang ACP$ if and only if $PA = PC$
(as $\triangle PAC$ is not degenerate)
and the second lemma turns our angle condition into $ABCD$ cyclic.
\ii Now assume $ABCD$ is quasi-harmonic and $P$ lies on line $AC$.
We ignore $(\spadesuit)$.
Instead, note that if $ABCD$ is also cyclic then
$\ol{BD}$ is a symmedian of $\triangle ABC$ and hence $P$ is the midpoint.
Conversely, suppose we know $\ol{BD}$ is a symmedian of $\triangle ABC$.
Let $D' \neq B$ be the point for which $ABCD'$ is cyclic and harmonic;
then $B$, $D$, $D'$ are collinear and $\frac{BD}{CD} = \frac{BD'}{CD'} = \frac{BA}{CA}$.
So $D' = D$ (the corresponding Apollonian circle only meets line $BD$ twice), as needed.
\end{itemize} |
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 $w \ge 2$,
\begin{itemize}
\ii there exists an even alternating multiple $g(w)$ of $2^{w+1}$
with exactly $w$ digits, and
\ii there exists an even alternating multiple $h(w)$ of $5^{w}$
with exactly $w$ digits.
\end{itemize}
\end{claim*}
(One might note this claim is implied by the problem, too.)
\begin{proof}
We prove the first point by induction on $w$.
If $w = 2$, take $g(2) = 32$.
In general, we can construct $g(w+2)$ from $g(w)$
by adding some element in
\[ 10^w \cdot \{10, 12, 14, 16, 18, 30, \dots, 98\} \]
to $g(w)$, corresponding to the digits
we want to append to the start.
This multiple is automatically divisible by $2^{w+1}$,
and also can take any of the four possible values modulo $2^{w+3}$.
The second point is a similar induction,
with base case $h(2) = 50$.
The same set above consists of numbers divisible by $5^w$,
and covers all residues modulo $5^{w+2}$.
Careful readers might recognize the second part
as essentially USAMO 2003/1.
\end{proof}
\begin{claim*}
[Head construction]
If $\gcd(n,10) = 1$, then for any $b$,
there exists an even alternating number $f(b \bmod n)$ which is
$b \pmod n$.
\end{claim*}
\begin{proof}
A standard argument shows that
\[ 10 \cdot \frac{100^m-1}{99}
= \underbrace{1010\dots10}_{m\ 10\text{'s}}
\equiv 0 \pmod n \]
for any $m$ divisible by $\varphi(99n)$.
Take a very large such $m$,
and then add on $b$ distinct numbers of the form $10^{\varphi(n)r}$
for various even values of $r$; these all are $1 \pmod n$
and change some of the $1$'s to $3$'s.
\end{proof}
Now, we can solve the problem.
Consider three cases:
\begin{itemize}
\ii If $n = 2^k m$ where $\gcd(m,10) = 1$ and $k \ge 2$ is even,
then the concatenated number
\[ 10^k f\left( -\frac{g(k)}{10^k} \bmod m \right) + g(k) \]
works fine.
\ii If $n = 5^k m$ where $\gcd(m,10) = 1$ and $k \ge 2$ is even,
then the concatenated number
\[ 10^k f\left( -\frac{h(k)}{10^k} \bmod m \right) + h(k) \]
works fine.
\ii If $n = 50m$ where $\gcd(m,10) = 1$,
then the concatenated number
\[ 100 f\left( -\frac{1}{2} \bmod m \right) + 50 \]
works fine.
\end{itemize}
Since every non-multiple of $20$ divides such a number, we are done. |
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 vanishing sum
must be rotations of each other by $120\dg$,
it follows they must also form an equilateral triangle.
\begin{center}
\begin{asy}
pair A_1 = origin;
pair A_2 = A_1+dir(0);
pair B_1 = A_2+dir(37);
pair B_2 = B_1+dir(120);
pair C_1 = B_2+dir(157);
pair C_2 = C_1+dir(240);
pair A = extension(B_1, B_2, C_1, C_2);
pair B = extension(C_1, C_2, A_1, A_2);
pair C = extension(A_1, A_2, B_1, B_2);
filldraw(A--B--C--cycle, opacity(0.1)+lightcyan, lightblue);
draw(A_1--A_2, red+1, EndArrow(TeXHead), Margins);
draw(B_1--B_2, red+1, EndArrow(TeXHead), Margins);
draw(C_1--C_2, red+1, EndArrow(TeXHead), Margins);
draw(A_2--B_1, blue+1, EndArrow(TeXHead), Margins);
draw(B_2--C_1, blue+1, EndArrow(TeXHead), Margins);
draw(C_2--A_1, blue+1, EndArrow(TeXHead), Margins);
filldraw(A_1--B_1--C_1--cycle, opacity(0.1)+lightgreen, heavygreen);
draw(A_1--B_2, dotted+heavycyan);
draw(B_1--C_2, dotted+heavycyan);
draw(C_1--A_2, dotted+heavycyan);
dot("$A_1$", A_1, dir(270));
dot("$A_2$", A_2, dir(270));
dot("$B_1$", B_1, dir(30));
dot("$B_2$", B_2, dir(30));
dot("$C_1$", C_1, dir(150));
dot("$C_2$", C_2, dir(150));
dot("$A$", A, dir(90));
dot("$B$", B, dir(210));
dot("$C$", C, dir(330));
/* TSQ Source:
A_1 = origin R270
A_2 = A_1+dir(0) R270
B_1 = A_2+dir(37) R30
B_2 = B_1+dir(120) R30
C_1 = B_2+dir(157) R150
C_2 = C_1+dir(240) R150
A = extension B_1 B_2 C_1 C_2 R90
B = extension C_1 C_2 A_1 A_2 R210
C = extension A_1 A_2 B_1 B_2 R330
A--B--C--cycle 0.1 lightcyan / lightblue
!draw(A_1--A_2, red+1, EndArrow(TeXHead), Margins);
!draw(B_1--B_2, red+1, EndArrow(TeXHead), Margins);
!draw(C_1--C_2, red+1, EndArrow(TeXHead), Margins);
!draw(A_2--B_1, blue+1, EndArrow(TeXHead), Margins);
!draw(B_2--C_1, blue+1, EndArrow(TeXHead), Margins);
!draw(C_2--A_1, blue+1, EndArrow(TeXHead), Margins);
A_1--B_1--C_1--cycle 0.1 lightgreen / heavygreen
A_1--B_2 dotted heavycyan
B_1--C_2 dotted heavycyan
C_1--A_2 dotted heavycyan
*/
\end{asy}
\end{center}
Consequently, triangles $A_1A_2B_1$, $B_1B_2C_1$, $C_1C_2A_1$
are congruent, as $\angle A_2 = \angle B_2 = \angle C_2$.
So triangle $A_1 B_1 C_1$ is equilateral and the
diagonals are concurrent at the center. |
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 $i,j \in [1,n]$ and $a_i \equiv a_j \pmod n$,
contradiction.
\end{proof}
\begin{claim*}
For any $n$, the set $\{a_1, \dots, a_n\}$
is of the form $\{k+1, \dots, k+n\}$ for some integer $k$.
\end{claim*}
\begin{proof}
By induction, with the base case $n=1$ being vacuous.
For the inductive step,
suppose $\{a_1, \dots, a_n\} = \{k+1, \dots, k+n\}$ are determined.
Then
\[ a_{n+1} \equiv k \pmod{n+1}. \]
Moreover by the earlier claim we have
\[ \left\lvert a_{n+1}-a_1 \right\rvert < n+1. \]
From this we deduce $a_{n+1} \in \{k, k+n+1\}$ as desired.
\end{proof}
This gives us actually a complete description
of all possible sequences satisfying the hypothesis:
choose any value of $a_1$ to start.
Then, for the $n$th term,
the set $S = \{a_1, \dots, a_{n-1}\}$
is (in some order) a set of $n-1$ consecutive integers.
We then let $a_n = \max S + 1$ or $a_n = \min S - 1$.
A picture of six possible starting terms is shown below.
\begin{center}
\begin{asy}
size(6cm);
defaultpen(fontsize(9pt));
int[] a = {0, 6, 5, 7, 4, 3, 8};
for (int i=1; i<=9; ++i) {
draw((0,i)--(7.2,i), gray+dotted);
}
for (int i=1; i<=7; ++i) {
draw((i,0)--(i,9.2), gray+dotted);
if (i <= 5) draw((i,a[i])--(i+1,a[i+1]), red+dashed);
if (i <= 6) dot("$"+(string) a[i] + "$", (i, a[i]), dir(15), red);
if (i <= 6) label("$a_{" + (string) i + "}$", (i, 0), dir(-90));
}
draw( (-0.2,0)--(7.2,0), black );
draw( (0,-0.2)--(0,9.2), black );
\end{asy}
\end{center}
Finally, we observe that the condition that
the sequence has infinitely many positive and negative terms
(which we have not used until now)
implies it is unbounded above and below.
Thus it must contain every integer. |
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 economically written as
\[ S = \sum_{\text{cyc}} x^2(x^2-yz)(y^4+x^3z+xz^3)(z^4+x^3y+xy^3)
\overset{?}{\ge} 0. \]
So, clearing all the denominators gives
\begin{align*}
S &= \sum_{\text{cyc}} x^2(x^2-yz)
\left[ y^4z^4 + x^3y^5 + xy^7 + x^3z^5 + x^6yz + x^4y^3z
+ xz^7 + x^4yz^3 + x^2y^3z^3 \right] \\
&= \sum_{\text{cyc}}
\left[ x^4y^4z^4 + x^7y^5 + x^5y^7 + x^7z^5 + x^{10}yz + x^8y^3z
+ x^5z^7 + x^8yz^3 + x^6y^3z^3 \right] \\
&- \sum_{\text{cyc}}
\left[ x^2y^5z^5 + x^5y^6z + x^3y^8z + x^5yz^6 + x^8y^2z^2 + x^6y^4z^2
+ x^3yz^8 + x^6y^2z^4 + x^4y^4z^4 \right] \\
&= \sum_{\text{cyc}}
\left[ x^7y^5 + x^5y^7 + x^7z^5 + x^{10}yz
+ x^5z^7 + x^6y^3z^3 \right] \\
&- \sum_{\text{cyc}}
\left[ x^2y^5z^5 + x^5y^6z + x^5yz^6 + x^8y^2z^2 + x^6y^4z^2
+ x^6y^2z^4 \right]
\end{align*}
In other words we need to show
\[
\sum_{\text{sym}} \left( 2x^7y^5
+\half x^{10}yz + \half x^6y^3z^3 \right)
\ge
\sum_{\text{sym}} \left( \half x^8 y^2 z^2
+ \half x^5 y^5 z^2 + x^6 y^4 z^2 + x^6 y^5 z \right).
\]
which follows by summing
\begin{align*}
\sum_{\text{sym}} \frac{x^{10} yz + x^6 y^3 z^3}{2}
&\ge \sum_{\text{sym}} x^8 y^2 z^2 \\
\half \sum_{\text{sym}} x^8 y^2 z^2
&\ge \half \sum_{\text{sym}} x^6 y^4 z^2 \\
\half \sum_{\text{sym}} x^7y^5
&\ge \half \sum_{\text{sym}} x^5 y^5 z^2 \\
\half \sum_{\text{sym}} x^7y^5
&\ge \half \sum_{\text{sym}} x^6 y^4 z^2 \\
\sum_{\text{sym}} x^7y^5
&\ge \sum_{\text{sym}} x^6 y^5 z.
\end{align*}
The first line here comes from AM-GM,
the rest come from Muirhead.
\begin{remark*}
More elegant approach is to use Cauchy in the form
\[ \frac{1}{x^5+y^2+z^2} \le \frac{x\inv+y^2+z^2}{(x^2+y^2+z^2)^2}. \]
\end{remark*} |
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 other larger prime works. |
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 circumcircles of the triangles $PQR$,
as $E$ and $F$ vary, have a common point other than $P$. | 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 be sent to each other under the spiral similarity too.
\begin{center}
\begin{asy}
size(10cm);
pair A = dir(120);
pair D = dir(210);
pair C = dir(330);
pair B = dir(80);
filldraw(A--D--B--C--cycle, opacity(0.2)+lightcyan, blue+1.2);
pair E = 0.8*B+0.2*C;
pair F = 0.8*D+0.2*A;
pair P = extension(A, C, B, D);
pair Q = extension(E, F, B, D);
pair R = extension(E, F, A, C);
draw(E--F, lightblue);
pair M = (A*B-C*D)/(A+B-C-D);
filldraw(circumcircle(P, Q, R), opacity(0.2)+yellow, red);
draw(circumcircle(A, P, D), lightblue);
draw(circumcircle(B, P, C), lightblue);
filldraw(circumcircle(F, A, R), opacity(0.1)+lightgreen, dotted+heavygreen);
filldraw(circumcircle(F, D, Q), opacity(0.1)+lightgreen, dotted+heavygreen);
filldraw(circumcircle(E, B, Q), opacity(0.1)+lightgreen, dotted+heavygreen);
filldraw(circumcircle(E, C, R), opacity(0.1)+lightgreen, dotted+heavygreen);
dot("$A$", A, dir(A));
dot("$D$", D, dir(240));
dot("$C$", C, dir(C));
dot("$B$", B, dir(B));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$P$", P, 1.8*dir(95));
dot("$Q$", Q, dir(Q));
dot("$R$", R, dir(R));
dot("$M$", M, 1.4*dir(-30));
/* TSQ Source:
!size(10cm);
A = dir 120
D = dir 210 R240
C = dir 330
B = dir 80
A--D--B--C--cycle 0.2 lightcyan / blue+1.2
E = 0.8*B+0.2*C
F = 0.8*D+0.2*A
P = extension A C B D 1.8R95
Q = extension E F B D
R = extension E F A C
E--F lightblue
M = (A*B-C*D)/(A+B-C-D) 1.4R-30
circumcircle P Q R 0.2 yellow / red
circumcircle A P D lightblue
circumcircle B P C lightblue
circumcircle F A R 0.1 lightgreen / dotted heavygreen
circumcircle F D Q 0.1 lightgreen / dotted heavygreen
circumcircle E B Q 0.1 lightgreen / dotted heavygreen
circumcircle E C R 0.1 lightgreen / dotted heavygreen
*/
\end{asy}
\end{center}
Thus $M$ is the Miquel point of complete quadrilateral $FACE$.
As $R = \ol{FE} \cap \ol{AC}$ we deduce $FARM$ is a cyclic quadrilateral
(among many others, but we'll only need one).
Now look at complete quadrilateral $AFQP$.
Since $M$ lies on $(DFQ)$ and $(RAF)$,
it follows that $M$ is in fact the Miquel point of $AFQP$ as well.
So $M$ lies on $(PQR)$.
Thus $M$ is the fixed point that we wanted.
\begin{remark*}
Naturally, the congruent length
condition can be relaxed to $DF/DA = BE/BC$.
\end{remark*} |
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 contestants who solved
$\{1,2,\dots,5\} \setminus \{i\}$ (and missed P6).
Let $b_{ij}$ denote the number of contestants who solved
$\{1,2,\dots,5,6\} \setminus \{i,j\}$, for $1 \le i < j \le 5$
(thus in particular they solved P6).
Thus
\[ n = 1 + \sum_{1 \le i \le 5} a_i + \sum_{1 \le i < j \le 5} b_{ij} \]
denotes the total number of contestants.
Considering contestants who solved P1/P6 we have
\[ t_1 \coloneq b_{23} + b_{24} + b_{25} + b_{34} + b_{35} + b_{45} \ge \frac25n + \frac15 \]
and we similarly define $t_2$, $t_3$, $t_4$, $t_5$.
(We have written $\frac25n+\frac15$ since we know
the left-hand side is an integer strictly larger than $\frac25n$.)
Also, by considering contestants who solved P1/P2 we have
\[ t_{12} = 1 + a_{3} + a_{4} + a_{5} + b_{34} + b_{35} + b_{45}
\ge \frac25n + \frac15 \]
and we similarly define $t_{ij}$ for $1 \le i < j \le 5$.
\begin{claim*}
The number $\frac{2n+1}{5}$ is equal to some integer $k$,
fourteen of the $t$'s are equal to $k$,
and the last one is equal to $k+1$.
\end{claim*}
\begin{proof}
First, summing all fifteen equations gives
\begin{align*}
6n+4 = 10 + 6(n-1) &= 10
+ \sum_{1 \le i \le 5} 6a_i + \sum_{1 \le i < j \le 5} 6b_{ij} \\
&= \sum_{1 \le i \le 5} t_i + \sum_{1 \le i < j \le 5} t_{ij}.
\end{align*}
Thus the sum of the $15$ $t$'s is $6n+4$.
But since all the $t$'s are integers at least
$\frac{2n+1}{5} = \frac{6n+3}{15}$, the conclusion follows.
\end{proof}
However, we will also manipulate the equations to get the following.
\begin{claim*}
We have
\[ t_{45} \equiv 1 + t_1 + t_2 + t_3 + t_{12} + t_{23} + t_{31} \pmod 3. \]
\end{claim*}
\begin{proof}
This follows directly by computing the coefficient of the $a$'s and $b$'s.
We will nonetheless write out a derivation of this equation, to
motivate it, but the proof stands without it.
Let $B = \sum_{1 \le i < j \le 5} b_{ij}$ be the sum of all $b$'s.
First, note that
\begin{align*}
t_1 + t_2 &= B + b_{34} + b_{45} + b_{35} - b_{12} \\
&= B + \left( t_{12}-1-a_3-a_4-a_5 \right) - b_{12} \\
\implies b_{12} &= B - (t_1 + t_2) + t_{12} - 1 - (a_3+a_4+a_5).
\end{align*}
This means we have more or less solved for each $b_{ij}$
in terms of only $t$ and $a$ variables.
Now
\begin{align*}
t_{45} &= 1 + a_1 + a_2 + a_3 + b_{12} + b_{23} + b_{31} \\
&= 1 + a_1 + a_2 + a_3 \\
&+ [B - (t_1 + t_2) + t_{12} - 1 - (a_3+a_4+a_5)] \\
&+ [B - (t_2 + t_3) + t_{23} - 1 - (a_1+a_4+a_5)] \\
&+ [B - (t_3 + t_1) + t_{13} - 1 - (a_2+a_4+a_5)] \\
&\equiv 1 + t_1 + t_2 + t_3 + t_{12} + t_{23} + t_{31} \pmod 3
\end{align*}
as desired.
\end{proof}
However, we now show the two claims are incompatible
(and this is easy, many ways to do this).
There are two cases.
\begin{itemize}
\ii Say $t_5 = k+1$ and the others are $k$.
Then the equation for $t_{45}$ gives that $k \equiv 6k+1 \pmod 3$.
But now the equation for $t_{12}$ give $k \equiv 6k \pmod 3$.
\ii Say $t_{45} = k+1$ and the others are $k$.
Then the equation for $t_{45}$ gives that $k+1 \equiv 6k \pmod 3$.
But now the equation for $t_{12}$ give $k \equiv 6k+1 \pmod 3$.
\end{itemize}
\begin{remark*}
It is significantly easier to prove that there
is at least one contestant who solved five problems.
One can see it by dropping the $+10$ in the proof of the claim,
and arrives at a contradiction.
In this situation it is not even necessary to set up
the many $a$ and $b$ variables;
just note that the expected number of contestants
solving any particular pair of problems is
$\frac{\binom42n}{\binom62} = \frac25n$.
The fact that $\frac{2n+1}{5}$
should be an integer also follows quickly,
since if not one can improve the bound to $\frac{2n+2}{5}$
and quickly run into a contradiction.
Again one can get here without setting up $a$ and $b$.
The main difficulty seems to be the precision required
in order to nail down the second $5$-problem solve.
\end{remark*}
\begin{remark*}
The second claim may look miraculous,
but the proof shows that it is not too unnatural
to consider $t_1 + t_2 - t_{12}$ to isolate $b_{12}$
in terms of $a$'s and $t$'s.
The main trick is: why mod $3$?
The reason is that if one looks closely, for a
fixed $k$ we have a system of $15$ equations in $15$ variables.
Unless the determinant $D$ of that system happens to be zero,
this means there will be a rational solution in $a$ and $b$,
whose denominators are bounded by $D$.
However if $p \mid D$ then we may conceivably run into mod $p$
issues.
This motivates the choice $p = 3$,
since it is easy to see the determinant is divisible by $3$,
since constant shifts of $\vec a$ and $\vec b$ are also solutions mod $3$.
(The choice $p = 2$ is a possible guess as well for this reason,
but the problem seems to have better $3$-symmetry.)
\end{remark*} |
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 $I$ are both inside $\triangle ABC$
that implies $P$ lies on the circumcircle of $\triangle BIC$.
It's well-known (by ``Fact 5'') that the circumcenter
of $\triangle BIC$ is the arc midpoint $M$ of $\widehat{BC}$.
Therefore
\[ AI + IM = AM \le AP + PM \implies AI \le AP \]
with equality holding iff $A$, $P$, $M$ are collinear, or $P=I$. |
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 point in the interior of $P$.
Find the maximum number of isosceles triangles having two good
sides that could appear in such a configuration. | 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 are congruent
(and not congruent to the third side).
\ii No two isosceles triangles share a good side.
\end{itemize}
\paragraph{Solution using bijections.}
Call a good diagonal \textbf{special} if it's part of a special triangle;
special diagonals come in pairs.
Consider the minor arc cut out by a special diagonal $d$,
which has an odd number of sides.
Since special diagonals come in pairs,
one can associate to $d$ a side of the polygon
not covered by any special diagonals from $d$.
Hence there are at most $2006$ special diagonals,
so at most $1003$ special triangles.
\paragraph{Solution using graph theory.}
Consider the tree $T$ formed by the $2004$
triangles in the dissection, with obvious adjacency.
Let $F$ be the forest obtained by deleting
any edge corresponding to a good diagonal.
Then the resulting graph $F$ has only degrees $1$ and $3$,
with special triangles only occurring at degree $1$ vertices.
If there are $k$ good diagonals drawn,
then this forest consists of $k+1$ trees.
A tree with $n_i$ vertices ($0 \le i \le k$)
consequently has $\frac{n_i+2}{2}$ leaves.
However by the earlier remark at least $k$ leaves
don't give special triangles
(one on each side of a special diagonal);
so the number of leaves that do give good triangles is at most
\[ -k + \sum_i \frac{n_i+2}{2}
= -k + \frac{2004 + 2(k+1)}{2} = 1003. \] |
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 them with the average
(this increases the LHS and decreases RHS).
So we can have $x=y$, $z=-2x$.
Now WLOG $x > 0$ to get
\[ 2x^3 \cdot s \le \frac{M}{9} \left( 6x^2+s^2 \right)^2. \]
After this routine calculation gives
$M = \frac{9}{32}\sqrt2$ works and is optimal
(by $6x^2+s^2 = 2x^2 + 2x^2 + 2x^2 + s^2$ and AM-GM). |
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 $y = 2^{x-1} m - 1$ for some odd $m$, so
\[ 1 + 2^{x+1} = m\left( 2^{x-2}m-1 \right)
\implies 2^x = \frac{4(1+m)}{m^2-8}. \]
\end{itemize}
In particular we need $4|1 \pm m| \ge 2^4 |m^2-8|$,
which is enough to imply $m < 5$.
From here easily recover $x = 4$, $m = 3$ as the last solution
(in the second case). |
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.
If any of the quantities are zero we are done.
Else, we must eventually have $x_i - x_{i+1} = -(x_{i+1} - x_{i+2})$,
so $x_i = x_{i+2}$ and we get $2$-periodicity.
\end{proof}
The tricky part is to study the $2$-orbits.
Suppose there exists a fixed pair $u \neq v$
with $P(u) = v$, $P(v) = u$.
(If no such pair exists, we are already done.)
Let $(a,b)$ be any other pair with $P(a) = b$, $P(b) = a$,
possibly even $a = b$, but $\{a,b\} \cap \{u,v\} = \varnothing$.
Then we should have
\[ u-a \mid P(u)-P(a) = v-b
\mid P(v) - P(b) = u-a \]
and so $u-a$ and $v-b$ divide each other (and are nonzero).
Similarly, $u-b$ and $v-a$ divide each other.
Hence $u-a = \pm (v-b)$ and $u-b = \pm (v-a)$.
We consider all four cases:
\begin{itemize}
\ii If $u-a = v-b$ and $u-b = v-a$
then $u-v = b-a = a-b$, contradiction.
\ii If $u-a = -(v-b)$ and $u-b = -(v-a)$
then $u+v = u-v = a+b$.
\ii If $u-a = -(v-b)$ and $u-b = v-a$,
we get $a+b = u+v$ from the first one
(discarding the second).
\ii If $u-a = v-b$ and $u-b = -(v-a)$,
we get $a+b = u+v$ from the second one
(discarding the first one).
\end{itemize}
Thus in all possible situations we have
\[ a+b = c \coloneq u+v \]
a fixed constant.
Therefore, any pair $(a,b)$ with $P(a) = b$
and $P(b) = a$ actually satisfies $P(a) = c-a$.
And since $\deg P > 1$,
this means there are at most $n$ roots to $a+P(a)=c$, as needed. |
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 point).
\begin{lemma*}
Let $N$ be an even integer.
Then any almost convex $N$-gon with area $S$
should have an inscribed triangle with area at least $2S/N$.
\end{lemma*}
The main work is the proof of the lemma.
Label the polygon $P_0 P_1 \dots P_{N-1}$.
Consider the $N/2$ major diagonals of the almost convex $N$-gon,
$P_0 P_{N/2}$, $P_1 P_{N/2+1}$, et cetera.
A \emph{butterfly} refers to a self-intersecting quadrilateral
$P_i P_{i+1} P_{i+1+N/2} P_{i+N/2}$.
An example of a butterfly is shown below for $N=8$.
\begin{center}
\begin{asy}
pair P_0 = (-1,-3);
pair P_1 = ( 0,-3);
pair P_2 = (2.5,-3);
pair P_3 = (3,0);
pair P_4 = (1.8,1.2);
pair P_5 = (1.1,1.9);
pair P_6 = (0.3,2.7);
pair P_7 = (-2,0.4);
filldraw(P_0--P_2--P_3--P_6--P_7--cycle, opacity(0.1)+lightcyan, blue);
filldraw(P_0--P_4--P_5--P_1--cycle, opacity(0.2)+lightred, red);
draw(P_1--P_5, red);
draw(P_2--P_6, red);
draw(P_3--P_7, red);
dot("$P_0$", P_0, dir(P_0));
dot("$P_1$", P_1, dir(P_1));
dot("$P_2$", P_2, dir(P_2));
dot("$P_3$", P_3, dir(P_3));
dot("$P_4$", P_4, dir(P_4));
dot("$P_5$", P_5, dir(P_5));
dot("$P_6$", P_6, dir(P_6));
dot("$P_7$", P_7, dir(P_7));
/* TSQ Source:
P_0 = (-1,-3)
P_1 = ( 0,-3)
P_2 = (2.5,-3)
P_3 = (3,0)
P_4 = (1.8,1.2)
P_5 = (1.1,1.9)
P_6 = (0.3,2.7)
P_7 = (-2,0.4)
P_0--P_2--P_3--P_6--P_7--cycle 0.1 lightcyan / blue
P_0--P_4--P_5--P_1--cycle 0.2 lightred / red
P_1--P_5 red
P_2--P_6 red
P_3--P_7 red
*/
\end{asy}
\end{center}
\begin{claim*}
Every point $X$ in the polygon is contained in the wingspan of some butterfly.
\end{claim*}
\begin{proof}
Consider a windmill-like process which
\begin{itemize}
\ii starts from some oriented red line $P_0 P_{N/2}$, oriented to face $P_0 P_{N/2}$
\ii rotates through $P_0 P_{N/2} \cap P_1 P_{N/2+1}$ to get line $P_1 P_{N/2+1}$,
\ii rotates through $P_1 P_{N/2+1} \cap P_2 P_{N/2+2}$ to get line $P_2 P_{N/2+2}$,
\ii \dots et cetera, until returning to line $P_{N/2} P_0$,
but in the reverse orientation.
\end{itemize}
At the end of the process, every point in the plane has switched sides with our moving line.
The moment that $X$ crosses the moving red line, we get it contained in a butterfly, as needed.
\end{proof}
\begin{claim*}
If $ABDC = P_i P_{i+1} P_{i+1+N/2} P_{i+N/2}$ is a butterfly,
one of the triangles $ABC$, $BCD$, $CDA$, $DAB$
has area at least that of the butterfly.
\end{claim*}
\begin{proof}
Let the diagonals of the butterfly meet at $O$,
and let $a = AO$, $b = BO$, $c = CO$, $d = DO$.
If we assume WLOG $d = \min(a,b,c,d)$
then it follows $[ABC] = [AOB] + [BOC] \ge [AOB] + [COD]$, as needed.
\end{proof}
Now, since the $N/2$ butterflies cover an area of $S$,
it follows that one of the butterflies
has area at least $S / (N/2) = 2S/N$,
and so that butterfly gives a triangle with area at least $2S/N$,
completing the proof of the lemma.
\paragraph{Main proof.}
Let $a_1$, \dots, $a_n$ be the numbers assigned to the sides.
Assume for contradiction $a_1 + \dots + a_n < 2S$.
We pick even integers $m_1$, $m_2$, \dots, $m_n$ such that
\begin{align*}
\frac{a_1}{S} &< \frac{2m_1}{m_1 + \dots + m_n} \\
\frac{a_2}{S} &< \frac{2m_2}{m_1 + \dots + m_n} \\
&\vdotswithin\le \\
\frac{a_n}{S} &< \frac{2m_n}{m_1 + \dots + m_n}.
\end{align*}
which is possible by rational approximation,
since the right-hand sides sum to $2$ and the left-hand sides sum to strictly less than $2$.
Now we break every side of $P$ into $m_i$ equal parts
to get an almost convex $N$-gon, where $N = m_1 + \dots + m_n$.
The main lemma then gives us a triangle $\Delta$ of the almost convex $N$-gon
which has area at least $\frac{2S}{N}$.
If $\Delta$ used the $i$th side then it then follows the label $a_i$ on that side should be
at least $m_i \cdot \frac{2S}{N}$, contradiction. |
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
\[
\max \left\{ \left\lvert x_i - a_i \right\rvert :
1 \le i \le n \right\}
\ge \half d.
\]
\ii Moreover, show that there exists some
choice of $x_1 \le \dots \le x_n$ which achieves equality.
\end{enumerate} | 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 longer consider this case.
Otherwise, consider any indices $i < j$ with $a_i > a_j$.
We first prove (a) by applying the following claim
with $p = a_i$ and $q = a_j$:
\begin{claim*}
For any $p \le q$, we have
either $|p - a_i| \ge \half(a_i-a_j)$
or $|q - a_j| \ge \half(a_i-a_j)$.
\end{claim*}
\begin{proof}
Assume for contradiction both are false.
Then $p > a_i - \half(a_i-a_j)
= a_j + \half(a_i-a_j) > q$, contradiction.
\end{proof}
As for (b), we let $i < j$ be any indices for which
$a_i - a_j = d > 0$ achieves the maximal difference.
We then define $x_\bullet$ in three steps:
\begin{itemize}
\ii We set $x_k = \frac{a_i + a_j}{2}$ for $k = i, \dots, j$.
\ii We recursively set $x_{k} = \max(x_{k-1}, a_k)$
for $k = j+1, j+2, \dots$.
\ii We recursively set $x_{k} = \min(x_{k+1}, a_k)$
for $k = i-1, i-2, \dots$.
\end{itemize}
By definition, these $x_\bullet$ are weakly increasing.
To prove this satisfies (b) we only need to check that
\[ \left\lvert x_k - a_k \right\rvert \le \frac{a_i-a_j}{2} \qquad
(\star) \]
for any index $k$ (as equality holds for $k = i$ or $k = j$).
We note $(\star)$ holds for $i < k < j$ by construction.
For $k > j$, note that $x_k \in \{a_j, a_{j+1}, \dots, a_k\}$
by construction, so $(\star)$ follows from our choice of $i$ and $j$
giving the largest possible difference; the case $k < i$ is similar. |
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 $\ell$ is the bisector of angle $ DAB$. | 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 = E*E/C;
pair B = C*K/D;
pair A = B+D-C;
pair M = foot(E, C, D);
pair N = foot(E, C, B);
pair P = foot(E, B, D);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
draw(B--C--D--cycle, blue);
filldraw(CP(E, C), opacity(0.1)+lightred, red+dotted);
pair F = 2*M-C;
pair G = 2*N-C;
draw(E--P, orange);
draw(E--M, orange);
draw(E--N, orange);
draw(C--G, blue);
draw(N--P, deepgreen);
draw(A--G, deepgreen);
draw(B--A--D, lightblue+dashed);
draw(A--C, lightblue+dashed);
dot("$C$", C, dir(180));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$B$", B, dir(B));
dot("$A$", A, dir(A));
dot("$M$", M, dir(135));
dot("$N$", N, dir(N));
dot("$P$", P, dir(P));
dot("$F$", F, dir(F));
dot("$G$", G, dir(G));
/* TSQ Source:
C = dir 200 R180
D = dir 340
E = dir 240
K := E*E/C
B = C*K/D
A = B+D-C
M = foot E C D R135
N = foot E C B
P = foot E B D
unitcircle 0.1 lightcyan / blue
B--C--D--cycle blue
CP E C 0.1 lightred / red dotted
F = 2*M-C
G = 2*N-C
E--P orange
E--M orange
E--N orange
C--G blue
N--P deepgreen
A--G deepgreen
B--A--D lightblue dashed
A--C lightblue dashed
*/
\end{asy}
\end{center}
However, since we also have $\ol{EM} \perp \ol{CF}$
and $\ol{EN} \perp \ol{CG}$ (from $EF=EG=EC$)
we conclude $\ol{PMN}$ is the Simson line of $E$ with respect to $\triangle BCD$,
which implies $\ol{EP} \perp \ol{BD}$.
In other words, $\ol{EP}$ is the perpendicular bisector of $\ol{BD}$,
so $E$ is the midpoint of arc $\widehat{BCD}$.
Finally,
\begin{align*}
\dang(\ol{AB}, \ell) &= \dang(\ol{CD}, \ol{MNP}) = \dang CMN = \dang CEN \\
&= 90\dg - \dang NCE = 90\dg + \dang ECB
\end{align*}
which means that $\ell$ is parallel to a bisector of $\angle BCD$,
and hence to one of $\angle BAD$.
(Moreover since $F$ lies on the interior of $\ol{CD}$,
it is actually the internal bisector.) |
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 same as the largest size of a clique contained in the other room. | 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 $r$ vertices of $A$ into $B$
to arrive at $\alpha(A) \le \alpha(B) \le \alpha(A)+1$.
\end{claim*}
\begin{proof}
This is actually obvious by discrete continuity.
