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sols-TST-IMO-2014_2
|
Let $a_1$, $a_2$, $a_3$, \dots be a sequence of integers,
with the property that every consecutive group of $a_i$'s
averages to a perfect square.
More precisely, for all positive integers $n$ and $k$, the quantity
\[ \frac{a_n + a_{n+1} + \dots + a_{n+k-1}}{k} \]
is always the square of an integer.
Prove that the sequence must be constant
(all $a_i$ are equal to the same perfect square).
|
Let $\nu_p(n)$ denote the largest exponent of $p$ dividing $n$.
The problem follows from the following proposition.
\begin{proposition*}
Let $(a_n)$ be a sequence of integers and let $p$ be a prime.
Suppose that every consecutive group of $a_i$'s
with length at most $p$ averages to a perfect square.
Then $\nu_p(a_i)$ is independent of $i$.
\end{proposition*}
We proceed by induction on the smallest value of $\nu_p(a_i)$ as $i$ ranges
(which must be even, as each of the $a_i$ are themselves a square).
First we prove two claims.
\begin{claim*}
If $j \equiv k \pmod p$ then $a_j \equiv a_k \pmod{p}$.
\end{claim*}
\begin{proof}
Taking groups of length $p$ in our given,
we find that $p \mid a_j +\dots + a_{j+p-1}$
and $p \mid a_{j+1} + \dots + a_{j+p}$ for any $j$.
So $a_j \equiv a_{j+p} \pmod p$ and the conclusion follows.
\end{proof}
\begin{claim*}
If some $a_i$ is divisible by $p$ then all of them are.
\end{claim*}
\begin{proof}
The case $p=2$ is trivial so assume $p \ge 3$.
Without loss of generality (via shifting indices)
assume that $a_1 \equiv 0 \pmod{p}$, and define
\[ S_n = a_1 + a_2 + \dots + a_n \equiv a_2 + \dots + a_n \pmod{p}. \]
Call an integer $k$ with $2 \le k < p$ a
\textbf{pivot} if $1-k^{-1}$ is a quadratic nonresidue modulo $p$.
We claim that for any pivot $k$, $S_k \equiv 0 \pmod{p}$.
If not, then
\[ \frac{a_1 + a_2 + \dots + a_k}{k}
\text{ and } \frac{a_2 + \dots + a_k}{k-1} \]
are both qudaratic residues.
Division implies that $\frac{k-1}{k} = 1 - k\inv$ is a
quadratic residue, contradiction.
Next we claim that there is an integer $m$
with $S_m \equiv S_{m+1} \equiv 0 \pmod{p}$,
which implies $p \mid a_{m+1}$.
If $2$ is a pivot, then we simply take $m=1$.
Otherwise, there are $\frac12(p-1)$ pivots,
one for each nonresidue (which includes neither $0$ nor $1$),
and all pivots lie in $[3,p-1]$,
so we can find an $m$ such that $m$ and $m+1$ are both pivots.
Repeating this procedure starting with $a_{m+1}$
shows that $a_{2m+1}, a_{3m+1}, \dots$ must all be divisible by $p$.
Combined with the first claim and the fact that $m < p$,
we find that all the $a_i$ are divisible by $p$.
\end{proof}
The second claim establishes the base case of our induction.
Now assume all $a_i$ are divisible by $p$ and hence $p^2$.
Then all the averages in our proposition (with length at most $p$)
are divisible by $p$ and hence $p^2$.
Thus the map $a_i \mapsto \frac{1}{p^2} a_i$ gives a new sequence
satisfying the proposition, and our inductive hypothesis completes the proof.
\begin{remark*}
There is a subtle bug that arises
if one omits the condition that $k \le p$ in the proposition.
When $k = p^2$ the average $\frac{a_1 + \dots + a_{p^2}}{p^2}$
is not necessarily divisible by $p$
even if all the $a_i$ are.
Hence it is not valid to divide through by $p$.
This is why the condition $k \le p$ was added.
\end{remark*}
|
sols-TST-IMO-2014_3
|
Let $n$ be an even positive integer,
and let $G$ be an $n$-vertex (simple) graph
with exactly $\frac{n^2}{4}$ edges.
An unordered pair of distinct vertices $\{x,y\}$
is said to be \emph{amicable} if they have a common neighbor
(there is a vertex $z$ such that $xz$ and $yz$ are both edges).
Prove that $G$ has at least $2\binom{n/2}{2}$
pairs of vertices which are amicable.
|
First, we prove the following lemma.
(\url{https://en.wikipedia.org/wiki/Friendship_paradox}).
\begin{lemma*}
[On average, your friends are more popular than you]
For a vertex $v$, let $a(v)$ denote
the average degree of the neighbors of $v$
(setting $a(v) = 0$ if $\deg v = 0$).
Then
\[ \sum_v a(v) \ge \sum_v \deg v = 2 \# E. \]
\end{lemma*}
\begin{proof}
Ignoring isolated vertices, we can write
\begin{align*}
\sum_v a(v)
&= \sum_v \frac{\sum_{w \sim v} \deg w}{\deg v} \\
&= \sum_{v} \sum_{w \sim v} \frac{\deg w}{\deg v} \\
&= \sum_{\text{edges } vw} \left(
\frac{\deg w}{\deg v} + \frac{\deg v}{\deg w} \right) \\
&\overset{\text{AM-GM}}{\ge} \sum_{\text{edges } vw} 2
= 2\#E = \sum_{v} \deg v
\end{align*}
as desired.
\end{proof}
\begin{corollary*}
[On average, your most popular friend is more popular than you]
For a vertex $v$, let $m(v)$ denote
the maximum degree of the neighbors of $v$
(setting $m(v) = 0$ if $\deg v = 0$).
Then
\[ \sum_v m(v) \ge \sum_v \deg v = 2 \# E. \]
\end{corollary*}
We can use this to count amicable pairs
by noting that any particular vertex $v$ is in
at least $m(v)-1$ amicable pairs.
So, the number of amicable pairs is at least
\[ \half \sum_v (m(v)-1) \ge \# E - \half \# V. \]
Note that up until now we haven't used any information about $G$.
But now if we plug in $\# E = n^2/4$, $\# V = n$,
then we get exactly the desired answer.
(Equality holds for $G = K_{n/2, n/2}$.)
|
sols-TST-IMO-2014_4
|
Let $n$ be a positive even integer,
and let $c_1$, $c_2$, \dots, $c_{n-1}$ be real numbers satisfying
\[ \sum_{i=1}^{n-1} \left\lvert c_i-1 \right\rvert < 1. \]
Prove that
\[ 2x^n - c_{n-1}x^{n-1} + c_{n-2}x^{n-2} - \dots - c_1x^1 + 2 \]
has no real roots.
|
We will prove the polynomial is positive for all $x \in \RR$.
As $c_i > 0$, the result is vacuous for $x \le 0$,
so we restrict attention to $x > 0$.
Then letting $c_i = 1 - d_i$ for each $i$,
the inequality we want to prove becomes
\[ x^n + 1 + \frac{x^{n+1}+1}{x+1} > \sum_1^{n-1} d_i x^i
\qquad\text{given } \sum |d_i| < 1. \]
But obviously $x^n+1 > x^i$ for any $1 \le i \le n-1$ and $x > 0$.
So in fact $x^n+1 > \sum_1^{n-1} |d_i| x^i$ holds for $x > 0$,
as needed.
|
sols-TST-IMO-2014_5
|
Let $ABCD$ be a cyclic quadrilateral,
and let $E$, $F$, $G$, and $H$ be the midpoints of $AB$, $BC$, $CD$, and $DA$ respectively.
Let $W$, $X$, $Y$ and $Z$ be the orthocenters of
triangles $AHE$, $BEF$, $CFG$ and $DGH$, respectively.
Prove that the quadrilaterals $ABCD$ and $WXYZ$ have the same area.
|
The following solution is due to Grace Wang.