We move one vertex at a time, noting $\alpha(A)$ decreases by one at each step,
while $\alpha(B)$ increases by either zero or one at each step.
We stop once $\alpha(B) \ge \alpha(A)$, which happens before we have moved $r$ vertices
(since then we have $\alpha(B) \ge r = \alpha(A)$).
The conclusion follows.
\end{proof}
So let's consider the situation
\[ \alpha(A) = k \ge r \qquad\text{and}\qquad \alpha(B) = k+1. \]
At this point $A$ is a set of $k$ red vertices,
while $B$ has the remaining $2r-k$ red vertices (and all the green ones).
An example is shown below with $k=4$ and $2r = 6$.
\begin{center}
\begin{asy}
size(10cm);
pair K1 = 2*dir(45);
pair K2 = 2*dir(135);
pair K3 = 2*dir(225);
pair K4 = 2*dir(315);
label("$A$", (0,2), dir(90), red);
label("$\alpha(A) = k$", (0,-2.5), dir(90), red);
label("$k$ red vertices", (0,-2.5), dir(-90), red);
pair K5 = (6,1);
pair K6 = (6,-1);
draw(K1--K5, palered);
draw(K2--K5, palered);
draw(K3--K5, palered);
draw(K4--K5, palered);
draw(K1--K6, palered);
draw(K2--K6, palered);
draw(K3--K6, palered);
draw(K4--K6, palered);
pair B1 = (4.8,1.3);
pair B2 = (3.7,0);
pair B3 = (4.8,-1.3);
pair C1 = (7.2,1.3);
pair C2 = (8.3,0);
pair C3 = (7.2,-1.3);
draw(K5--B1--K6, paleblue);
draw(K5--B2--K6, paleblue);
draw(K5--B3--K6, paleblue);
draw(B1--B2--B3--cycle, paleblue);
draw(K5--C1--K6, paleblue);
draw(K5--C2--K6, paleblue);
draw(K5--C3--K6, paleblue);
draw(C1--C2--C3--cycle, paleblue);
draw(K1--K2--K3--K4--cycle, red+1);
draw(K1--K3, red+1);
draw(K2--K4, red+1);
draw(K5--K6, red+1);
dot(K1, red);
dot(K2, red);
dot(K3, red);
dot(K4, red);
dot(K5, red);
dot(K6, red);
dot(B1, deepgreen);
dot(B2, deepgreen);
dot(B3, blue);
dot(C1, deepgreen);
dot(C2, deepgreen);
dot(C3, blue);
draw(circle(B3, 0.15), blue);
draw(circle(C3, 0.15), blue);
label("$B$", (6,2), dir(90), deepgreen);
label("$\alpha(B) = k+1$", (6,-2.5), dir(90), deepgreen);
label("$2r-k$ red vertices", (6,-2.5), dir(-90), deepgreen);
draw((3,2.8)--(3,-2.8), black+1.5);
\end{asy}
\end{center}
Now, if we can move any red vertex from $B$ back to $A$
without changing the clique number of $B$, we do so, and win.
Otherwise, it must be the case that \emph{every}
$(k+1)$-clique in $B$ uses \emph{every} red vertex in $B$.
For each $(k+1)$-clique in $B$ (in arbitrary order), we do the following procedure.
\begin{itemize}
\ii If all $k+1$ vertices are still green, pick one and re-color it blue.
This is possible since $k+1 > 2r-k$.
\ii Otherwise, do nothing.
\end{itemize}
Then we move all the blue vertices from $B$ to $A$,
one at a time, in the same order we re-colored them.
This forcibly decreases the clique number of $B$ to $k$,
since the clique number is $k+1$ just before the last blue vertex is moved,
and strictly less than $k+1$ (hence equal to $k$) immediately after that.
\begin{claim*}
After this, $\alpha(A) = k$ still holds.
\end{claim*}
\begin{proof}
Assume not, and we have a $(k+1)$-clique
which uses $b$ blue vertices and $(k+1)-b$ red vertices in $A$.
Together with the $2r-k$ red vertices already in $B$ we then get a clique of size
\[ b + \left( (k+1-b) \right) + \left( 2r-k \right) = 2r + 1 \]
which is a contradiction.
\end{proof}
\begin{remark*}
Dragomir Grozev posted the following motivation on \href{https://dgrozev.wordpress.com/2019/12/05/splitting-the-cliques-of-a-graph-imo-2007-p3/}{his blog}:
\begin{quote}
I think, it's a natural idea to place all students in one room and begin
moving them one by one into the other one.
Then the max size of the cliques in the first and second room
increase (resp.\ decrease) at most with one.
So, there would be a moment both sizes are almost the same.
At that moment we may adjust something.
Trying the idea, I had some difficulties
keeping track of the maximal cliques in the both rooms.
It seemed easier all the students in one of the rooms to comprise a clique.
It could be achieved by moving only the members of the maximal clique.
Following this path the remaining obstacles can be overcome naturally.
\end{quote}
\end{remark*} |
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 $OP = OQ$.
Since $OC = OR$ as well, we get the conclusion.
\end{proof}
Denote by $X$ and $Y$ the feet from $R$ to $\ol{CA}$
and $\ol{CB}$, so $\triangle CXR \cong \triangle CYR$.
Then, let $t = \frac{CQ}{CR} = 1 - \frac{CP}{CR}$.
\begin{center}
\begin{asy}
pair C = dir(90);
pair R = -C;
pair X = dir(205);
pair Y = -conj(X);
pair Q = 0.41*R+0.59*C;
pair P = R+C-Q;
pair L = foot(Q, C, X);
pair K = foot(P, C, Y);
pair A = 2*L-C;
pair B = 2*K-C;
pair O = extension(L, Q, P, K);
filldraw(L--Q--X--cycle, opacity(0.1)+orange, dotted+orange);
filldraw(P--K--Y--cycle, opacity(0.1)+orange, dotted+orange);
draw(C--A--B--cycle, lightblue);
draw(A--X, lightblue);
draw(circumcircle(C, A, B), lightblue+dashed);
draw(C--R, deepgreen);
draw(X--R--Y, deepgreen);
draw(L--O, blue);
draw(P--K, blue);
dot("$C$", C, dir(C));
dot("$R$", R, dir(R));
dot("$X$", X, dir(180));
dot("$Y$", Y, dir(0));
dot("$Q$", Q, dir(45));
dot("$P$", P, dir(215));
dot("$L$", L, dir(L));
dot("$K$", K, dir(K));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$O$", O, dir(315));
/* TSQ Source:
C = dir 90
R = -C
X = dir 205 R180
Y = -conj(X) R0
Q = 0.41*R+0.59*C R45
P = R+C-Q R215
L = foot Q C X
K = foot P C Y
A = 2*L-C
B = 2*K-C
O = extension L Q P K R315
L--Q--X--cycle 0.1 orange / dotted orange
P--K--Y--cycle 0.1 orange / dotted orange
C--A--B--cycle lightblue
A--X lightblue
circumcircle C A B lightblue dashed
C--R deepgreen
X--R--Y deepgreen
L--O blue
P--K blue
*/
\end{asy}
\end{center}
Then it follows that
\[ [RQL] = [XQL] = t(1-t) \cdot [XRC]
= t(1-t) \cdot [YCR] = [YKP] = [RKP] \]
as needed.
\begin{remark*}
Trigonometric approaches are very possible
(and easier to find) as well:
both areas work out to be $\frac 18 ab \tan \half C$.
\end{remark*} |
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 with $a > b$
(we have $a \neq b$ since $k \neq 0$).
By Vieta jumping, $(b, \frac{b^2+k}{a})$ is also such a solution.
But now
\begin{align*}
\frac{b^2+k}{a} \ge a &\implies k \ge a^2 - b^2 \\
&\implies \frac{(a-b)^2}{4ab-1} \ge a^2-b^2 \\
&\implies a-b \ge (4ab-1)(a+b)
\end{align*}
which is absurd for $a,b \in \ZZ_{>0}$.
(In the last step we divided by $a-b > 0$.) |
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 polynomials
\begin{align*}
A(x,y,z) &= \prod_{i=1}^n (x-i) \prod_{i=1}^n (y-i) \prod_{i=1}^n (z-i) \\
B(x,y,z) &= \prod_{i=1}^N \left( a_i x + b_i y + c_i z + 1 \right).
\end{align*}
Note that $A(0,0,0) = (-1)^n (n!)^3 \neq 0$
and $B(0,0,0) = 1 \neq 0$,
but $A(x,y,z) = B(x,y,z) = 0$ for any $(x,y,z) \in S$.
Also, the coefficient of $x^n y^n z^n$ in $A$ is $1$,
while the coefficient of $x^n y^n z^n$ in $B$ is $0$.
Now, define
\[ P(x,y,z) \coloneq A(x,y,z) - \lambda B(x,y,z). \]
where $\lambda = \frac{A(0,0,0)}{B(0,0,0)} = (-1)^{n} (n!)^3$.
We now have that
\begin{itemize}
\ii $P(x,y,z) = 0$ for any $x,y,z \in \left\{ 0,1,\dots,n \right\}^3$.
\ii But the coefficient of $x^n y^n z^n$ is $1$.
\end{itemize}
This is a contradiction to Alon's combinatorial nullstellensatz. |
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 $A_1$, $A_2$, $B_1$, $B_2$, $C_1$, $C_2$ are concyclic. | 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 circles $\Gamma_B$ and $\Gamma_C$.
\begin{center}
\begin{asy}
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
draw(A--B--C--cycle);
pair D = midpoint(B--C);
pair E = midpoint(C--A);
pair F = midpoint(A--B);
pair H = orthocenter(A, B, C);
draw(E--F);
pair X = -H+2*foot(H, E, F);
path w1 = Drawing(CP(E,H));
path w2 = Drawing(CP(F,H));
pair B_1 = IP(w1, A--C);
pair B_2 = OP(w1, A--C);
pair C_1 = IP(w2, A--B);
pair C_2 = OP(w2, A--B);
pair T = foot(A, B, C);
draw(A--T, dotted);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$H$", H, 1.4*dir(-10));
dot("$X$", X, 1.4*dir(10));
dot("$B_1$", B_1, 1.5*dir(90));
dot("$B_2$", B_2, 1.2*dir(-10));
dot("$C_1$", C_1, 1.2*dir(120));
dot("$C_2$", C_2, dir(C_2));
/* TSQ Source:
A = dir 110
B = dir 210
C = dir 330
A--B--C--cycle
D = midpoint B--C
E = midpoint C--A
F = midpoint A--B
H = orthocenter A B C 1.4R-10
E--F
X = -H+2*foot H E F 1.4R10
! path w1 = Drawing(CP(E,H));
! path w2 = Drawing(CP(F,H));
B_1 = IP w1 A--C 1.5R90
B_2 = OP w1 A--C 1.2R-10
C_1 = IP w2 A--B 1.2R120
C_2 = OP w2 A--B
T := foot A B C
A--T dotted
*/
\end{asy}
\end{center}
Let $X$ be the second intersection of $\Gamma_B$ and $\Gamma_C$.
Clearly $\ol{XH}$ is perpendicular to the line
joining the centers of the circles, namely $\ol{EF}$.
But $\ol{EF} \parallel \ol{BC}$, so $\ol{XH} \perp \ol{BC}$.
Since $\ol{AH} \perp \ol{BC}$ as well,
we find that $A$, $X$, $H$ are collinear, as needed.
Thus, $B_1$, $B_2$, $C_1$, $C_2$ are concyclic.
Similarly, $C_1$, $C_2$, $A_1$, $A_2$ are concyclic,
as are $A_1$, $A_2$, $B_1$, $B_2$.
Now if any two of these three circles coincide, we are done;
else the pairwise radical axii are not concurrent, contradiction.
(Alternatively, one can argue directly that $O$ is the center of all
three circles, by taking the perpendicular bisectors.)
\paragraph{Second solution using length chase (Ritwin Narra).}
We claim the circumcenter $O$ of $\triangle ABC$
is in fact the center of $(A_1A_2B_1B_2C_1C_2)$.
Define $D$, $E$, $F$ as before.
Then since $\ol{OD} \perp \ol{A_1A_2}$ and $DA_1 = DA_2$,
which means $OA_1 = OA_2$. Similarly, we have $OB_1 = OB_2$ and $OC_1 = OC_2$.
Now since $DA_1 = DA_2 = DH$, we have $OA_1^2 = OD^2 + HD^2$.
We seek to show
\[ OD^2 + HD^2 = OE^2 + HE^2 = OF^2 + HF^2. \]
This is clear by Appollonius's Theorem
since $D$, $E$, and $F$ lie on the nine-point circle,
which is centered at the midpoint of $\ol{OH}$. |
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 are in such a situation,
given that there is an infinite curve of rationals).
For equality, it suffices to show there are infinitely many integer solutions to
\[ a^2b+b^2c+c^2a=3abc \iff \frac ac + \frac ba + \frac ca = 3 \]
or equivalently that there are infinitely many rational solutions to
\[ u + v + \frac{1}{uv} = 3. \]
For any $0 \neq u \in \QQ$ the real solution for $u$ is
\[ v = \frac{-u + (u-1)\sqrt{1-4/u} + 3}{2} \]
and there are certainly infinitely many rational numbers $u$
for which $1-4/u$ is a rational square
(say, $u = \frac{-4}{q^2-1}$ for $q \neq \pm 1$ a rational number). |
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 $p \equiv 1 \pmod 8$ henceforth.
Then there exists some $\frac 12 p < x < p$
such that $x^2 \equiv -1 \pmod p$,
and we set \[ x = \frac{p+1}{2} + t. \]
\begin{claim*}
We have $t \ge \frac{h-1}{2}$ and hence may take $n = p-x$.
\end{claim*}
\begin{proof}
Assume for contradiction this is false; then
\begin{align*}
0 &\equiv 4(x^2+1) \pmod{p} \\
&= \left( p+1+2t \right)^2 + 4 \\
&\equiv (2t+1)^2 + 4 \pmod{p} \\
&< h^2+4
\end{align*}
So we have that $(2t+1)^2+4$ is positive and divisible by $p$,
yet at most $\left\lceil \sqrt{p} \right\rceil^2 + 4 < 2p$.
So it must be the case that $(2t+1)^2+4 = p$,
but this has no solutions modulo $8$.
\end{proof} |
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 $f(t) \in \{t, 1/t\}$.
Now assume $f(a) = a$ and $f(b) = 1/b$.
Setting $(\sqrt a, \sqrt b, 1, \sqrt{ab})$ gives
\[ \frac{a + 1/b}{f(ab) + 1} = \frac{a+b}{ab+1}. \]
One can check the two cases on $f(ab)$ each imply
$a=1$ and $b=1$ respectively.
Hence the only answers are those claimed. |
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 $N$ be the number of such sequences consisting of $k$ steps
and resulting in the state where lamps $1$ through $n$ are all on,
and lamps $n + 1$ through $2n$ are all off.
Let $M$ be number of such sequences consisting of $k$ steps,
resulting in the state where lamps $1$ through $n$ are all on,
and lamps $n + 1$ through $2n$ are all off,
but where none of the lamps $n + 1$ through $2n$ is ever switched on.
Determine $\frac{N}{M}$. | 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 $144225253 \mapsto 111222223$.)
Clearly this is map is well-defined and surjective.
So all that remains is:
\begin{claim*}
Every $M$-sequence has exactly $2^{k-n}$ pre-images under $\Psi$.
\end{claim*}
\begin{proof}
Indeed, suppose that there are $c_1$ instances of lamp $1$.
Then we want to pick an odd subset of the $1$'s to change to $n+1$'s,
so $2^{c_1 - 1}$ ways to do this.
And so on.
Hence the number of pre-images is
\[ \prod_i 2^{c_i - 1} = 2^{k-n}. \qedhere \]
\end{proof} |
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 common external tangents to
$\omega_1$ and $\omega_2$ intersect on $\omega$.
\end{enumerate} | 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 orient $AC$ horizontally and let $K$ be the ``uppermost'' point of $\omega$, as shown.
\begin{center}
\begin{asy}
size(12cm);
pair W = dir(48.4);
pair X = dir(68.4);
pair Y = dir(138.4);
pair Z = dir(173.4);
pair A = 2*X*Z/(X+Z);
pair B = 2*W*Z/(W+Z);
pair C = 2*W*Y/(W+Y);
pair D = 2*X*Y/(X+Y);
draw(arc(origin, 1, 40, 180), dashed+orange);
filldraw(incircle(A, B, C), opacity(0.1)+lightred, red);
filldraw(incircle(A, D, C), opacity(0.1)+lightred, red);
filldraw(A--B--C--D--cycle, opacity(0.1)+yellow, orange);
draw(A--C, red);
pair I_B = incenter(A, B, C);
pair I_D = incenter(A, D, C);
pair P = foot(I_B, A, C);
pair Q = 2*I_B-P;
pair T = foot(I_D, A, C);
pair S = 2*I_D-T;
draw(A--Z, orange);
draw(C--W, orange);
pair E = 2*Y*Z/(Y+Z);
pair F = 2*X*W/(X+W);
draw(E--D--F, orange);
pair K = extension(B, Q, D, P);
draw(B--K--P, dashed+red);
draw(P--Q, red);
draw(S--T, red);
dot("$W$", W, dir(W));
dot("$X$", X, dir(95));
dot("$Y$", Y, dir(Y));
dot("$Z$", Z, dir(Z));
dot("$A$", A, dir(135));
dot("$B$", B, dir(B));
dot("$C$", C, dir(45));
dot("$D$", D, dir(-D));
dot("$P$", P, dir(270));
dot("$Q$", Q, 0.8*dir(95));
dot("$T$", T, dir(45));
dot("$S$", S, dir(225));
dot("$K$", K, dir(315));
/* TSQ Source:
!size(12cm);
W = dir 48.4
X = dir 68.4 R95
Y = dir 138.4
Z = dir 173.4
A = 2*X*Z/(X+Z) R135
B = 2*W*Z/(W+Z)
C = 2*W*Y/(W+Y) R45
D = 2*X*Y/(X+Y) R-D
!draw(arc(origin, 1, 40, 180), dashed+orange);
incircle A B C 0.1 lightred / red
incircle A D C 0.1 lightred / red
A--B--C--D--cycle 0.1 yellow / orange
A--C red
I_B := incenter A B C
I_D := incenter A D C
P = foot I_B A C R270
Q = 2*I_B-P 0.8R95
T = foot I_D A C R45
S = 2*I_D-T R225
A--Z orange
C--W orange
E := 2*Y*Z/(Y+Z)
F := 2*X*W/(X+W)
E--D--F orange
K = extension B Q D P R315
B--K--P dashed red
P--Q red
S--T red
*/
\end{asy}
\end{center}
Consequently, a homothety at $B$ maps $Q$, $T$, $K$ to each other
(since $T$ is the uppermost of the excircle, $Q$ of the incircle).
Similarly, a homothety at $D$ maps $P$, $S$, $K$ to each other.
As $\ol{PQ}$ and $\ol{ST}$ are parallel diameters
it then follows $K$ is the exsimilicenter of $\omega_1$ and $\omega_2$. |
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 $a_2-1$.
\begin{itemize}
\ii If $p \mid a_1$, then $p \nmid a_1 - 1$. Then
\[ a_k (a_1-1) \equiv 0 \pmod q \implies a_k \equiv 0 \pmod q. \]
In particular, $p \mid a_k$.
So repeating this argument,
we get $a_{k-1} \equiv 0 \pmod q$, $a_{k-2} \equiv 0 \pmod q$, and so on.
\ii Similarly, if $p \mid a_2 - 1$ then $p \nmid a_2$, and
\[ a_2 (a_3-1) \equiv 0 \pmod q \implies a_3 \equiv 1 \pmod q. \]
In particular, $p \mid a_3 - 1$.
So repeating this argument,
we get $a_4 \equiv 0 \pmod q$, $a_5 \equiv 0 \pmod q$, and so on.
\end{itemize}
Either way, we find $a_i \pmod q$ is constant (and either $0$ or $1$).
Since $q$ was an arbitrary prime power dividing $n$,
by Chinese remainder theorem we conclude that $a_i \pmod n$ is constant as well.
But this contradicts the assumption of distinctness. |
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)+lightred, red);
pair M = midpoint(P--Q);
pair K = midpoint(P--B);
pair L = midpoint(Q--C);
filldraw(circumcircle(M, K, L), opacity(0.3)+lightcyan, blue);
draw(P--Q, heavygreen+1);
draw(B--P, heavygreen);
draw(C--Q, heavygreen);
filldraw(K--M--L--cycle, opacity(0.4)+cyan, heavycyan);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(120));
dot("$M$", M, dir(M));
dot("$K$", K, dir(K));
dot("$L$", L, dir(20));
/* TSQ Source:
A = dir 70
B = dir 210
C = dir 330
P = 0.4*A+0.6*C
Q = 0.25*B+0.75*A R120
A--B--C--cycle 0.2 lightred / red
M = midpoint P--Q
K = midpoint P--B
L = midpoint Q--C R20
circumcircle M K L 0.3 lightcyan / blue
P--Q heavygreen+1
B--P heavygreen
C--Q heavygreen
K--M--L--cycle 0.4 cyan / heavycyan
*/
\end{asy}
\end{center}
As $\ol{ML} \parallel \ol{AC}$ and $\ol{MK} \parallel \ol{AB}$ we have that
\begin{align*}
\dang APQ &= \dang LMP = \dang LKM \\
\dang PQA &= \dang KMQ = \dang MLK
\end{align*}
and consequently we have the (opposite orientation) similarity
\[ \triangle APQ \overset{-}{\sim} \triangle MKL. \]
Therefore
\[ \frac{AQ}{AP} = \frac{ML}{MK} = \frac{2ML}{2MK} = \frac{PC}{QB} \]
id est $AQ \cdot QB = AP \cdot PC$, which is what we wanted to prove. |
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 arithmetic progression. | 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 $\sup d_n < \infty$ since $d_n$ is bounded above by $A$.
Then we let
\[ m \coloneq \min d_n, \qquad M \coloneq \max d_n. \]
Now suppose $a$ achieves the maximum, meaning $s(a+1)-s(a) = M$.
Then
\begin{align*}
\underbrace{d_{s(s(a))} + \dots + d_{s(s(a+1))-1}}_{D \text{ terms}}
&= \boxed{s(s(s(a+1))) - s(s(s(a)))} \\
&= (D \cdot s(a+1) + A) - (D \cdot s(a) + A) = DM.
\end{align*}
Now $M$ was maximal hence $M = d_{s(s(a))} = \dots = d_{s(s(a+1))-1}$.
But $d_{s(s(a))} = B-A$ is a constant.
Hence $M = B-A$.
In the same way $m = B-A$ as desired.
\paragraph{Second solution.}
We retain the notation $D$, $A$, $B$ above,
as well as $m = \min_n s(n+1)-s(n) \ge B-A$.
We do the involution trick first as:
\[ D = \boxed{s(s(s(n)+1)) - s(s(s(n)))} = s(Dn+B) - s(Dn+A) \]
and hence we recover $D \ge m(B-A)$.
The edge case $D = B-A$ is easy since then $m=1$
and $D = s(Dn+B)-s(Dn+A)$ forces $s$ to be a constant shift.
So henceforth assume $D > B-A$.
The idea is that right now the $B$ terms are ``too big'',
so we want to use the involution trick in a way that makes as many
``$A$ minus $B$'' shape expressions as possible.
This motivates considering $s(s(s(n+1))) - s(s(s(n)+1)+1) > 0$,
since the first expression will have all $A$'s
and the second expression will have all $B$'s.
Calculation gives:
\begin{align*}
s(D(n+1)+A) - s(Dn+B+1)
&= \boxed{s(s(s(n+1))) - s(s(s(n)+1)+1)} \\
&= (Ds(n+1) + A) - (D(s(n)+1) + B) \\
&= D\left( s(n+1)-s(n) \right) + A-B-D.
\end{align*}
Then by picking $n$ achieving the minimum $m$,
\[ \underbrace{m(D+A-B-1)}_{>0} \le s(s(s(n+1))) - s(s(s(n)+1)+1) \le Dm + A-B-D \]
which becomes
\[ \left( D-m(B-A) \right) + \left( (B-A)-m \right) \le 0. \]
Since both of these quantities were supposed to be nonnegative,
we conclude $m = B-A$ and $D = m^2$.
Now the estimate $D = s(Dn+B) - s(Dn+A) \ge m(B-A)$ is actually sharp,
so it follows that $s(n)$ is arithmetic. |
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 chased all the angles we want, we need a relationship.
We can find it by considering the side ratio $\frac{IK}{KC}$.
Using the angle bisector theorem,
we can express this in terms of triangle $IDC$;
however we can also express it in terms of triangle $IEC$.
\begin{center}
\begin{asy}
size(7cm);
pair A = Drawing("A", (0,7), dir(90));
pair C = Drawing("C", (3,0), dir(-45));
pair B = Drawing("B", -C, dir(225));
draw(A--B--C--cycle);
pair I = incenter(A,B,C);
pair D = Drawing("D", origin, dir(-90));
pair E = Drawing("E", extension(B,I,A,C), dir(45));
pair K = Drawing("K", incenter(A,D,C), dir(-30));
draw(A--D--K);
draw(B--E--K);
draw(incircle(A,D,C));
markangle(9.0,B,E,K);
\end{asy}
\quad
\begin{asy}
pair A = Drawing("A", (0,5), dir(90));
pair C = Drawing("C", (3,0), dir(-45));
pair B = -C;
pair D = Drawing("D", origin, dir(-90));
pair I = Drawing("I", incenter(A,B,C), dir(180));
pair E = Drawing("E", extension(B,I,A,C), dir(45));
pair K = Drawing("K", incenter(A,D,C), 1.3*dir(-90));
draw(A--D--C--cycle);
draw(C--I--E--K);
draw(D--K);
MP("2x", A, dir(K-A)*8);
MP(rotate(-35)*"45^{\circ}-x", C+dir(130)*1.2);
MP(rotate(-10)*"45^{\circ}-x", C+dir(155));
MP("45^{\circ}", E, dir(240)*6);
MP("3x", E, dir(-80)*6);
markangle(9.0,D,I,C);
markangle(9.0,E,I,A);
\end{asy}
\end{center}
By the law of sines, we obtain
\[ \frac{IK}{KC} = \frac%
{\sin 45\dg \cdot \frac{EK}{\sin \left( 90\dg - 2x \right)}}%
{\sin \left( 3x \right) \cdot \frac{EK}{\sin \left( 45\dg-x \right)}}
= \frac{\sin 45\dg \sin \left( 45\dg - x \right)}%
{\sin \left( 3x \right) \sin \left( 90\dg - 2x \right)}. \]
Also, by the angle bisector theorem on $\triangle IDC$,
we have
\[ \frac{IK}{KC} = \frac{ID}{DC}
= \frac{\sin \left( 45\dg-x \right)}{\sin\left( 45\dg+x \right)}. \]
Equating these and cancelling $\sin \left( 45\dg-x \right) \neq 0$ gives
\[ \sin 45\dg \sin \left( 45\dg + x \right)
= \sin 3x \sin \left( 90\dg - 2x \right). \]
Applying the product-sum formula
(again, we are just trying to break down things as much as possible),
this just becomes
\[ \cos\left( x \right) - \cos \left( 90\dg + x \right)
= \cos \left( 5x-90\dg \right) - \cos \left( 90\dg+x \right) \]
or $\cos x = \cos \left( 5x-90\dg \right)$.
At this point we are basically done;
the rest is making sure we do not miss any solutions
and write up the completion nicely.
One nice way to do this is by using product-sum in reverse as
\[ 0 = \cos \left( 5x-90\dg \right) - \cos x
= 2 \sin \left( 3x - 45\dg \right) \sin \left( 2x-45\dg \right). \]
This way we merely consider the two cases
\[ \sin \left( 3x-45\dg \right) = 0 \text{ and }
\sin \left( 2x - 45\dg \right) = 0. \]
Notice that $\sin\theta = 0$ if and only $\theta$
is an integer multiple of $180\dg$.
Using the bound $0\dg < x < 45\dg$,
it is easy to see that that the permissible values of $x$
are $x = 15\dg$ and $x = \frac{45}{2}\dg$.
As $\angle A = 4x$, this corresponds to $\angle A = 60\dg$
and $\angle A = 90\dg$, which can be seen to work. |
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, the function $f$ is periodic modulo $N = f(1)-1 \ge 1$.
This is impossible since we can fix $b$ and let $a$ be arbitrarily
large in some residue class modulo $N$.
Hence $f(1)=1$, so taking $P(n,1)$ gives $f(f(n)) = n$.
\end{proof}
\begin{claim*}
Let $\delta = f(2)-1 > 0$.
Then for every $n$,
\[ f(n+1) = f(n) + \delta
\quad\text{ or }\quad f(n-1) = f(n) + \delta \]
\end{claim*}
\begin{proof}
Use
\[ P(2, f(n)) \implies n-2 < f( f(n) + \delta ) < n+2. \]
Let $y = f(f(n)+\delta)$, hence $n-2 < y < n+2$
and $f(y) = f(n)+\delta$.
But, remark that if $y = n$, we get $\delta = 0$, contradiction.
So $y \in \{n+1, n-1\}$ and that is all.
\end{proof}
We now show $f$ is an arithmetic progression
with common difference $+\delta$.
Indeed we already know $f(1) = 1$ and $f(2) = 1+\delta$.
Now suppose $f(1)=1$, \dots, $f(n) = 1 + (n-1)\delta$.
Then by induction for any $n \ge 2$,
the second case can't hold,
so we have $f(n+1) = f(n)+\delta$, as desired.
Combined with $f(f(n)) = n$, we recover that $f$ is the identity. |
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 order can be chosen in such a way that
the grasshopper never lands on any point in $M$.
\end{enumerate} | 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 $[0, x]$ and so we use induction to reach $x$,
then leap from $x$ to $s$ and win.
\ii If $x$ has no mine but there is also no mine to the right of $x$,
then let $m$ be the maximal mine.
By induction hypothesis on $M \setminus \{m\}$, there is a path to $x$
using $\{a_1, \dots, a_{n-1}\}$ which avoids mines except possibly $m$.
If the path hits the mine $m$ on the hop of length $a_k$,
we then swap that hop with $a_n$, and finish.
\ii If $x$ has a mine, but there are no mines to the right of $x$,
we can repeat the previous case with $m = x$.
\ii Now suppose $x$ has a mine, \emph{and} there is a mine past $x$.
There should exist an integer $1 \le i \le n-1$
such that $s-a_i$ and $y = s-a_i-a_n$ both have no mine.
By induction hypothesis, we can then reach $y$ in $n-2$ steps
(as there are two mines to the right of $y$); then $y \to s-a_i \to s$ finishes.
\end{itemize}
\begin{remark*}
It seems much of the difficulty of the problem is realizing induction will actually work.
Attempts at induction are, indeed, a total minefield (ha!),
and given the position P6 of the problem, it is expected that many contestants
will abandon induction after some cursory attempts fail.
\end{remark*} |
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 $f(x) \left\lfloor f(0) \right\rfloor = f(0)$,
thus $f$ is constant.
Thus assume henceforth $f(0) = 0$.
Now set $x=y=1$ to get
\[ f(1) = f(1) \left\lfloor f(1) \right\rfloor \]
so either $f(1) = 0$ or $1 \le f(1) < 2$.
We split into cases:
\begin{itemize}
\ii If $f(1) = 0$, pick $x=1$ to get $f(y) \equiv 0$.
\ii If $1 \le f(1) < 2$,
then $y=1$ gives
\[ f(\left\lfloor x \right\rfloor) = f(x) \]
from $y=1$, in particular $f(x) = 0$ for $0 \le x < 1$.
Choose $(x,y) = \left( 2, \half \right)$ to get
$f(1) = f(2) \left\lfloor f\left( \half \right) \right\rfloor = 0$.
\end{itemize} |
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 $\ol{DG}$ and $\ol{EI}$ intersect on $\Gamma$. | Let $\ol{EI}$ meet $\Gamma$ again at $K$.
Then it suffices to show that $\ol{KD}$ bisects $\ol{IF}$.
Let $\ol{AF}$ meet $\Gamma$ again at $H$, so $\ol{HE} \parallel \ol{BC}$.
By Pascal theorem on \[ AHEKDD \]
we then obtain that $P = \ol{AH} \cap \ol{KD}$ lies on a line through $I$
parallel to $\ol{BC}$.
Let $I_A$ be the $A$-excenter,
and set $Q = \ol{I_AF} \cap \ol{IP}$, and $T = \ol{AIDI_A} \cap \ol{BFC}$.
Then
\[ -1 = (AI;TI_A) \overset{F} = (IQ;\infty P) \]
where $\infty$ is the point at infinity along $\ol{IPQ}$.
Thus $P$ is the midpoint of $\ol{IQ}$.
Since $D$ is the midpoint of $\ol{II_A}$ by ``Fact 5'',
it follows that $\ol{DP}$ bisects $\ol{IF}$.
\begin{center}
\begin{asy}
size(9cm);
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
pair D = dir(-90);
pair I = incenter(A, B, C);
pair I_A = 2*D-I;
pair E = dir(310);
pair K = -E+2*foot(origin, E, I);
pair H = B*C/E;
pair F = extension(B, C, A, H);
pair G = extension(D, K, I, F);
pair P = extension(A, F, D, G);
pair Q = extension(I, P, I_A, F);
draw(A--B--C--cycle, lightblue);
filldraw(unitcircle, opacity(0.1)+lightcyan, lightblue);
draw(A--I_A, lightblue);
draw(A--H--E--K--D--cycle, orange);
draw((D+0.3)--(D-0.3), orange);
draw(I--Q--I_A, red);
draw(I--F, blue);
pair T = extension(B, C, A, D);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(220));
dot("$I$", I, dir(10));
dot("$I_A$", I_A, dir(I_A));
dot("$E$", E, dir(E));
dot("$K$", K, dir(K));
dot("$H$", H, dir(H));
dot("$F$", F, dir(F));
dot("$G$", G, dir(350));
dot("$P$", P, dir(45));
dot("$Q$", Q, dir(Q));
dot("$T$", T, dir(T));
/* TSQ Source:
!size(9cm);
A = dir 110
B = dir 210
C = dir 330
D = dir -90 R220
I = incenter A B C R10
I_A = 2*D-I
E = dir 310
K = -E+2*foot origin E I
H = B*C/E
F = extension B C A H
G = extension D K I F R350
P = extension A F D G R45
Q = extension I P I_A F
A--B--C--cycle lightblue
unitcircle 0.1 lightcyan / lightblue
A--I_A lightblue
A--H--E--K--D--cycle orange
(D+0.3)--(D-0.3) orange
I--Q--I_A red
I--F blue
T = extension B C A D
*/
\end{asy}
\end{center} |
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
\[ \nu_p(M+g(n)), \quad \nu_p(M+g(n')) \quad \text{are both odd}. \]
(It's not hard to see this is always possible.)
Now, since each of
\begin{align*}
\left( M + g(n) \right)&\left( n + g(M) \right) \\
\left( M + g(n') \right)&\left( n' + g(M) \right)
\end{align*}
is a square, we get $n \equiv n' \equiv -g(M) \pmod p$.
\end{proof}
This claim implies that
\begin{itemize}
\ii The numbers $g(n)$ and $g(n+1)$ differ by $\pm 1$ for any $n$, and
\ii The function $g$ is injective.
\end{itemize}
It follows $g$ is a linear function with slope $\pm 1$, hence done. |
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 $\triangle CDP$
is orthogonal to $\Gamma$
(passing through the orthocenter of $\triangle MPN$).
\begin{center}
\begin{asy}
pair N = dir(100);
pair M = -N;
pair L = dir(150);
pair K = N*N/L;
pair C = dir(122);
pair D = conj(C);
pair P = extension(D, N, M, C);
pair A = -K+2*foot(origin, K, P);
pair B = -L+2*foot(origin, L, P);
pair S = circumcenter(C, P, D);
filldraw(unitcircle, opacity(0.1)+lightgreen, heavygreen);
draw(C--S--D, heavygreen);
draw(A--C--B--S, lightblue);
draw(A--D--B, lightblue);
filldraw(L--N--K--M--cycle, opacity(0.1)+lightred, lightred);
draw(M--N, lightred);
draw(K--L, lightred);
draw(A--K, dashed+orange);
draw(B--L, dashed+orange);
draw(C--M, dashed+orange);
draw(D--N, dashed+orange);
draw(S--P, heavygreen);
draw(arc(S,abs(S-D),-60,60), heavygreen);
dot("$N$", N, dir(N));
dot("$M$", M, dir(M));
dot("$L$", L, dir(170));
dot("$K$", K, dir(K));
dot("$C$", C, dir(80));
dot("$D$", D, dir(280));
dot("$P$", P, dir(P));
dot("$A$", A, dir(220));
dot("$B$", B, dir(B));
dot("$S$", S, dir(S));
/* TSQ Source:
N = dir 100
M = -N
L = dir 150 R170
K = N*N/L
C = dir 122 R80
D = conj(C) R280
P = extension D N M C
A = -K+2*foot origin K P R220
B = -L+2*foot origin L P
S = circumcenter C P D
unitcircle 0.1 lightgreen / heavygreen
C--S--D heavygreen
A--C--B--S lightblue
A--D--B lightblue
L--N--K--M--cycle 0.1 lightred / lightred
M--N lightred
K--L lightred
A--K dashed orange
B--L dashed orange
C--M dashed orange
D--N dashed orange
S--P heavygreen
!draw(arc(S,abs(S-D),-60,60), heavygreen);
*/
\end{asy}
\end{center}
Now, we are given that point $S$ is such that $\ol{SC}$
is tangent to $\Gamma$, and $SC = SP$.
It follows that $S$ is the circumcenter of $\triangle CDP$,
and hence $\ol{SC}$ and $\ol{SD}$ are tangents to $\Gamma$.
Then $-1 = (AB;CD) \overset{P}{=} (KL;MN)$.
Since $\ol{MN}$ is a diameter, this implies $MK = ML$.
\begin{remark*}
I think it's more natural to come up with
this solution in reverse.
Namely, suppose we define the points the other way:
let $\ol{SD}$ be the other tangent, so $(AB;CD) = -1$.
Then project through $P$ to get $(KL;MN) = -1$,
where $N$ is the second intersection of $\ol{DP}$.
However, if $ML = MK$ then $KMLN$ must be a kite.
Thus one can recover the solution in reverse.
\end{remark*}
\paragraph{Second solution (Sebastian Jeon).}
We have \[ SP^2 = SC^2 = SA \cdot SB
\implies
\dang SPA = \dang PBA = \dang LBA = \dang LKA = \dang LKP \]
(the latter half is Reim's theorem).
Therefore $\ol{SP}$ and $\ol{LK}$ are \emph{parallel}.
Now, let $\ol{SP}$ meet $\Gamma$ again at $X$ and $Y$,
and let $Q$ be the antipode of $P$ on $(S)$.
Then
\[ SP^2 = SQ^2 = SX \cdot SY
\implies (PQ;XY) = -1 \implies \angle QCP = 90\dg \]
that $\ol{CP}$ bisects $\angle XCY$.
Since $\ol{XY} \parallel \ol{KL}$,
it follows $\ol{CP}$ bisects to $\angle LCK$ too. |
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 $B_k$, $1\leq k \leq 4$,
remove one coin from $B_k$ and swap the contents
(possibly empty) of the boxes $B_{k+1}$ and $B_{k+2}$.
\end{enumerate}
Determine if there exists a finite sequence of operations of the allowed types,
such that the five boxes $B_1$, $B_2$, $B_3$, $B_4$, $B_5$ become empty,
while box $B_6$ contains exactly $2010^{2010^{2010}}$ coins. | First,
\begin{align*}
(1,1,1,1,1,1) &\to (0,3,1,0,3,1) \to (0,0,7,0,0,7) \\
&\to (0,0,6,2,0,7) \to (0,0,6,1,2,7) \to (0,0,6,1,0,11) \\
&\to (0,0,6,0,11,0) \to (0,0,5,11,0,0).
\end{align*}
and henceforth we ignore boxes $B_1$ and $B_2$,
looking at just the last four boxes;
so we write the current position as $(5,11,0,0)$.
We prove a lemma:
\begin{claim*}
Let $k \ge 0$ and $n > 0$.
From $(k,n,0,0)$ we may reach $(k-1,2^n,0,0)$.
\end{claim*}
\begin{proof}
Working with only the last three boxes for now,
\begin{align*}
(n,0,0) &\to (n-1, 2, 0) \to (n-1, 0, 4) \\
&\to (n-2, 4, 0) \to (n-2, 0, 8) \\
&\to (n-3, 8, 0) \to (n-3, 0, 16) \\
&\to \dots \to (1, 2^{n-1}, 0) \to (1, 0, 2^n) \to (0, 2^n, 0).
\end{align*}
Finally we have $(k,n,0,0) \to (k,0,2^n,0) \to (k-1,2^n, 0,0)$.
\end{proof}
Now from $(5,11,0,0)$ we go as follows:
\begin{align*} (5,11,0,0) &\to (4, 2^{11}, 0, 0)
\to \left(3, 2^{2^{11}}, 0, 0\right)
\to \left(2, 2^{2^{2^{11}}}, 0, 0\right) \\
&\to \left( 1, 2^{2^{2^{2^{11}}}}, 0, 0\right)
\to \left(0, 2^{2^{2^{2^{2^{11}}}}}, 0, 0\right).
\end{align*}
Let $A = 2^{2^{2^{2^{2^{11}}}}} > 2010^{2010^{2010}} = B$.
Then by using move 2 repeatedly on the fourth box
(i.e., throwing away several coins by swapping the empty $B_5$ and $B_6$),
we go from $(0,A,0,0)$ to $(0,B/4,0,0)$.
From there we reach $(0,0,0,B)$. |
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
\[
a_n =
a_{\ell} + a_{n - \ell} \text{ for all $n \ge N$}.
\]
\end{enumerate} | Let \[ w_1 = \frac{a_1}{1}, \quad w_2 = \frac{a_2}{2},
\quad \dots, \quad w_s = \frac{a_s}{s}. \]
(The choice of the letter $w$ is for ``weight''.)
We claim the right choice of $\ell$
is the one maximizing $w_\ell$.
Our plan is to view each $a_n$ as a linear combination
of the weights $w_1, \dots, w_s$ and track their coefficients.
To this end, let's define an \emph{$n$-type}
to be a vector $T = \left< t_1, \dots, t_s\right>$
of nonnegative integers such that
\begin{itemize}
\ii $n = t_1 + \dots + t_s$; and
\ii $t_i$ is divisible by $i$ for every $i$.
\end{itemize}
We then define its \emph{valuation} as $v(T) = \sum_{i=1}^s w_i t_i$.
Now we define a $n$-type to be \emph{valid}
according to the following recursive rule.
For $1 \le n \le s$ the only valid $n$-types are
\begin{align*}
T_1 &= \left< 1, 0, 0, \dots, 0 \right> \\
T_2 &= \left< 0, 2, 0, \dots, 0 \right> \\
T_3 &= \left< 0, 0, 3, \dots, 0 \right> \\
&\vdotswithin= \\
T_s &= \left< 0, 0, 0, \dots, s \right>
\end{align*}
for $n = 1, \dots, s$, respectively.
Then for any $n > s$, an $n$-type is valid
if it can be written as the sum of a valid $k$-type
and a valid $(n-k)$-type, componentwise.
These represent the linear combinations possible in the recursion;
in other words the recursion in the problem is phrased as
\[ a_n = \max_{T \text{ is a valid $n$-type}} v(T). \]
In fact, we have the following description of valid $n$-types:
\begin{claim*}
Assume $n > s$.
Then an $n$-type $\left< t_1, \dots, t_s \right>$ is valid
if and only if either
\begin{itemize}
\ii there exist indices $i < j$ with $i+j > s$,
$t_i \ge i$ and $t_j \ge j$; or
\ii there exists an index $i > s/2$
with $t_i \ge 2i$.
\end{itemize}
\end{claim*}
\begin{proof}
Immediate by forwards induction on $n > s$
that all $n$-types have this property.
The reverse direction is by downwards induction on $n$.
Indeed if $\sum_i \frac{t_i}{i} > 2$,
then we may subtract off on of $\{T_1, \dots, T_s\}$
while preserving the condition;
and the case $\sum_i \frac{t_i}{i} = 2$
is essentially by definition.
\end{proof}
\begin{remark*}
The claim is a bit confusingly stated in its two cases;
really the latter case should be thought of as the situation
$i=j$ but requiring that $t_i/i$ is counted with multiplicity.
\end{remark*}
Now, for each $n > s$ we pick a valid $n$-type $T_n$
with $a_n = v(T_n)$;
if there are ties, we pick one for which the $\ell$th
entry is as large as possible.
\begin{claim*}
For any $n > s$ and index $i \neq \ell$,
the $i$th entry of $T_n$
is at most $2s + \ell i$.
\end{claim*}
\begin{proof}
If not, we can go back $i\ell$ steps to get
a valid $(n-i\ell)$-type $T$
achieved by decreasing the $i$th entry of $T_n$ by $i \ell$.
But then we can add $\ell$ to the $\ell$th entry $i$
times to get another $n$-type $T'$ which obviously
has valuation at least as large,
but with larger $\ell$th entry.
\end{proof}
Now since all other entries in $T_n$ are bounded,
eventually the sequence $(T_n)_{n > s}$
just consists of repeatedly
adding $1$ to the $\ell$th entry, as required.
\begin{remark*}
One big step is to consider $w_k = a_k / k$.
You can get this using wishful thinking
or by examining small cases.
(In addition this normalization makes it easier
to see why the largest $w$ plays an important role,
since then in the definition of type,
the $n$-types all have a sum of $n$.
Unfortunately, it makes the characterization
of valid $n$-types somewhat clumsier too.)
\end{remark*} |
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 value of $n_A$. | There are two curves of solutions,
namely $\{x,5x,7x,11x\}$ and $\{x,11x,19x,29x\}$, for any positive integer $x$,
achieving $n_A = 4$ (easy to check).
We'll show that $n_A \le 4$ and equality holds only in one of the curves.
Let $A = \{a < b < c < d\}$.
\begin{claim*}
We have $n_A \le 4$ with equality iff
\[ a+b \mid c+d, \quad a+c \mid b+d, \quad a+d = b+c. \]
\end{claim*}
\begin{proof}
Note $a+b \mid s_A \iff a+b \mid c+d$ etc.
Now $c+d \nmid a+b$ and $b+d \nmid a+c$ for size reasons, so we already have $n_A \le 4$;
moreover $a+d \mid b+c$ and $b+c \mid a+d$ if and only if $a+d = b+c$.
\end{proof}
We now show the equality curve is the one above.
\[ a+c \mid b+d \iff a+c \mid -a+2b+c \iff a+c \mid 2(b-a). \]
Since $a+c > |b-a|$, so we must have $a+c=2(b-a)$.
So we now have
\begin{align*}
c &= 2b-3a \\
d &= b+c-a = 3b+c-4a.
\end{align*}
The last condition is
\[ a+b \mid c+d = 5b-7a \iff a+b \mid 12a. \]
Now, let $x = \gcd(a,b)$.
The expressions for $c$ and $d$ above imply that
$x \mid c,d$ so we may scale down so that $x = 1$.
Then $\gcd(a+b,a) = \gcd(a,b) = 1$ and so $a+b \mid 12$.
We have $c > b$, so $3a < b$.
The only pairs $(a,b)$ with $3a < 2b$, $\gcd(a,b) = 1$ and $a+b \mid 12$
are $(a,b) \in \left\{ (1,5), (1,11) \right\}$ which give the solutions earlier. |
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 the line meets some other point belonging to $\mathcal S$.
This point, $Q$, takes over as the new pivot,
and the line now rotates clockwise about $Q$,
until it next meets a point of $\mathcal S$.
This process continues indefinitely.
Show that we can choose a point $P$ in $\mathcal S$ and
a line $\ell$ going through $P$ such that the resulting windmill
uses each point of $\mathcal S$ as a pivot infinitely many times. | Orient $\ell$ in some direction,
and color the plane such that its left half is red
and right half is blue.
The critical observation is that:
\begin{claim*}
The number of points on the red side of $\ell$ does not change,
nor does the number of points on the blue side
(except at a moment when $\ell$ contains two points).
\end{claim*}
Thus, if $|\mathcal S| = n+1$,
it suffices to pick the initial configuration
so that there are $\left\lfloor n/2 \right\rfloor$
red and $\left\lceil n/2 \right\rceil$ blue points.
Then when the line $\ell$ does a full $180\dg$ rotation,
the red and blue sides ``switch'',
so the windmill has passed through every point.
(See official shortlist for verbose write-up;
this is deliberately short to make a point.) |
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{symmetry trick} to write
\[ f(f(x)) \le (f(x)-w)f(w) + f(f(w)) \]
so that when we sum we get
\[ wf(w) + xf(x) \le 2f(x)f(w). \]
%(We remark now that if we set $w = 0$,
%we find that $xf(x) \le 2f(x)f(0)$ for all $x$;
%since we know $f(0)$ should be zero, this is pretty good.)
Next we use \emph{cancellation trick}: set $w = 2f(x)$ in the above to get
\[ xf(x) \le 0 \quad \forall x \in \RR. \qquad (\spadesuit) \]
\begin{claim*}
For every $p \in \RR$, we have $f(p) \le 0$.
\end{claim*}
\begin{proof}
Assume $f(p) > 0$ for some $p \in \RR$.
Then for any negative number $z$,
\[ 0 \overset{(\spadesuit)}{\le} f(z)
\overset{(\heartsuit)}{\le} (z-p)f(p) + f(f(p)). \]
which is false if we let $z \to -\infty$.
\end{proof}
Together with $(\spadesuit)$ we derive $f(x) = 0$ for $x < 0$.
Finally, letting $x$ and $z$ be any negative numbers
in $(\heartsuit)$, we get $f(0) \ge 0$, so $f(0) = 0$ too.
\begin{remark*}
As another corollary of the claim, $f(f(x)) = 0$ for all $x$.
\end{remark*}
\begin{remark*}
A nontrivial example of a working $f$ is to take
\[ f(x) = \begin{cases}
-\exp(\exp(\exp(x))) & x > 0 \\
0 & x \le 0.
\end{cases} \]
or some other negative function growing rapidly in absolute value for $x > 0$.
\end{remark*} |
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 $f(x) = f(-x)$ and $f$ is even.
\end{itemize}
\begin{claim*}
Let $x$, $y$, $z$ be integers with $x+y+z=0$.
Then among $f(x)$, $f(y)$, $f(z)$,
two of them are equal and divide the third.
\end{claim*}
\begin{proof}
Let $a = f(\pm x)$, $b = f(\pm y)$, $c = f(\pm z)$
be positive integers.
Note that
\begin{align*}
a &\mid b-c \\
b &\mid c-a \\
c &\mid a-b
\end{align*}
from $P(y,-z)$ and similarly.
WLOG $c = \max(a,b,c)$; then $c > |a-b|$
so $a=b$. Thus $a=b \mid c$ from the first two.
\end{proof}
This implies the problem,
by taking $x$ and $y$ in the previous claim
to be the integers $m$ and $n$.
\begin{remark*}
At \url{https://aops.com/community/c6h418981p2381909},
Davi Medeiros gives the following characterization
of functions $f$ satisfying the hypothesis.
Pick $f(0)$, $k$ positive integers,
a chain $d_1 \mid d_2 \mid \dots \mid d_k$ of divisors of $f(0)$,
and positive integers $a_1,a_2,\dots,a_{k-1}$,
greater than $1$ (if $k=1$, $a_i$ doesn't exist, for every $i$).
We'll define $f$ as follows:
\begin{itemize}
\ii $f(n)=d_1$, for every integer $n$ that is not divisible by $a_1$;
\ii $f(a_1n)=d_2$, for every integer $n$ that is not divisible by $a_2$;
\ii $f(a_1a_2n)=d_3$, for every integer $n$ that is not divisible by $a_3$;
\ii $f(a_1a_2a_3n)=d_4$, for every integer $n$ that is not divisible by $a_4$;
\ii \dots
\ii $f(a_1a_2 \dots a_{k-1}n)=d_k$, for every integer $n$;
\end{itemize}
\end{remark*} |
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 $\ell_c$
is tangent to the circle $\Gamma$.
\end{enumerate} | This is a hard problem with many beautiful solutions.
The following solution is not very beautiful but not too hard to find during an olympiad,
as the only major insight it requires is the construction of $A_2$, $B_2$, and $C_2$.
\begin{center}
\begin{asy}
size(11cm);
pair A = dir(110);
dot("$A$", A, dir(A));
pair B = dir(195);
dot("$B$", B, dir(160));
pair C = dir(325);
dot("$C$", C, 1.4*dir(30));
pair P = dir(270);
dot("$P$", P, dir(P));
draw(unitcircle);
draw(A--B--C--cycle);
pair U = P+(2,0);
pair V = 2*P-U;
pair X_1 = reflect(B,C)*P;
pair Y_1 = reflect(C,A)*P;
pair Z_1 = reflect(A,B)*P;
pair X_2 = extension(B, C, U, V);
dot(X_2);
pair Y_2 = extension(C, A, U, V);
dot(Y_2);
pair Z_2 = extension(A, B, U, V);
dot(Z_2);
draw(B--Z_2, dotted);
draw(C--Y_2, dotted);
draw(C--X_2, dotted);
draw(X_2--Z_2);
pair A_1 = extension(Y_1, Y_2, Z_1, Z_2);
dot("$A_1$", A_1, dir(A_1));
pair B_1 = extension(Z_1, Z_2, X_1, X_2);
dot("$B_1$", B_1, dir(B_1));
pair C_1 = extension(X_1, X_2, Y_1, Y_2);
dot("$C_1$", C_1, dir(50));
draw(A_1--B_1--C_1--cycle, black+1);
draw(C_1--X_2, dotted);
pair O_1 = circumcenter(A_1, B_1, C_1);
draw(arc(O_1, abs(O_1-A_1), -80, 140));
pair A_2 = A*A/P;
dot("$A_2$", A_2, dir(-20));
pair B_2 = B*B/P;
dot("$B_2$", B_2, dir(130));
pair C_2 = C*C/P;
dot("$C_2$", C_2, dir(C_2));
draw(A_2--B_2--C_2--cycle, black+1);
pair T = extension(A_1, A_2, B_1, B_2);
dot("$T$", T, dir(T));
draw(T--A_1, dashed);
draw(T--B_1, dashed);
draw(T--C_1, dashed);
/*
A = dir 110
B = dir 195
C = dir 325
P = dir 270
unitcircle
A--B--C--cycle blue
U := P+(2,0)
V := 2*P-U
X = reflect(B,C)*P
Y = reflect(C,A)*P
Z = reflect(A,B)*P
Y--Z heavygreen
X1 := extension B C U V
Y1 := extension C A U V
Z1 := extension A B U V
Line Y1 Z1
A_1 = extension Y Y1 Z Z1
B_1 = extension Z Z1 X X1
C_1 = extension X X1 Y Y1
A_1--B_1--C_1--cycle red
circumcircle A_1 B_1 C_1
circumcircle B Z X dotted
circumcircle C X Y dotted
circumcircle A Y Z dotted
M = extension A_1 A*A/P B_1 B*B/P
*/
\end{asy}
\end{center}
We apply complex numbers with $\omega$ the unit circle and $p=1$. Let $A_1 = \ell_B \cap \ell_C$, and let $a_2 = a^2$ (in other words, $A_2$ is the reflection of $P$ across the diameter of $\omega$ through $A$). Define the points $B_1$, $C_1$, $B_2$, $C_2$ similarly.
We claim that $\ol{A_1A_2}$, $\ol{B_1B_2}$, $\ol{C_1C_2}$ concur at a point on $\Gamma$.
We begin by finding $A_1$. If we reflect the points $1+i$ and $1-i$ over $\ol{AB}$, then we get two points $Z_1$, $Z_2$ with
\begin{align*}
z_1 &= a+b-ab(1-i) = a+b-ab+abi \\
z_2 &= a+b-ab(1+i) = a+b-ab-abi. \\
\intertext{Therefore,}
z_1 - z_2 &= 2abi \\
\ol{z_1}z_2 - \ol{z_2}{z_1}
&= -2i \left( a+b+\frac1a+\frac1b-2 \right).
\end{align*}
Now $\ell_C$ is the line $\ol{Z_1Z_2}$,
so with the analogous equation $\ell_B$ we obtain:
\begin{align*}
a_1 &= \frac{ -2i\left( a+b+\frac1a+\frac1b-2 \right)\left( 2ac i \right) +
2i\left( a+c+\frac1a+\frac1c-2 \right)(2abi) }
{ \left( -\frac{2}{ab}i \right)
\left( 2ac i \right) - \left( -\frac{2}{ac}i \right) \left( 2abi \right)} \\
&= \frac{\left[ c-b \right]a^2 + \left[ \frac cb - \frac bc - 2c + 2b \right]a + (c-b) }{\frac cb - \frac bc} \\
&= a + \frac{(c-b)\left[ a^2-2a+1 \right]}{(c-b)(c+b)/bc} \\
&= a + \frac{bc}{b+c} (a-1)^2.
\end{align*}
Then the second intersection of $\ol{A_1A_2}$ with $\omega$ is given by
\begin{align*}
\frac{a_1-a_2}{1-a_2\ol{a_1}}
&= \frac{a+\frac{bc}{b+c}(a-1)^2-a^2}{1-a-a^2 \cdot \frac{(1-1/a)^2}{b+c}} \\
&= \frac{a + \frac{bc}{b+c}(1-a)}{1 - \frac{1}{b+c}(1-a)} \\
&= \frac{ab+bc+ca - abc}{a+b+c-1}.
\end{align*}
Thus, the claim is proved.
Finally, it suffices to show $\ol{A_1B_1} \parallel \ol{A_2B_2}$.
One can also do this with complex numbers;
it amounts to showing $a^2-b^2$, $a-b$, $i$
(corresponding to $\ol{A_2 B_2}$, $\ol{A_1 B_1}$, $\ol{PP}$)
have their arguments an arithmetic progression, equivalently
\[ \frac{(a-b)^2}{i(a^2-b^2)} \in \RR
\iff
\frac{(a-b)^2}{i(a^2-b^2)}
= \frac{\left( \frac 1a-\frac1b \right)^2}
{\frac1i\left(\frac{1}{a^2}-\frac{1}{b^2}\right)}
\]
which is obvious.
\begin{remark*}
One can use directed angle chasing for this last part too.
Let $\ol{BC}$ meet $\ell$ at $K$ and $\ol{B_2C_2}$ meet $\ell$ at $L$.
Evidently
\begin{align*}
-\dang B_2LP &= \dang LPB_2 + \dang PB_2L \\
&= 2 \dang KPB + \dang PB_2C_2 \\
&= 2 \dang KPB + 2\dang PBC \\
&= -2\dang PKB \\
&= \dang PKB_1
\end{align*}
as required.
\end{remark*} |
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 $BC$,
and let $T$ be the point of intersection of the lines $AG$ and $BC$.
Prove that $M$ is the midpoint of $\ol{ST}$. | 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
\[ T = \left( 0 : b : \frac{-as + (s-c)}{s-b} \right) = ( 0 : b(s-b) : b(s-c) - as). \]
Normalizing, we see that $T = \left( 0, -\frac{b}{a}, 1 + \frac{b}{a} \right)$,
from which we quickly obtain $MT = s$.
Similarly, $MS = s$, so we're done. |
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 these all gives the result.
The inequality is strict since it's not possible
that $a_2 = 1$, $a_3 = \half$, et cetera. |
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 player $B$.
Player $B$ now tries to obtain information about $x$
by asking player $A$ questions as follows:
each question consists of $B$ specifying an arbitrary set $S$
of positive integers (possibly one specified in some previous question),
and asking $A$ whether $x$ belongs to $S$.
Player $B$ may ask as many questions as he wishes.
After each question, player $A$ must immediately answer
it with yes or no, but is allowed to lie as many times as she wants;
the only restriction is that, among any $k+1$ consecutive answers,
at least one answer must be truthful.
After $B$ has asked as many questions as he wants,
he must specify a set $X$ of at most $n$ positive integers.
If $x$ belongs to $X$, then $B$ wins;
otherwise, he loses.
Prove that:
\begin{enumerate}[(a)]
\ii If $n \ge 2^k$, then $B$ can guarantee a win.
\ii For all sufficiently large $k$,
there exists an integer $n \ge (1.99)^k$
such that $B$ cannot guarantee a win.
\end{enumerate} | 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''
or until Bob has asked $k+1$ questions.
If Alice answers ``no'' to all of these then Bob rules out $2^k+1$.
So let's assume Alice just said ``yes''.
Now let $T = \{1, \dots, 2^k\}$.
Then, he asks $k$-follow up questions $S_1$, \dots, $S_k$
defined as follows:
\begin{itemize}
\ii $S_1 = \{1, 3, 5, 7, \dots, 2^k-1\}$ consists of all numbers
in $T$ whose least significant digit in binary is $1$.
\ii $S_2 = \{ 2, 3, 6, 7, \dots, 2^k-2, 2^k-1\}$
consists of all numbers in $T$ whose second least
significant digit in binary is $1$.
\ii More generally $S_i$
consists of all numbers in $T$ whose $i$th least
significant digit in binary is $1$.
\end{itemize}
WLOG Alice answers these all as ``yes'' (the other cases are similar).
Among the last $k+1$ answers at least one must be truthful,
and the number $2^k$ (having zeros in all relevant digits)
does not appear in any of $S_0$, \dots, $S_k$ and is ruled out.
\end{proof}
Thus in this way Bob can repeatedly find non-possibilities for $x$
(and then relabel the remaining candidates $1$, \dots, $N-1$)
until he arrives at a set of at most $2^k$ numbers.
\textbf{Part (b)}:
It suffices to consider $n = \left\lceil 1.99^k \right\rceil$
and $N = n+1$ for large $k$.
At the $t$th step, Bob asks some question $S_t$;
we phrase each of Alice's answers in the form ``$x \notin B_t$'',
where $B_t$ is either $S_t$ or its complement.
(You may think of these as ``bad sets'';
the idea is to show we can avoid having any number
appear in $k+1$ consecutive bad sets,
preventing Bob from ruling out any numbers.)
Main idea: for every number $1 \le x \le N$,
at time step $t$ we define its \emph{weight}
to be \[ w(x) = 1.998^e \]
where $e$ is the largest number
such that $x \in B_{t-1} \cap B_{t-2} \cap \dots \cap B_{t-e}$.
\begin{claim*}
Alice can ensure the total weight never exceeds $1.998^{k+1}$
for large $k$.
\end{claim*}
\begin{proof}
Let $W_{t}$ denote the sum of weights after the $t$th question.
We have $W_0 = N < 1000n$.
We will prove inductively that $W_t < 1000n$ always.
At time $t$, Bob specifies a question $S_t$.
We have Alice choose $B_t$ as whichever of $S_t$ or $\ol{S_t}$
has lesser total weight, hence at most $W_t/2$.
The weights of for $B_t$ increase by a factor of $1.998$,
while the weights for $\ol{B_t}$ all reset to $1$.
So the new total weight after time $t$ is
\[ W_{t+1} \le 1.998 \cdot \frac{W_t}{2}
+ \# \ol{B_t} \le 0.999 W_t + n. \]
Thus if $W_t < 1000n$ then $W_{t+1} < 1000n$.
To finish, note that
$1000n < 1000 \left( 1.99^k + 1 \right) < 1.998^{k+1}$
for $k$ large.
\end{proof}
In particular, no individual number can have weight $1.998^{k+1}$.
Thus for every time step $t$ we have
\[ B_t \cap B_{t+1} \cap \dots \cap B_{t+k} = \varnothing. \]
Then once Bob stops, if he declares a set of $n$ positive integers,
and $x$ is an integer Bob did not choose,
then Alice's question history is consistent with $x$ being Alice's number,
as among any $k+1$ consecutive answers
she claimed that $x \in \ol{B_t}$ for some $t$ in that range.
\begin{remark*}
[Motivation]
In our $B_t$ setup, let's think backwards.
The problem is equivalent to avoiding $e = k+1$ at any time step $t$,
for any number $x$.
That means
\begin{itemize}
\ii have at most two elements with $e = k$ at time $t-1$,
\ii thus have at most four elements with $e = k-1$ at time $t-2$,
\ii thus have at most eight elements with $e = k-2$ at time $t-3$,
\ii and so on.
\end{itemize}
We already exploited this in solving part (a).
In any case it's now natural to try letting $w(x) = 2^e$,
so that all the cases above sum to ``equally bad'' situations:
since $8 \cdot 2^{k-2} = 4 \cdot 2^{k-1} = 2 \cdot 2^k$, say.
However, we then get $W_{t+1} \le \half (2W_t) + n$,
which can increase without bound due to contributions
from numbers resetting to zero.
The way to fix this is to change the weight to $w(x) = 1.998^e$,
taking advantage of the little extra space we have
due to having $n \ge 1.99^k$ rather than $n \ge 2^k$.
\end{remark*} |
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 more natural to think of these as
$f(x) = x^2$, $f(x) = x^2 \pmod 4$, $f(x) = x^2 \pmod 8$,
and multiples thereof.)
Set $a=b=c=0$ to get $f(0)=0$.
Then set $c=0$ to get $f(a) = f(-a)$, so $f$ is even.
Now \[ f(a)^2 + f(b)^2 + f(a+b)^2
= 2f(a+b)\left( f(a)+f(b) \right) + 2f(a)f(b) \]
or
\[ \left( f(a+b) - \left( f(a)+f(b) \right) \right)^2
= 4f(a)f(b). \]
Hence $f(a)f(b)$ is a perfect square for all $a,b \in \ZZ$.
So there exists a $\lambda$ such that $f(n) = \lambda g(n)^2$, where $g(n) \ge 0$.
From here we recover
\[ \boxed{g(a+b) = \pm g(a) \pm g(b)}. \]
Also $g(0) = 0$.
Let $k = g(1) \neq 0$.
We now split into cases on $g(2)$:
\begin{itemize}
\ii $g(2) = 0$. Put $b = 2$ in original to get
$g(a+2) = \pm g(a) = +g(a)$.
\ii $g(2) = 2k$. Cases on $g(4)$:
\begin{itemize}
\ii $g(4) = 0$, then we get
$(g(n))_{n\ge0} = (0,1,2,1,0,1,2,1,\dots)$. This works.
\ii $g(4) = 4k$. This only happens when
$g(1) = k$, $g(2) = 2k$, $g(3) = 3k$, $g(4) = 4k$.
Then
\begin{itemize}
\ii $g(5) = \pm 3k \pm 2k = \pm 4k \pm k$.
\ii $g(6) = \pm4k \pm 2k = \pm5k \pm k$.
\ii \dots
\end{itemize}
and so by induction $g(n) = nk$.
\end{itemize}
\end{itemize} |
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 $M = \ol{AL} \cap \ol{BK}$.
Prove that $MK = ML$. | 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 A = Drawing("A", dir(180), dir(225));
pair B = Drawing("B", dir(0), dir(-45));
pair C = Drawing("C", dir(140), 1.2*dir(110));
draw(A--B--C--cycle);
pair D = Drawing("D", (C.x,0), dir(-90));
pair X = Drawing("X", 0.55*C+0.45*D, dir(-80));
pair K = Drawing("K", IP(A--X, Drawing(CP(B,C))), dir(-50));
pair L = Drawing("L", IP(B--X, Drawing(CP(A,C))), dir(240));
pair K1= Drawing("K^\ast", 2*foot(B,A,K)-K, dir(60));
pair L1= Drawing("L^\ast", 2*foot(A,B,L)-L, dir(L-B));
draw(A--K1);
draw(B--L1);
draw(circumcircle(K,L,K1), dashed);
/*
pair O = Drawing("O", circumcenter(K,L,K1), dir(90));
draw(A--O--B, dotted);
draw(O--D, dotted);
*/
pair M = Drawing("M", extension(A,L,B,K), dir(270));
draw(A--L, dotted);
draw(B--K, dotted);
clip((1.2,-0.4)--(1.2,2.3)--(-2.4,2.3)--(-2.4,-0.4)--cycle);
\end{asy}
\end{center}
By orthogonality of $(A)$ and $(B)$ we find that
$\ol{AL}$, $\ol{AL^\ast}$,
$\ol{BK}$, $\ol{BK^\ast}$ are tangents to $\omega$
(in particular, $KLK^{\ast}L^{\ast}$ is harmonic).
In particular $\ol{MK}$ and $\ol{ML}$ are tangents to $\omega$,
so $MK = ML$. |
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 the user \texttt{cfheolpiixn}.
\begin{claim*}
If $n = 2k-1$ works then so does $n = 2k$.
\end{claim*}
\begin{proof}
Replace
\[ \frac{k}{3^r} = \frac{k}{3^{r+1}} + \frac{2k}{3^{r+1}}.
\qquad(\ast) \qedhere \]
\end{proof}
\begin{claim*}
If $n = 4k+2$ works then so does $n = 4k + 13$.
\end{claim*}
\begin{proof}
First use the identity
\[
\frac{k+2}{3^r} = \frac{k+2}{3^{r+2}}
+ \frac{4k+ 3}{3^{r+3}}
+ \frac{4k+ 5}{3^{r+3}}
+ \frac{4k+ 7}{3^{r+3}}
+ \frac{4k+ 9}{3^{r+3}}
+ \frac{4k+11}{3^{r+3}}
+ \frac{4k+13}{3^{r+3}}
\]
to fill in the odd numbers.