We begin with:
\begin{claim*}
Point $W$ has coordinates $\half(2a+b+d)$.
\end{claim*}
\begin{proof}
The orthocenter of $\triangle DAB$ is $d+a+b$,
and $\triangle AHE$ is homothetic to $\triangle DAB$
through $A$ with ratio $1/2$.
Hence $w = \half(a+(d+a+b))$ as needed.
\end{proof}
By symmetry, we have
\begin{align*}
w &= \half(2a+b+d) \\
x &= \half(2b+c+a) \\
y &= \half(2c+d+b) \\
z &= \half(2d+a+c).
\end{align*}
We see that $w-y = a-c$, $x-z = b-d$.
So the diagonals of $WXYZ$ have the same length as those
of $ABCD$ as well as the same directed angle between them.
This implies the areas are equal, too.
|
sols-TST-IMO-2014_6
|
For a prime $p$, a subset $S$ of residues modulo $p$
is called a \emph{sum-free multiplicative subgroup}
of $\FF_p$ if
\begin{itemize}
\ii there is a nonzero residue $\alpha$ modulo $p$
such that $S = \left\{ 1, \alpha^1, \alpha^2, \dots \right\}$
(all considered mod $p$), and
\ii there are no $a,b,c \in S$
(not necessarily distinct) such that $a+b \equiv c \pmod p$.
\end{itemize}
Prove that for every integer $N$,
there is a prime $p$ and a sum-free multiplicative subgroup $S$
of $\FF_p$ such that $\left\lvert S \right\rvert \ge N$.
\end{enumerate}
|
We first prove the following general lemma.
\begin{lemma*}
If $f,g \in \ZZ[X]$ are relatively prime nonconstant polynomials,
then for sufficiently large primes $p$,
they have no common root modulo $p$.
\end{lemma*}
\begin{proof}
By B\'{e}zout Lemma,
there exist polynomials $a(X)$ and $b(X)$ in $\ZZ[X]$
and a nonzero constant $c \in \ZZ$ satisfying the identity
\[ a(X) f(X) + b(X) g(X) \equiv c. \]
So, plugging in $X = r$ we get $p \mid c$,
so the set of permissible primes $p$ is finite.
\end{proof}
With this we can give the construction.
\begin{claim*}
Suppose that
\begin{itemize}
\ii $n$ is a positive integer with $n \not\equiv 0 \pmod 3$;
\ii $p$ is a prime which is $1 \bmod n$; and
\ii $\alpha$ is a primitive $n$'th root of unity modulo $p$.
\end{itemize}
Then $|S| = n$ and, if $p$ is sufficiently large in $n$,
is also sum-free.
\end{claim*}
\begin{proof}
The assertion $|S|=n$ is immediate from the choice of $\alpha$.
As for sum-free, assume for contradiction that
\[ 1 + \alpha^k \equiv \alpha^m \pmod p \]
for some integers $k,m \in \ZZ$.
This means $(X+1)^n-1$ and $X^n-1$ have common root $X=\alpha^k$.
But
\[ \gcd_{\ZZ[x]} \Big( (X+1)^n-1, \; X^n-1 \Big)
= 1 \qquad \forall n \not\equiv 0 \pmod 3 \]
because when $3 \nmid n$ the two polynomials
have no common complex roots.
(Indeed, if $|\omega| = |1+\omega| = 1$
then $\omega = -\half \pm \frac{\sqrt3}{2}i$.)
Thus $p$ is bounded by the lemma, as desired.
\end{proof}
|
sols-TST-IMO-2015_1
|
Let $ABC$ be a scalene triangle with incenter $I$ whose incircle is
tangent to $\ol{BC}$, $\ol{CA}$, $\ol{AB}$ at $D$, $E$, $F$,
respectively. Denote by $M$ the midpoint of $\ol{BC}$ and
let $P$ be a point in the interior of $\triangle ABC$
so that $MD = MP$ and $\angle PAB = \angle PAC$.
Let $Q$ be a point on the incircle such that $\angle AQD = 90^{\circ}$.
Prove that either $\angle PQE = 90^{\circ}$ or $\angle PQF = 90^{\circ}$.
|
We present two solutions.
\paragraph{Official solution.}
Assume without loss of generality that $AB < AC$; we show $\angle PQE=90^{\circ}$.
\begin{center}
\begin{asy}
size(9cm);
pair A = dir(130);
pair B = dir(210);
pair C = dir(330);
filldraw(A--B--C--cycle, opacity(0.2)+palecyan, blue);
pair I = incenter(A, B, C);
filldraw(incircle(A, B, C), opacity(0.3)+palered, red);
pair D = foot(I, B, C);
pair E = foot(I, C, A);
pair F = foot(I, A, B);
pair M = midpoint(B--C);
pair P = extension(A, I, D, E);
draw(A--P, orange);
pair D1 = 2*M-D;
pair Q = foot(D, A, D1);
draw(P--Q--E, heavyred+1);
draw(M--Q, red);
markrightangle(E,Q,P,heavyred+1);
pair N = midpoint(A--B);
draw(N--M, dashed+orange);
draw(D--E, dashed+orange);
filldraw(CP(M, P), opacity(0.2)+palegreen, heavygreen+dashed);
pair T = extension(A, Q, B, C);
pair S = extension(D, I, A, T);
draw(A--T, heavygreen);
draw(D--Q, red);
draw(D--S, dashed+red);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$I$", I, dir(I));
dot("$D$", D, dir(D));
dot("$E$", E, dir(P-D));
dot("$F$", F, dir(F));
dot("$M$", M, dir(M));
dot("$P$", P, dir(200));
dot("$Q$", Q, dir(25));
dot("$N$", N, dir(N));
dot("$T$", T, dir(T));
dot("$S$", S, dir(70));
/* Source generated by TSQ
A = dir 130
B = dir 210
C = dir 330
A--B--C--cycle
I = incenter A B C
incircle A B C
D = foot I B C
E = foot I C A RP-D
F = foot I A B
M = midpoint B--C
P = extension A I D E R200
A--P
D1 := 2*M-D
Q = foot D A D1 R25
E--Q--P--M--Q
!markrightangle(E,Q,P,heavyred+1);
N = midpoint A--B
N--M dotted
D--E dotted
CP M P dashed
T = extension A Q B C
S = extension D I A T R70
A--T
D--Q
D--S dashed
*/
\end{asy}
\end{center}
First, we claim that $D$, $P$, $E$ are collinear.
Let $N$ be the midpoint of $\ol{AB}$.
It is well-known that the three lines $MN$, $DE$, $AI$ are concurrent at a point
(see for example problem 6 of USAJMO 2014).
Let $P'$ be this intersection point, noting that $P'$ actually lies on segment $DE$.
Then $P'$ lies inside $\triangle ABC$ and moreover
\[ \triangle DP'M \sim \triangle DEC \]
so $MP' = MD$.
Hence $P'=P$, proving the claim.
Let $S$ be the point diametrically opposite $D$ on the incircle,
which is also the second intersection of $\ol{AQ}$ with the incircle.
Let $T = \ol{AQ} \cap \ol{BC}$.
Then $T$ is the contact point of the $A$-excircle;
consequently, \[ MD = MP = MT \] and we obtain a circle with diameter $\ol{DT}$.
Since $\angle DQT = \angle DQS = 90^{\circ}$ we have $Q$ on this circle as well.
As $\ol{SD}$ is tangent to the circle with diameter $\ol{DT}$,
we obtain \[ \angle PQD = \angle SDP = \angle SDE = \angle SQE. \]
Since $\angle DQS = 90^{\circ}$, $\angle PQE = 90^{\circ}$ too.
\paragraph{Solution using spiral similarity.}
We will ignore for now the point $P$.
As before define $S$, $T$ and note $\ol{AQST}$ collinear,
as well as $DPQT$ cyclic on circle $\omega$ with diameter $\ol{DT}$.