The even numbers can then be instantiated with $(\ast)$ too.
\end{proof}
Thus it suffices to construct base cases
for $n = 1$, $n = 5$, $n = 9$.
They are
\begin{align*}
1 &= \frac{1}{3^0} \\
&= \frac{1}{3^2} + \frac{2}{3^2} + \frac{3}{3^2}
+ \frac{4}{3^3} + \frac{5}{3^3} \\
&= \frac{1}{3^2} + \frac{2}{3^3} + \frac{3}{3^3}
+ \frac{4}{3^3} + \frac{5}{3^3} + \frac{6}{3^4}
+ \frac{7}{3^4} + \frac{8}{3^4} + \frac{9}{3^4}.
\end{align*} |
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 write
\[
\frac{n + (2^k-1)}{n}
= \left( 1 + \frac{1}{n} \right) \cdot \frac{\frac{n+1}{2} + (2^{k-1}-1)}{\frac{n+1}2}
\]
and use inductive hypothesis on the second term. |
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 $ABC$.
Prove that triangle $ABC$ is right-angled. | 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 similarity $\triangle XBC_1 \sim \triangle XCB_1$
which must actually be a spiral congruence
since $BC_1 = B_1C$.
\end{proof}
We define the arc midpoints $Y$ and $Z$ similarly,
which lie on the perpendicular bisectors of
$\ol{A_1 C_1}$, $\ol{A_1 B_1}$.
\begin{center}
\begin{asy}
pair A = dir(110);
pair B = dir(180);
pair C = dir(0);
pair X = dir(90);
pair Y = dir(235);
pair Z = dir(325);
pair A_1 = foot(Y+Z-X, B, C);
pair B_1 = foot(Z+X-Y, C, A);
pair C_1 = foot(X+Y-Z, A, B);
filldraw(unitcircle, opacity(0.1)+gray, gray);
draw(C_1--X--B_1, red);
draw(A--B--C--cycle, gray);
filldraw(A_1--B_1--C_1--cycle, opacity(0.1)+cyan, heavycyan);
draw(Y--X--Z, orange);
draw(circumcircle(X, B_1, C_1), dashed+gray);
draw(X--A_1, red+dashed);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
dot("$Z$", Z, dir(Z));
dot("$A_1$", A_1, dir(270));
dot("$B_1$", B_1, dir(B_1));
dot("$C_1$", C_1, dir(C_1));
/* TSQ Source:
A = dir 110
B = dir 180
C = dir 0
X = dir 90
Y = dir 235
Z = dir 325
A_1 = foot Y+Z-X B C R270
B_1 = foot Z+X-Y C A
C_1 = foot X+Y-Z A B
unitcircle 0.1 gray / gray
C_1--X--B_1 red
A--B--C--cycle gray
A_1--B_1--C_1--cycle 0.1 cyan / heavycyan
Y--X--Z orange
circumcircle X B_1 C_1 dashed gray
X--A_1 red dashed
*/
\end{asy}
\end{center}
We now turn to the problem condition
which asserts the circumcenter $W$ of $\triangle A_1B_1C_1$
lies on $(ABC)$.
\begin{claim*}
We may assume WLOG that $W = X$.
\end{claim*}
\begin{proof}
This is just configuration analysis,
since we already knew that the arc midpoints
both lie on $(ABC)$ and the relevant perpendicular bisectors.
Point $W$ lies on $(ABC)$ and hence outside $\triangle ABC$,
hence outside $\triangle A_1 B_1 C_1$.
Thus we may assume WLOG that $\angle B_1 A_1 C_1 > 90\dg$.
Then $A$ and $X$ lie on the same side of line $\ol{B_1 C_1}$,
and since $W$ is supposed to lie both on $(ABC)$
and the perpendicular bisector of $\ol{B_1C_1}$ it follows $W = X$.
\end{proof}
Consequently, $\ol{XY}$ and $\ol{XZ}$
are exactly the perpendicular bisectors
of $\ol{A_1 C_1}$, $\ol{A_1 B_1}$.
The rest is angle chase, the fastest one is
\begin{align*}
\angle A &= \angle C_1 X B_1
= \angle C_1 X A_1 + \angle A_1 X B_1
= 2 \angle YXA_1 + 2 \angle A_1 X Z \\
&= 2 \angle YXZ = 180\dg - \angle A
\end{align*}
which solves the problem.
\begin{remark*}
Angle chasing is also possible even without
the points $Y$ and $Z$, though it takes much longer.
Introduce the Bevan point $V$ and use the fact
that $VA_1B_1C$ is cyclic (with diameter $\ol{VC}$)
and similarly $VA_1C_1B$ is cyclic;
a calculation then gives $\angle CVB = 180\dg - \half \angle A$.
Thus $V$ lies on the circle with diameter $\ol{I_b I_c}$.
\end{remark*} |
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 of $\omega_1$.
Similarly, $\omega_2$ is the circumcircle of triangle $CWM$
and $Y$ is a point such that $\ol{WY}$ is a diameter of $\omega_2$.
Show that the points $X$, $Y$, and $H$ are collinear. | 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{center}
\begin{asy}
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
pair M = foot(B, A, C);
pair N = foot(C, A, B);
draw(B--M);
draw(C--N);
filldraw(A--B--C--cycle, opacity(0.1)+lightcyan, lightblue);
draw(B--M, lightblue);
draw(C--N, lightblue);
pair W = 0.55*C+0.45*B;
pair H = A+B+C;
pair P = foot(H, A, W);
pair X = extension(P, H, B, B+A-H);
pair Y = extension(P, H, C, C+A-H);
filldraw(circumcircle(B, N, W), opacity(0.1)+lightgreen, heavygreen);
filldraw(circumcircle(C, M, W), opacity(0.1)+lightgreen, heavygreen);
draw(X--W--Y, heavygreen);
draw(B--X, lightblue);
draw(C--Y, lightblue);
draw(X--Y, red+dotted);
draw(A--W, red+dotted);
draw(circumcircle(A, M, N), red+dashed);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$M$", M, dir(M));
dot("$N$", N, dir(N));
dot("$W$", W, dir(W));
dot("$H$", H, dir(H));
dot("$P$", P, dir(P));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
/* TSQ Source:
A = dir 110
B = dir 210
C = dir 330
M = foot B A C
N = foot C A B
B--M
C--N
A--B--C--cycle 0.1 lightcyan / lightblue
B--M lightblue
C--N lightblue
W = 0.55*C+0.45*B
H = A+B+C
P = foot H A W
X = extension P H B B+A-H
Y = extension P H C C+A-H
circumcircle B N W 0.1 lightgreen / heavygreen
circumcircle C M W 0.1 lightgreen / heavygreen
X--W--Y heavygreen
B--X lightblue
C--Y lightblue
X--Y red dotted
A--W red dotted
circumcircle A M N red dashed
*/
\end{asy}
\end{center}
We now contend:
\begin{claim*}
Points $P$, $H$, $X$ collinear.
(Similarly, points $P$, $H$, $Y$ are collinear.)
\end{claim*}
\begin{proof}
[Proof using power of a point]
By radical axis on $BNMC$, $\omega_1$, $\omega_2$,
it follows that $A$, $P$, $W$ are collinear.
We know that $\angle APH = 90\dg$, and also
$\angle XPW = 90\dg$ by construction.
Thus $P$, $H$, $X$ are collinear.
\end{proof}
\begin{proof}
[Proof using angle chasing]
This is essentially Reim's theorem:
\[ \dang NPH = \dang NAH = \dang BAH = \dang ABX = \dang NBX = \dang NPX \]
as desired.
Alternatively, one may prove $A$, $P$, $W$ are collinear by
$\dang NPA = \dang NMA = \dang NMC = \dang NBC = \dang NBW = \dang NPW$.
\end{proof}
\paragraph{Second solution, by moving points.}
Fix $\triangle ABC$ and vary $W$.
Let $\infty$ be the point at infinity perpendicular to $\ol{BC}$
for brevity.
By spiral similarity, the point $X$ moves linearly on $\ol{B\infty}$
as $W$ varies linearly on $\ol{BC}$.
Similarly, so does $Y$.
So in other words, the map \[ X \mapsto W \mapsto Y \]
is linear.
However, the map \[ X \mapsto Y' \coloneq \ol{XH} \cap \ol{C\infty} \]
is linear too.
To show that these maps are the same,
it suffices to check it thus at two points.
\begin{itemize}
\ii When $W = B$, the circle $(BNW)$
degenerates to the circle through $B$ tangent to $\ol{BC}$,
and $X = \ol{CN} \cap \ol{B\infty}$.
We have $Y = Y' = C$.
\ii When $W = C$, the result is analogous.
\ii Although we don't need to do so,
it's also easy to check the result if $W$
is the foot from $A$
since then $XHWB$ and $YHWC$ are rectangles.
\end{itemize} |
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 positive rational numbers $x$. | 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 $f$ takes only positive values,
and hence by (ii) is strictly increasing.
\end{claim*}
\begin{proof}[Proof, suggested by Gopal Goel]
Let $p,q > 0$ be integers.
Then $f(q) f(p/q) \ge f(p)$, and since both $\min(f(p), f(q)) > 0$
it follows $f(p/q) > 0$.
\end{proof}
\begin{claim*}
For any $x > 1$ we have $f(x) \ge x$.
\end{claim*}
\begin{proof}
Note that
\[ f(x)^N \ge f(x^N) \ge f\left( \left\lfloor x^N \right\rfloor \right)
\ge \left\lfloor x^N \right\rfloor > x^N-1 \]
for any integer $N$.
Since $N$ can be arbitrarily large, we conclude $f(x) \ge x$ for $x > 1$.
\end{proof}
On the other hand, $f$ has arbitrarily large fixed points
(namely powers of $a$) so from (ii) we're essentially done.
First, for $x > 1$ pick a large $m$ and note
\[ a^m = f(a^m) \ge f(a^m-x) + f(x) \ge (a^m-x)+x = a^m. \]
Finally, for $x \le 1$ use
\[ nf(x) = f(n)f(x) \ge f(nx) \ge nf(x) \]
for large $n$.
\begin{remark*}
Note that $a > 1$ is essential;
if $b \ge 1$ then $f(x) = bx^2$ works with unique fixed point $1/b \le 1$.
\end{remark*} |
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 of the circle.
A labelling is called \emph{beautiful} if, for any four labels $a < b < c < d$ with $a + d = b + c$,
the chord joining the points labelled $a$ and $d$
does not intersect the chord joining the points labelled $b$ and $c$.
Let $M$ be the number of beautiful labellings,
and let $N$ be the number of ordered pairs $(x, y)$ of positive integers
such that $x + y \le n$ and $\gcd(x, y) = 1$.
Prove that $M = N + 1$.
\end{enumerate} | 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(r), dotted+blue);
draw(pic, g(a)--g(e), dotted+blue);
draw(pic, g(b)--g(d), dotted+blue);
dot(pic, "$0$", dir(90), dir(90));
dot(pic, "$1$", g(a), g(a));
dot(pic, "$2$", g(b), g(b));
dot(pic, "$3$", g(c), g(c));
dot(pic, "$4$", g(d), g(d));
dot(pic, "$5$", g(e), g(e));
dot(pic, "$6$", dir(r), dir(r), red);
return pic;
}
add(shift(3,6)*ring(2,4,5,1,3,0));
add(shift(0,6)*ring(3,1,4,2,5,240));
add(shift(3,3)*ring(4,2,5,3,1,0));
add(shift(0,3)*ring(2,3,4,5,1,240));
add(shift(3,0)*ring(1,2,3,4,5,50));
add(shift(0,0)*ring(1,2,3,4,5,130));
\end{asy}
\end{center}
Abbreviate ``beautiful labelling of points around a circle'' to ring.
Moreover, throughout the solution we will allow degenerate
chords that join a point to itself;
this has no effect on the problem statement.
The idea is to proceed by induction in the following way.
A ring of $[0,n]$ is called \emph{linear}
if it is an arithmetic progression modulo $n+1$.
For example, the first two rings in the diagram
and the last one are linear for $n = 6$,
while the other three are not.
Of course we can move from any ring on $[0,n]$
to a ring on $[0,n-1]$ by deleting $n$.
We are going to prove that:
\begin{itemize}
\ii Each linear ring on $[0,n-1]$ yields exactly two
rings of $[0,n]$, and
\ii Each nonlinear ring on $[0,n-1]$ yields exactly one
rings of $[0,n]$.
\end{itemize}
In light of the fact there are obviously $\varphi(n)$ linear rings on $[0,n]$,
the conclusion will follow by induction.
We say a set of chords (possibly degenerate) is \emph{pseudo-parallel}
if for any three of them, one of them separates the two.
(Pictorially, one can perturb the endpoints along the circle
in order to make them parallel in Euclidean sense.)
The main structure lemma is going to be:
\begin{lemma*}
In any ring, the chords of sum $k$
(even including degenerate ones) are pseudo-parallel.
\end{lemma*}
\begin{proof}
By induction on $n$.
By shifting, we may assume that one of the chords is $\{0,k\}$
and discard all numbers exceeding $k$; that is, assume $n = k$.
Suppose the other two chords are $\{a, n-a\}$ and $\{b, n-b\}$.
\begin{center}
\begin{asy}
size(5cm);
draw(unitcircle);
pair O = dir(210);
pair N = dir(330);
draw(O--N);
pair A1 = dir(100);
pair A2 = dir(140);
pair B1 = dir( 80);
pair B2 = dir( 40);
draw(A1--A2);
draw(B1--B2);
pair U = dir(200);
pair V = dir(340);
draw(U--V, blue);
pair X = dir(250);
pair Y = dir(310);
draw(X--Y, red);
draw(O--X, blue+dashed);
dot("$a$", A1, A1);
dot("$b$", B1, B1);
dot("$n-a$", A2, A2);
dot("$n-b$", B2, B2);
dot("$u$", U, U, blue);
dot("$v$", V, V, blue);
dot("$u+v$", X, X, red);
dot("$n-(u+v)$", Y, Y, red);
dot("$0$", O, O);
dot("$n$", N, N);
\end{asy}
\end{center}
We consider the chord $\{u,v\}$ directly above $\{0,n\}$, drawn in blue.
There are now three cases.
\begin{itemize}
\ii If $u+v = n$, then delete $0$ and $n$
and decrease everything by $1$.
Then the chords $\{a-1, n-a-1\}$, $\{b-1, n-b-1\}$, $\{u-1, v-1\}$
contradict the induction hypothesis.
\ii If $u+v < n$, then search for the chord
$\{u+v, n-(u+v)\}$.
It lies on the other side of $\{0, n\}$
in light of chord $\{0,u+v\}$.
Now again delete $0$ and $n$ and decrease everything by $1$.
Then the chords $\{a-1, n-a-1\}$, $\{b-1, n-b-1\}$, $\{u+v-1, n-(u+v)-1\}$
contradict the induction hypothesis.
%% MAX LU: this case can't occur at all;
%% nowhere to put $n-u$ without $0+n = u+(n-u)
%% and $v + (n-(u+v))$.
\ii If $u+v > n$, apply the map $t \mapsto n-t$ to the entire ring.
This gives the previous case as now $(n-u)+(n-v) < n$.
\qedhere
\end{itemize}
\end{proof}
Next, we give another characterization of linear rings.
\begin{lemma*}
A ring on $[0,n-1]$ is linear if and only if the point $0$
does not lie between two chords of sum $n$.
\end{lemma*}
\begin{proof}
It's obviously true for linear rings.
Conversely, assume the property holds for some ring.
Note that the chords with sum $n-1$ are pseudo-parallel and encompass every point,
so they are \emph{actually} parallel.
Similarly, the chords of sum $n$ are \emph{actually} parallel
and encompass every point other than $0$.
So the map
\[ t \mapsto n-t \mapsto (n-1)-(n-t) = t-1 \pmod n \]
is rotation as desired.
\end{proof}
\begin{lemma*}
Every nonlinear ring on $[0,n-1]$ induces exactly one ring on $[0,n]$.
\end{lemma*}
\begin{proof}
Because the chords of sum $n$ are pseudo-parallel,
there is at most one possibility for the location $n$.
Conversely, we claim that this works.
The chords of sum $n$ (and less than $n$) are OK by construction, so
assume for contradiction that there exists $a,b,c \in \{1,\dots,n-1\}$
such that $a + b = n + c$.
Then, we can ``reflect'' them using the (pseudo-parallel)
chords of length $n$ to find that $(n-a) + (n-b) = 0 + (n-c)$,
and the chords joining $0$ to $n-c$ and $n-a$ to $n-b$ intersect,
by definition.
\begin{center}
\begin{asy}
size(5cm);
draw(unitcircle);
draw(dir(140)--dir(220), blue+dotted);
draw(dir(250)--dir(110), blue+dotted);
draw(dir(340)--dir(20), blue+dotted);
draw(dir(60)--dir(300), blue+dotted);
draw(dir(220)--dir(60), lightred);
draw(dir(340)--dir(110), lightred);
draw(dir(140)--dir(300), orange+dashed);
draw(dir(250)--dir(20), orange+dashed);
dot("$0$", dir(140), dir(140));
dot("$n$", dir(220), dir(220), red);
dot("$n-a$", dir(250), dir(250));
dot("$n-c$", dir(300), dir(300));
dot("$b$", dir(340), dir(340));
dot("$n-b$", dir(20), dir(20));
dot("$c$", dir(60), dir(60));
dot("$a$", dir(110), dir(110));
\end{asy}
\end{center}
This is a contradiction that the original numbers on $[0,n-1]$ form a ring.
\end{proof}
\begin{lemma*}
Every linear ring on $[0,n-1]$ induces
exactly two rings on $[0,n]$.
\end{lemma*}
\begin{proof}
Because the chords of sum $n$ are pseudo-parallel,
the point $n$ must lie either directly to the left or right of $0$.
For the same reason as in the previous proof, both of them work.
\end{proof} |
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
\[ b_n < a_0 \le b_{n+1} \]
which obviously exists and is unique. |
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
\[ \angle CHS - \angle CSB = 90^{\circ},
\quad \angle THC - \angle DTC = 90^{\circ}. \]
Prove that line $BD$ is tangent to the circumcircle of triangle $TSH$. | \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{align*}
\dang ATH &= \dang DTH \\
&= \dang DTC + \dang CTH \\
&= \dang DTC - \dang THC - \dang HCT \\
&= 90\dg - \dang HCT = 90\dg + \dang TCH.
\end{align*}
which implies conclusion.
\end{proof}
\begin{center}
\begin{asy}
pair A = dir(110);
pair B = dir(200);
pair D = dir(-20);
pair C = -A;
pair H = foot(A, B, D);
pair Cs = -C+2*foot(origin, H, C);
pair As = -A+2*foot(origin, A, H);
filldraw(unitcircle, opacity(0.1)+lightcyan, lightblue);
draw(A--B--C--D--cycle, lightblue);
pair M = midpoint(B--As);
pair N = midpoint(D--As);
pair U = IP(Cs--N, circumcircle(D, H, As));
pair V = IP(Cs--M, circumcircle(B, H, As));
pair Ss = D+As-U;
pair Ts = B+As-V;
pair T = extension(Ts, H, A, D);
pair S = extension(Ss, H, A, B);
pair O_D = circumcenter(C, H, T);
filldraw(circumcircle(C, H, T), opacity(0.1)+yellow, red);
pair P = extension(A, H, O_D, midpoint(T--H));
draw(A--H, lightblue);
draw(B--D, lightblue);
draw(H--C--T--cycle, red);
draw(H--O_D, lightblue);
draw(P--O_D, heavygreen);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$D$", D, dir(D));
dot("$C$", C, dir(C));
dot("$H$", H, dir(H));
dot("$T$", T, dir(80));
dot("$S$", S, dir(S));
dot("$O_D$", O_D, dir(45));
dot("$P$", P, dir(45));
/* TSQ Source:
A = dir 110
B = dir 200
D = dir -20
C = -A
H = foot A B D
C* := -C+2*foot origin H C R80
A* := -A+2*foot origin A H
unitcircle 0.1 lightcyan / lightblue
A--B--C--D--cycle lightblue
M := midpoint B--As
N := midpoint D--As
U := IP Cs--N circumcircle D H As
V := IP Cs--M circumcircle B H As
S* := D+As-U
T* := B+As-V
T = extension Ts H A D R80
S = extension Ss H A B
O_D = circumcenter C H T R45
circumcircle C H T 0.1 yellow / red
P = extension A H O_D midpoint T--H R45
A--H lightblue
B--D lightblue
H--C--T--cycle red
H--O_D lightblue
P--O_D heavygreen
*/
\end{asy}
\end{center}
Let the perpendicular bisector of $TH$ meet $AH$ at $P$ now.
It suffices to show that $\frac{AP}{PH}$ is symmetric in $b = AD$ and $d=AB$,
because then $P$ will be the circumcenter of $\triangle TSH$.
To do this, set $AH = \frac{bd}{2R}$ and $AC=2R$.
Let $O$ denote the circumcenter of $\triangle CHT$.
Use the Law of Cosines on $\triangle ACO$ and $\triangle AHO$,
using variables $x=AO$ and $r=HO$.
We get that
\[ r^2 = x^2 + AH^2 - 2x \cdot AH \cdot \frac{d}{2R} = x^2 + (2R)^2 - 2bx. \]
By the angle bisector theorem,
$\frac{AP}{PH} = \frac{AO}{HO}$.
The rest is computation: notice that
\[ r^2 - x^2 = h^2 - 2xh \cdot \frac{d}{2R} = (2R)^2 - 2bx \]
where $h = AH = \frac{bd}{2R}$, whence
\[ x = \frac{(2R)^2-h^2}{2b - 2h \cdot \frac{d}{2R}}. \]
Moreover,
\[ \frac{1}{2} \left( \frac{r^2}{x^2}-1 \right)
= \frac{1}{x} \left( \frac 2x R^2 - b \right). \]
Now, if we plug in the $x$ in the
right-hand side of the above, we obtain
\[ \frac{2b-2h \cdot \frac{d}{2R}}{4R^2-h^2}
\left( \frac{2b-2h \cdot \frac{d}{2R}}{4R^2-h^2} \cdot 2R^2 - b\right)
= \frac{2h}{(4R^2-h^2)^2} \left( b- h \cdot \frac{d}{2R} \right)
\left( -2hdR + bh^2 \right). \]
Pulling out a factor of $-2Rh$ from the rightmost term,
we get something that is symmetric in $b$ and $d$, as required.
\paragraph{Second solution (Victor Reis).}
Here is the fabled solution using inversion at $H$.
First, we rephrase the angle conditions in the following ways:
\begin{itemize}
\ii $\ol{AD} \perp (THC)$,
which is equivalent to the claim from the first solution.
\ii $\ol{AB} \perp (SHC)$, by symmetry.
\ii $\ol{AC} \perp (ABCD)$, by definition.
\end{itemize}
Now for concreteness we will use a negative inversion at $H$
which swaps $B$ and $D$ and overlay it on the original diagram.
As usual we denote inverses with stars.
Let us describe the inverted problem.
We let $M$ and $N$ denote the midpoints of $\ol{A^\ast B^\ast}$
and $\ol{A^\ast D^\ast}$, which are the centers of
$(HA^\ast B^\ast)$ and $(HA^\ast D^\ast)$.
From $\ol{T^\ast C^\ast} \perp (HA^\ast D^\ast)$,
we know have $C^\ast$, $M$, $T^\ast$ collinear.
Similarly, $C^\ast$, $N$, $S^\ast$ are collinear.
We have that $(A^\ast H C^\ast)$ is orthogonal to $(ABCD)$ which remains fixed.
We wish to show $\ol{T^\ast S^\ast}$ and $\ol{MN}$ are parallel.
\begin{center}
\begin{asy}
pair A = dir(100);
pair B = dir(200);
pair D = dir(-20);
pair C = -A;
pair H = foot(A, B, D);
pair Cs = -C+2*foot(origin, H, C);
pair As = -A+2*foot(origin, A, H);
filldraw(unitcircle, opacity(0.1)+lightcyan, lightblue);
draw(A--B--C--D--cycle, lightblue);
filldraw(circumcircle(B, H, As), opacity(0.1)+lightgreen, dotted+heavygreen);
filldraw(circumcircle(D, H, As), opacity(0.1)+lightgreen, dotted+heavygreen);
pair M = midpoint(B--As);
pair N = midpoint(D--As);
pair U = IP(Cs--N, circumcircle(D, H, As));
pair V = IP(Cs--M, circumcircle(B, H, As));
pair Ss = D+As-U;
pair Ts = B+As-V;
pair T = extension(Ts, H, A, D);
pair S = extension(Ss, H, A, B);
draw(Ts--Cs--Ss, heavygreen);
draw(Ts--Ss, red);
draw(B--D, lightblue);
draw(A--As, lightblue);
draw(A--B--As--D--cycle, lightblue);
draw(C--Cs--A--cycle, lightblue);
draw(S--Ss, dotted+blue);
draw(T--Ts, dotted+blue);
draw(M--N, red);
filldraw(circumcircle(As, M, N), opacity(0.1)+lightred, red+dashed);
filldraw(circumcircle(As, H, Cs), opacity(0.1)+pink, purple+dashed);
clip(box((-1.4,-1.4),(2,2)));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$D$", D, dir(D));
dot("$C$", C, dir(C));
dot("$H$", H, dir(H));
dot("$C^\ast$", Cs, dir(80));
dot("$A^\ast$", As, dir(As));
dot("$M$", M, dir(M));
dot("$N$", N, dir(N));
dot("$S^\ast$", Ss, dir(Ss));
dot("$T^\ast$", Ts, dir(Ts));
dot("$T$", T, dir(50));
dot("$S$", S, dir(S));
/* TSQ Source:
A = dir 100
B = dir 200
D = dir -20
C = -A
H = foot A B D
C* = -C+2*foot origin H C R80
A* = -A+2*foot origin A H
unitcircle 0.1 lightcyan / lightblue
A--B--C--D--cycle lightblue
circumcircle B H As 0.1 lightgreen / dotted heavygreen
circumcircle D H As 0.1 lightgreen / dotted heavygreen
M = midpoint B--As
N = midpoint D--As
U := IP Cs--N circumcircle D H As
V := IP Cs--M circumcircle B H As
S* = D+As-U
T* = B+As-V
T = extension Ts H A D R50
S = extension Ss H A B
Ts--Cs--Ss heavygreen
Ts--Ss red
B--D lightblue
A--As lightblue
A--B--As--D--cycle lightblue
C--Cs--A--cycle lightblue
S--Ss dotted blue
T--Ts dotted blue
M--N red
circumcircle As M N 0.1 lightred / red dashed
circumcircle As H Cs 0.1 pink / purple dashed
!clip(box((-1.4,-1.4),(2,2)));
*/
\end{asy}
\end{center}
Lot $\omega$ denote the circumcircle of $\triangle A^\ast H C^\ast$,
which is orthogonal to the original circle $(ABCD)$.
It would suffices to show $(A^\ast H C^\ast)$
is an $H$-Apollonius circle with respect to $\ol{MN}$,
from which we would get $C^\ast M / H M = C^\ast N / H N$.
However, $\omega$ through $H$ and $A$,
hence it center lies on line $MN$.
Moreover $\omega$ is orthogonal to $(A^\ast MN)$
(since $(A^\ast MN)$ and $(A^\ast BD)$ are homothetic).
This is enough (for example, if we let $O$ denote the center of $\omega$,
we now have $\mathrm{r}(\omega)^2 = OH^2 = OM \cdot ON$).
(Note in this proof that the fact that $C^\ast$ lies on $(ABCD)$
is not relevant.) |
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
circumcircle of $\triangle ABC$. | 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, blue);
pair T = 2*B*C/(B+C);
pair P = extension(A, B+A-T, B, C);
pair Q = extension(A, C+A-T, B, C);
pair M = 2*P-A;
pair N = 2*Q-A;
pair X = extension(B, M, C, N);
draw(B--M, lightblue);
draw(C--N, lightblue);
draw(A--M, lightred);
draw(A--N, orange);
real r = 0.4;
draw((B+r*dir(B-T))--(B+r*dir(T-B)), lightred);
draw((C+r*dir(C-T))--(C+r*dir(T-C)), orange);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$P$", P, dir(225));
dot("$Q$", Q, dir(225));
dot("$M$", M, dir(M));
dot("$N$", N, dir(N));
dot("$X$", X, dir(280));
/* TSQ Source:
A = dir 70
B = dir 190
C = dir 350
unitcircle 0.1 palecyan / lightblue
A--B--C--cycle 0.1 heavycyan / blue
T := 2*B*C/(B+C)
P = extension A B+A-T B C R225
Q = extension A C+A-T B C R225
M = 2*P-A
N = 2*Q-A
X = extension B M C N R280
B--M lightblue
C--N lightblue
A--M lightred
A--N orange
! real r = 0.4;
(B+r*dir(B-T))--(B+r*dir(T-B)) lightred
(C+r*dir(C-T))--(C+r*dir(T-C)) orange
*/
\end{asy}
\end{center}
The angle conditions imply that the tangent to $(ABC)$ at $B$
is parallel to $\ol{AP}$.
Let $\infty$ be the point at infinity along line $AP$.
Then \[ -1 = (AM;P\infty) \overset{B}{=} (AX;BC). \]
Similarly, if $\ol{CN}$ meets the circumcircle at $Y$
then $(AY;BC) = -1$ as well.
Hence $X=Y$, which implies the problem.
\paragraph{Second solution by similar triangles.}
Once one observes $\triangle CAQ \sim \triangle CBA$,
one can construct $D$ the reflection of $B$ across $A$,
so that $\triangle CAN \sim \triangle CBD$.
Similarly, letting $E$ be the reflection of $C$ across $A$,
we get $\triangle BAP \sim \triangle BCA
\implies \triangle BAM \sim \triangle BCE$.
Now to show $\angle ABM + \angle ACN = 180\dg$
it suffices to show $\angle EBC + \angle BCD = 180\dg$,
which follows since $BCDE$ is a parallelogram.
\paragraph{Third solution by barycentric coordinates.}
Since $PB = c^2/a$ we have
\[ P = (0 : a^2-c^2 : c^2) \]
so the reflection $\vec M = 2\vec P - \vec A$ has coordinates
\[ M = (-a^2 : 2(a^2-c^2) : 2c^2). \]
Similarly $N = (-a^2 : 2b^2 : 2(b^2-a^2))$. Thus
\[ \ol{BM} \cap \ol{CN} = (-a^2 : 2b^2 : 2c^2) \]
which clearly lies on the circumcircle,
and is in fact the point identified in the first solution. |
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 has total value at most $1$. | 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,
group it into a single group and induct downwards.
\end{itemize}
Apply this operation repeatedly until it cannot be done anymore.
Now construct boxes $B_0$, $B_1$, \dots, $B_{k-1}$.
In box $B_0$ put any coins of size $\tfrac 12$ (clearly there is at most one).
More generally for $m \ge 0$, $B_m$, put coins of size
$\frac{1}{2m+1}$ and $\frac{1}{2m+2}$
(there at most $2m$ of the former and at most one of the latter).
Note that
\[ \text{total weight in $B_m$} \le 2m \cdot \frac{1}{2m+1} + \frac{1}{2m+2} < 1. \]
Finally, place the remaining ``light'' coins of size at most $\frac{1}{2k+1}$ in a pile.
\begin{remark*}
One way to explain where this boxing comes from is to imagine
what happens after applying the operation repeatedly until it can't be done any more.
We expect the most troublesome situation would be if the leftover coins are
as large as possible, which looks like
\[ \half, \qquad \frac13 \times 2, \qquad \frac14, \qquad \frac15 \times 4,
\qquad \frac16, \qquad \frac17 \times 6, \qquad \frac 18, \qquad \dots \]
Seeing this the boxing $B_i$ is quite reasonable because it groups these
into groups of almost $1$ without too much effort:
\begin{align*}
B_0 &\longrightarrow \half \\
B_1 &\longrightarrow \frac13 \cdot 2 + \frac 14 = \frac{11}{12} \\
B_2 &\longrightarrow \frac15 \cdot 4 + \frac 16 = \frac{29}{30} \\
B_3 &\longrightarrow \frac17 \cdot 6 + \frac 18 = \frac{55}{56} \\
&\vdotswithin\longrightarrow
\end{align*}
\end{remark*}
Then just toss coins from the pile into the boxes arbitrarily,
other than the proviso that no box should have its weight exceed $1$.
We claim this uses up all coins in the pile.
Assume not, and that some coin remains in the pile
when all the boxes are saturated.
Then all the boxes must have at least $1 -\frac{1}{2k+1}$,
meaning the total amount in the boxes is strictly greater than
\[ k \left( 1 - \frac{1}{2k+1} \right) > k - \tfrac 12 \]
which is a contradiction.
\begin{remark*}
This gets a stronger bound $k - \frac{k}{2k+1}$ than the requested $k-\tfrac 12$.
\end{remark*} |
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 $n$ lines in general position
it is possible to colour at least $\sqrt{n}$ lines blue
in such a way that none of its finite regions
has a completely blue boundary.
\end{enumerate} | 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;
this is the intersection of two blue lines $v$.
We'll say $\ell$ is the \emph{eyelid} of $v$.
\begin{center}
\begin{asy}
size(2cm);
pair A = dir( 0); dot(A);
pair B = dir( 72); dot(B);
pair C = dir(144); dot(C);
pair D = dir(216); dot(D);
pair E = dir(288); dot(E);
draw(D--E--A--B--C, blue);
draw(C--D, red);
label("$\ell$", 0.5*C+0.5*D, dir(180));
label("$v$", E, dir(E));
\end{asy}
\end{center}
You can prove without too much difficulty that every intersection of two blue lines
has at most two eyelids.
Since there are $\binom k2$ such intersections, we see that
\[ n-k \le 2 \binom k2 = k^2 - k\]
so $n \le k^2$, as required.
\begin{remark*}
In fact, $k = \sqrt n$ is ``sharp for greedy algorithms'',
as illustrated below for $k=3$:
\begin{center}
\begin{asy}
size(2.5cm);
real R = 7;
real r = 3;
real e = 2;
pair A = R * dir(90);
pair B = R * dir(210);
pair C = R * dir(330);
draw( (A-r*dir(B-A)) -- (B-r*dir(A-B)), blue );
draw( (B-r*dir(C-B)) -- (C-r*dir(B-C)), blue );
draw( (C-r*dir(A-C)) -- (A-r*dir(C-A)), blue );
void yanpi(pair P) {
dot(P, blue);
draw( (P+e*dir(P)*dir(70)) -- (P+e*dir(P)*dir(-70)), red);
draw( (P+e*dir(P)*dir(110)) -- (P+e*dir(P)*dir(-110)), red);
}
yanpi(A);
yanpi(B);
yanpi(C);
\end{asy}
\end{center}
\end{remark*} |
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 $\mathcal{S}$,
there are no points $P$ in $\mathcal{S}$ such that $PA=PB=PC$.
\begin{enumerate}
\item[(a)] Show that for all integers $n\ge 3$,
there exists a balanced set consisting of $n$ points.