Let $\tau$ be the spiral similarity at $Q$
sending $\omega$ to the incircle.
We have $\tau(T) = D$, $\tau(D) = S$, $\tau(Q) = Q$.
Now
\[ I = \ol{DD} \cap \ol{QQ} \implies
\tau(I) = \ol{SS} \cap \ol{QQ} \]
and hence we conclude $\tau(I)$ is the pole of $\ol{ASQT}$
with respect to the incircle, which lies on line $EF$.
Then since $\ol{AI} \perp \ol{EF}$ too,
we deduce $\tau$ sends line $AI$ to line $EF$,
hence $\tau(P)$ must be either $E$ or $F$ as desired.
\paragraph{Authorship comments.}
Written April 2014.
I found this problem while playing with GeoGebra.
Specifically, I started by drawing in the points $A$, $B$, $C$,
$I$, $D$, $M$, $T$, common points.
I decided to add in the circle with diameter $DT$,
because of the synergy it had with the rest of the picture.
After a while of playing around,
I intersected ray $AI$ with the circle to get $P$,
and was surprised to find that $D$, $P$, $E$ were collinear,
which I thought was impossible since the setup should have been symmetric.
On further reflection, I realized it was because $AI$
intersected the circle twice, and set about trying to prove this.
I noticed the relation $\angle PQE = 90\dg$ in my attempts
to prove the result, even though this ended up being a corollary
rather than a useful lemma.
|
sols-TST-IMO-2015_2
|
Prove that for every positive integer $n$, there exists a set $S$ of $n$ positive integers
such that for any two distinct $a,b \in S$, $a-b$ divides $a$ and $b$
but none of the other elements of $S$.
|
The idea is to look for a
sequence $d_1, \dots, d_{n-1}$ of ``differences''
such that the following two conditions hold.
Let $s_i = d_1 + \dots + d_{i-1}$,
and $t_{i,j} = d_i + \dots + d_{j-1}$ for $i \le j$.
\begin{enumerate}
\item[(i)] No two of the $t_{i,j}$ divide each other.
\item[(ii)] There exists an integer $a$
satisfying the CRT equivalences
\[ a \equiv -s_i \pmod{t_{i,j}} \quad \forall i \le j \]
\end{enumerate}
Then the sequence $a+s_1$, $a+s_2$, \dots, $a+s_n$ will work.
For example, when $n=3$ we can take $(d_1,d_2)=(2,3)$ giving
\[ 10 \overbrace{\underbrace{\qquad}_2 12 \underbrace{\qquad}_3}^5 15 \]
because the only conditions we need satisfy are
\begin{align*}
a &\equiv 0 \pmod 2 \\
a &\equiv 0 \pmod 5 \\
a &\equiv -2 \pmod 3.
\end{align*}
But with this setup we can just construct the $d_i$ inductively.
To go from $n$ to $n+1$, take a $d_1, \dots, d_{n-1}$
and let $p$ be a prime
not dividing any of the $d_i$.
Moreover, let $M$ be a multiple of $\prod_{i \le j} t_{i,j}$ coprime to $p$.
Then we claim that $d_1M, d_2M, \dots, d_{n-1}M, p$
is such a difference sequence.
For example, the previous example extends as follows with $M = 300$ and $p=7$.
\[
a \overbrace{\underbrace{\qquad}_{600} b
\overbrace{\underbrace{\qquad}_{900} c
\underbrace{\qquad}_{7}}^{907}}^{1507} d
\]
The new numbers $p$, $p+Mt_{n-1,n}$, $p+Mt_{n-2,n}$, \dots\
are all relatively prime to everything else.
Hence (i) still holds.
To see that (ii) still holds,
just note that we can still get a family of solutions
for the first $n$ terms, and then the last $(n+1)$st term
can be made to work by Chinese Remainder Theorem
since all the new $p+Mt_{n-2,n}$ are coprime to everything.
|
sols-TST-IMO-2015_3
|
A physicist encounters $2015$ atoms called usamons.
Each usamon either has one electron or zero electrons,
and the physicist can't tell the difference.
The physicist's only tool is a diode.
The physicist may connect the diode from any usamon $A$
to any other usamon $B$. (This connection is directed.)
When she does so, if usamon $A$ has an electron and usamon $B$
does not, then the electron jumps from $A$ to $B$.
In any other case, nothing happens. In addition,
the physicist cannot tell whether an electron jumps during any given step.
The physicist's goal is to isolate two usamons that she is 100\%
sure are currently in the same state.
Is there any series of diode usage that makes this possible?
|
The answer is no.
Call the usamons $U_1, \dots, U_m$ (here $m=2015$).
Consider models $M_k$ of the following form:
$U_1, \dots, U_k$ are all charged for some $0 \le k \le m$
and the other usamons are not charged.
Note that for any pair there's a model
where they are different states, by construction.
We can consider the physicist as acting
on these $m+1$ models simultaneously,
and trying to reach a state where there's a pair
in all models which are all the same charge.
(This is a necessary condition for a winning strategy to exist.)
But we claim that any diode operation $U_i \to U_j$
results in the $m+1$ models being an isomorphic copy
of the previous set.
If $i < j$ then the diode operation can be interpreted
as just swapping $U_i$ with $U_j$, which doesn't change anything.
Moreover if $i > j$ the operation never does anything.
The conclusion follows from this.
\begin{remark*}
This problem is not a ``standard'' olympiad problem,
so I can't say it's trivial.
But the idea is pretty natural I think.
You can motivate it as follows:
there's a sequence of diode operations
you can do which forces the situation to be one of the $M_k$ above:
first, use the diode into $U_1$ for all other $U_i$'s,
so that either no electrons exist at all or $U_1$ has an electron.
Repeat with the other $U_i$.
This leaves us at the situation described at the start of the problem.
Then you could guess the answer was ``no'' just based on the fact
that it's impossible for $n=2,3$
and that there doesn't seem to be a reasonable strategy.
In this way it's possible to give a pretty good description
of what it's possible to do.
One possible phrasing:
``the physicist can arrange the usamons in a line such that
all the charged usamons are to the left of the un-charged usamons,
but can't determine the \emph{number} of charged usamons''.
\end{remark*}
|
sols-TST-IMO-2015_4
|
Let $f \colon \QQ \to \QQ$ be a function such that for any $x,y \in \QQ$,
the number $f(x+y)-f(x)-f(y)$ is an integer.
Decide whether there must exist a constant $c$
such that $f(x) - cx$ is an integer for every rational number $x$.
|
No, such a constant need not exist.
One possible solution is as follows:
define a sequence by $x_0 = 1$ and
\begin{align*}
2x_1 &= x_0 \\
2x_2 &= x_1 + 1 \\
2x_3 &= x_2 \\
2x_4 &= x_3 + 1 \\
2x_5 &= x_4 \\
2x_6 &= x_5 + 1 \\
&\vdotswithin=
\end{align*}
Set $f(2^{-k}) = x_k$ and $f(2^k)=2^k$ for $k=0,1,\dots$.
Then, let
\[ f\left( a \cdot 2^k + \frac bc \right)
= a f\left( 2^k \right) + \frac bc \]
for odd integers $a$, $b$, $c$.
One can verify this works.
A second shorter solution (given by the proposer) is to set,
whenever $\gcd(p,q) = 1$ and $q > 0$,
\[ f\left( \frac pq \right)
= \frac pq \left( 1! + 2! + \dots + q! \right). \]
\begin{remark*}
Silly note: despite appearances,
$f(x) = \left\lfloor x \right\rfloor$
is not a counterexample since one can take $c = 0$.
\end{remark*}
|
sols-TST-IMO-2015_5
|
Fix a positive integer $n$.
Find the smallest positive integer $\chi$ for which
there exists a tournament on $n$ vertices,
and a coloring of each of the tournament's edges by one of $\chi$ colors, such that:
any two directed edges $u \to v$ and $v \to w$ have different colors.
|
The answer is
\[ \chi = \left\lceil \log_2 n \right\rceil. \]
First, we prove by induction on $n$
that $\chi \ge \log_2 n$ for any coloring and any tournament.
The base case $n = 1$ is obvious.
Now given any tournament, consider any used color $c$.
Then it should be possible to divide the tournament
into two subsets $A$ and $B$ such that all $c$-colored edges point from $A$ to $B$
(for example by letting $A$ be all vertices
which are the starting point of a $c$-edge).
\begin{center}
\begin{asy}
size(5cm);
filldraw(box( (-2,-2), (-0.6,2) ), opacity(0.1)+lightcyan, lightblue );
filldraw(box( (0.6,-2), (2,2) ), opacity(0.1)+lightcyan, lightblue );
label("$A$", (-2,0), dir(180), blue);
label("$B$", ( 2,0), dir( 0), blue);
label("all edges colored $c$", (0,2), dir(90), deepgreen);
draw( (-1.3,1.2)--(1.5,0.2), deepgreen, EndArrow);
draw( (-0.9,-0.2)--(1.1, -0.1), deepgreen, EndArrow);
draw( (-1.4, 0.4)--(0.9, 0.8), deepgreen, EndArrow);
draw( (-1.4, -0.8)--(1.3, -0.6), deepgreen, EndArrow);
draw( (-1.3, -0.5)--(1.2, -1.6), deepgreen, EndArrow);
\end{asy}
\end{center}
One of $A$ and $B$ has size at least $n/2$, say $A$.
Since $A$ has no $c$ edges,
and uses at least $\log_2 |A|$ colors other than $c$, we get
\[ \chi \ge 1 + \log_2 (n/2) = \log_2 n \]
completing the induction.
One can read the construction off from the argument above,
but here is a concrete description.
For each integer $n$, consider the tournament whose vertices are
the binary representations of $S =\{0, \dots, n-1\}$.
Instantiate colors $c_1$, $c_2$, \dots.
Then for $v, w \in S$,
we look at the smallest order bit for which they differ; say the $k$th one.
If $v$ has a zero in the $k$th bit, and $w$ has a one in the $k$th bit,
we draw $v \to w$.
Moreover we color the edge with color $c_k$.
This works and uses at most $\left\lceil \log_2 n \right\rceil$ colors.
\begin{remark*}
[Motivation]
The philosophy ``combinatorial optimization'' applies here.
The idea is given any color $c$,
we can find sets $A$ and $B$ such that all $c$-edges point $A$ to $B$.
Once you realize this, the next insight is to realize
that you may as well color \emph{all} the edges from $A$ to $B$ by $c$;
after all, this doesn't hurt the condition and makes your life easier.
Hence, if $f$ is the answer, we have already a proof that
$f(n) = 1 + \max \left( f(|A|), f(|B|) \right)$
and we choose $|A| \approx |B|$.
This optimization also gives the inductive construction.
\end{remark*}
|
sols-TST-IMO-2015_6
|
Let $ABC$ be a non-equilateral triangle
and let $M_a$, $M_b$, $M_c$ be the midpoints
of the sides $BC$, $CA$, $AB$, respectively.
Let $S$ be a point lying on the Euler line.
Denote by $X$, $Y$, $Z$ the second intersections
of $M_a S$, $M_b S$, $M_c S$ with the nine-point circle.
Prove that $AX$, $BY$, $CZ$ are concurrent.
\end{enumerate}
|
We assume now and forever that $ABC$ is scalene since the problem
follows by symmetry in the isosceles case.
We present four solutions.
\paragraph{First solution by barycentric coordinates (Evan Chen).}
Let $AX$ meet $M_b M_c$ at $D$,
and let $X$ reflected over $M_b M_c$'s midpoint be $X'$.
Let $Y'$, $Z'$, $E$, $F$ be similarly defined.
\begin{center}
\begin{asy}
pair M_a = dir(110);
pair M_b = dir(210);
pair M_c = dir(330);
pair A = M_b+M_c-M_a;
pair B = M_c+M_a-M_b;
pair C = M_a+M_b-M_c;
pair O = circumcenter(A, B, C);
pair S = 0.65*O;
pair N = origin;
pair X = -M_a+2*foot(N,M_a,S);
pair Y = -M_b+2*foot(N,M_b,S);
pair Z = -M_c+2*foot(N,M_c,S);
filldraw(unitcircle, opacity(0.1)+lightcyan, lightblue);
draw(M_a--M_b--M_c--cycle, lightblue);
draw(A--B--C--cycle, lightblue);
draw(X--M_a, cyan);
draw(Y--M_b, cyan);
draw(Z--M_c, cyan);
pair D = extension(X, A, M_b, M_c);
pair Dp = M_b+M_c-D;
pair E = extension(Y, B, M_c, M_a);
pair Ep = M_c+M_a-E;
pair F = extension(Z, C, M_a, M_b);
pair Fp = M_a+M_b-F;
draw(A--D, deepgreen);
draw(B--E, deepgreen);
draw(C--F, deepgreen);
draw(M_a--Dp, lightred);
draw(M_b--Ep, lightred);
draw(M_c--Fp, lightred);
draw(M_a--D, lightred+dotted);
draw(M_b--E, lightred+dotted);
draw(M_c--F, lightred+dotted);
dot("$X$", X, dir(250));
dot("$Y$", Y, dir(80));
dot("$Z$", Z, dir(200));
dot("$M_a$", M_a, dir(M_a));
dot("$M_b$", M_b, dir(M_b));
dot("$M_c$", M_c, dir(M_c));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$S$", S, dir(310));
dot("$D$", D, dir(D));
dot("$D'$", Dp, dir(Dp));
dot("$E$", E, dir(E));
dot("$E'$", Ep, dir(Ep));
dot("$F$", F, dir(220));
dot("$F'$", Fp, dir(Fp));
/* TSQ Source:
M_a = dir 110
M_b = dir 210
M_c = dir 330
A = M_b+M_c-M_a
B = M_c+M_a-M_b
C = M_a+M_b-M_c
O := circumcenter A B C
S = 0.65*O R310
N := origin R45
X := -M_a+2*foot(N,M_a,S)
Y := -M_b+2*foot(N,M_b,S)
Z := -M_c+2*foot(N,M_c,S)
unitcircle 0.1 lightcyan / lightblue
M_a--M_b--M_c--cycle lightblue
A--B--C--cycle lightblue
X--M_a cyan
Y--M_b cyan
Z--M_c cyan
D = extension X A M_b M_c
D' = M_b+M_c-D
E = extension Y B M_c M_a
E' = M_c+M_a-E
F = extension Z C M_a M_b R220
F' = M_a+M_b-F
A--D deepgreen
B--E deepgreen
C--F deepgreen
M_a--Dp lightred
M_b--Ep lightred
M_c--Fp lightred
M_a--D lightred dotted
M_b--E lightred dotted
M_c--F lightred dotted
!dot("$X$", X, dir(250));
!dot("$Y$", Y, dir(80));
!dot("$Z$", Z, dir(200));
*/
\end{asy}
\end{center}
By Cevian Nest Theorem it suffices to
prove that $M_a D$, $M_b E$, $M_c F$ are concurrent.
Taking the isotomic conjugate and recalling that
$M_a M_b A M_c$ is a parallelogram,
we see that it suffices to prove
$M_a X'$, $M_b Y'$, $M_c Z'$ are concurrent.
We now use barycentric coordinates on $\triangle M_a M_b M_c$.
Let \[ S = \left( a^2S_A + t : b^2S_B + t : c^2S_C + t \right) \]
(possibly $t=\infty$ if $S$ is the centroid).
Let $v = b^2S_B+t$, $w=c^2S_C+t$.