\item[(b)] Determine all integers $n\ge 3$ for which
there exists a balanced center-free set consisting of $n$ points.
\end{enumerate} | 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, dir(-90), blue);
pair A = dir(37);
pair B = A*dir(-60);
pair C = B*dir(-60);
pair X = dir(117);
pair Y = X*dir(60);
draw(O--X--Y--cycle, deepgreen+dashed);
dot(X, deepgreen);
dot(Y, deepgreen);
draw(B--O--A--B--C--O, red+dashed);
dot(A, red);
dot(B, red);
dot(C, red);
\end{asy}
\end{center}
For part (b), the answer is odd $n$, achieved by taking a regular $n$-gon.
To show even $n$ fail, note that some point is on the perpendicular bisector of
\[ \left\lceil \frac 1n \binom n2 \right\rceil = \frac{n}{2} \]
pairs of points, which is enough.
(This is a standard double-counting argument.)
\begin{remark*}
As an aside, there is a funny joke about this problem.
There are two types of people in the world:
\begin{itemize}
\ii Those who solve (b) quickly and then take forever to solve (a),
\ii those who solve (a) quickly and then can't solve (b) at all.
\end{itemize}
(Empirically true when the Taiwan IMO 2014 team was working on it.)
\end{remark*} |
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 even, then
\begin{align*}
ca-b &= \gcd (ab-c, ca-b) \le \gcd (ab-c, a(ca-b)+ab-c) \\
&= \gcd\left( ab-c, c(a^2-1) \right).
\end{align*}
As $a^2-1$ is odd, we conclude $ca-b \le c$.
This implies $a=b=c=2$.
\ii If $a$, $b$, $c$ are all odd, then $a > b > c > 1$ follows.
Then as before
\[ ca-b \le \gcd (ab-c, c(a^2-1))
\le 2^{\nu_2(a^2-1)} \le 2a+2 \le 3a-b \]
so $c = 3$ and $a = b+2$.
As $3a-b = ca-b \ge 2(bc-a) = 6b-2a$ we then conclude $a=7$ and $b=5$.
\ii If $a$ is odd and $b$, $c$ are even, then $bc-a=1$
and hence $bc^2 - b - c = ca - b$.
Then from the miraculous identity
\[ c^3-b-c = (1-c^2)(ab-c) + a(\underbrace{bc^2-b-c}_{=ca-b}) + (ca-b) \]
so we conclude $\gcd(ab-c, ca-b) = \gcd(ab-c, c^3-b-c)$, in other words
\[ bc^2-b-c = ca-b = \gcd(ab-c, ca-b) = \gcd(ab-c, c^3-b-c). \]
We thus consider two more cases:
\begin{itemize}
\ii If $c^3-b-c \neq 0$ then
the above implies $|c^3-b-c| \ge bc^2-b-c$.
As $b \ge c > 1$, we must actually have $b = c$,
thus $a = c^2-1$.
Finally $ab-c = c(c^2-2)$ is a power of $2$, hence $b=c=2$, so $a=3$.
\ii In the second case, assume $c^3-b-c = 0$, hence $c^3-c$.
From $bc-a=1$ we obtain $a=c^4-c^2-1$,
hence
\[ ca-b = c^5-2c^3 = c^3(c^2-2) \]
is a power of $2$, hence again $c = 2$.
Thus $a=11$ and $b=6$.
\end{itemize}
\end{itemize}
This finishes all cases, so the proof is done. |
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 $\angle HKQ = 90\dg$.
Assume that the points $A$, $B$, $C$, $K$ and $Q$ are all different and lie on $\Gamma$ in this order.
Prove that the circumcircles of triangles $KQH$ and $FKM$ are tangent to each other. | 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{center}
\begin{asy}
size(10cm);
pair A = dir(80);
pair B = dir(220);
pair C = dir(-40);
filldraw(unitcircle, opacity(0.1)+yellow, orange);
pair O = circumcenter(A, B, C);
pair H = orthocenter(A, B, C);
pair Q = IP(CP(midpoint(A--H), H), unitcircle);
pair N = midpoint(H--Q);
pair T = midpoint(A--H);
pair N_9 = midpoint(O--H);
pair M = midpoint(B--C);
pair F = foot(A, B, C);
filldraw(circumcircle(M, F, N), opacity(0.1)+yellow, heavyred);
pair L = M+T-N;
pair K = foot(Q, L, H);
draw(A--B--C--cycle, red+1);
draw(A--F, red+1);
draw(T--M--L--N, red);
draw(M--Q--A, brown);
draw(Q--K--L, brown);
filldraw(A--Q--L--cycle, opacity(0.1)+cyan, cyan);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$O$", O, dir(O));
dot("$H$", H, dir(H));
dot("$Q$", Q, dir(Q));
dot("$N$", N, dir(N));
dot("$T$", T, dir(T));
dot("$N_9$", N_9, dir(N_9));
dot("$M$", M, dir(M));
dot("$F$", F, dir(F));
dot("$L$", L, dir(L));
dot("$K$", K, dir(K));
/* TSQ Source:
!size(10cm);
A = dir 80
B = dir 220
C = dir -40
unitcircle 0.1 yellow / orange
O = circumcenter A B C
H = orthocenter A B C
Q = IP CP midpoint A--H H unitcircle
N = midpoint H--Q
T = midpoint A--H
N_9 = midpoint O--H
M = midpoint B--C
F = foot A B C
circle M F N 0.1 yellow / heavyred
L = M+T-N
K = foot Q L H
A--B--C--cycle red+1
A--F red+1
T--M--L--N red
M--Q--A brown
Q--K--L brown
A--Q--L--cycle 0.1 cyan / cyan
*/
\end{asy}
\end{center}
\begin{claim*}
$\ol{LM} \parallel \ol{AQ}$.
\end{claim*}
\begin{proof}
Both are perpendicular to $\ol{MHQ}$.
\end{proof}
\begin{claim*}
$LA = LQ$.
\end{claim*}
\begin{proof}
Let $N$ and $T$ be the midpoints of $\ol{HQ}$ and $\ol{AH}$,
and $O$ the circumcenter.
As $\ol{MT}$ is a diameter, we know $LTNM$ is a rectangle,
so $\ol{LT}$ passes through $O$.
Since $\ol{LOT} \perp \ol{AQ}$ and $OA=OQ$, the proof is complete.
\end{proof}
Together these two claims solve the problem. |
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 $A$, $F$, $B$, $C$, and $G$ lie on $\Omega$ in this order.
Let $K = (BDF) \cap \ol{AB} \neq B$
and $L = (CGE) \cap \ol{AC} \neq C$
and assume these points do not lie on line $FG$.
Define $X = \ol{FK} \cap \ol{GL}$.
Prove that $X$ lies on the line $AO$. | 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(A, r), unitcircle);
pair G = OP(CR(A, r), unitcircle);
pair D = IP(CR(A, r), B--C);
pair E = OP(CR(A, r), B--C);
pair K = IP(circumcircle(B, D, F), A--B);
pair L = IP(circumcircle(C, E, G), A--C);
draw(unitcircle, orange);
draw(arc(A,r,160,380), orange);
filldraw(A--B--C--cycle, opacity(0.1)+lightred, red);
pair X = extension(F, K, G, L);
draw(F--G, red);
draw(F--X--G, lightblue);
pair F_2 = IP(circumcircle(B, F, D), F--G);
pair G_2 = IP(F--G, circumcircle(C, E, G));
filldraw(circumcircle(B, F, D), opacity(0.1)+yellow, lightblue);
filldraw(circumcircle(C, E, G), opacity(0.1)+yellow, lightblue);
draw(B--F_2--D--F--cycle, lightcyan);
draw(C--G_2--E--G--cycle, lightcyan);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$F$", F, dir(F));
dot("$G$", G, dir(G));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$K$", K, dir(100));
dot("$L$", L, dir(L));
dot("$X$", X, dir(X));
dot("$F_2$", F_2, dir(70));
dot("$G_2$", G_2, dir(G_2));
/* TSQ Source:
A = dir 105
B = dir 200
C = dir 340
! real r = 1.337;
F = IP CR A r unitcircle
G = OP CR A r unitcircle
D = IP CR A r B--C
E = OP CR A r B--C
K = IP circumcircle B D F A--B R100
L = IP circumcircle C E G A--C
unitcircle orange
!draw(arc(A,r,160,380), orange);
A--B--C--cycle 0.1 lightred / red
X = extension F K G L
F--G red
F--X--G lightblue
F_2 = IP circumcircle B F D F--G R70
G_2 = IP F--G circumcircle C E G
circumcircle B F D 0.1 yellow / lightblue
circumcircle C E G 0.1 yellow / lightblue
B--F_2--D--F--cycle lightcyan
C--G_2--E--G--cycle lightcyan
*/
\end{asy}
\end{center}
\begin{claim*}
Quadrilaterals $FBDF_2$ and $G_2ECG$ are similar,
actually homothetic through $\ol{FG} \cap \ol{BC}$.
\end{claim*}
\begin{proof}
This is essentially a repeated application
of being ``anti-parallel'' through $\angle(FG, BC)$.
Note the four angle relations
\begin{align*}
\dang(FD, FG) = \dang(BC,GE) = \dang(G_2C,FG)
&\implies \ol{FD} \parallel \ol{G_2C} \\
\dang(F_2B, FG) = \dang(BC,FD) = \dang(GE,FG)
&\implies \ol{F_2B} \parallel \ol{GE} \\
\dang(FB, FG) = \dang(BC,GC) = \dang(G_2E,FG)
&\implies \ol{FB} \parallel \ol{G_2E} \\
\dang(F_2D, FG) = \dang(BC,FB) = \dang(GC,FG)
&\implies \ol{F_2D} \parallel \ol{GC}.
\end{align*}
This gives the desired homotheties.
\end{proof}
To finish the angle chase,
\begin{align*}
\dang GFK = \dang F_2BK &= \dang F_2BF - \dang ABF
= \dang F_2DF - \dang ABF \\
&= \dang F_2DF - \dang GCA
= \dang GCG_2 - \dang GCA \\
&= \dang LCG_2 = \dang LGF
\end{align*}
as needed.
(Here $\dang ABF = \dang GCA$ since $AF = AG$.) |
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 $f(0) \in \{0,2\}$.
\ii From $P(x,1)$
we find that $x+f(x+1) \in S$ for all $x$.
\end{itemize}
We now solve the case $f(0) = 2$.
\begin{claim*}
If $f(0) = 2$ then $f(x) \equiv 2-x$.
\end{claim*}
\begin{proof}
Let $t \in S$ be any fixed point.
Then $P(0,t)$ gives $2 = 2t$ or $t = 1$;
so $S = \{1\}$.
But we also saw $x+f(x+1) \in S$,
which implies $f(x) \equiv 2-x$.
\end{proof}
Henceforth, assume $f(0) = 0$.
\begin{claim*}
If $f(0) = 0$ then $f$ is odd.
\end{claim*}
\begin{proof}
Note that $P(1,-1) \implies f(1) + f(-1) = 1 - f(1)$
and $P(-1,1) \implies f(-1) + f(-1) = -1 + f(1)$,
together giving $f(1) = 1$ and $f(-1) = -1$.
To prove $f$ odd we now obtain more fixed points:
\begin{itemize}
\ii From $P(x,0)$ we find that $x+f(x) \in S$ for all $x \in \RR$.
\ii From $P(x-1,1)$ we find that $x-1+f(x) \in S$ for all $x \in \RR$.
\ii From $P(1, f(x)+x-1)$ we find $x+1+f(x) \in S$ for all $x \in \RR$.
\end{itemize}
Finally $P(x, -1)$ gives $f$ odd.
\end{proof}
To finish from $f$ odd, notice that
\begin{align*}
P(x,-x) &\implies f(x) + f(-x^2) = x - xf(x) \\
P(-x,x) &\implies f(-x) + f(-x^2) = -x + xf(-x)
\end{align*}
which upon subtracting gives $f(x) \equiv x$. |
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
\[ \left\lvert\sum_{j=m+1}^n (a_j-b) \right\rvert \le 1007^2 \]
for all integers $m$ and $n$ such that $n > m\ge N$.
\end{enumerate} | 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 more ascending chains
(which skip by at most $2015$).
There are at most $2015$ such chains,
since among any $2015$ consecutive points in $\NN$
every chain must have an element.
We claim we may take $b$ to be the number of such chains,
and $N$ to be the largest of the start-points of all the chains.
% TODO picture here would be nice
Consider an interval $I = [m+1, n]$.
We have that
\[ \sum_{m<j\le n} a_j = \sum_{\text{chain } c} \left[ \min \left\{ x > n, x \in c \right\}
- \min \left\{ x > m, x \in c \right\} \right]. \]
Thus the upper bound is proved by the calculation
\begin{align*}
\sum_{m<j\le n} (a_j-b)
&= \sum_{\text{chain } c} \left[ (\min \left\{ x > n, x \in c \right\} - n)
- (\min \left\{ x > m, x \in c \right\} - m) \right] \\
&= \sum_{\text{chain } c} \left[ (\min \left\{ x > n, x \in c \right\} - n) \right]
- \sum_{\text{chain } c} \left[
\min \left\{ x > m, x \in c \right\} - m \right] \\
&\le (1+2015+2014+\dots+(2015-(b-2)))-(1+2+\dots+b) \\
&= (b-1)(2015-b)
\end{align*}
from above (noting that $n+1$ has to belong to some chain).
The lower bound is similar.
\paragraph{Second solution (juggling).}
This solution is essentially the same, but phrased as a juggling problem.
Here is a solution in this interpretation:
we will consider several balls thrown in the air,
which may be at heights $0$, $1$, $2$, \dots, $2014$.
The process is as follows:
\begin{itemize}
\ii Initially, at time $t = 0$, there are no balls in the air.
\ii Then at each integer time $t$ thereafter,
if there is a ball at height $0$, it is caught;
otherwise a ball is added to the juggler's hand.
This ball (either caught or added) is then thrown to a height of $a_t$.
\ii Immediately afterwards, all balls have their height decreased by one.
\end{itemize}
The condition $a_k + k \neq \ell + a_\ell$ thus ensures that
no two balls are ever at the same height.
In particular, there will never be more than $2016$ balls,
since there are only $2015$ possible heights.
We claim we may set.
\begin{align*}
b &= \text{number of balls in entire process} \\
N &= \text{last moment in time at which a ball is added}.
\end{align*}
Indeed, the key fact is that if we let $S_t$ denote
the sum of the height of all the balls just after time $t+\half$, then
\[ S_{t+1} - S_t = a_{t+1} - b \]
After all, at each time step $t$, the caught ball is thrown to height $a_t$,
and then all balls have their height decreased by $1$,
from which the conclusion follows.
Hence the quantity in the problem is exactly equal to
\[ \left\lvert\sum_{j=m+1}^n (a_j-b) \right\rvert
= \left\lvert S_m - S_n \right\rvert. \]
For a fixed $b$, we easily have the inequalities
$0 + 1 + \dots + (b-1) \le S_t \le 2014 + 2013 + \dots + (2015-b)$.
Hence $|S_m - S_n| \le (b-1)(2015-b) \le 1007^2$ as desired. |
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 $\ol{FX}$, $\ol{EM}$, $\ol{BD}$ are concurrent. | 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
\[ \theta = \angle FAB = \angle FBA = \angle DAC = \angle DCA
= \angle EAD = \angle EDA. \]
We begin by focusing just on $ABCD$ with point $F$,
ignoring for now the points $E$ and $X$
(and to some extent even point $M$).
It turns out this is a very familiar configuration.
\begin{lemma*}
[Central lemma]
The points $F$ and $C$
are the incenter and $A$-excenter of $\triangle DAB$.
Moreover, $\triangle DAB$ is isosceles with $DA = DB$.
\end{lemma*}
\begin{proof}
The proof uses three observations:
\begin{itemize}
\ii We already know that $\ol{FAC}$ is the angle bisector of $\angle ABD$.
\ii We were given $\angle FBC = 90\dg$.
\ii Next, note that $\triangle AFB \sim \triangle ADC$
(they are similar isosceles triangles).
From this it follows that $AF \cdot AC = AB \cdot AD$.
\end{itemize}
These three facts, together with $F$ lying inside $\triangle ABD$,
are enough to imply the result.
\end{proof}
\begin{center}
\begin{asy}
pair F = dir(157);
pair C = -F;
pair B = conj(F);
pair M = origin;
pair A = dir(F-C)*abs(F-B)+F;
pair t(pair X) { return A + (X-A) * (F-A) / (B-A); }
pair D = t(C);
filldraw(A--F--B--cycle, opacity(0.1)+red, gray);
filldraw(A--C--D--cycle, opacity(0.1)+orange, gray);
filldraw(F--B--C--cycle, opacity(0.1)+lightcyan, gray);
draw(circumcircle(A, B, D), orange+dashed);
draw(circumcircle(B, C, F), orange+dashed);
draw(B--D, gray);
draw(F--D, gray);
draw(A--B--D--cycle, black+0.8);
dot("$F$", F, dir(115));
dot("$C$", C, dir(C));
dot("$B$", B, dir(B));
dot("$M$", M, dir(280));
dot("$A$", A, dir(A));
dot("$D$", D, dir(60));
/* TSQ Source:
F = dir 157 R115
C = -F
B = conj(F)
M = origin R280
A = dir(F-C)*abs(F-B)+F
!pair t(pair X) { return A + (X-A) * (F-A) / (B-A); }
D = t(C) R60
A--F--B--cycle 0.1 red / gray
A--C--D--cycle 0.1 orange / gray
F--B--C--cycle 0.1 lightcyan / gray
circumcircle A B D orange dashed
circumcircle B C F orange dashed
B--D gray
F--D gray
A--B--D--cycle black+0.8
*/
\end{asy}
\end{center}
\begin{corollary*}
The point $M$ is the midpoint of arc $\widehat{BD}$ of $(DAB)$,
and the center of cyclic quadrilateral $FDCB$.
\end{corollary*}
\begin{proof}
Fact 5.
\end{proof}
Using these observations as the anchor
for everything that follows,
we now prove several claims about $X$ and $E$ in succession.
\begin{center}
\begin{asy}
pair F = dir(157);
pair C = -F;
pair B = conj(F);
pair M = origin;
pair A = dir(F-C)*abs(F-B)+F;
pair t(pair X) { return A + (X-A) * (F-A) / (B-A); }
pair D = t(C);
pair E = t(D);
pair X = E+M-A;
draw(E--F, lightgreen);
draw(D--M, lightgreen);
draw(B--D, lightblue);
draw(F--X, lightblue);
draw(M--E, lightblue);
filldraw(A--F--B--cycle, opacity(0.1)+red, gray);
filldraw(A--C--D--cycle, opacity(0.1)+orange, gray);
filldraw(A--E--D--cycle, opacity(0.1)+yellow, gray);
filldraw(F--B--C--cycle, opacity(0.1)+lightcyan, gray);
draw(A--E--X--M--cycle, heavygreen);
draw(circumcircle(E, F, X), lightgreen+dashed);
draw(circumcircle(A, B, D), orange);
draw(circumcircle(B, C, F), orange);
dot("$F$", F, dir(45));
dot("$C$", C, dir(C));
dot("$B$", B, dir(B));
dot("$M$", M, dir(260));
dot("$A$", A, dir(A));
dot("$D$", D, dir(60));
dot("$E$", E, dir(E));
dot("$X$", X, dir(X));
/* TSQ Source:
F = dir 157 R45
C = -F
B = conj(F)
M = origin R260
A = dir(F-C)*abs(F-B)+F
!pair t(pair X) { return A + (X-A) * (F-A) / (B-A); }
D = t(C) R60
E = t(D)
X = E+M-A
E--F lightgreen
D--M lightgreen
B--D lightblue
F--X lightblue
M--E lightblue
A--F--B--cycle 0.1 red / gray
A--C--D--cycle 0.1 orange / gray
A--E--D--cycle 0.1 yellow / gray
F--B--C--cycle 0.1 lightcyan / gray
A--E--X--M--cycle heavygreen
circumcircle E F X lightgreen dashed
circumcircle A B D orange
circumcircle B C F orange
*/
\end{asy}
\end{center}
\begin{claim*}
Point $E$ is the midpoint of arc $\widehat{AD}$
in $(ABMD)$, and hence lies on ray $BF$.
\end{claim*}
\begin{proof}
This follows from
$\angle EDA = \theta = \angle EBA$.
\end{proof}
\begin{claim*}
Points $X$ is the second intersection of ray $\ol{ED}$
with $(BFDC)$.
\end{claim*}
\begin{proof}
First, $\ol{ED} \parallel \ol{AC}$ already since
$\angle AED = 180\dg - 2\theta$
and $\angle CAE = 2\theta$.
Now since $DB = DA$, we get $MB = MD = ED = EA$.
Thus, $MX = AE = MB$,
so $X$ also lies on the circle $(BFDC)$ centered at $M$.
\end{proof}
\begin{claim*}
The quadrilateral $EXMF$ is an isosceles trapezoid.
\end{claim*}
\begin{proof}
We already know $\ol{EX} \parallel \ol{FM}$.
Since $\angle EFA = 180\dg - \angle AFB = 2\theta = \angle FAE$,
we have $EF = EA$ as well (and $F \neq A$).
As $EXMA$ was a parallelogram,
it follows $EXMF$ is an isosceles trapezoid.
\end{proof}
The problem then follows by radical axis theorem
on the three circles $(AEDMB)$, $(BFDXC)$ and $(EXMF)$. |
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 diagonal is a multiple of three,
then one third of the entries are $I$, one third are $M$ and one third are $O$.
\end{itemize}
Note that an $n \times n$ table has $4n-2$ diagonals. | The answer is $n$ divisible by $9$.
First we construct $n=9$ and by extension every multiple of $9$.
\[
\begin{array}{|ccc|ccc|ccc|} \hline
I & I & I & M & M & M & O & O & O \\
M & M & M & O & O & O & I & I & I \\
O & O & O & I & I & I & M & M & M \\\hline
I & I & I & M & M & M & O & O & O \\
M & M & M & O & O & O & I & I & I \\
O & O & O & I & I & I & M & M & M \\\hline
I & I & I & M & M & M & O & O & O \\
M & M & M & O & O & O & I & I & I \\
O & O & O & I & I & I & M & M & M \\\hline
\end{array}
\]
We now prove $9 \mid n$ is necessary.
Let $n = 3k$, which divides the given grid into $k^2$ sub-boxes
(of size $3 \times 3$ each).
We say a multiset of squares $S$ is \emph{clean} if
the letters distribute equally among them;
note that unions of clean multisets are clean.
Consider the following clean sets (given to us by problem statement):
\begin{itemize}
\ii All columns indexed $2 \pmod 3$,
\ii All rows indexed $2 \pmod 3$, and
\ii All $4k-2$ diagonals mentioned in the problem.
\end{itemize}
Take their union.
This covers the center of each box four times,
and every other cell exactly once.
We conclude the set of $k^2$ center squares
are clean, hence $3 \mid k^2$ and so $9 \mid n$,
as desired.
Shown below is the sums over all diagonals only,
and of the entire union.
\[
\begin{array}{|ccc|ccc|ccc|} \hline
1 & & 1 & 1 & & 1 & 1 & & 1 \\
& 2 & & & 2 & & & 2 & \\
1 & & 1 & 1 & & 1 & 1 & & 1 \\\hline
1 & & 1 & 1 & & 1 & 1 & & 1 \\
& 2 & & & 2 & & & 2 & \\
1 & & 1 & 1 & & 1 & 1 & & 1 \\\hline
1 & & 1 & 1 & & 1 & 1 & & 1 \\
& 2 & & & 2 & & & 2 & \\
1 & & 1 & 1 & & 1 & 1 & & 1 \\\hline
\end{array}
\qquad
\begin{array}{|ccc|ccc|ccc|} \hline
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 4 & 1 & 1 & 4 & 1 & 1 & 4 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\\hline
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 4 & 1 & 1 & 4 & 1 & 1 & 4 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\\hline
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 4 & 1 & 1 & 4 & 1 & 1 & 4 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\\hline
\end{array}
\] |
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 divisible by $n$. | 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 diagonal and induct down on $k$ by using inversion.
Let the polygon be $A_1 A_2 \dots A_{k+1}$
and suppose for contradiction that all sides are divisible by $p^e$
but no diagonals are.
Let $O = A_{k+1}$ for notational convenience.
By applying inversion around $O$ with radius $1$,
we get the ``generalized Ptolemy theorem''
\[
\frac{A_1A_2}{OA_1 \cdot OA_2}
+ \frac{A_2A_3}{OA_2 \cdot OA_3}
+ \dots
+ \frac{A_{k-1} A_k}{OA_{k-1} \cdot OA_k}
= \frac{A_1 A_k}{OA_1 \cdot OA_k}
\]
or, making use of square roots,
\[
\sqrt{\frac{A_1A_2^2}{OA_1^2 \cdot OA_2^2}}
+ \sqrt{\frac{A_2A_3^2}{OA_2^2 \cdot OA_3^2}}
+ \dots
+ \sqrt{\frac{A_{k-1} A_k^2}{OA_{k-1}^2 \cdot OA_k^2}}
= \sqrt{\frac{A_1 A_k^2}{OA_1^2 \cdot OA_k^2}}
\]
Suppose $\nu_p$ of all diagonals is strictly less than $e$.
Then the relation becomes
\[ \sqrt{q_1} + \dots + \sqrt{q_{k-1}} = \sqrt q \]
where $q_i$ are positive rational numbers.
Since there are no nontrivial relations between square roots
(see \href{https://qchu.wordpress.com/2009/07/02/square-roots-have-no-unexpected-linear-relationships/}{this link})
there is a positive rational number $b$
such that $r_i = \sqrt{q_i/b}$ and $r = \sqrt{q/b}$
are all rational numbers.
Then
\[ \sum r_i = r. \]
However, the condition implies that $\nu_p(q_i^2) > \nu_p(q^2)$ for all $i$
(check this for $i=1$, $i=k-1$ and $2 \le i \le k-2$),
and hence $\nu_p(r_i) > \nu_p(r)$.
This is absurd.
\begin{remark*}
I think you basically have to use some Ptolemy-like geometric property,
and also all correct solutions I know of for $n = p^e$
depend on finding a diagonal and inducting down.
(Actually, the case $k=4$ is pretty motivating;
Ptolemy implies one can cut in two.)
\end{remark*} |
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 which the set
\[ \{P(a+1),P(a+2),\dots,P(a+b)\} \]
is fragrant? | 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 any GCD's among $P(a+1)$, \dots, $P(a+b)$ must be among $\{3, 7, 19\}$.
Consider a multi-graph on $\{a+1, \dots, a+b\}$ where we join two elements with nontrivial GCD
and label the edge with the corresponding prime.
Then we readily see there is at most one edge each of $\{3, 7, 19\}$:
id est at most one edge of gap $2$, $3$, $4$ (and no edges of gap $1$).
(By the gap of an edge $e = \{u,v\}$ we mean $|u - v|$.)
But one can see that it's now impossible for every vertex to have nonzero degree, contradiction.
To construct $b = 6$ we use the Chinese remainder theorem: select $a$ with
\begin{align*}
a+1 & \equiv 7 \pmod{19} \\
a+5 & \equiv 11 \pmod{19} \\
a+2 & \equiv 2 \pmod{7} \\
a+4 & \equiv 4 \pmod{7} \\
a+3 & \equiv 1 \pmod{3} \\
a+6 & \equiv 1 \pmod{3}
\end{align*}
which does the trick. |
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 equation
has no real solutions? | 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) \\
(x-5)(x-8) &< (x-6)(x-7) \\
(x-9)(x-12) &< (x-10)(x-11) \\
&\vdots \\
(x-2013)(x-2016) &< (x-2014)(x-2015).
\end{align*}
Notice that in all these inequalities, at most one of them
has non-positive numbers in it, and we never have both zero.
If there is exactly one negative term among the $1008 \cdot 2 = 2016$ sides,
it is on the left and we can multiply all together.
Thus the only case that remains is if $x \in (4m-2, 4m-1)$ for some $m$,
say the $m$th inequality.
In that case, the two sides of that inequality differ by a factor of at least $9$.
\begin{claim*}
We have \[ \prod_{k \ge 0} \frac{(4k+2)(4k+3)}{(4k+1)(4k+4)} < e. \]
\end{claim*}
\begin{proof}
[Proof of claim using logarithms]
It's equivalent to prove
\[ \sum_{k \ge 0} \log \left( 1 + \frac{2}{(4k+1)(4k+4)} \right) < 1. \]
To this end, we use the deep fact that $\log(1+t) \le t$,
and thus it follows from
$\sum_{k \ge 0} \frac{1}{(4k+1)(4k+4)} < \half$,
which one can obtain for example by noticing
it's less than $\frac14\frac{\pi^2}{6}$.
\end{proof}
\begin{remark*}
[Elementary proof of claim, given by Espen Slettnes]
To avoid calculus as above, for each $N \ge 0$, note the partial product is bounded by
\begin{align*}
\prod_{k=0}^N \frac{(4k+2)(4k+3)}{(4k+1)(4k+4)}
&=
\frac 21 \cdot \left( \frac 34 \cdot \frac 65 \right)
\cdot \left( \frac 78 \cdot \frac{10}{9} \right) \cdot \dots
\cdot \frac{4N+3}{4N+4} \\
&< 2 \cdot 1 \cdot 1 \cdot \dots \cdot \frac{4N+3}{4N+4}
< 2 < e. \qedhere
\end{align*}
\end{remark*}
This solves the problem, because then the factors being multiplied
on by the positive inequalities before the
$m$th one are both less than $e$, and $e^2 < 9$.
In symbols, for $4m-2 < x < 4m-1$ we should have
\[ \frac{(x-(4m-6))(x-(4m-5))}{(x-(4m-7))(x-(4m-4))}
\times \dots \times \frac{(x-2)(x-3)}{(x-1)(x-4)} < e \]
and
\[ \frac{(x-(4m+2))(x-(4m+3))}{(x-(4m+1))(x-(4m+4))}
\times \dots \times \frac{(x-2014)(x-2015)}{(x-2013)(x-2016)} < e \]
because the $(k+1)$st term of each left-hand side
is at most $\frac{(4k+2)(4k+3)}{(4k+1)(4k+4)}$, for $k \ge 0$.
As $e^2 < 9$, we're okay. |
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
to the next intersection point on its segment.
Frogs never change the direction of their jumps.
Geoff wishes to place the frogs in such a way that no two of them
will ever occupy the same intersection point at the same time.
\begin{enumerate}[(a)]
\ii Prove that Geoff can always fulfill his wish if $n$ is odd.
\ii Prove that Geoff can never fulfill his wish if $n$ is even.
\end{enumerate}
\end{enumerate} | 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 every pair of segments to meet,
each line segment must be of the form $P_i P_{i+n}$.
\begin{center}
\begin{asy}
draw(unitcircle);
pair P_1 = dir(100);
pair P_2 = dir(150);
pair P_3 = dir(190);
pair P_4 = dir(210);
pair P_5 = dir(280);
pair P_6 = dir(340);
pair P_7 = dir(2);
pair P_8 = dir(55);
dot("$1$", P_1, dir(P_1));
dot("$2$", P_2, dir(P_2));
dot("$3$", P_3, dir(P_3));
dot("$4$", P_4, dir(P_4));
dot("$5$", P_5, dir(P_5));
dot("$6$", P_6, dir(P_6));
dot("$7$", P_7, dir(P_7));
dot("$8$", P_8, dir(P_8));
draw(P_1--P_5);
draw(P_2--P_6);
draw(P_3--P_7);
draw(P_4--P_8);
dot(extension(P_1, P_5, P_2, P_6), red);
dot(extension(P_1, P_5, P_3, P_7), red);
dot(extension(P_1, P_5, P_4, P_8), red);
dot(extension(P_2, P_6, P_3, P_7), red);
dot(extension(P_2, P_6, P_4, P_8), red);
dot(extension(P_3, P_7, P_4, P_8), red);
\end{asy}
\end{center}
Then:
\begin{enumerate}[(a)]
\ii Place the frogs on $P_1$, $P_3$, \dots, $P_{2n-1}$.
A simple parity arguments shows this works.
\ii Observe that we cannot place frogs on consecutive $P_i$,
so the $n$ frogs must be placed on alternating points.
But since we also are supposed to not place frogs on
diametrically opposite points,
for even $n$ we immediately get a contradiction.
\end{enumerate}
\begin{remark*}
Yang says: this is easy to guess if you just do a
few small cases and notice that the pairs of
``violating points'' just forms a large cycle around.
\end{remark*} |
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 $a_n = A$ for infinitely many values of $n$. | 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 $c \not\equiv 2 \pmod 3$.
As $c$ is not itself a square,
the next perfect square after $c$ in the sequence
is one of $\left( \left\lfloor \sqrt c \right\rfloor + 1\right)^2$,
$\left( \left\lfloor \sqrt c \right\rfloor + 2\right)^2$,
or $\left( \left\lfloor \sqrt c \right\rfloor + 3\right)^2$.
So by minimality we require
\[ c \le \left\lfloor \sqrt c \right\rfloor + 3 \le \sqrt c + 3 \]
which requires $c \le 5$.
Since $c \neq 1,2,4,5$ we conclude $c = 3$.
\end{proof}
Now we split the problem into two cases:
\begin{itemize}
\ii If $a_0 \equiv 0 \pmod 3$, then all terms of the sequence are $0 \pmod 3$.
The smallest term of the sequence is thus $3$ by the lemma
and we have \[ 3 \to 6 \to 9 \to 3 \]
so $A = 3$ works fine.
\ii If $a_0 \not\equiv 0 \pmod 3$,
then no term of the sequence is $0 \pmod 3$,
and so in particular $3$ does not appear in the sequence.
So the smallest term of the sequence is $2 \pmod 3$ by lemma.
But since no squares are $2 \pmod 3$,
the sequence $a_k$ grows without bound forever after,
so no such $A$ can exist.
\end{itemize}
Hence the answer is $a_0 \equiv 0 \pmod 3$ only.
\paragraph{Second solution.}
We clean up the argument by proving the following lemma.
\begin{lemma*}
If $a_n$ is constant modulo $3$ and not $2 \pmod 3$,
then $a_n$ must eventually cycle in the form
$(m, m+3, m+6, \dots, m^2)$,
with no squares inside the cycle except $m^2$.
\end{lemma*}
\begin{proof}
Observe that $a_n$ must eventually hit a square, say $a_k = c^2$;
the next term is $a_{k+1} = c$.