Hence \[ X = \left( -a^2 vw : (b^2w+c^2v)v : (b^2w+c^2v)w \right). \]
Consequently,
\[ X' = \left( a^2vw : -a^2vw + (b^2w+c^2v)w : -a^2vw + (b^2w+c^2v)v \right) \]
We can compute
\[ b^2w+c^2v = (bc)^2(S_B+S_C)+ (b^2+c^2)t = (abc)^2 + (b^2+c^2)t. \]
Thus
\[ -a^2v+b^2w+c^2v
= (b^2+c^2)t + (abc)^2 - (ab)^2S_B - a^2t =
S_A( (ab)^2 +t ). \]
Finally
\[ X' = \left( a^2vw :
S_A (c^2S_C+t) \left( (ab)^2 +2t \right) :
S_A (b^2S_B+t) \left( (ac)^2 +2t \right)
\right) \]
and from this it's evident that $AX'$, $BY'$, $CZ'$ are concurrent.
\paragraph{Second solution by moving points (Anant Mudgal).}
Let $H_a$, $H_b$, $H_c$ be feet of altitudes,
and let $\gamma$ denote the nine-point circle.
The main claim is that:
\begin{claim*}
Lines $XH_a$, $YH_b$, $ZH_c$ are concurrent,
\end{claim*}
\begin{proof}
In fact, we claim that the concurrence point lies on the Euler line $\ell$.
This gives us a way to apply the moving points method:
fix triangle $ABC$ and animate $S \in \ell$;
then the map
\begin{align*}
\ell & \to \gamma \to \ell \\
S &\mapsto X \mapsto S_a \coloneq \ell \cap \ol{H_a X}
\end{align*}
is projective, because it consists of two perspectivities.
So we want the analogous maps $S \mapsto S_b$, $S \mapsto S_c$ to coincide.
For this it suffices to check three positions of $S$;
since you're such a good customer here are four.
\begin{itemize}
\ii If $S$ is the orthocenter of $\triangle M_a M_b M_c$
(equivalently the circumcenter of $\triangle ABC$)
then $S_a$ coincides with the circumcenter of $M_a M_b M_c$
(equivalently the nine-point center of $\triangle ABC$).
By symmetry $S_b$ and $S_c$ are too.
\ii If $S$ is the circumcenter of $\triangle M_a M_b M_c$
(equivalently the nine-point center of $\triangle ABC$)
then $S_a$ coincides with the de Longchamps point of $\triangle M_a M_b M_c$
(equivalently orthocenter of $\triangle ABC$).
By symmetry $S_b$ and $S_c$ are too.
\ii If $S$ is either of the intersections of the Euler
line with $\gamma$, then $S = S_a = S_b = S_c$
(as $S = X = Y = Z$).
\end{itemize}
This concludes the proof.
\end{proof}
\begin{center}
\begin{asy}
pair M_a = dir(110);
pair M_b = dir(210);
pair M_c = dir(330);
pair A = M_b+M_c-M_a;
pair B = M_c+M_a-M_b;
pair C = M_a+M_b-M_c;
pair O = circumcenter(A, B, C);
pair S = 0.75*O;
pair N = origin;
pair X = -M_a+2*foot(N,M_a,S);
pair Y = -M_b+2*foot(N,M_b,S);
pair Z = -M_c+2*foot(N,M_c,S);
filldraw(unitcircle, opacity(0.1)+lightcyan, lightblue);
draw(M_a--M_b--M_c--cycle, lightblue);
draw(A--B--C--cycle, lightblue);
draw(X--M_a, cyan);
draw(Y--M_b, cyan);
draw(Z--M_c, cyan);
pair H_a = M_b*M_c/M_a;
pair H_b = M_c*M_a/M_b;
pair H_c = M_a*M_b/M_c;
draw(X--A, deepgreen);
draw(Y--B, deepgreen);
draw(Z--C, deepgreen);
filldraw(H_a--H_b--H_c--cycle, opacity(0.1)+lightred, lightred);
draw(H_a--X, lightred);
draw(H_b--Y, lightred);
draw(H_c--Z, lightred);
dot("$X$", X, dir(250));
dot("$Y$", Y, dir(80));
dot("$Z$", Z, dir(200));
dot("$M_a$", M_a, dir(M_a));
dot("$M_b$", M_b, dir(M_b));
dot("$M_c$", M_c, dir(M_c));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$S$", S, dir(310));
dot("$H_a$", H_a, dir(H_a));
dot("$H_b$", H_b, dir(H_b));
dot("$H_c$", H_c, dir(H_c));
/* TSQ Source:
M_a = dir 110
M_b = dir 210
M_c = dir 330
A = M_b+M_c-M_a
B = M_c+M_a-M_b
C = M_a+M_b-M_c
O := circumcenter A B C
S = 0.75*O R310
N := origin R45
X := -M_a+2*foot(N,M_a,S)
Y := -M_b+2*foot(N,M_b,S)
Z := -M_c+2*foot(N,M_c,S)
unitcircle 0.1 lightcyan / lightblue
M_a--M_b--M_c--cycle lightblue
A--B--C--cycle lightblue
X--M_a cyan
Y--M_b cyan
Z--M_c cyan
H_a = M_b*M_c/M_a
H_b = M_c*M_a/M_b
H_c = M_a*M_b/M_c
X--A deepgreen
Y--B deepgreen
Z--C deepgreen
H_a--H_b--H_c--cycle 0.1 lightred / lightred
H_a--X lightred
H_b--Y lightred
H_c--Z lightred
!dot("$X$", X, dir(250));
!dot("$Y$", Y, dir(80));
!dot("$Z$", Z, dir(200));
*/
\end{asy}
\end{center}
We now use Trig Ceva to carry over the concurrence.
By sine law,
\[ \frac{\sin \angle M_{c}AX}{\sin \angle AM_{c}X}
=\frac{M_{c}X}{AX} \]
and a similar relation for $M_b$ gives that
\[ \frac{\sin \angle M_{c}AX}{\sin \angle M_{b}AX}
= \frac{\sin \angle AM_{c}X}{\sin \angle AM_{b}X} \cdot \frac{M_{c}X}{M_{b}X}
= \frac{\sin \angle AM_{c}X}{\sin \angle AM_{b}X}
\cdot \frac{\sin \angle XM_{a}M_{c}}{\sin \angle XM_{a}M_{b}}. \]
Thus multiplying cyclically gives
\[ \prod_{\text{cyc}} \frac{\sin \angle M_{c}AX}{\sin \angle M_{b}AX}
= \prod_{\text{cyc}} \frac{\sin \angle AM_{c}X}{\sin \angle AM_{b}X}
\prod_{\text{cyc}} \frac{\sin \angle XM_{a}M_{c}}{\sin \angle XM_{a}M_{b}}. \]
The latter product on the right-hand side equals $1$
by Trig Ceva on $\triangle M_a M_b M_c$ with cevians
$\ol{M_a X}$, $\ol{M_b Y}$, $\ol{M_c Z}$.
The former product also equals $1$
by Trig Ceva for the concurrence in the previous claim
(and the fact that $\angle AM_{c}X=\angle H_{c}H_{a}X$).
Hence the left-hand side equals $1$, implying the result.
\paragraph{Third solution by moving points (Gopal Goel).}
In this solution,
we will instead use barycentric coordinates with resect to $\triangle ABC$
to bound the degrees suitably,
and then verify for seven distinct choices of $S$.
We let $R$ denote the radius of $\triangle ABC$,
and $N$ the nine-point center.
First, imagine solving for $X$ in the following way.
Suppose $\vec X = (1-t_a) \vec M_a + t_a \vec S$.
Then, using the dot product
(with $\left\lvert \vec v \right\rvert^2 = \vec v \cdot \vec v$ in general)
\begin{align*}
\frac14 R^2
&= \left\lvert \vec X - \vec N \right\rvert^2 \\
&= \left\lvert t_a(\vec S - \vec M_a) + \vec M_a - \vec N \right\rvert^2 \\
&= \left\lvert t_a(\vec S - \vec M_a) \right\rvert^2
+ 2t_a \left( \vec S - \vec M_a \right) \cdot \left( \vec M_a - \vec N \right)
+ \left\lvert \vec M_a - \vec N \right\rvert^2 \\
&= t_a^2 \left\lvert (\vec S - \vec M_a) \right\rvert^2
+ 2t_a \left( \vec S - \vec M_a \right) \cdot \left( \vec M_a - \vec N \right)
+ \frac 14 R^2
\end{align*}
Since $t_a \neq 0$ we may solve to obtain
\[ t_a = -\frac{2(\vec M_a - \vec N) \cdot (\vec S - \vec M_a)}%
{\left\lvert \vec S - \vec M_a \right\rvert^2}. \]
Now imagine $S$ varies along the Euler line,
meaning there should exist linear functions
$\alpha, \beta, \gamma \colon \RR \to \RR$ such that
\[ S = \left( \alpha(s) , \beta(s) , \gamma(s) \right)
\qquad s \in \RR \]
with $\alpha(s) + \beta(s) + \gamma(s) = 1$.
Thus $t_a = \frac{f_a}{g_a} = \frac{f_a(s)}{g_a(s)}$
is the quotient of a linear function
$f_a(s)$ and a quadratic function $g_a(s)$.
So we may write:
\begin{align*}
X &= (1-t_a) \left( 0, \half, \half \right)
+ t_a \left( \alpha, \beta, \gamma \right) \\
&= \left( t_a \alpha, \half(1-t_a) + t_a \beta,
\half(1-t_a) + t_a \gamma \right) \\
&= \left( 2f_a \alpha : g_a - f_a + 2f_a \beta : g_a - f_a + 2 f_a \gamma \right).
\end{align*}
Thus the coordinates of $X$ are quadratic polynomials in $s$
when written in this way.
In a similar way, the coordinates of $Y$ and $Z$
should be quadratic polynomials in $s$.
The Ceva concurrence condition
\[ \prod_{\text{cyc}} \frac{g_a-f_a+2f_a\beta}{g_a-f_a+2f_a\gamma}
= 1 \]
is thus a polynomial in $s$ of degree at most six.
Our goal is to verify it is identically zero,
thus it suffices to check seven positions of $S$.
\begin{itemize}
\ii If $S$ is the circumcenter of $\triangle M_a M_b M_c$
(equivalently the nine-point center of $\triangle ABC$)
then $\ol{AX}$, $\ol{BY}$, $\ol{CZ}$
are altitudes of $\triangle ABC$.
\ii If $S$ is the centroid of $\triangle M_a M_b M_c$
(equivalently the centroid of $\triangle ABC$),
then $\ol{AX}$, $\ol{BY}$, $\ol{CZ}$
are medians of $\triangle ABC$.
\ii If $S$ is either of the intersections of the Euler
line with $\gamma$, then $S = X = Y = Z$
and all cevians concur at $S$.
\ii If $S$ lies on the $\ol{M_a M_b}$,
then $Y = M_a$, $X = M_c$, and thus $\ol{AX} \cap \ol{BY} = C$,
which is of course concurrent with $\ol{CZ}$
(regardless of $Z$).
Similarly if $S$ lies on the other sides of $\triangle M_a M_b M_c$.
\end{itemize}
Thus we are also done.
\paragraph{Fourth solution using Pascal (official one).}
We give a different proof of the claim that
$\ol{XH_a}$, $\ol{YH_b}$, $\ol{ZH_c}$ are concurrent
(and then proceed as in the end of the second solution).
Let $H$ denote the orthocenter, $N$ the nine-point center,
and moreover let $N_a$, $N_b$, $N_c$ denote the midpoints
of $\ol{AH}$, $\ol{BH}$, $\ol{CH}$,
which also lie on the nine-point circle
(and are the antipodes of $M_a$, $M_b$, $M_c$).
\begin{itemize}
\ii By Pascal's theorem on $M_b N_b H_b M_c N_c H_c$,
the point $P = \ol{M_c H_b} \cap \ol{M_b H_c}$
is collinear with $N = \ol{M_b N_b} \cap \ol{M_c N_c}$,
and $H = \ol{N_b H_b} \cap \ol{N_c H_c}$.
So $P$ lies on the Euler line.
\ii By Pascal's theorem on $M_b Y H_b M_c Z H_c$,
the point $\ol{YH_b} \cap \ol{ZH_c}$ is collinear
with $S = \ol{M_b Y} \cap \ol{M_c Z}$
and $P = \ol{M_b H_c} \cap \ol{M_c H_b}$.
Hence $YH_b$ and $ZH_c$ meet on the Euler line, as needed.
\end{itemize}
|
sols-TST-IMO-2016_1
|
Let $S = \{1, \dots, n\}$.
Given a bijection $f \colon S \to S$ an \emph{orbit} of $f$ is a
set of the form $\{x, f(x), f(f(x)), \dots \}$ for some $x \in S$.
We denote by $c(f)$ the number of distinct orbits of $f$.
For example, if $n=3$ and $f(1)=2$, $f(2)=1$, $f(3)=3$,
the two orbits are $\{1,2\}$ and $\{3\}$, hence $c(f)=2$.
Given $k$ bijections $f_1$, \dots, $f_k$ from $S$ to itself, prove that
\[ c(f_1) + \dots + c(f_k) \le n(k-1) + c(f) \]
where $f \colon S \to S$ is the composed function $f_1 \circ \dots \circ f_k$.
|
Most motivated solution is to consider $n - c(f)$ and show this is the transposition distance.
Dumb graph theory works as well.
|
sols-TST-IMO-2016_2
|
Let $ABC$ be a scalene triangle with circumcircle $\Omega$,
and suppose the incircle of $ABC$ touches $BC$ at $D$.
The angle bisector of $\angle A$ meets $BC$ and $\Omega$ at $K$ and $M$.
The circumcircle of $\triangle DKM$ intersects the $A$-excircle
at $S_1$, $S_2$, and $\Omega$ at $T \neq M$.
Prove that line $AT$ passes through either $S_1$ or $S_2$.
|
We present an angle-chasing solution,
and then a more advanced alternative finish.
\paragraph{First solution (angle chasing).}
Assume for simplicity $AB < AC$.
Let $E$ be the contact point of the $A$-excircle on $BC$;
also let ray $TD$ meet $\Omega$ again at $L$.
From the fact that $\angle MTL = \angle MTD = 180^{\circ} - \angle MKD$,
we can deduce that $\angle MTL = \angle ACM$,
meaning that $L$ is the reflection of $A$ across the
perpendicular bisector $\ell$ of $BC$.
If we reflect $T$, $D$, $L$ over $\ell$, we deduce $A$, $E$ and the reflection of $T$ across
$\ell$ are collinear, which implies that $\angle BAT = \angle CAE$.
Now, consider the reflection point $E$ across line $AI$, say $S$.
Since ray $AI$ passes through the $A$-excenter, $S$ lies on the $A$-excircle.
Since $\angle BAT = \angle CAE$, $S$ also lies on ray $AT$.
But the circumcircles of triangles $DKM$ and $KME$ are congruent (from $DM = EM$),
so $S$ lies on the circumcircle of $\triangle DKM$ too.