Then it is forever impossible to exceed $c^2$ again,
by what is essentially discrete intermediate value theorem.
Indeed, suppose $a_\ell > c^2$ and take $\ell > k$ minimal
(in particular $a_{\ell} \neq \sqrt{a_{\ell-1}}$).
% We have $a_{\ell-1} \neq a_{\ell}^2$, since otherwise.
Thus $a_{\ell-1} \in \{c^2-2, c^2-1, c^2\}$
and thus for modulo $3$ reasons we have $a_{\ell-1} = c^2$.
But that should imply $a_\ell = c < c^2$, contradiction.
We therefore conclude $\sup \{a_n, a_{n+1}, \dots \}$ is a
decreasing integer sequence in $n$.
It must eventually stabilize, say at $m^2$.
Now we can't hit a square between $m$ and $m^2$,
and so we are done.
\end{proof}
Now, we contend that all $a_0 \equiv 0 \pmod 3$ work.
Indeed, for such $a_0$ we have $a_n \equiv 0 \pmod 3$ for all $n$,
so the lemma implies that the problem statement is valid.
Next, we observe that if $a_i \equiv 2 \pmod 3$,
then the sequence grows without bound afterwards
since no squares are $2 \pmod 3$.
In particular, if $a_0 \equiv 2 \pmod 3$ the answer is no.
Finally, we claim that if $a_0 \equiv 1 \pmod 3$,
then eventually some term is $2 \pmod 3$.
Assume for contradiction this is not so;
then $a_n \equiv 1 \pmod 3$ must hold forever,
and the lemma applies to give us a cycle of the form
$(m, m+3, \dots, m^2)$ where $m \equiv 1 \pmod 3$.
In particular $m \ge 4$ and
\[ m \le (m-2)^2 < m^2 \]
but $(m-2)^2 \equiv 1 \pmod 3$ which is a contradiction. |
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, $f(0)=1$ and $f(1)=0$.
\end{claim*}
\begin{proof}
For the forwards direction, if $f(z)=0$ and $z \neq 1$
one may put $(x,y) = \left( z, z(z-1)\inv \right)$
(so that $x+y=xy$) we deduce $f(0) = 0$ which is a contradiction.
For the reverse, $f(f(0)^2)=0$ by setting $x=y=0$, and use the previous part.
We also conclude $f(1) = 0$, $f(0) = 1$.
\end{proof}
\begin{claim*}
If $f$ is injective, we are done.
\end{claim*}
\begin{proof}
Setting $y=0$ in the original equation gives $f(f(x)) = 1-f(x)$.
We apply this three times on the expression $f^3(x)$:
\[ f(1-f(x)) = f(f(f(x))) = 1 - f(f(x)) = f(x). \]
Hence $1-f(x) = x$ or $f(x) = 1-x$.
\end{proof}
\begin{remark*}
The result $f(f(x)) + f(x) = 1$ also implies that surjectivity
would solve the problem.
\end{remark*}
\begin{claim*}
$f$ is injective.
\end{claim*}
\begin{proof}
Setting $y=1$ in the original equation gives $f(x+1) = f(x)-1$, and by induction
\begin{equation}
f(x+n) = f(x)-n.
\label{eq:linshift}
\end{equation}
Assume now $f(a) = f(b)$.
By using \eqref{eq:linshift} we may shift $a$ and $b$ to be large enough that
we may find $x$ and $y$ obeying $x+y=a+1$, $xy=b$.
Setting these gives
\begin{align*}
f(f(x)f(y)) &= f(xy) - f(x+y) = f(b) - f(a+1) \\
&= f(b) + 1 - f(a) = 1
\end{align*}
from which we conclude
\[ f\left( f(x)f(y) + 1 \right) = 0. \]
Hence by the first claim
we have $f(x)f(y) + 1 = 1$, so $f(x)f(y) = 0$.
Applying the first claim again gives $1 \in \{x,y\}$.
But that implies $a=b$.
\end{proof}
\begin{remark*}
Jessica Wan points out that for any $a \neq b$,
at least one of $a^2 > 4(b-1)$ and $b^2 > 4(a-1)$ is true.
So shifting via \eqref{eq:linshift} is actually unnecessary for this proof.
\end{remark*}
\begin{remark*}
One can solve the problem over $\QQ$ using only \eqref{eq:linshift} and the easy parts.
Indeed, that already implies $f(n) = 1-n$ for all $n$.
Now we induct to show $f(p/q) = 1-p/q$ for all $0 < p < q$ (on $q$).
By choosing $x = 1+p/q$, $y = 1+q/p$, we cause $xy = x+y$,
and hence $0 = f\left( f(1+p/q)f(1+q/p) \right)$
or $1 = f(1+p/q)f(1+q/p)$.
By induction we compute $f(1+q/p)$ and this gives $f(p/q+1) = f(p/q)-1$.
\end{remark*} |
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 has a tracking device (e.g.\ dog)
which reports an approximate location $P_n$ of the rabbit,
such that $P_n A_n \le 1$.
\ii The hunter moves visibly from $B_{n-1}$ to a point $B_n$
such that $B_{n-1} B_n = 1$.
\end{enumerate}
Let $N = 10^9$. Can the hunter guarantee that $A_N B_N < 100$? | 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 about the rabbit.
\end{claim*}
\begin{proof}
Consider a positive integer $n > d$, to be chosen later.
Let the hunter start at $B$ and the rabbit at $A$, as shown.
Let $\ell$ denote line $AB$.
Now, we may assume the rabbit reveals its location $A$,
so that all previous information becomes irrelevant.
The rabbit chooses two points $X$ and $Y$ symmetric about $\ell$
such that $XY = 2$ and $AX = AY = n$, as shown.
The rabbit can then hop to either $X$ or $Y$,
pinging the point $P_n$ on the $\ell$ each time.
This takes $n$ hops.
\begin{center}
\begin{asy}
pair A = 2*dir(0);
pair B = dir(180);
pair H = B+4*A;
pair X = (2+63**0.5,1);
pair Y = (2+63**0.5,-1);
draw(B--H, blue);
draw(A--X, red, EndArrow(TeXHead), Margins);
draw(A--Y, red, EndArrow(TeXHead), Margins);
pair M = midpoint(X--Y);
draw(H--M, dotted);
draw(X--Y, dotted);
draw(X--H--Y, dashed+heavygreen);
dot("$A$", A, dir(270));
label("rabbit", A, dir(90));
dot("$B$", B, dir(270));
label("hunter", B, dir(90));
dot("$H$", H, dir(270));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
dot("$M$", M, dir(M));
label("$n$", A--X, dir(90), red);
label("$n$", A--Y, dir(-90), red);
/* TSQ Source:
A = 2*dir(0) R270
B = dir(180) R270
H = B+4*A R270
X = (2+63**0.5,1)
Y = (2+63**0.5,-1)
B--H blue
X--A--Y red
M = midpoint X--Y
H--M dotted
X--Y dotted
X--H--Y dashed heavygreen
*/
\end{asy}
\end{center}
Now among all points $H$ the hunter can go to,
$\min \max \{HX, HY\}$ is clearly minimized with $H \in \ell$ by symmetry.
So the hunter moves to a point $H$ such that $BH = n$ as well.
In that case the new distance is $HX = HY$.
We now compute
\begin{align*}
HX^2 &= 1 + HM^2 = 1 + \left( \sqrt{AX^2-1}-AH \right)^2 \\
&= 1 + \left( \sqrt{n^2-1}-(n-d) \right)^2 \\
&\ge 1 + \left( \left( n-\frac1n \right) - (n-d) \right)^2 \\
&= 1 + (d-1/n)^2
\end{align*}
which exceeds $d^2 + \half$ whenever $n \ge 4d$.
\end{proof}
In particular we can always take $n = 400$ even very crudely;
applying the lemma $2 \cdot 100^2$ times,
this gives a bound of $400 \cdot 2 \cdot 100^2 < 10^9$, as desired.
\begin{remark*}
The step of revealing the location of the rabbit seems
critical because as far as I am aware it is basically
impossible to keep track of ping locations in the problem.
\end{remark*}
\begin{remark*}
Reasons to believe the answer is ``no'':
the $10^9$ constant,
and also that ``follow the last ping'' is losing for the hunter.
\end{remark*}
\begin{remark*}
I think there are roughly two ways you can approach the problem
once you recognize the answer.
\begin{enumerate}[(i)]
\ii Try and control the location of the pings
\ii Abandon the notion of controlling possible locations,
and try to increase the distance by a little bit,
say from $d$ to $\sqrt{d^2+\varepsilon}$.
This involves revealing the location of the rabbit
before each iteration of several jumps.
\end{enumerate}
I think it's clear that the difficulty of
my approach is realizing that (ii) is possible;
once you do, the two-point approach is more or less the only one possible.
My opinion is that (ii) is not that magical;
as I said it was the first idea I had.
But I am biased, because when I test-solved the problem
at the IMO it was called ``C5'' and not ``IMO3'';
this effectively told me it was unlikely that the official solution
was along the lines of (i),
because otherwise it would have been placed much later in the shortlist.
\end{remark*} |
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 $\ell$ at two distinct points.
Let $A$ be the common point of $\Gamma$ and $\ell$ closer to $R$.
Line $AJ$ meets $\Omega$ again at $K$.
Prove that line $KT$ is tangent to $\Gamma$. | \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 $-\dang RKS = \dang SKR = \dang SRA = \dang TRA$.
\ii Similarly, $\ol{TK}$ is tangent at $T$
iff $\triangle ATS \sim \triangle TRK$.
\ii The two similarities are equivalent because $RS = ST$
the SAS gives $KR \cdot TA = RS \cdot RT = TS \cdot TR$.
\end{itemize}
\begin{center}
\begin{asy}
size(10cm);
pair T = dir(100);
pair S = dir(165);
pair R = 2*S-T;
pair E = dir(0);
pair A = IP(unitcircle, R--(R+100*E));
pair B = OP(unitcircle, R--(R+100*E));
pair K = extension(T, T+dir(90)*T, R, R+T-A);
draw(R--B, blue);
filldraw(unitcircle, opacity(0.05)+lightcyan, blue);
filldraw(circumcircle(R, S, K), opacity(0.05)+paleblue, heavycyan);
pair J = -A+2*foot(origin, A, K);
filldraw(K--R--A--T--cycle, opacity(0.1)+lightred, red);
draw(A--K, orange);
draw(R--T, orange);
// invert
pair Js = extension(T, T+A-B, R, J);
pair Ks = extension(T, T+A-B, R, K);
draw(K--Ks, heavygreen);
draw(Ks--Js, heavygreen);
draw(Js--B, dotted+blue);
draw(R--Js, dotted+blue);
dot("$T$", T, dir(T));
dot("$S$", S, dir(355));
dot("$R$", R, dir(270));
dot("$A$", A, dir(225));
dot("$B$", B, dir(315));
dot("$K$", K, dir(200));
dot("$J$", J, dir(180));
dot("$J^\ast$", Js, dir(Js));
dot("$K^\ast$", Ks, dir(Ks));
/* TSQ Source:
!size(10cm);
T = dir 100
S = dir 165 R355
R = 2*S-T R270
E := dir 0
A = IP unitcircle R--(R+100*E) R225
B = OP unitcircle R--(R+100*E) R315
K = extension T T+dir(90)*T R R+T-A R200
T--K blue
R--B blue
unitcircle 0.05 lightcyan / blue
circumcircle R S K 0.05 paleblue / heavycyan
J = -A+2*foot origin A K R180
K--R--A--T--cycle 0.1 lightred / red
A--K orange
R--T orange
// invert
J* = extension T T+A-B R J
K* = extension T T+A-B R K
K--Ks heavygreen
Ks--Js heavygreen
Js--B dotted blue
R--Js dotted blue
*/
\end{asy}
\end{center}
\begin{remark*}
The problem is actually symmetric with respect to two circles;
$\ol{RA}$ is tangent at $R$ if and only if $\ol{TK}$ at $T$.
\end{remark*}
\paragraph{Second solution (inversion).}
Consider an inversion at $R$ fixing the circumcircle $\Gamma$ of $TSJA$.
Then:
\begin{itemize}
\ii $T$ and $S$ swap,
\ii $A$ and $B$ swap, where $B$ is the second intersection
of $\ell$ with $\Gamma$.
\ii Circle $\Omega$ inverts to the line through $T$
parallel to $\ol{RAB}$, call it $\ell$.
\ii $J^\ast$ is the second intersection of $\ell$ with $\Gamma$.
\ii $K^\ast$ is the intersection of $\ell$ with the circumcircle
of $RBJ^\ast$; this implies $RK^\ast J^\ast B$ is an isosceles trapezoid.
In particular, one reads $\ol{RK^\ast} \parallel \ol{AT}$ from this,
hence $RK^\ast TA$ is a parallelogram.
\end{itemize}
Thus we wish to show the circumcircle of $RSK^\ast$ is tangent to $\Gamma$.
But that follows from the final parallelogram observed:
$S$ is the center of the parallelogram since it is the midpoint of the diagonal.
\begin{remark*}
This also implies $RKTB$ is cyclic, from $\ol{K^\ast SA}$ collinear.
Moreover, quadrilateral $KK^\ast TS$ is cyclic (by power of a point);
this leads to the second official solution to the problem.
\end{remark*} |
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 fourth tallest players, \dots,
no one stands between the two shortest players.
Prove that this is possible. | 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}
size(10cm);
int[] ys = {7,11,2,5,0,10,9,0,12,1,6,8,4,3};
real r = 0.2;
fill(box((-1, 8.5),(4,4.5)), opacity(0.3)+lightred);
fill(box((-1,12.5),(7,8.5)), opacity(0.3)+lightcyan);
fill(box((-1, 4.5),(14,0.5)), opacity(0.3)+lightgreen);
fill(box((4, 8.5),(14,4.5)), opacity(0.3)+gray);
fill(box((7,12.5),(14,8.5)), opacity(0.3)+gray);
draw( (-1,4.5)--(14,4.5) );
draw( (-1,8.5)--(14,8.5) );
pair P(int x) {
return (x,ys[x]);
}
pair O;
int y;
pen c;
for (int x=0; x<=13; ++x) {
y = ys[x];
if (y==0) continue;
O = (x, y);
if ((y==7) || (y==5)) c = red;
else if ((y==10) || (y==9)) c = blue;
else if ((y==1) || (y==4)) c = heavygreen;
else c = gray;
filldraw(circle(O, r), c, black+1);
label("$"+(string)y+"$", O+(r,0), 0.5*dir(0), c);
}
pen border = black+2;
pen dash = dotted+1;
draw( (4,12.5)--(4,0.5), dash );
draw( (7,12.5)--(7,0.5), dash );
path bracket(real x0, real x1) {
return (x0,0.7)--(x0,0.5)--(x1,0.5)--(x1,0.7);
}
draw(bracket(-0.8,3.8), border);
draw(bracket(4.2,6.8), border);
draw(bracket(7.2,13.8), border);
\end{asy}
\end{center}
Take a partition of $N$ groups in order by height:
$G_1 = \{1,\dots,N+1\}$, $G_2 = \{N+2, \dots, 2N+2\}$, and so on.
We will pick two people from each group $G_k$.
Scan from the left until we find two people in the same group $G_k$.
Delete all people scanned and also everyone in $G_k$.
All the groups still have at least $N$ people left,
so we can induct down with the non-deleted people;
the chosen pair is to the far left anyways.
\begin{remark*}
The important bit is to \emph{scan by position}
but \emph{group by height},
and moreover not change the groups as we scan.
Dually, one can have a solution which scans by height
but groups by position.
\end{remark*} |
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{enumerate} | 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*}
[Standard]
By a suitable linear transformation we may assume
\[ (a_{m+1},b_{m+1}) = (1,0). \]
\end{claim*}
\begin{proof}
[Outline of proof]
It would be sufficient to show there exists a $2 \times 2$ matrix
$T = \left[ \begin{smallmatrix} u & v \\ s & t \end{smallmatrix} \right]$
with integer entries such that $\det T = 1$ and
$T \cdot \left[ \begin{smallmatrix} a_{m+1} \\ b_{m+1} \end{smallmatrix} \right]
= \left[ \begin{smallmatrix} 1 \\ 0 \end{smallmatrix} \right]$.
Then we could apply $T$ to all the ordered pairs in $S$ (viewed as column vectors);
if $f$ was a polynomial that works on the transformed ordered pairs,
then $f(ux+vy, sx+ty)$ works for the original ordered pairs.
(Here, the condition that $\det T = 1$ ensures that $T\inv$ has integer entries,
and hence that $T$ maps irreducible lattice points to irreducible lattice points.)
To generate such a matrix $T$, choose
$T = \left[ \begin{smallmatrix} u & v \\ -b_{m+1} & a_{m+1} \end{smallmatrix} \right]$
where $u$ and $v$ are chosen via B\'{e}zout lemma so that $ua_{m+1} + vb_{m+1} = 1$.
This matrix $T$ is rigged so that $\det T = 1$ and the leftmost column of $T\inv$ is
$\left[ \begin{smallmatrix} a_{m+1} \\ b_{m+1} \end{smallmatrix} \right]$.
\end{proof}
\begin{remark*}
This transformation claim is not necessary to proceed;
the solution below can be rewritten to avoid it with only cosmetic edits.
However, it makes things a bit easier to read.
\end{remark*}
Let $g(x,y)$ be a polynomial which works on the latter set.
We claim we can choose the new polynomial $f$ of the form
\[ f(x,y) = g(x,y)^{M} - C x^{\deg g \cdot M-m} \prod_{i=1}^m (b_i x - a_i y). \]
where $C$ and $M$ are integer parameters we may adjust.
Since $f(a_i, b_i) = 1$ by construction we just need
\[ 1 = f(1,0) = g(1,0)^M - C \prod b_i. \]
If $\prod b_i = 0$ we are done,
since $b_i = 0 \implies a_i = \pm 1$ in that case
and so $g(1, 0) = \pm 1$, thus take $M = 2$.
So it suffices to prove:
\begin{claim*}
We have $\gcd\left( g(1,0), b_i \right) = 1$ when $b_i \neq 0$.
\end{claim*}
\begin{proof}
Fix $i$. If $b_i = 0$ then $a_i = \pm 1$ and $g(\pm 1,0) = \pm 1$.
Otherwise know
\[ 1 = g(a_i, b_i) \equiv g(a_i, 0) \pmod{b_i} \]
and since the polynomial is homogeneous with $\gcd(a_i, b_i) = 1$
it follows $g(1,0) \not\equiv 0 \pmod{b_i}$ as well.
\end{proof}
Then take $M$ a large multiple of $\varphi(\prod |b_i|)$ and we're done.
\paragraph{Second solution (Lagrange).}
The main claim is that:
\begin{claim*}
For every positive integer $N$,
there is a homogeneous polynomial $P(x,y)$ such that
$P(x,y) \equiv 1 \pmod N$ whenever $\gcd(x,y) = 1$.
\end{claim*}
(This claim is actually implied by the problem.)
\begin{proof}
For $N = p^e$ a prime take $(x^{p-1} + y^{p-1})^{\varphi(N)}$
when $p$ is odd, and $(x^2+xy+y^2)^{\varphi(N)}$ for $p=2$.
Now, if $N$ is a product of primes,
we can collate coefficient by coefficient using the
Chinese remainder theorem.
% Now suppose $N = q_1 q_2 \dots q_k$ where $q_i$ are prime powers.
% Look at the polynomial $Q_i$ described above for $i=1, \dots, k$.
% Now \[ \frac{N}{q_i} Q_i(x,y) \equiv \frac{N}{q_i} \pmod{N} \]
% for all $x$ and $y$;
% so we can put together the polynomials $\frac{N}{q_i} Q_i$ by B\'{e}zout lemma.
\end{proof}
Let $S = \left\{ (a_i, b_i) \mid i=1, \dots, m \right\}$.
We have the natural homogeneous ``Lagrange polynomials''
\[ L_k(x,y) \coloneq \prod_{i \neq k} (b_i x - a_i y) \]
Now let
\[ N \coloneq \prod_k L_k(x_k, y_k) \]
and take $P$ as in the claim.
Then we can take a large power of $P$,
and for each $i$ subtract an appropriate multiple of $L_i(x,y)$; that is, choose
\[
f(x,y)
= P(x,y)^{C} - \sum_i L_i(x,y) \cdot Q_i(x,y)
\]
where $C$ is large enough that $C \deg P > \max_i \deg L_i$,
and $Q_i(x,y)$ is any homogeneous polynomial of degree $C \deg P - \deg L_i$
such that $L_k(x_k, y_k) Q_k(x_k, y_k) = \frac{P(x_k, y_k)^C - 1}{N} \cdot L_k(x_k, y_k)$
(which is an integer). |
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 $FG$ are parallel. | 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 $\Gamma$.
As $\ol{XF} \parallel \ol{AB}$ we can let $M$
denote the midpoint of minor arcs $\widehat{XF}$ and $\widehat{AB}$
(which coincide). Define $N$ similarly.
\begin{center}
\begin{asy}
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
pair F = dir(190);
pair M = dir(160);
pair N = dir(40);
pair G = M*N/F;
pair K = foot(F, A, B);
pair L = foot(G, A, C);
pair D = -B+2*K;
pair E = -C+2*L;
pair X = M*M/F;
pair Y = N*N/G;
filldraw(unitcircle, opacity(0.1)+lightcyan, lightblue);
draw(D--B--C--E--cycle, lightblue);
draw(X--A, red);
draw(A--Y, pink);
draw(F--X, heavygreen);
draw(Y--G, heavygreen);
draw(F--G, dotted+blue);
draw(M--N, dotted+blue);
filldraw(A--X--F--D--cycle, opacity(0.1)+lightgreen, heavygreen);
filldraw(A--E--G--Y--cycle, opacity(0.1)+lightgreen, heavygreen);
draw(B--F, heavygreen);
draw(C--G, heavygreen);
draw(X--F, heavycyan+1);
draw(D--A--E, heavycyan+1);
draw(Y--G, heavycyan+1);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$F$", F, dir(F));
dot("$M$", M, dir(M));
dot("$N$", N, dir(N));
dot("$G$", G, dir(G));
dot("$D$", D, dir(160));
dot("$E$", E, dir(80));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
/* TSQ Source:
A = dir 110
B = dir 210
C = dir 330
F = dir 190
M = dir 160
N = dir 40
G = M*N/F
K := foot F A B
L := foot G A C
D = -B+2*K R160
E = -C+2*L R80
X = M*M/F
Y = N*N/G
unitcircle 0.1 lightcyan / lightblue
D--B--C--E--cycle lightblue
X--A red
A--Y pink
F--X heavygreen
Y--G heavygreen
F--G dotted blue
M--N dotted blue
A--X--F--D--cycle 0.1 lightgreen / heavygreen
A--E--G--Y--cycle 0.1 lightgreen / heavygreen
B--F heavygreen
C--G heavygreen
X--F heavycyan+1
D--A--E heavycyan+1
Y--G heavycyan+1
*/
\end{asy}
\end{center}
Observe that $XF = AD = AE = YG$,
so arcs $\widehat{XF}$ and $\widehat{YG}$ have equal measure;
hence arcs $\widehat{MF}$ and $\widehat{NG}$ have equal measure;
therefore $\ol{MN} \parallel \ol{FG}$.
Since $\ol{MN}$ and $\ol{DE}$ are both perpendicular
to the $\angle A$ bisector, we're done.
\paragraph{Complex numbers solution.}
Let $b$, $c$, $f$, $g$, $a$ be as usual.
Note that
\begin{align*}
d-a &= \left( 2 \cdot \frac{f+a+b-ab\ol f}{2} -b \right)-a
= f - \frac{ab}{f} \\
e-a &= g - \frac{ac}{g}
\end{align*}
We are given $AD = AE$ from which one deduces
\begin{align*}
\left( \frac{e-a}{d-a} \right)^2 &= \frac cb
\implies \frac{(g^2-ac)^2}{(f^2-ab)^2} = \frac{g^2 c}{f^2 b} \\
\implies bc(bg^2-cf^2)a^2 &= g^2f^4c - f^2g^4b = f^2g^2(f^2c-g^2b) \\
\implies bc \cdot a^2 &= (fg)^2 \implies \left( -\frac{fg}{a} \right)^2 = bc.
\end{align*}
Since $\frac{-fg}{a}$ is the point $X$ on the circle
with $\ol{AX} \perp \ol{FG}$,
we conclude $\ol{FG}$ is either parallel or perpendicular
to the $\angle A$-bisector; it must the latter
since the $\angle A$-bisector separates the two minor arcs.
\paragraph{Trig solution (outline).}
Let $\ell$ denote the $\angle A$ bisector.
Fix $D$ and $F$.
We define the phantom point $G'$ such that $\ol{FG'} \perp \ell$
and $E'$ on side $\ol{AC}$ such that $GE'=GC$.
\begin{claim*}
[Converse of the IMO problem]
We have $AD = AE'$, so that $E = E'$.
\end{claim*}
\begin{proof}
Since $\ol{FG'} \perp \ell$,
one can deduce $\angle FBD = \half C + x$
and $\angle GCA = \half B + x$ for some $x$.
(One fast way to see this is to note that $\ol{FG} \parallel \ol{MN}$
where $M$ and $N$ are in the first solution.)
Then $\angle FAB = \half C -x$ and $\angle GAC = \half B - x$.
Let $R$ be the circumradius.
Now, by the law of sines,
\[ BF = 2R \sin\left( \half C - x \right). \]
From there we get
\begin{align*}
BD &= 2 \cdot BF \cos\left(\half C + x\right)
= 4R \cos\left(\half C+x\right) \sin \left(\half C-x\right) \\
DA &= AB - BD = 2R\sin C
- 4R\cos\left(\half C+x\right) \sin\left(\half C -x\right) \\
&= 2R\left[ \sin C - 2\cos\left( \half C + x \right) \sin \left( \half C -x \right) \right] \\
&= 2R\left[ \sin C - \left( \sin C - \sin 2x \right) \right]
= 2R \sin 2x.
\end{align*}
A similar calculation gives $AE' = 2R \sin 2x$ as needed.
\end{proof}
Thus, $\ol{FG'} \parallel \ol{DE}$, so $G = G'$ as well.
This concludes the proof.
\paragraph{Synthetic solution from Derek Liu.}
Let lines $FD$ and $GE$ intersect $\Gamma$ again at $J$ and $K$, respectively.
\begin{center}
\begin{asy}
unitsize(0.5cm);
import graph;
pair A=(-2.5,6), B=(-5.2,-3.9), C=(5.2,-3.9), D=(-4,0.5), E=(1,1.5), F=(-6.39,-1.21), G=(6.36,1.33), J=(3.19,5.66), K=(-6.27,1.72);
draw(Circle((0,0),6.5)); draw(A--B--C--A); draw(B--F--J--A--K--G--C); draw(Circle(A,5.7),dashed);
dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(J); dot(K);
label("$A$",A,N);
label("$B$",B,SW);
label("$C$",C,SE);
label("$D$",D,SSE);
label("$E$",E,S);
label("$F$",F,W);
label("$G$",G,E);
label("$J$",J,ENE);
label("$K$",K,WSW);
\end{asy}
\end{center}
Notice that $\triangle BFD\sim\triangle JAD$; as $FB=FD$, it follows that $AJ=AD$.
Likewise, $\triangle CGE\sim\triangle KAE$ and $GC=GE$, so $AK=AE$.
Hence,
\[ AK=AE=AD=AJ, \]
so $DEJK$ is cyclic with center $A$.
It follows that
\[ \dang KED=\dang KJD=\dang KJF=\dang KGF, \]
so we're done.
\begin{remark*}
Note that $K$ and $J$ must be distinct for this solution to work.
Since $G$ and $K$ lie on opposite sides of $AC$, $K$ is on major arc $ABC$.
As $AK=AD=AE\le \min(AB,AC)$, $K$ lies on minor arc $AB$.
Similarly, $J$ lies on minor arc $AC$, so $K\neq J.$
\end{remark*} |
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*}
Cyclically summing $a_{i+2}^2 - a_{i+2} = a_i a_{i+3} - a_i$ then gives
\[ \sum_i a_{i+2}^2 = \sum_i a_i a_{i+3}
\iff \sum_{\text{cyc}} \left( a_i - a_{i+3} \right)^2 = 0. \]
This means for inequality reasons the sequence is $3$-periodic.
Since the sequence is clearly not $1$-periodic,
as $x^2 + 1 = x$ has no real solutions.
Thus $3 \mid n$.
\paragraph{Second solution by sign counting.}
Extend $a_n$ to be a periodic sequence.
The idea is to look at the signs, and show the sequence
of the signs must be $--+$ repeated.
This takes several steps:
\begin{itemize}
\ii The pattern $---$ is impossible. Obvious, since the third term should be $ > 1$.
\ii The pattern $++$ is impossible. Then the sequence becomes strictly increasing,
hence may not be periodic.
\ii Zeros are impossible. If $a_1 = 0$, then $a_2 = 0$, $a_3 > 0$, $a_4 > 0$,
which gives the impossible $++$.
\ii The pattern $--+-+$ is impossible.
Compute the terms:
\begin{align*}
a_1 &= -x < 0 \\
a_2 &= -y < 0 \\
a_3 &= 1 + xy > 1 \\
a_4 &= 1 - y(1+xy) < 0 \\
a_5 &= 1 + (1+xy)(1-y(1+xy)) < 1.
\end{align*}
But now
\[ a_6 - a_5 = (1 + a_5 a_4) - (1 + a_3 a_4)
= a_4 (a_5 - a_3) > 0 \]
since $a_5 > 1 > a_3$.
This means we have the impossible $++$ pattern.
\ii The infinite alternating pattern $-+-+-+-+\dots$ is impossible.
Note that
\[ a_1 a_2 + 1 = a_3 < 0 < a_4 = 1 + a_2 a_3 \implies a_1 < a_3 \]
since $a_2 > 0$;
extending this we get $a_1 < a_3 < a_5 < \dots$
which contradicts the periodicity.
\end{itemize}
We finally collate the logic of sign patterns.
Since the pattern is not alternating, there must be $--$ somewhere.
Afterwards must be $+$, and then after that must be two minus signs
(since one minus sign is impossible by impossibility of $--+-+$
and $---$ is also forbidden);
thus we get the periodic $--+$ as desired. |
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 integer from $1$ to $10$.
\begin{center}
\begin{tikzpicture}[scale = 0.8]
\node at (1.5,2.58) {$4$};
\node at (1,1.72) {$2$};
\node at (2,1.72) {$6$};
\node at (0.5,0.86) {$5$};
\node at (1.5,0.86) {$7$};
\node at (2.5,0.86) {$1$};
\node at (0,0) {$8$};
\node at (1,0) {$3$};
\node at (2,0) {$10$};
\node at (3,0) {$9$};
\end{tikzpicture}
\end{center}
Does there exist an anti-Pascal triangle with $2018$ rows
which contains every integer from $1$ to $1+2+\dots +2018$? | 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 like lightning strikes).
Consider the directed path starting from the top vertex $A$.
Starting from the first number, it increments by at least $1+2+\dots+n$,
since the increments at each step in the path are distinct;
therefore equality must hold
and thus the path from the top ends at $N = 1+2+\dots+n$
with all the numbers $\{1,2,\dots,n\}$ being close by.
Let $B$ be that position.
\begin{center}
\begin{asy}
size(8cm);
pair P(int x, int y) {
return x*dir(60)+y*dir(0);
}
pair A = P(12,0);
draw(A--P(0,0)--P(0,12)--cycle, gray+0.4);
draw(A--P(11,1)--P(9,1)--P(8,2)--P(5,5)--P(3,7)--P(1,7)--P(0,8),
orange+1.1, EndArrow(TeXHead));
draw(P(0,7)--P(7,0), blue);
draw(P(0,9)--P(3,9), blue);
draw(P(7,0)--P(6,0)--P(5,1)--P(3,1)--P(2,2)--P(0,2),
orange+1.1, EndArrow(TeXHead));
dot("$A$", A, dir(90));
dot("$B$", P(0,8), dir(-90));
dot("$C$", P(7,0), dir(150));
dot("$D$", P(0,2), dir(-90));
dot("$X$", P(0,7), dir(-90));
dot("$Y$", P(0,9), dir(-90));
dot(P(3,9));
\end{asy}
\end{center}
Consider the two left/right neighbors $X$ and $Y$ of the endpoint $B$.
Assume that $B$ is to the right of the midpoint of the bottom side,
and complete the equilateral triangle as shown to an apex $C$.
Consider the lightning strike from $C$ hitting the bottom at $D$.
It travels at least $\left\lfloor n/2-1 \right\rfloor$ steps, by construction.
But the increases must be at least $n+1$, $n+2$, \dots since $1,2,\dots,n$
are close to the $A \to B$ lightning path.
Then the number at $D$ is at least
\[ (n+1) + (n+2) + \dots +
\left( n+\left( \left\lfloor n/2-1 \right\rfloor \right) \right)
> 1 + 2 + \dots + n \]
for $n \ge 2018$, contradiction. |
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 $\sqrt5$.
They stop once either player cannot move.
Find the greatest $K$ such that Amy can ensure that
she places at least $K$ stones. | 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 $\half \cdot 200 = 100$ stones.
Second, we show Ben can prevent Amy from placing more than $100$ stones.
Divide into several $4 \times 4$ squares and then further partition
each $4 \times 4$ squares as shown in the grid below.
\[
\left[
\begin{array}{cccc}
1 & 2 & 3 & 4 \\
3 & 4 & 1 & 2 \\
2 & 1 & 4 & 3 \\
4 & 3 & 2 & 1
\end{array}
\right]
\]
The squares with each label form $4$-cycles by knight jumps.
For each such cycle, whenever Amy plays in the cycle,
Ben plays in the opposite point of the cycle,
preventing Amy from playing any more stones in that original cycle.
Hence Amy can play at most in $1/4$ of the stones, as desired. |
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 $\nu_p(a_n) < \nu_p(a_{n+1})$, then $\nu_p(a_{n+1}) = \nu_p(a_1)$.
\ii If $\nu_p(a_n) = \nu_p(a_{n+1})$, no conclusion.
\ii If $\nu_p(a_n) > \nu_p(a_{n+1})$,
then $\nu_p(a_{n+1}) \ge \nu_p(a_1)$.
\end{itemize}
In other words:
\begin{claim*}
Let $p$ be a prime. Consider the sequence
$\nu_p(a_{N+1})$, $\nu_p(a_{N+2})$, \dots.
Then either:
\begin{itemize}
\ii We have $\nu_p(a_{N+1}) \ge \nu_p(a_{N+2}) \ge \dots$
and so on, i.e.\ the sequence is weakly decreasing immediately; or
\ii For some index $K > N$ we have
$\nu_p(a_K) < \nu_p(a_{K+1}) = \nu_p(a_{K+2}) = \dots = \nu_p(a_1)$,
i.e.\ the sequence ``jumps'' to $\nu_p(a_1)$
at some point and then stays there forever after.