Hence $S$ is the desired intersection point.
\begin{center}
\begin{asy}
size(12cm);
pair A = dir(40);
pair B = dir(210);
pair C = dir(330);
pair I = incenter(A, B, C);
pair D = foot(I, B, C);
pair K = extension(A, I, B, C);
pair M = dir(-90);
pair E = B+C-D;
pair I_A = 2*M-I;
pair S = -E+2*foot(E, K, M);
pair L = dir(90);
filldraw(A--B--C--cycle, opacity(0.2)+lightblue, blue);
filldraw(unitcircle, opacity(0.1)+lightblue, blue);
filldraw(circumcircle(D, K, M), opacity(0.2)+lightgreen, green);
filldraw(circumcircle(E, K, M), opacity(0.2)+lightred, red);
draw(A--I_A, blue);
draw(arc(I_A, abs(E-I_A), -20, 120), heavycyan);
draw(E--S, orange);
pair T = extension(I, L, A, S);
draw(E--A--S, orange+dashed);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(90));
dot("$K$", K, dir(100));
dot("$M$", M, dir(M));
dot("$E$", E, dir(E));
dot("$I_A$", I_A, dir(I_A));
dot("$S$", S, dir(10));
dot("$T$", T, dir(T));
/* Source generated by TSQ
!size(12cm);
A = dir 40
B = dir 210
C = dir 330
I := incenter A B C
D = foot I B C R90
K = extension A I B C R100
M = dir -90
E = B+C-D
I_A = 2*M-I
S = -E+2*foot E K M R10
L := dir 90
A--B--C--cycle 0.2 lightblue / blue
unitcircle 0.1 lightblue / blue
circumcircle D K M 0.2 lightgreen / green
circumcircle E K M 0.2 lightred / red
A--I_A blue
!draw(arc(I_A, abs(E-I_A), -20, 120), heavycyan);
E--S orange
T = extension I L A S
E--A--S orange dashed
*/
\end{asy}
\end{center}
\paragraph{Second solution (advanced).}
It's known that $T$ is the touch-point of the $A$-mixtilinear incircle.
Let $E$ be contact point of $A$-excircle on $BC$.
Now the circumcircles of $\triangle DKM$ and $\triangle KME$ are congruent,
since $DM = ME$ and the angles at $K$ are supplementary.
Let $S$ be the reflection of $E$ across line $KM$, which by
the above the above comment lies on the circumcircle of $\triangle DKM$.
Since $KM$ passes through the $A$-excenter, $S$ also lies on the $A$-excircle.
But $S$ also lies on line $AT$, since lines $AT$ and $AE$ are isogonal
(the mixtilinear cevian is isogonal to the Nagel line).
Thus $S$ is the desired intersection point.
\paragraph{Authorship comments.}
This problem comes from an observation of mine:
let $ABC$ be a triangle,
let the $\angle A$ bisector meet $\ol{BC}$ and $(ABC)$ at $E$ and $M$.
Let $W$ be the tangency point of the $A$-mixtilinear excircle
with the circumcircle of $ABC$.
Then $A$-Nagel line passed through a common intersection
of the circumcircle of $\triangle MEW$
and the $A$-mixtilinear incircle.
This problem is the inverted version of this observation.
|
sols-TST-IMO-2016_3
|
Let $p$ be a prime number. Let $\FF_p$ denote the integers modulo $p$,
and let $\FF_p[x]$ be the set of polynomials with coefficients in $\FF_p$.
Define $\Psi \colon \FF_p[x] \to \FF_p[x]$ by
\[ \Psi\left( \sum_{i=0}^n a_i x^i \right) = \sum_{i=0}^n a_i x^{p^i}. \]
Prove that for nonzero polynomials $F,G \in \FF_p[x]$,
\[ \Psi(\gcd(F,G)) = \gcd(\Psi(F), \Psi(G)). \]
%Here, a polynomial $Q$ divides $P$ if there exists $R \in \FF_p[x]$
%such that $P(x) - Q(x) R(x)$ is the polynomial with all coefficients $0$
%(with all addition and multiplication in the coefficients taken modulo $p$),
%and the gcd of two polynomials is the highest degree polynomial
%with leading coefficient $1$ which divides both of them.
%A non-zero polynomial is a polynomial with not all coefficients $0$.
%As an example of multiplication, $(x+1)(x+2)(x+3) = x^3+x^2+x+1$
%in $\FF_5[x]$.
|
Observe that $\Psi$ is also a linear map of $\FF_p$ vector spaces,
and that $\Psi(xP) = \Psi(P)^p$ for any $P \in \FF_p[x]$.
(In particular, $\Psi(1) = x$, not $1$, take caution!)
\paragraph{First solution (Ankan Bhattacharya).}
We start with:
\begin{claim*}
If $P \mid Q$ then $\Psi(P) \mid \Psi(Q)$.
\end{claim*}
\begin{proof}
Set $Q = PR$, where $R = \sum_{i=0}^k r_i x^i$.
Then
\[ \Psi(Q) = \Psi\left( P\sum_{i=0}^k r_i x^i \right)
= \sum_{i=0}^k \Psi\left( P \cdot r_i x^i \right)
= \sum_{i=0}^k r_i \Psi(P)^{p^i} \]
which is divisible by $\Psi(P)$.
\end{proof}
This already implies
\[ \Psi(\gcd(F,G)) \mid \gcd(\Psi(F), \Psi(G)). \]
For the converse, by Bezout there exists $A,B \in \FF_p[x]$
such that $AF + BG = \gcd(F,G)$, so taking $\Psi$ of both sides gives
\[ \Psi(AF) + \Psi(BG) = \Psi\left( \gcd(F,G) \right). \]
The left-hand side is divisible by $\gcd(\Psi(F), \Psi(G))$
since the first term is divisible by $\Psi(F)$
and the second term is divisible by $\Psi(G)$.
So $\gcd(\Psi(F), \Psi(G)) \mid \Psi(\gcd(F,G))$
and noting both sides are monic we are done.
\paragraph{Second solution.}
Here is an alternative (longer but more conceptual) way to finish without Bezout lemma.
Let $\beth \subseteq \FF_p[x]$ denote the set of polynomials in the image of $\Psi$,
thus $\Psi \colon \FF_p[x] \to \beth$ is a bijection on the level of sets.
\begin{claim*}
If $A,B \in \beth$ then $\gcd(A,B) \in \beth$.
\end{claim*}
\begin{proof}
It suffices to show that if $A$ and $B$ are monic,
and $\deg A > \deg B$,
then the remainder when $A$ is divided by $B$ is in $\beth$.
Suppose $\deg A = p^k$ and $B = x^{p^{k-1}} - c_2x^{p^{k-2}} - \dots - c_k$.
Then
\begin{align*}
x^{p^k} &\equiv \left( c_2x^{p^{k-2}} + c_3x^{p^{k-3}}
+ \dots + c_k \right)^p \pmod B \\
&\equiv c_2x^{p^{k-1}} + c_3x^{p^{k-2}} \dots + c_k \pmod B
\end{align*}
since exponentiation by $p$ commutes with addition in $\FF_p$.
This is enough to imply the conclusion.
The proof if $\deg B$ is smaller less than $p^{k-1}$ is similar.
\end{proof}
Thus, if we view $\FF_p[x]$ and $\beth$ as partially ordered sets
under polynomial division, then $\gcd$ is the
``greatest lower bound'' or ``meet'' in both partially ordered sets.
We will now prove that $\Psi$ is an \emph{isomorphism} of the posets.
We have already seen that $P \mid Q \implies \Psi(P) \mid \Psi(Q)$
from the first solution. For the converse:
\begin{claim*}
If $\Psi(P) \mid \Psi(Q)$ then $P \mid Q$.
\end{claim*}
\begin{proof}
Suppose $\Psi(P) \mid \Psi(Q)$, but $Q=PA+B$ where $\deg B < \deg P$.
Thus $\Psi(P) \mid \Psi(PA) + \Psi(B)$, hence $\Psi(P) \mid \Psi(B)$,
but $\deg \Psi(P) > \deg \Psi(B)$ hence $\Psi(B) = 0 \implies B = 0$.
\end{proof}
This completes the proof.
\begin{remark*}
In fact $\Psi \colon \FF_p[x] \to \beth$ is a ring isomorphism
if we equip $\beth$ with function composition as the ring multiplication.