Note this requires $\nu_p(a_1) > 0$.
\end{itemize}
\end{claim*}
A cartoon of the situation is drawn below.
\begin{center}
\begin{asy}
label("$\nu_p(a_1)$", (0,3), dir(180), deepcyan);
draw( (0,3)--(11,3), blue+dotted );
draw( (0,9)--(0,-1)--(12,-1), black, Arrows );
dot( (1,6), red );
dot( (2,5), red );
dot( (3,5), red );
dot( (4,5), red );
dot( (5,4), red );
dot( (6,4), red );
dot( (7,4), red );
dot( (8,3), red );
dot( (1,9), darkred );
dot( (2,7), darkred );
dot( (3,6), darkred );
dot( (4,6), darkred );
dot( (5,6), darkred );
dot( (6,6), darkred );
dot( (7,6), darkred );
dot( (8,6), darkred );
dot( (9,6), darkred );
draw( (1,6)--(2,5)--(4,5)--(5,4)--(7,4)--(8,3), red );
draw( (1,9)--(2,7)--(3,6)--(4,6)--(9,6), darkred );
dot( (1,1), brown );
dot( (2,1), brown );
dot( (3,1), brown );
dot( (4,1), brown );
dot( (5,1), brown );
dot( (6,3), brown );
dot( (6,3), brown );
draw( (1,1)--(5,1)--(6,3), brown );
dot( (1,0), orange );
dot( (2,0), orange );
dot( (3,0), orange );
dot( (4,0), orange );
dot( (5,0), orange );
dot( (6,0), orange );
dot( (7,0), orange );
dot( (8,0), orange );
dot( (9,0), orange );
draw( (1,0)--(9,0), orange );
draw( (1,-0.8)--(1,-1.2) );
draw("$n > N$", (1,-1.2), dir(-90) );
\end{asy}
\end{center}
As only finitely many primes $p$ divide $a_1$,
after some time $\nu_p(a_n)$ is fixed for all such $p \mid a_1$.
Afterwards, the sequence satisfies $a_{n+1} \mid a_n$ for each $n$,
and thus must be eventually constant.
\begin{remark*}
This solution is almost completely $p$-adic,
in the sense that I think a similar result
holds if one replaces $a_n \in \ZZ$
by $a_n \in \ZZ_p$ for any particular prime $p$.
In other words, the primes almost do not talk to each other.
There is one caveat: if $x_n$ is an integer sequence
such that $\nu_p(x_n)$ is eventually constant for each prime
then $x_n$ may not be constant.
For example, take $x_n$ to be the $n$th prime!
That's why in the first claim (applied to co-finitely many of the primes),
we need the stronger non-decreasing condition,
rather than just eventually constant.
\end{remark*}
\begin{remark*}
An alternative approach is to show that, when the fractions $a_n / a_1$
is written in simplest form for $n = N+1, N+2, \dots$,
the numerator and denominator are both weakly decreasing.
Hence it must eventually be constant; in which case it equals $\frac11$.
\end{remark*} |
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-harmonic quadrilateral is determined
up to similarity by its angles.
\end{claim*}
\begin{proof}
Do some inequalities.
\end{proof}
\begin{remark*}
This could be expected by degrees of freedom;
a quadrilateral has four degrees of freedom up to similarity;
the pseudo-harmonic condition is one
while the measures of angles $\angle A$, $\angle B$, $\angle C$, $\angle D$
(summing to $360\dg$) provide three degrees of freedom.
(Note that the point $X$ plays no role in this comment.)
\end{remark*}
Performing an inversion at $X$, one obtains a
second quasi-harmonic quadrilateral
$A^\ast B^\ast C^\ast D^\ast$ which has the same angles
as the original one, $\angle D^\ast = \angle A$,
$\angle A^\ast = \angle B$, and so on.
Thus by the claim we obtain similarity
\[ D^\ast A^\ast B^\ast C^\ast \sim ABCD. \]
If one then maps $D^\ast A^\ast B^\ast C^\ast$,
onto $ABCD$, the image of $X^\ast$
becomes a point isogonally conjugate to $X$.
In other words, $X$ has an isogonal conjugate in $ABCD$.
It is well-known that this is equivalent to
$\angle BXA + \angle DXC = 180\dg$,
for example by inscribing an ellipse with foci $X$ and $X^\ast$.
\paragraph{Second solution: ``rhombus inversion'', by Michael Ren.}
Since
\[ \frac{AB}{AD} = \frac{CB}{CD} \]
and
\[ \frac{BA}{BC} = \frac{DA}{DC} \]
it follows that $B$ and $D$ lie on an Apollonian circle $\omega_{AC}$
through $A$ and $C$,
while $A$ and $C$ lie on an Apollonian circle $\omega_{BD}$
through $B$ and $D$.
We let these two circles intersect at a point $P$ inside $ABCD$.
The main idea is then to
perform an inversion about $P$ with radius $1$.
We obtain:
%\paragraph{Rhombus construction.}
\begin{lemma*}
The image of $ABCD$ is a rhombus.
\end{lemma*}
\begin{proof}
By the inversion distance formula, we have
\[ \frac{1}{A'B'} = \frac{PA}{AB} \cdot PB = \frac{PC}{BC} \cdot PB = \frac{1}{B'C'} \]
and so $A'B' = B'C'$.
In a similar way, we derive $B'C' = C'D' = D'A'$,
so the image is a rhombus as claimed.
\end{proof}
Let us now translate the angle conditions.
We were given that $\dang XAB = \dang XCD$, but
\begin{align*}
\dang XAB &= \dang XAP + \dang PAB = \dang PX'A' + \dang A'B'P \\
\dang XCD &= \dang XCP + \dang PCD = \dang PX'C' + \dang C'D'P
\intertext{so subtracting these gives}
\dang A'X'C' &= \dang A'B'P + \dang PD'C' = \dang (A'B', B'P) + \dang (PD', C'D') \\
&= \dang (A'B', B'P) + \dang (PD', A'B') = \dang D' P B'. \tag{1}
\end{align*}
since $\ol{A'B'} \parallel \ol{C'D'}$.
Similarly, we obtain
\[ \dang B'X'D' = \dang A'PC' \tag{2}. \]
We now translate the desired condition.
Since
\begin{align*}
\dang AXB &= \dang AXP + \dang PXB = \dang PA'X' + \dang X'B'P \\
\dang CXD &= \dang CXP + \dang PXD = \dang PC'X' + \dang X'DP'
\end{align*}
we compute
\begin{align*}
\dang AXB + \dang CXD &= (\dang PA'X' + \dang X'B'P) + (\dang PC'X' + \dang X'D'P) \\
&= -\left[ \left( \dang A'X'P + \dang X'PA' \right)
+ \left( \dang PX'B' + \dang B'PX' \right) \right] \\
&\quad- \left[ \left( \dang C'X'P + \dang X'PC' \right)
+ \left( \dang PX'D' + \dang D'PX' \right) \right] \\
&= \left[ \dang PX'A' + \dang BX'P + \dang PX'C' + \dang D'X'P \right] \\
&\quad+ \left[ \dang A'PX' + \dang X'PB' + \dang C'PX' + \dang X'PD' \right] \\
&= \dang A'PB' + \dang C'PD' + \dang B'X'C + \dang D'X'A
\end{align*}
and we wish to show this is equal to zero, i.e.\
the desired becomes
\[ \dang A'PB' + \dang C'PD' + \dang B'X'C + \dang D'X'A = 0. \tag{3} \]
In other words, the problem is to show (1) and (2) implies (3).
Henceforth drop apostrophes.
Here is the inverted diagram (with apostrophes dropped).
\begin{center}
\begin{asy}
size(12cm);
pair A = (0,3);
pair B = (-8,0);
pair C = (0,-3);
pair D = (8,0);
pair X = (-5.6,7.8);
pair Y = reflect(circumcenter(A,C,X), circumcenter(D,B,X))*X;
pair Q = IP(circumcircle(A, Y, D), circumcircle(B, Y, C));
pair P = conj(Q);
filldraw(A--B--C--D--cycle, opacity(0.1)+lightred, red);
draw(A--C, red);
draw(B--D, red);
draw(B--X--D, lightgreen);
draw(A--P--C, lightgreen);
draw(C--X--A, lightblue);
draw(B--P--D, lightblue);
draw(circumcircle(A, X, C), heavycyan);
draw(circumcircle(B, X, D), heavycyan);
draw(circumcircle(B, Q, C), lightcyan);
draw(circumcircle(A, Q, D), lightcyan);
draw(Q--P, red);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
dot("$Q$", Q, dir(Q));
dot("$P$", P, dir(P));
/* TSQ Source:
A = (0,3)
B = (-8,0)
C = (0,-3)
D = (8,0)
X = (-5.6,7.8)
Y = OP circumcircle A X C circumcircle B X D
Q = IP circumcircle A Y D circumcircle B Y C
P = conj(Q)
A--B--C--D--cycle 0.1 lightred / red
A--C red
B--D red
B--X--D lightgreen
A--P--C lightgreen
C--X--A lightblue
B--P--D lightblue
circumcircle A X C heavycyan
circumcircle B X D heavycyan
circumcircle B Q C lightcyan
circumcircle A Q D lightcyan
Q--P red
*/
\end{asy}
\end{center}
Let $Q$ denote the reflection of $P$
and let $Y$ denote the second intersection of $(BQC)$ and $(AQD)$.
Then
\begin{align*}
-\dang AXC &= -\dang DPB = \dang BQD = \dang BQY + \dang YQD = \dang BCY + \dang YAD \\
&= \dang(BC,CY) + \dang(YA,AD) = \dang YCA = -\dang AYC.
\end{align*}
% again using $\ol{AB} \parallel \ol{CD}$.
Hence $XACY$ is concyclic; similarly $XBDY$ is concyclic.
\begin{claim*}
$X \neq Y$.
\end{claim*}
\begin{proof}
To see this: Work pre-inversion assuming $AB < AC$.
Then $Q$ was the center of $\omega_{BD}$.
If $T$ was the second intersection of $BA$ with $(QBC)$,
then $QB = QD = QT = \sqrt{QA \cdot QC}$, by shooting lemma.
Since $\angle BAD < 180\dg$,
it follows $(QBCY)$ encloses $ABCD$ (pre-inversion).
(This part is where the hypothesis that
$ABCD$ is convex with $X$ inside is used.)
\end{proof}
Finally, we do an angle chase to finish:
\begin{align*}
\dang DXA &= \dang DXY + \dang YXA = \dang DBY + \dang YCA \\
&= \dang (DB, YB) + \dang (CY, CA) = \dang CYB + 90\dg \\
&= \dang CQB + 90\dg = -\dang APB + 90\dg \tag{4}.
\end{align*}
Similarly,
\[ \dang BXC = \dang DPC + 90\dg. \tag{5} \]
%% \dang AXB = \dang DYC + 90\dg, \quad
%% \dang CXD = \dang BYC + 90\dg.
Summing (4) and (5) gives (3).
% $\dang BXC + \dang DXA = (\dang BPA + 90\dg) + (\dang DPC + 90\dg) = -(\dang APB + \dang CPD)$.
\begin{remark*}
A difficult part of the problem in many solutions
is that the conclusion is false in the directed sense,
if the point $X$ is allowed to lie outside the quadrilateral.
We are saved in the first solution because the equivalence
of the isogonal conjugation requires $X$ inside the quadrilateral.
On the other hand, in the second solution,
the issue appears in the presence of the second point $Y$.
\end{remark*} |
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, $P(a,b)$ and $P(0,a+b)$ give
\begin{align*}
f(f(a+b)) &= f(2a) + 2f(b) = f(0) + 2f(a+b) \\
\implies [2f(a) - f(0)] + 2f(b) &= f(0) + 2f(a+b) \\
\implies \left( f(a)-f(0) \right) + \left( f(b)-f(0) \right)
&= \left( f(a+b)-f(0) \right).
\end{align*}
Thus the map $x \mapsto f(x) - f(0)$ is additive,
therefore linear.
\begin{remark*}
The same proof works on the functional equation
\[ f(2a) + 2f(b) = g(a+b) \]
where $g$ is an arbitrary function (it implies that $f$ is linear).
\end{remark*} |
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 $QA_1$ beyond $A_1$
such that $\angle CQ_1Q = \angle CBA$.
Prove that points $P_1$, $Q_1$, $P$, $Q$ are cyclic. | 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 $T = \ol{PX} \cap \ol{QY}$ (possibly at infinity),
it suffices to show that the
radical axis of $\triangle CXA$ and $\triangle CYB$ passes through $T$,
because that would imply $P_1XYQ_1$ is cyclic
(by power of a point when $T$ is Euclidean,
and because it is an isosceles trapezoid if $T$ is at infinity).
\begin{center}
\begin{asy}
pair C = dir(110);
pair A = dir(210);
pair B = dir(330);
pair A_1 = 0.45*C+0.55*B;
pair P = 0.53*A+0.47*A_1;
pair B_1 = midpoint(A--C);
pair Q = extension(B, B_1, P, P+A-B);
pair T = extension(A_1, Q, B_1, P);
pair X = extension(A, B, P, T);
pair Y = extension(A, B, Q, T);
filldraw(A--B--C--cycle, opacity(0.1)+lightcyan, lightblue);
filldraw(circumcircle(A, C, X), opacity(0.05)+yellow, lightred);
filldraw(circumcircle(B, C, Y), opacity(0.05)+yellow, lightred);
draw(A--A_1, lightblue);
draw(B--B_1, lightblue);
draw(P--Q, blue);
draw(C--T, deepgreen);
pair P_1 = -X+2*foot(circumcenter(A, X, C), P, X);
pair Q_1 = -Y+2*foot(circumcenter(B, Y, C), Q, Y);
draw(P_1--T--Q_1, orange);
dot("$C$", C, dir(C));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$A_1$", A_1, dir(A_1));
dot("$P$", P, dir(160));
dot("$B_1$", B_1, dir(B_1));
dot("$Q$", Q, dir(20));
dot("$T$", T, dir(T));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
dot("$P_1$", P_1, dir(P_1));
dot("$Q_1$", Q_1, dir(Q_1));
/* TSQ Source:
C = dir 110
A = dir 210
B = dir 330
A_1 = 0.45*C+0.55*B
P = 0.53*A+0.47*A_1 R160
B_1 = midpoint A--C
Q = extension B B_1 P P+A-B R20
T = extension A_1 Q B_1 P
X = extension A B P T
Y = extension A B Q T
A--B--C--cycle 0.1 lightcyan / lightblue
circumcircle A C X 0.05 yellow / lightred
circumcircle B C Y 0.05 yellow / lightred
A--A_1 lightblue
B--B_1 lightblue
P--Q blue
C--T deepgreen
P_1 = -X+2*foot circumcenter A X C P X
Q_1 = -Y+2*foot circumcenter B Y C Q Y
P_1--T--Q_1 orange
*/
\end{asy}
\end{center}
To this end we use barycentric coordinates on $\triangle ABC$.
We begin by writing
\[ P = (u+t : s : r), \quad Q = (t : u+s : r) \]
from which it follows that
$A_1 = (0 : s : r)$ and $B_1 = (t : 0 : r)$.
Next, compute $X = \left(
\det \left[ \begin{smallmatrix} u+t & r \\ t & r \end{smallmatrix} \right]
: \det \left[ \begin{smallmatrix} s & r \\ 0 & r \end{smallmatrix} \right]
: 0 \right) = (u : s : 0)$.
Similarly, $Y = (t : u : 0)$.
So we have computed all points.
\begin{claim*}
Line $B_1X$ has equation
$-rs \cdot x + ru \cdot y + st \cdot z = 0$,
while line $C_1 Y$ has equation
$ru \cdot x - rt \cdot y + st \cdot z = 0$.
\end{claim*}
\begin{proof}
Line $B_1X$ is $0 = \det(B_1, X, -)
= \det \left[ \begin{smallmatrix}
t & 0 & r \\
u & s & 0 \\
x & y & z
\end{smallmatrix} \right]$.
Line $C_1Y$ is analogous.
\end{proof}
\begin{claim*}
The radical axis $(u+t) y - (u+s) x = 0$.
\end{claim*}
\begin{proof}
Circle $(AXC)$ is given by
$-a^2yz - b^2zx - c^2xy + (x+y+z) \cdot \frac{c^2 \cdot u}{u+s} y = 0$.
Similarly, circle $(BYC)$ has equation
$-a^2yz - b^2zx - c^2xy + (x+y+z) \cdot \frac{c^2 \cdot u}{u+t} x = 0$.
Subtracting gives the radical axis.
\end{proof}
Finally, to see these three lines are concurrent, we now compute
\begin{align*}
\det \begin{bmatrix}
-rs & ru & st \\
ru & -rt & st \\
-(u+s) & u+t & 0
\end{bmatrix}
&= rst \left[ [u(u+t)-t(u+s)] + [s(u+t)-u(u+s)] \right] \\
&= rst \left[ (u^2-st) + (st-u^2) \right] = 0.
\end{align*}
This completes the proof.
\paragraph{Second official solution by tricky angle chasing.}
Let lines $AA_1$ and $BB_1$ meet at the circumcircle of $\triangle ABC$
again at points $A_2$ and $B_2$.
By Reim's theorem, $PQA_2B_2$ are cyclic.
\begin{center}
\begin{asy}
pair C = dir(110);
pair A = dir(210);
pair B = dir(330);
pair A_1 = 0.45*C+0.55*B;
pair P = 0.53*A+0.47*A_1;
pair B_1 = midpoint(A--C);
pair Q = extension(B, B_1, P, P+A-B);
pair T = extension(A_1, Q, B_1, P);
pair X = extension(A, B, P, T);
pair Y = extension(A, B, Q, T);
filldraw(A--B--C--cycle, opacity(0.1)+lightcyan, lightblue);
pair P_1 = -X+2*foot(circumcenter(A, X, C), P, X);
pair Q_1 = -Y+2*foot(circumcenter(B, Y, C), Q, Y);
pair A_2 = -A+2*foot(origin, A, P);
pair B_2 = -B+2*foot(origin, B, Q);
filldraw(unitcircle, opacity(0.1)+lightcyan, lightblue);
draw(A--A_2, lightblue);
draw(B--B_2, lightblue);
filldraw(circumcircle(C, B_1, B_2), opacity(0.05)+orange, red);
filldraw(circumcircle(C, A_1, A_2), opacity(0.05)+orange, red);
draw(P_1--P--Q--Q_1, blue);
filldraw(circumcircle(P, Q, Q_1), opacity(0.1)+yellow, deepgreen+dashed);
dot("$C$", C, dir(C));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$A_1$", A_1, dir(A_1));
dot("$P$", P, dir(270));
dot("$B_1$", B_1, dir(B_1));
dot("$Q$", Q, dir(270));
dot("$P_1$", P_1, dir(P_1));
dot("$Q_1$", Q_1, dir(Q_1));
dot("$A_2$", A_2, dir(A_2));
dot("$B_2$", B_2, dir(B_2));
/* TSQ Source:
C = dir 110
A = dir 210
B = dir 330
A_1 = 0.45*C+0.55*B
P = 0.53*A+0.47*A_1 R270
B_1 = midpoint A--C
Q = extension B B_1 P P+A-B R270
T := extension A_1 Q B_1 P
X := extension A B P T
Y := extension A B Q T
A--B--C--cycle 0.1 lightcyan / lightblue
P_1 = -X+2*foot circumcenter A X C P X
Q_1 = -Y+2*foot circumcenter B Y C Q Y
A_2 = -A+2*foot origin A P
B_2 = -B+2*foot origin B Q
unitcircle 0.1 lightcyan / lightblue
A--A_2 lightblue
B--B_2 lightblue
circumcircle C B_1 B_2 0.05 orange / red
circumcircle C A_1 A_2 0.05 orange / red
P_1--P--Q--Q_1 blue
circumcircle P Q Q_1 0.1 yellow / deepgreen dashed
*/
\end{asy}
\end{center}
\begin{claim*}
The points $P$, $Q$, $A_2$, $Q_1$ are cyclic.
Similarly the points $P$, $Q$, $B_2$, $P_1$ are cyclic.
\end{claim*}
\begin{proof}
Note that $CA_1A_2Q_1$ is cyclic since
$\dang CQ_1A_1 = \dang CQ_1Q = \dang CBA = \dang CA_2A = \dang CA_2A_1$.
Then $\dang QQ_1A_2 = \dang A_1Q_1A_2 = \dang A_1 C A_2
= \dang B C A_2 = \dang B A A_2 = \dang Q P A_2$.
\end{proof}
This claim obviously solves the problem. |
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 $AC$ are no longer friends.
Initially, $1009$ users have $1010$ friends and $1010$ users have $1009$ friends.
Prove that the administrator can make a sequence of operations
such that all users have at most $1$ friend. | 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 the smallest cycle $C$; by hypothesis $C \neq G$.
If $C$ is not a triangle (equivalently, $G$ is triangle-free),
then let $b \notin C$ be a vertex adjacent to $C$, say at $a$.
Take a vertex $c$ of the cycle adjacent to $a$ (hence not to $b$).
Then we can toggle $abc$.
Now assume there exists a triangle; let $K$ be the maximal clique.
By hypothesis, $K \neq G$.
We take an edge $e = ab$ dangling off the clique,
with $a \in K$ and $b \notin K$.
Note some vertex $c$ of $K$ is not adjacent to $b$; now toggle $abc$.
\end{proof}
Back to the original problem;
let $G_{\text{imo}}$ be the given graph.
The point is that we can apply toggles (by the claim) repeatedly,
without disconnecting the graph, until we get a tree.
This is because
\begin{itemize}
\ii $G_{\text{imo}}$ is connected,
since any two vertices which are not adjacent
have a common neighbor by pigeonhole
($1009 + 1009 + 2 > 2019$).
\ii $G_{\text{imo}}$ cannot become a cycle,
because it initially has an odd-degree vertex,
and toggles preserve parity of degree!
\ii $G_{\text{imo}}$ is obviously not a clique initially
(and hence not afterwards).
\end{itemize}
So, we can eventually get $G_{\text{imo}}$ to be a tree.
Once $G_{\text{imo}}$ is a tree the problem follows by repeatedly applying
toggles arbitrarily until no more are possible;
the graph (although now disconnected) remains acyclic
(in particular having no triangles)
and therefore can only terminate in the desired situation.
\begin{remark*}
The above proof in fact shows the following better result:
\begin{quote}
The task is possible if and only if
$G_{\text{imo}}$ is a connected graph which is not a clique
and has any vertex of odd degree.
\end{quote}
The ``only if'' follows from the observation
that toggles preserve parity of degree.
Thus the given condition about the degrees of vertices
being $1009$ and $1010$ is largely a red herring;
it's a somewhat strange way of masking the correct and more natural
both-sufficient-and-necessary condition.
\end{remark*} |
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_2(k!) &= \nu_2(A) = 1 + 2 + \dots + (n-1) = \frac{n(n-1)}{2} \\
\left\lfloor \frac k3 \right\rfloor
\le \nu_3(k!) &= \nu_3(A) = \left\lfloor \frac n2 \right\rfloor
+ \left\lfloor \frac n6 \right\rfloor + \dots < \frac 34n.
\end{align*}
where the first inequality follows by Legendre's formula $\nu_2(k!) = k - s_2(k)$.
In this way, we get
\[ \frac 94 n + 3 > k > \frac{n(n-1)}{2} \]
which means $n \le 6$; a manual check then shows the
solutions we claimed earlier are the only ones.
\begin{remark*}
An amusing corollary of the problem pointed out in the shortlist
is that the symmetric group $S_k$ cannot be isomorphic to the group $\GL_n(\FF_2)$
unless $(n,k) = (1,1)$ or $(n,k) = (2,3)$, which indeed produce isomorphisms.
\end{remark*} |
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
$THT \to HHT \to HTT \to TTT$, which takes three steps.)
Prove the process will always terminate, and determine the average number of steps
this takes over all $2^n$ configurations. | 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 binary strings as a sequence of $n$ symbols.
The main claim is that $G_n$
can be described explicitly in terms of $G_{n-1}$:
\begin{itemize}
\ii We take two copies $X$ and $Y$ of $G_{n-1}$.
\ii In $X$, we take each string of length $n-1$
and just append a $0$ to it. In symbols,
we replace $s_1 \dots s_{n-1} \mapsto s_1 \dots s_{n-1} 0$.
\ii In $Y$, we toggle every bit, then reverse the order,
and then append a $1$ to it.
In symbols, we replace
$s_1 \dots s_{n-1} \mapsto \ol s_{n-1} \ol s_{n-2} \dots \ol s_{1} 1$.
\ii Finally, we add one new edge from $Y$ to $X$ by
$11 \dots 1 \to 11\dots110$.
\end{itemize}
An illustration of $G_4$ is given below.
\begin{center}
\begin{asy}
unitsize(0.8cm);
label("$0000$", (0, 0), red);
label("$1000$", (0, 1), red);
label("$1100$", (0, 2), red);
label("$1110$", (0, 3), red);
label("$0100$", (2, 2), red);
label("$1010$", (2, 3), red);
label("$0010$", (4, 3), red);
label("$0110$", (6, 3), red);
label("$1111$", (0, 4), blue);
label("$1101$", (0, 5), blue);
label("$1001$", (0, 6), blue);
label("$0001$", (0, 7), blue);
label("$1011$", (2, 6), blue);
label("$0101$", (2, 7), blue);
label("$0111$", (4, 7), blue);
label("$0011$", (6, 7), blue);
label(scale(1.5)*"$\downarrow$", (0, 0.5), red);
label(scale(1.5)*"$\downarrow$", (0, 1.5), red);
label(scale(1.5)*"$\downarrow$", (0, 2.5), red);
label(scale(1.2)*"$\leftarrow$", (1, 2), red);
label(scale(1.2)*"$\leftarrow$", (1, 3), red);
label(scale(1.2)*"$\leftarrow$", (3, 3), red);
label(scale(1.2)*"$\leftarrow$", (5, 3), red);
label(scale(1.5)*"$\downarrow$", (0, 4.5), blue);
label(scale(1.5)*"$\downarrow$", (0, 5.5), blue);
label(scale(1.5)*"$\downarrow$", (0, 6.5), blue);
label(scale(1.2)*"$\leftarrow$", (1, 6), blue);
label(scale(1.2)*"$\leftarrow$", (1, 7), blue);
label(scale(1.2)*"$\leftarrow$", (3, 7), blue);
label(scale(1.2)*"$\leftarrow$", (5, 7), blue);
label(scale(1.7)*"$\Downarrow$", (0, 3.5), deepgreen);
\end{asy}
\end{center}
To prove this claim, we need only show
the arrows of this directed graph remain valid.
The graph $X$ is correct as a subgraph of $G_n$,
since the extra $0$ makes no difference.
As for $Y$, note that if $s = s_1 \dots s_{n-1}$ had $k$ ones,
then the modified string has $(n-1-k)+1 = n-k$ ones, ergo
$\ol s_{n-1} \dots \ol s_1 1
\mapsto \ol s_{n-1} \dots \ol s_{k+1} s_k \ol s_{k-1} \dots \ol s_1 1$
which is what we wanted.
Finally, the one edge from $Y$ to $X$ is obviously correct.
To finish, let $E_n$ denote the desired expected value.
Since $1 \dots 1$ takes $n$ steps to finish we have
\[ E_n = \half \left[ E_{n-1} + (E_{n-1}+n) \right] \]
based on cases on whether the chosen string is in $X$ or $Y$ or not.
By induction, we have $E_n = \half (1 + \dots + n) = \frac14 n(n+1)$,
as desired.
\begin{remark*}
Actually, the following is true:
if the indices of the $1$'s are $1 \le i_1 < \dots < i_\ell \le n$,
then the number of operations required is
\[ 2(i_1 + \dots + i_\ell) - \ell^2. \]
This problem also has an interpretation as a Turing machine:
the head starts at a position on the tape (the binary string).
If it sees a $1$, it changes the cell to a $0$ and moves left;
if it sees a $0$, it changes the cell to a $1$ and moves right.
\end{remark*} |
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 $R$).
Suppose the circumcircles of $\triangle PCE$ and $\triangle PBF$ meet again at $Q$ (other than $P$).
Prove that lines $DI$ and $PQ$ meet on the external $\angle A$-bisector.
\end{enumerate} | 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)+lightcyan, blue);
pair A = 2*E*F/(E+F);
pair B = 2*F*D/(F+D);
pair C = 2*D*E/(D+E);
draw(A--B--C--cycle, blue);
pair R = -E*F/D;
pair P = -R+2*foot(I, A, R);
filldraw(circumcircle(P, C, E), opacity(0.1)+lightcyan, deepgreen);
filldraw(circumcircle(P, B, F), opacity(0.1)+lightcyan, deepgreen);
pair Q = -P+2*foot(P, circumcenter(C, P, E), circumcenter(B, P, F));
pair T = extension(P, Q, D, I);
draw(A--T, deepcyan);
draw(D--T--P, deepcyan);
draw(D--R, lightblue);
draw(A--P, lightblue);
dot("$I$", I, dir(45));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$R$", R, dir(R));
dot("$P$", P, dir(200));
dot("$Q$", Q, dir(70));
dot("$T$", T, dir(T));
/* TSQ Source:
I = origin R45
D = dir 110
E = dir 210
F = dir 330
D--E--F--cycle blue
unitcircle 0.1 lightcyan / blue
A = 2*E*F/(E+F)
B = 2*F*D/(F+D)
C = 2*D*E/(D+E)
A--B--C--cycle blue
R = -E*F/D
P = -R+2*foot I A R R200
circumcircle P C E 0.1 lightcyan / deepgreen
circumcircle P B F 0.1 lightcyan / deepgreen
Q = -P+2*foot P circumcenter C P E circumcenter B P F R70
T = extension P Q D I
A--T deepcyan
D--T--P deepcyan
D--R lightblue
A--P lightblue
*/
\end{asy}
\end{center}
Then $A = \frac{2yz}{y+z}$,
$R = \frac{-yz}{x}$ and so
\[ P = \frac{A-R}{1-R\ol{A}}
= \frac{\frac{2yz}{y+z} + \frac{yz}{x}}
{1 + \frac{yz}{x} \cdot \frac{2}{y+z}}
= \frac{yz(2x+y+z)}{2yz+x(y+z)}. \]
We now compute
\begin{align*}
O_B &= \det \begin{bmatrix}
P & P \ol P & 1 \\
F & F \ol F & 1 \\
B & B \ol B & 1
\end{bmatrix}
\div \det \begin{bmatrix}
P & \ol P & 1 \\
F & \ol F & 1 \\
B & \ol B & 1
\end{bmatrix}
= \det \begin{bmatrix}
P & 1 & 1 \\
z & 1 & 1 \\
\frac{2xz}{x+z} & \frac{4xz}{(x+z)^2} & 1
\end{bmatrix}
\div \det \begin{bmatrix}
P & 1/P & 1 \\
z & 1/z & 1 \\
\frac{2xz}{x+z} & \frac{2}{x+z} & 1
\end{bmatrix} \\
&= \frac{1}{x+z} \det \begin{bmatrix}
P & 0 & 1 \\
z & 0 & 1 \\
2xz(x+z) & -(x-z)^2 & (x+z)^2
\end{bmatrix}
\div \det \begin{bmatrix}
P & 1/P & 1 \\
z & 1/z & 1 \\
2xz & 2 & x+z
\end{bmatrix} \\
&= \frac{(x-z)^2}{x+z} \cdot \frac{P-z}{(x+z)(P/z-z/P)+2z-2x + \frac{2xz}{P}-2P} \\
&= \frac{(x-z)^2}{x+z} \cdot \frac{P-z}{
(\frac xz-1) P - 2(x-z) + (xz-z^2) \frac 1P } \\
&= \frac{x-z}{x+z} \cdot \frac{P-z}{P/z + z/P - 2}
= \frac{x-z}{x+z} \cdot \frac{P-z}{\frac{(P-z)^2}{Pz}}
= \frac{x-z}{x+z} \cdot \frac{1}{\frac 1z - \frac 1P} \\
&= \frac{x-z}{x+z} \cdot \frac{y(2x+y+z)}{y(2x+y+z) - (2yz+xy+xz)}
= \frac{x-z}{x+z} \cdot \frac{yz(2x+y+z)}{xy+y^2-yz-xz} \\
&= \frac{x-z}{x+z} \cdot \frac{yz(2x+y+z)}{(y-z)(x+y)}.
\end{align*}
Similarly
\[ O_C = \frac{x-y}{x+y} \cdot \frac{yz(2x+y+z)}{(z-y)(x+z)}. \]
Therefore, subtraction gives
\[ O_B-O_C
=
\frac{yz(2x+y+z)}{(x+y)(x+z)(y-z)}
\left[ (x-z) + (x-y) \right]
= \frac{yz(2x+y+z)(2x-y-z)}{(x+y)(x+z)(z-y)}.
\]
It remains to compute $T$.
Since $T \in \ol{ID}$ we have $t/x \in \RR$
so $\ol t = t/x^2$.
Also,
\begin{align*}
\frac{t - \frac{2yz}{y+z}}{y+z} \in i \RR
\implies 0 &= \frac{t-\frac{2yz}{y+z}}{y+z}
+ \frac{\frac{t}{x^2}-\frac{2}{y+z}}{\frac1y+\frac1z} \\
&= \frac{1+\frac{yz}{x^2}}{y+z} t - \frac{2yz}{(y+z)^2} - \frac{2yz}{(y+z)^2} \\
\implies t &= \frac{x^2}{x^2+yz} \cdot \frac{4yz}{y+z}
\end{align*}
Thus
\begin{align*}
P-T &= \frac{yz(2x+y+z)}{2yz+x(y+z)} - \frac{4x^2yz}{(x^2+yz)(y+z)} \\
&= yz \cdot \frac{(2x+y+z)(x^2+yz)(y+z) - 4x^2(2yz+xy+xz)}
{(y+z)(x^2+yz)(2yz+xy+xz)} \\
&= -yz \cdot \frac{(2x-y-z)(x^2y+x^2z+4xyz+y^2z+yz^2)}
{(y+z)(x^2+yz)(2yz+xy+xz)}.
\end{align*}
This gives $\ol{PT} \perp \ol{O_B O_C}$ as needed.
\paragraph{Second solution by tethered moving points, with optimization (Evan Chen).}
Fix $\triangle DEF$ and $\omega$, with $B = \ol{DD} \cap \ol{FF}$
and $C = \ol{DD} \cap \ol{EE}$.
We consider a variable point $M$ on $\omega$
and let $X$, $Y$ be on $\ol{EF}$ with
$\ol{CY} \cap \parallel \ol{ME}$, $\ol{BX} \cap \parallel \ol{MF}$.
We define $W = \ol{CY} \cap \ol{BX}$.
Also, let line $MW$ meet $\omega$ again at $V$.
\begin{center}
\begin{asy}
pair D = dir(105);
pair E = dir(200);
pair F = dir(340);
pair M = dir(275);
pair N = extension(D, M, E, F);
pair Rp = -D;
pair B = 2*D*F/(D+F);
pair C = 2*D*E/(D+E);
pair X = extension(E, F, B, B+M-F);
pair Y = extension(E, F, C, C+M-E);
pair W = extension(B, X, C, Y);
pair V = extension(N, Rp, M, W);
filldraw(unitcircle, opacity(0.1)+lightred, red);
draw(D--E--F--cycle, red);
draw(E--C--D--B--F, orange);
draw(V--M);
draw(X--B, deepgreen);
draw(C--Y, deepgreen);
draw(E--M--F, heavygreen);
draw(D--M, lightred);
draw(V--M, blue);
filldraw(circumcircle(C, W, V), opacity(0.1)+lightcyan, blue);
filldraw(circumcircle(B, W, V), opacity(0.1)+lightcyan, blue);
draw(V--Rp, gray+dotted);
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$M$", M, dir(M));
dot("$N$", N, dir(N));
dot("$R'$", Rp, dir(Rp));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$X$", X, dir(270));
dot("$Y$", Y, dir(270));
dot("$W$", W, dir(W));
dot("$V$", V, dir(200));
/* TSQ Source:
D = dir 105
E = dir 200
F = dir 340
M = dir 275
N = extension D M E F
R' = -D
B = 2*D*F/(D+F)
C = 2*D*E/(D+E)
X = extension E F B B+M-F R270
Y = extension E F C C+M-E R270
W = extension B X C Y
V = extension N Rp M W R200
unitcircle 0.1 lightred / red
D--E--F--cycle red
E--C--D--B--F orange
V--M
X--B deepgreen
C--Y deepgreen
E--M--F heavygreen
D--M lightred
V--M blue
circumcircle C W V 0.1 lightcyan / blue
circumcircle B W V 0.1 lightcyan / blue
V--Rp gray dotted
*/
\end{asy}
\begin{asy}
pair D = dir(115);
pair E = dir(210);
pair F = dir(330);
pair N = midpoint(E--F);
pair M = -D+2*foot(origin, D, N);
pair Rp = -D;
pair B = 2*D*F/(D+F);
pair C = 2*D*E/(D+E);
pair X = extension(E, F, B, B+M-F);
pair Y = extension(E, F, C, C+M-E);
pair Q = extension(B, X, C, Y);
pair P = extension(N, Rp, M, Q);
pair R = E*F/Rp;
pair Mp = E*F/M;
pair A = 2*E*F/(E+F);
pair Z = extension(P, M, D, Rp);
filldraw(unitcircle, opacity(0.1)+lightred, red);
draw(D--E--F--cycle, red);
draw(E--A--F, orange);
draw(P--M);
draw(D--M, lightred);
// E--D--C orange
draw(P--Rp, deepgreen);
draw(A--Z, lightblue);
draw(P--Z--D, blue);
draw(P--A--D, deepcyan);
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$N$", N, dir(N));
dot("$M$", M, dir(M));
dot("$R'$", Rp, dir(Rp));
dot("$Q$/$W$", Q, dir(Q));
dot("$P$/$V$", P, dir(200));
dot("$R$", R, dir(R));
dot("$M'$", Mp, dir(Mp));
dot("$A$", A, dir(A));
/* TSQ Source:
D = dir 115
E = dir 210
F = dir 330
N = midpoint E--F
M = -D+2*foot origin D N
R' = -D
B := 2*D*F/(D+F)
C := 2*D*E/(D+E)
X := extension E F B B+M-F R270
Y := extension E F C C+M-E R270
Q = extension B X C Y
P = extension N Rp M Q R200
R = E*F/Rp
M' = E*F/M
A = 2*E*F/(E+F)
Z := extension P M D Rp
unitcircle 0.1 lightred / red
D--E--F--cycle red
E--A--F orange
P--M
D--M lightred
// E--D--C orange
P--Rp deepgreen
A--Z lightblue
P--Z--D blue
P--A--D deepcyan
*/
\end{asy}
\end{center}
\begin{claim*}
[Angle chasing]
Pentagons $CVWXE$ and $BVWYF$ are cyclic.
\end{claim*}
\begin{proof}
By $\dang EVW = \dang EVM = \dang EFM = \dang CEM = \dang ECW$
and $\dang EXW = \dang EFM = \dang CEM = \dang ECW$.
\end{proof}
Let $N = \ol{DM} \cap \ol{EF}$ and $R'$ be the $D$-antipode on $\omega$.
\begin{claim*}
[Black magic]
The points $V$, $N$, $R'$ are collinear.
\end{claim*}
\begin{proof}
We use tethered moving points with $\triangle DEF$ fixed.
Obviously the map $\omega \mapsto \ol{EF} \mapsto \omega$ by $M \mapsto N \mapsto \ol{R'N} \cap \omega$ is projective.
Also, the map $\omega \mapsto \ol{EF} \mapsto \omega$ by $M \mapsto X \mapsto V$ is also projective
(the first by projection to the line at infinity at back;
the second say by inversion at $E$).
So it suffices to check for three points.
When $M=E$ we get $N=E$ so $\ol{R'N} \cap \omega = E$, while $W=E$ and thus $V=E$.
The case $M=F$ is similar.
Finally, if $M = R'$, then $W$ is the center of $\omega$ and so $V = \ol{R'N} \cap \ol{EF} = D$.
\end{proof}
We now address the original problem by specializing $M$:
choose it so that $N$ is the midpoint of $\ol{EF}$.
Let $M' = \ol{DA} \cap (DEF)$.
\begin{claim*}
After this specialization, $V=P$ and $W=Q$.
\end{claim*}
\begin{proof}
Thus $\ol{RR'}$ and $\ol{MM'}$ are parallel to $\ol{EF}$.
From $(EF;PR) = -1 = (EF;N\infty) \overset{R'}{=} (EF;NV)$,
we derive that $P=V$ and $Q=R$, proving (i).
\end{proof}
Finally, the concurrence requested follows by Pascal theorem on $M'MDR'PR$.
\paragraph{Third solution by power of a point linearity (Luke Robitaille).}
Let us define
\[ f(\bullet) = \opname{Pow}(\bullet, (CPE))
- \opname{Pow}(\bullet, (BPF)) \]
which is a linear function from the plane to $\RR$.
Define $W = \ol{BA} \cap \ol{PE}$, $V = \ol{AC} \cap \ol{PF}$.
Also, let $W_1 = \ol{ER} \cap \ol{AB}$, $V_1 = \ol{FR} \cap \ol{AC}$.
Note that
\[ -1 = (PR;EF) \overset{E}{=} (WA; W_1F) \]
and similarly $(VA; V_1E) = -1$.
\begin{claim*}
We have
\begin{align*}
f(F) &= \frac{|EF| \cdot (s-c) \sin C/2}{\sin B/2} \\
f(E) &= -\frac{|EF| \cdot (s-b) \sin B/2}{\sin C/2}.
\end{align*}
\end{claim*}
\begin{proof}
We have
\[ f(W) = WF^2 - WB \cdot WF = WF \cdot BF \]
where lengths are directed.
Next,
\begin{align*}
f(F) &= \frac{AF \cdot f(W) + FW \cdot f(A)}{AW} \\
&= \frac{AF \cdot WF \cdot BF + FW \cdot
\left( AE \cdot AC - AF \cdot AB \right)}{AW} \\
&= \frac{WF(AF \cdot BF + AF \cdot AB) + FW \cdot AE \cdot AC}{AW} \\
&= \frac{WF \cdot AF^2 - WF \cdot AE \cdot AC}{AW}
= \frac{WF}{AW} \cdot (AE^2 - AE \cdot AC) \\
&= \frac{WF}{AW} \cdot AE \cdot CE
= -\frac{W_1F}{AW_1} \cdot AE \cdot CE.
\end{align*}
Since $\triangle DEF$ is acute,
the point $R$ lies inside $\triangle AEF$.
Thus $W_1$ lies inside segment $\ol{AF}$
and the ratio $\frac{W_1F}{AW_1}$ is positive.
We now determine its value: by the ratio lemma
\begin{align*}
\frac{|W_1F|}{|AW_1|}
&= \frac{|EF| \sin \angle W_1 E F}{|AE| \sin \angle A E W_1} \\
&= \frac{|EF| \sin \angle REF}{|AE| \sin \angle AER} \\
&= \frac{|EF| \sin \angle RDF}{|AE| \sin \angle EDR} \\
&= \frac{|EF| \sin C/2}{|AE| \sin B/2}.
\end{align*}
Also, we have $AE \cdot CE < 0$ since $E$ lies inside $\ol{AC}$.
Hence
\[ f(F) = -\frac{|EF| \sin C/2}{|AE| \sin B/2}.
\cdot AE \cdot CE
= |EF| \cdot \frac{|CE| \sin B/2}{\sin C/2}
= |EF| \cdot \frac{(s-c) \sin B/2}{\sin C/2}. \]
The calculation for $f(E)$ is similar,
(noting the sign flips since $f$ is anti-symmetric
in terms of $B$ and $C$).
\end{proof}
Let $Z \in \ol{DI}$ with $\angle ZAI = 90\dg$ be the point requested in the problem now.
Our goal is to show $f(Z) = 0$.
We assume WLOG that $AB < AC$, so $\frac{ZA}{EF} > 0$.
Then
\begin{align*}
|ZA| &= |AI| \cdot \tan \angle AIZ \\
&= |AI| \cdot \tan \angle(\ol{AI}, \ol{DI}) \\
&= \frac{s-a}{\cos A/2} \cdot \tan (\ol{BC}, \ol{EF}) \\
&= \frac{s-a}{\cos A/2} \tan (B/2-C/2).
\end{align*}
To this end we compute
\begin{align*}
f(Z) &= f(A) + \left[ f(Z) - f(A) \right]
= f(A) + \frac{ZA}{EF} \left[ f(E)-f(F) \right] \\
&= f(A) - \frac{ZA}{EF}
\left[ \frac{|EF| \cdot (s-b) \sin B/2}{\sin C/2}
+ \frac{|EF| \cdot (s-c) \sin C/2}{\sin B/2} \right] \\
&= f(A) - |ZA| \left[ \frac{(s-b) \sin B/2}{\sin C/2}
+ \frac{(s-c) \sin C/2}{\sin B/2} \right] \\
&= \left[ b(s-a) - c(s-a) \right]
- |ZA| \left[ \frac{(s-b) \sin B/2}{\sin C/2}
+ \frac{(s-c) \sin C/2}{\sin B/2} \right] \\
&= (b-c)(s-a) - \frac{s-a}{\cos A/2} \tan (B/2-C/2)
\left[ \frac{(s-b) \sin B/2}{\sin C/2}
+ \frac{(s-c) \sin C/2}{\sin B/2} \right].
\end{align*}
Dividing out,
\begin{align*}
\frac{f(Z)}{s-a}
&= (b-c) - \frac{1}{\cos A/2} \tan (B/2-C/2)
\left[ \frac{r \cos B/2}{\sin C/2}
+ \frac{r \cos C/2}{\sin B/2} \right] \\
&= (b-c) - \frac{r \tan(B/2-C/2)}{\cos A/2}
\cdot \frac{\cos B/2 \sin B/2 + \cos C/2 \sin C/2}%
{\sin C/2 \sin B/2} \\
&= (b-c) - \frac{r \tan(B/2-C/2)}{\cos A/2}
\cdot \frac{\sin B + \sin C}%
{2\sin C/2 \sin B/2} \\
&= (b-c) - \frac{r \tan(B/2-C/2)}{\cos A/2}
\cdot \frac{\sin(B/2+C/2)\cos(B/2-C/2)}
{\sin C/2 \sin B/2} \\
&= (b-c) - r \frac{\sin(B/2-C/2)}{\sin B/2 \sin C/2} \\
&= (b-c) - r(\cot C/2 - \cot B/2)
= (b-c) - \left( (s-c) - (s-b) \right) = 0.
\end{align*}
\paragraph{Fourth solution by incircle inversion (USA IMO live stream, led by Andrew Gu).}
Let $T$ be the intersection of line $DI$ and the external $\angle A$-bisector.
Also, let $G$ be the antipode of $D$ on $\omega$.
We perform inversion around $\omega$, using $\bullet^\ast$ for the inverse.
Then $\triangle A^\ast B^\ast C^\ast$ is the medial triangle of $\triangle DEF$,
and $T^\ast$ is the foot from $A^\ast$ on to $\ol{DI}$.
If we denote $Q^\ast$ as the second intersection
of $(PC^\ast E)$ and $(PB^\ast F)$,
then the goal is to show that $Q^\ast$ lies on $(PIT^\ast)$.
\begin{center}
\begin{asy}
pair D = dir(115);
pair E = dir(210);
pair F = dir(330);
pair As = midpoint(E--F);
pair G = -D;
pair I = origin;
pair A = 2*E*F/(E+F);
pair R = -E*F/D;
pair P = -R+2*foot(I, R, A);
pair T = extension(D, I, A, E-F+A);
pair Bs = midpoint(D--F);
pair Cs = midpoint(D--E);
pair M = midpoint(As--D);
pair Qs = OP(circumcircle(Cs, P, E), circumcircle(Bs, P, F));
filldraw(unitcircle, opacity(0.1)+lightcyan, deepcyan);
draw(D--E--F--cycle, deepcyan);
draw(E--A--F, lightblue);
draw(P--G, deepgreen);
draw(D--T, lightblue);
draw(A--T, lightblue);
draw(A--P, lightblue);
pair Ts = foot(As, D, T);
draw(As--Ts, lightred);
draw(Qs--Bs, lightred+dashed);
draw(D--As, lightblue);
filldraw(circumcircle(P, Cs, E), opacity(0.05)+yellow, lightred);
filldraw(circumcircle(P, Bs, F), opacity(0.05)+yellow, lightred);
draw(CP(M, D), lightgreen);
draw(M--I, deepgreen);
clip(box((-1.8,-2.2), (2,2)));
filldraw(circumcircle(P, I, Ts), opacity(0.1)+orange, red+dashed);
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$A^\ast$", As, dir(225));
dot("$G$", G, dir(245));
dot("$I$", I, dir(45));
dot("$A$", A, dir(A));
dot("$R$", R, dir(R));
dot("$P$", P, dir(P));
dot("$T$", T, dir(T));
dot("$B^\ast$", Bs, dir(Bs));
dot("$C^\ast$", Cs, dir(45));
dot("$M$", M, dir(225));
dot("$Q^\ast$", Qs, dir(Qs));
dot("$T^\ast$", Ts, dir(45));
/* TSQ Source:
D = dir 115
E = dir 210
F = dir 330
A* = midpoint E--F R225
G = -D R245
I = origin R45
A = 2*E*F/(E+F)
R = -E*F/D
P = -R+2*foot I R A
T = extension D I A E-F+A
B* = midpoint D--F
C* = midpoint D--E R45
M = midpoint As--D R225
Q* = OP circumcircle Cs P E circumcircle Bs P F
unitcircle 0.1 lightcyan / deepcyan
D--E--F--cycle deepcyan
E--A--F lightblue
P--G deepgreen
D--T lightblue
A--T lightblue
A--P lightblue
T* = foot As D T R45
As--Ts lightred
Qs--Bs lightred dashed
D--As lightblue
circumcircle P Cs E 0.05 yellow / lightred
circumcircle P Bs F 0.05 yellow / lightred
CP M D lightgreen
M--I deepgreen
!clip(box((-1.8,-2.2), (2,2)));
circumcircle P I Ts 0.1 orange / red dashed
*/
\end{asy}
\end{center}
\begin{claim*}
Points $Q^\ast$, $B^\ast$, $C^\ast$ are collinear.
\end{claim*}
\begin{proof}
$\dang PQ^\ast C^\ast = \dang PEC^\ast = \dang PED = \dang PFD
= \dang PFB^\ast = \dang PQ^\ast B^\ast$.
\end{proof}
\begin{claim*}
[cf Brazil 2011/5]
Points $P$, $A^\ast$, $G$ are collinear.
\end{claim*}
\begin{proof}
Project harmonic quadrilateral $PERF$ through $G$,
noting $\ol{GR} \parallel \ol{EF}$.
\end{proof}
Denote by $M$ the center of parallelogram $DC^\ast A^\ast B^\ast$.
Note that it is the center of the circle with
diameter $\ol{DA^\ast}$, which passes through $P$ and $T^\ast$.
Also, $\ol{MI} \parallel \ol{PA^\ast G}$.
\begin{claim*}
Points $P$, $M$, $I$, $T^\ast$ are cyclic.
\end{claim*}
\begin{proof}
$\dang IT^\ast P = \dang D T^\ast P = \dang DA^\ast P
= \dang M A^\ast P = \dang A^\ast P M = \dang IMP$.
\end{proof}
\begin{claim*}
Points $P$, $M$, $I$, $Q^\ast$ are cyclic.
\end{claim*}
\begin{proof}
$\dang MQ^\ast P = \dang C^\ast Q^\ast P = \dang C^\ast E P
= \dang D E P = \dang D G P = \dang GPI = \dang MIP$.
\end{proof}
\paragraph{Fifth solution by double inversion (Brandon Wang, Luke Robitaille, Michael Ren, Evan Chen).}
We outline one final approach.
After inverting about $\omega$ as in the previous approach,
we then apply another inversion around $P$.
Dropping the apostrophes/stars/etc now one can check
that the problem we arrive at becomes the following.
\begin{proposition*}
[Doubly inverted problem]
In $\triangle PEF$, the $P$-symmedian meets $\ol{EF}$
and $(PEF)$ at $K$, $L$.
Let $D \in \ol{EF}$ with $\angle DPK = 90\dg$,
and let $T$ be the foot from $K$ to $\ol{DL}$.
Denote by $I$ the reflection of $P$ about $\ol{EF}$.
Finally, let $PDNE$ and $PDMF$ be cyclic harmonic quadrilaterals.
Then lines $EN$, $MF$, $TI$, are concurrent.
\end{proposition*}
The proof proceeds in three steps.
Suppose the line through $L$ perpendicular to $\ol{EF}$
meets $\ol{EF}$ at $W$ and $(PEF)$ at $Z$.
\begin{center}
\begin{asy}
size(10cm);
pair P = dir(190);
pair E = dir(220);
pair F = dir(-40);
pair T = 2*E*F/(E+F);
pair K = extension(E, F, P, T);
pair L = -P+2*foot(origin, P, K);
pair W = foot(L, E, F);
pair Z = -L+2*foot(origin, W, L);
pair D = extension(E, F, P, P+dir(90)*(P-K));
pair T = foot(K, D, L);
pair I = -P+2*foot(P, E, F);
pair H = extension(P, Z, W, I);
pair N = -E+2*foot(circumcenter(P, D, E), H, E);
pair M = -F+2*foot(circumcenter(P, D, F), H, F);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
draw(P--E--F--cycle, blue);
draw(D--E, blue);
draw(D--P--L--cycle, deepcyan);
draw(T--K, deepcyan);
draw(E--Z, orange);
draw(F--Z, orange);
draw(H--Z, orange);
draw(L--Z, orange);
filldraw(circumcircle(P, D, E), opacity(0.1)+lightgreen, deepgreen);
filldraw(circumcircle(P, D, F), opacity(0.1)+lightgreen, deepgreen);
draw(E--H, gray);
draw(W--H, gray);
draw(F--M, gray);
clip(box((-3,-2.3), (1.5,1.5)));
dot("$P$", P, dir(110));
dot("$E$", E, dir(135));
dot("$F$", F, dir(350));
dot("$K$", K, dir(70));
dot("$L$", L, dir(270));
dot("$W$", W, dir(45));
dot("$Z$", Z, dir(Z));
dot("$D$", D, dir(D));
dot("$T$", T, dir(270));
dot("$I$", I, dir(315));
dot(H);
dot("$N$", N, dir(100));
dot("$M$", M, dir(M));
/* TSQ Source:
P = dir 190 R110
E = dir 220 R135
F = dir -40 R350
T := 2*E*F/(E+F)
K = extension E F P T R70
L = -P+2*foot origin P K R270
W = foot L E F R45
Z = -L+2*foot origin W L
D = extension E F P P+dir(90)*(P-K)
T = foot K D L R270
I = -P+2*foot P E F R315
H .= extension P Z W I
N = -E+2*foot circumcenter P D E H E R100
M = -F+2*foot circumcenter P D F H F
unitcircle 0.1 lightcyan / blue
P--E--F--cycle blue
D--E blue
D--P--L--cycle deepcyan
T--K deepcyan
E--Z orange
F--Z orange
H--Z orange
L--Z orange
circumcircle P D E 0.1 lightgreen / deepgreen
circumcircle P D F 0.1 lightgreen / deepgreen
E--H gray
W--H gray
F--M gray
! clip(box((-3,-2.3), (1.5,1.5)));
*/
\end{asy}
\end{center}
\begin{enumerate}
\ii Since $\dang ZEP = \dang WLP = \dang WDP$,
it follows $\ol{ZE}$ is tangent to $(PDNE)$.
Similarly, $\ol{ZF}$ is tangent to $(PDMF)$.
\ii $\triangle WTP$ is the orthic triangle of $\triangle DKL$,
so $\ol{WD}$ bisects $\angle PWT$ and $\ol{WTI}$ collinear.
\ii $-1 = E(PN;DZ) = F(PM;DZ) = W(PI;DZ)$, so
$\ol{EN}$, $\ol{FM}$, $\ol{WI}$ meet on $\ol{PZ}$.
\end{enumerate} |
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 $\angle PCB$
and the perpendicular bisector of segment $AB$. | 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, opacity(0.1)+lightcyan, blue+dotted);
filldraw(A--B--C--D--cycle, opacity(0.1)+yellow, gray);
filldraw(P--A--B--cycle, opacity(0.1)+lightcyan, blue);
draw(circumcircle(P, O, B), lightred);
draw(circumcircle(P, O, D), lightred);
draw(D--P--C, gray);
draw(D--O--C, dashed+red);
dot("$O$", O, dir(270));
dot("$P$", P, dir(P));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
/* TSQ Source:
O = origin R270
P = dir 118
A = dir 190
B = dir 350
C = extension P/dir(64) B P B*dir(192)
D = extension P*dir(36) A P A/dir(108)
unitcircle 0.1 lightcyan / blue dotted
A--B--C--D--cycle 0.1 yellow / gray
P--A--B--cycle 0.1 lightcyan / blue
circumcircle P O B lightred
circumcircle P O D lightred
D--P--C gray
D--O--C dashed red
*/
\end{asy}
\end{center}
Indeed, $O$ obviously lies on the perpendicular bisector of $AB$.
Now
\begin{align*}
\dang BCP &= \dang CBP + \dang BPC \\
&= 2\dang BAP = \dang BOP
\end{align*}
it follows $BOPC$ are cyclic.
And since $OP = OB$, it follows that $O$ is on
the bisector of $\angle PCB$, as needed.
\begin{remark*}
The angle equality is only used insomuch $\angle BAP$
is the average of $\angle CBP$ and $\angle BPC$,
i.e.\ only $\frac{1+3}{2} = 2$ matters.
\end{remark*} |
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
\[
\begin{array}{cccc}
+a^3 &+ b^2a &+ c^2a & +d^2a \\
+2a^2b &+ 2b^3 &+ 2bc^2 & +2d^2b \\
+3a^2c & + 3b^2c & + 3c^3 & + 3d^2c \\
+4a^2d &+ 4b^2d & + 4c^2d & + 4d^3
\end{array}
<
\begin{array}{cccc}
+a^3 &+ 3b^2a &+ 3c^2a & +3d^2a \\
+3a^2b &+ b^3 &+ 3bc^2 & +3d^2b \\
+3a^2c &+ 3b^2c &+ c^3 &+ 3d^2c \\
+3a^2d &+ 3b^2d &+ 3c^2d &+ d^3 \\
+6abc &+ 6bcd &+ 6cda &+ 6dab.
\end{array}
\]
In other words, we need to prove that
\[
\begin{array}{cccc}
& && \\
&+ b^3 & & \\
& & +2c^3 & \\
+a^2d &+ b^2d & + c^2d & + 3d^3 \\
\end{array}
<
\begin{array}{cccc}
&+ 2b^2a &+ 2c^2a & +2d^2a \\
+a^2b & &+ bc^2 & +d^2b \\
&&& \\
&&& \\
+6abc &+ 6bcd &+ 6cda &+ 6dab
\end{array}
\]
This follows since
\begin{align*}
2b^2a &\ge b^3 + b^2d \\
2c^2a &\ge 2c^3 \\
2d^2a &\ge 2d^3 \\
a^2b &\ge a^2d \\
bc^2 &\ge c^2d \\
d^2b &\ge d^3
\end{align*}
and $6(abc+bcd+cda+dab) > 0$.
\begin{remark*}
Some students complained this problem was ``unfair'' and they couldn't solve it
because they didn't think an IMO problem would be solved only by expansion.
I don't agree with this.
Fedor Petrov provides the following motivational comments
for why the existence of this solution should not be that surprising:
\begin{quote}
Better to think about mathematics.
You have to bound from above a product $(a+2b+3c+4d)(a^2+b^2+c^2+d^2)$,
the coefficients $1,2,3,4$ are increasing and so play on your side,
so plausibly $(a+b+c+d)^3$ should majorize this term-wise,
you check it and this appears to be true.
\end{quote}
He also gave the \href{https://aops.com/community/p18331252}{following advice}:
\begin{quote}
The general advice is to study mathematics, not \emph{olympiad problems in past years}.
If the IMO problems set the students and their teachers on this path, I am more than satisfied.
\end{quote}
\end{remark*} |
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 such that each box has two blue strings
and two green strings
(where a string between two pebbles in the same box counts double).
\begin{center}
\begin{asy}
size(12cm);
// picture box;
pair A = 0.5*dir(135);
pair C = 0.5*dir(225);
pair D = 0.5*dir(315);
pair B = 0.5*dir(45);
path boxoutline = (dir(135)+dir(200)*0.4)--dir(135)--dir(225)
--dir(315)--dir(45)--(dir(45)+dir(-20)*0.4);
transform[] t = new transform[5];
for (int i=0; i<5; ++i) {
t[i] = shift(3.4*dir(90+72*i));
draw(t[i]*boxoutline, brown+1.5);
label("Box " + array("ABCDE")[i], t[i]*dir(-90), brown);
}
void rope(pair X, pair Y, string s1, string s2, pen p) {
draw(X--Y, p);
dot(s1, X, dir(-90), p + fontsize(9pt));
dot(s2, Y, dir(-90), p + fontsize(9pt));
}
rope(t[0]*A, t[0]*B, "1", "20", blue);
rope(t[0]*C, t[1]*A, "4", "17", deepgreen);
rope(t[0]*D, t[3]*D, "3", "18", deepgreen);
rope(t[1]*B, t[4]*A, "8", "13", blue);
rope(t[1]*D, t[4]*C, "7", "14", deepgreen);
rope(t[4]*B, t[2]*A, "15", "6", deepgreen);
rope(t[1]*C, t[2]*B, "11", "10", blue);
rope(t[4]*D, t[3]*B, "9", "12", blue);
rope(t[2]*C, t[3]*A, "19", "2", blue);
rope(t[2]*D, t[3]*C, "5", "16", deepgreen);
/*
for (int i=0; i<=3; ++i) {
dot(box, 0.5*dir(45+90*i));
}
*/
\end{asy}
\end{center}
We can therefore rephrase the problem as follows,
if we view boxes as vertices and strings as edges:
\begin{claim*}
Given a $4$-regular multigraph on $n$ vertices
(where self-loops are allowed and have degree $2$),
one can color the edges blue and green
such that each vertex has two blue and two green edges.
\end{claim*}
\begin{proof}
Each connected component of the graph can be decomposed
into an Eulerian circuit, since $4$ is even.
A connected component with $k$ vertices has $2k$
edges in its Eulerian circuit,
so we may color the edges in this circuit alternating green and blue.
This may be checked to work.
\end{proof} |
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 $k$ different starting points
and $k$ different finishing points, and a cable car which starts higher also finishes higher.
The same conditions hold for $B$.
We say that two stations are linked by a company if one can start from the lower station
and reach the higher one by using one or more cars of that company
(no other movements between stations are allowed).
Determine the smallest positive integer $k$ for which one can guarantee
that there are two stations that are linked by both companies. | 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) );
if (j!=0) {
draw(P(i,j-1)--P(i,j), red, EndArrow, Margin(4,4));
}
if (i != 0) {
draw(P(i-1,j)--P(i,j), blue, EndArrow, Margin(4,4));
}
}
}
\end{asy}
\end{center}
To see this is sharp, view $A$ and $B$ as graphs
whose connected components are paths (possibly with $0$ edges;
the direction of these edges is irrelevant).
Now, if $k = n^2-n+1$ it follows that $A$ and $B$
each have exactly $n-1$ connected components.
But in particular some component of $A$ has at least $n+1$ vertices.
This component has two vertices in the same component of $B$, as desired.
\begin{remark*}
The main foothold for this problem is the hypothesis
that the number of stations should be $n^2$ rather than, say, $n$.
This gives a big hint towards finding the construction
which in turn shows how the bound can be computed.
On the other hand, the hypothesis that
``a cable car which starts higher
also finishes higher'' appears to be superfluous.
\end{remark*} |
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 equal? | 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 assume
\[ \gcd(a_1, a_2, \dots, a_n) = 1. \]
WLOG we also assume \[ a_1 \ge a_2 \ge \dots \ge a_n. \]
\bigskip
\textbf{Main proof.}
As $a_1 \ge 2$, let $p$ be a prime divisor of $a_1$.
Let $k$ be smallest index such that $p \nmid a_k$ (which must exist).
In particular, note that $a_1 \neq a_k$.
Consider the mean $x = \frac{a_1+a_k}{2}$; by assumption,
it equals some geometric mean, hence
\[ \sqrt[m]{a_{i_1} \dots a_{i_m}} = \frac{a_1 + a_k}{2} > a_k. \]
Since the arithmetic mean is an integer not divisible by $p$,
all the indices $i_1$, $i_2$, \dots, $i_m$
must be at least $k$.
But then the GM is at most $a_k$, contradiction.
\begin{remark*}
A similar approach could be attempted by using
the smallest numbers rather than the largest ones,
but one must then handle the edge case $a_n = 1$
separately since no prime divides $1$.
\end{remark*}
\begin{remark*}
Since $\frac{27+9}{2} = 18 = \sqrt[3]{27 \cdot 27 \cdot 8}$,
it is not true that in general the AM of two largest different cards
is not the GM of other numbers in the sequence
(say the cards are $27, 27, 9, 8, \dots$).
\end{remark*} |
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 line $\ell$ separates a set of points $S$
if some segment joining two points in $\mathcal S$ crosses $\ell$.)
\end{enumerate} | 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 make the following the definitions:
\begin{itemize}
\ii Define $A$ and $B$ as the two points farthest apart in $\mathcal S$.
This means that all points lie in the intersections
of the circles centered at $A$ and $B$ with radius $R = AB \ge 1$.
\ii We pick chord $\ol{XY}$ of $\odot(B)$
such that $\ol{XY} \perp \ol{AB}$ and the distance
from $A$ to $\ol{XY}$ is exactly $\half$.
\ii We denote by $\mathcal T$
the smaller region bound by $\odot(B)$ and chord $\ol{XY}$.
\end{itemize}
The figure is shown below with $\mathcal T$ drawn in yellow,
and points of $\mathcal S$ drawn in blue.
\begin{center}
\begin{asy}
size(10cm);
pair A = (-2,0);
pair B = (5,0);
pen auxpen = brown+linetype("4 2");
pair X = (0, 24^0.5);
pair Y = (0, -24^0.5);
pair[] S = {
(-1.3, 2.7),
(-0.57, -2.61),
(-0.21, 1.07),
(0.33,-1.06),
(0.83, 3.71),
(1.57, -3.18),
(2.14, 1.66),
(2.72, 3.32),
(3.37, -1.17),
(3.97, 1.34),
(4.37, -2.35),
};
fill(Y..A..X--cycle, paleyellow);
for (int i=0; i<S.length; ++i) {
draw(S[i]--(S[i].x,0), gray);
dot(S[i], blue);
}
draw(A--B, blue+1.3);
draw(arc(B,abs(B-A), 110, 250), auxpen);
draw(arc(A,abs(B-A), -70, 70), auxpen);
draw(B--X--Y--cycle, auxpen);
dot("$A$", A, dir(180), blue);
dot("$B$", B, dir(0), blue);
dot("$X$", X, dir(90), auxpen);
dot("$Y$", Y, dir(270), auxpen);
draw( (-1.9,-0.1)--(-1.9,-0.2)--(-0.1,-0.2)--(-0.1, -0.1) );
label("$\frac12$", (-1, -0.2), dir(-90));
draw( (2.14,-0.1)--(2.14,-0.2)--(2.72,-0.2)--(2.72,-0.1) );
label("$< \delta$", (2.43, -0.2), dir(-90), fontsize(9pt));
label("$\mathcal T$", (-0.7,3), fontsize(14pt));
\end{asy}
\end{center}
\begin{claim*}
[Length of $AB$ + Pythagorean theorem]
We have $XY < 2\sqrt{n\delta}$.
\end{claim*}
\begin{proof}
First, note that we have $R = AB < (n-1) \cdot \delta$,
since the $n$ projections of points onto $AB$
are spaced at most $\delta$ apart.
The Pythagorean theorem gives
\[ XY = 2\sqrt{R^2 - \left(R-\half\right)^2}
= 2\sqrt{R - \frac14} < 2\sqrt{n\delta}. \qedhere \]
\end{proof}
\begin{claim*}
[$|\mathcal T|$ lower bound + narrowness]
We have $XY > \frac{\sqrt3}{2} \left( \half \delta\inv - 1 \right)$.
\end{claim*}
\begin{proof}
Because $\mathcal T$ is so narrow (has width $\half$ only),
the projections of points in $\mathcal T$ onto line $XY$
are spaced at least $\frac{\sqrt3}{2}$ apart (more than just $\delta$).
This means
\[ XY > \frac{\sqrt3}{2}
\left( \left\lvert \mathcal T \right\rvert - 1 \right). \]
But projections of points in $\mathcal T$
onto the segment of length $\half$ are spaced at most $\delta$ apart,
so apparently
\[ \left\lvert \mathcal T \right\rvert > \half \cdot \delta\inv. \]
This implies the result.
\end{proof}
Combining these two this implies $\delta \ge \Omega(n^{-1/3})$ as needed.
\begin{remark*}
The constant $1/3$ in the problem is actually optimal
and cannot be improved;
the constructions give an example showing $\Theta(n^{-1/3} \log n)$.
\end{remark*} |
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