Indeed in the proof of the first claim
(that $P \mid Q \implies \Psi(P) \mid \Psi(Q)$) we saw that
\[ \Psi(RP) = \sum_{i=0}^k r_i \Psi(P)^{p^i} = \Psi(R) \circ \Psi(P). \]
\end{remark*}
|
sols-TST-IMO-2016_4
|
Let $\sqrt{3}=1.b_1b_2b_3\dots_{(2)}$ be the binary representation
of $\sqrt 3$. Prove that for any positive integer $n$, at least one of
the digits $b_n, b_{n+1}, \dots, b_{2n}$ equals 1.
|
Assume the contrary, so that for some integer $k$ we have
\[k < 2^{n-1} \sqrt 3 < k + \frac{1}{2^{n+1}}. \]
Squaring gives
\begin{align*}
k^2 < 3 \cdot 2^{2n-2} &< k^2 + \frac{k}{2^n} + \frac{1}{2^{2n+2}} \\
&\le k^2 + \frac{2^{n-1} \sqrt 3}{2^n} + \frac{1}{2^{2n+2}} \\
&= k^2 + \frac{\sqrt3}{2} + \frac{1}{2^{2n+2}} \\
&\le k^2 + \frac{\sqrt3}{2} + \frac{1}{16} \\
&< k^2+1
\end{align*}
and this is a contradiction.
|
sols-TST-IMO-2016_5
|
Let $n \ge 4$ be an integer.
Find all functions $W \colon \{1, \dots, n\}^2 \to \RR$ such that
for every partition $[n] = A \cup B \cup C$ into disjoint sets,
\[ \sum_{a \in A} \sum_{b \in B} \sum_{c \in C} W(a,b) W(b,c)
= |A| |B| |C|. \]
|
Of course, $W(k,k)$ is arbitrary for $k \in [n]$.
We claim that $W(a,b) = \pm 1$ for any $a \neq b$, with the sign fixed.
(These evidently work.)
First, let $X_{abc} = W(a,b)W(b,c)$ for all distinct $a$, $b$, $c$,
so the given condition is
\[ \sum_{a,b,c \in A \times B \times C} X_{abc} = |A| |B| |C|. \]
Consider the given equation with the particular choices
\begin{itemize}
\ii $A = \{1\}$, $B = \{2\}$, $C = \{3,4,\dots,n\}$.
\ii $A = \{1\}$, $B = \{3\}$, $C = \{2,4,\dots,n\}$.
\ii $A = \{1\}$, $B = \{2,3\}$, $C = \{4,\dots,n\}$.
\end{itemize}
This gives
\begin{align*}
X_{123} + X_{124} + \dots + X_{12n} &= n-2 \\
X_{132} + X_{134} + \dots + X_{13n} &= n-2 \\
(X_{124} + \dots + X_{12n})
+ (X_{134} + \dots + X_{13n}) &= 2(n-3).
\end{align*}
Adding the first two and
subtracting the last one gives $X_{123} + X_{132} = 2$.
Similarly, $X_{123} + X_{321} = 2$,
and in this way we have $X_{321} = X_{132}$.
Thus $W(3,2)W(2,1) = W(1,3)W(3,2)$,
and since $W(3,2) \neq 0$ (clearly) we get $W(2,1) = W(3,2)$.
Analogously, for any distinct $a$, $b$, $c$ we have $W(a,b) = W(b,c)$.
For $n \ge 4$ this is enough to imply $W(a,b) = \pm 1$ for $a \neq b$
where the choice of sign is the same for all $a$ and $b$.
\begin{remark*}
Surprisingly, the $n = 3$ case has ``extra'' solutions for
$W(1,2) = W(2,3) = W(3,1) = \pm1$,
$W(2,1) = W(3,2) = W(1,3) = \mp1$.
\end{remark*}
\begin{remark*}
[Intuition]
It should still be possible to solve the problem
with $X_{abc}$ in place of $W(a,b) W(b,c)$,
because we have about far more equations than variables $X_{a,b,c}$
so linear algebra assures us we almost certainly have a unique solution.
\end{remark*}
|
sols-TST-IMO-2016_6
|
Let $ABC$ be an acute scalene triangle
and let $P$ be a point in its interior.
Let $A_1$, $B_1$, $C_1$ be projections of $P$ onto
triangle sides $BC$, $CA$, $AB$, respectively.
Find the locus of points $P$ such that
$AA_1$, $BB_1$, $CC_1$ are concurrent
and $\angle PAB + \angle PBC + \angle PCA = 90\dg$.
\end{enumerate}
|
In complex numbers with $ABC$ the unit circle,
it is equivalent to solving the following two cubic equations
in $p$ and $q = \ol p$:
\begin{align*}
(p-a)(p-b)(p-c) &= (abc)^2 (q -1/a)(q - 1/b)(q - 1/c) \\
0 &= \prod_{\text{cyc}} (p+c-b-bcq) + \prod_{\text{cyc}} (p+b-c-bcq).
\end{align*}
Viewing this as two cubic curves in $(p,q) \in \CC^2$,
by \emph{B\'ezout's Theorem} it follows there are at most nine solutions
(unless both curves are not irreducible,
but it's easy to check the first one cannot be factored).
Moreover it is easy to name nine solutions (for $ABC$ scalene):
the three vertices, the three excenters, and $I$, $O$, $H$.
Hence the answer is just those three triangle centers $I$, $O$ and $H$.
\begin{remark*}
On the other hand it is not easy to solve the cubics by hand;
I tried for an hour without success.
So I think this solution is only feasible
with knowledge of algebraic geometry.
\end{remark*}
\begin{remark*}
These two cubics have names:
\begin{itemize}
\ii The locus of $\angle PAB + \angle PBC + \angle PCA = 90\dg$
is the \textbf{McCay cubic},
which is the locus of points $P$
for which $P$, $P^\ast$, $O$ are collinear.
\ii The locus of the pedal condition
is the \textbf{Darboux cubic},
which is the locus of points $P$
for which $P$, $P^\ast$, $L$ are collinear,
$L$ denoting the de Longchamps point.
\end{itemize}
Assuming $P \neq P^\ast$,
this implies $P$ and $P^\ast$
both lie on the Euler line of $\triangle ABC$,
which is possible only if $P=O$ or $P=H$.
Some other synthetic solutions are posted at
\url{https://aops.com/community/c6h1243902p6368189}.
\end{remark*}
|
sols-TST-IMO-2017_1
|
In a sports league, each team uses a set of at most $t$ signature colors.
A set $S$ of teams is \emph{color-identifiable} if one can assign
each team in $S$ one of their signature colors,
such that no team in $S$ is assigned
\emph{any} signature color of a different team in $S$.
For all positive integers $n$ and $t$,
determine the maximum integer $g(n,t)$ such that:
In any sports league with exactly $n$ distinct colors
present over all teams, one can always
find a color-identifiable set of size at least $g(n,t)$.
|
Answer: $\left\lceil n/t \right\rceil$.
To see this is an upper bound, note that one can easily construct
a sports league with that many teams anyways.
A quick warning:
\begin{remark*}
[Misreading the problem]
It is common to misread the problem by ignoring the word ``any''.
Here is an illustration.
Suppose we have two teams, MIT and Harvard;
the colors of MIT are red/gray/black, and the colors of Harvard are red/white.
(Thus $n=4$ and $t=3$.)
The assignment of MIT to gray and Harvard to red is \emph{not} acceptable
because red is a signature color of MIT, even though not the one assigned.
\end{remark*}
We present two proofs of the lower bound.
\paragraph{Approach by deleting teams (Gopal Goel).}
Initially, place all teams in a set $S$.
Then we repeat the following algorithm:
\begin{quote}
If there is a team all of whose signature colors
are shared by some other team in $S$ already, then we delete that team.
\end{quote}
(If there is more than one such team, we pick arbitrarily.)
At the end of the process, all $n$ colors are still present at least once,
so at least $\left\lceil n/t \right\rceil$ teams remain.
Moreover, since the algorithm is no longer possible,
the remaining set $S$ is already color-identifiable.
\begin{remark*}
[Gopal Goel]
It might seem counter-intuitive that we are \emph{deleting} teams from the full set
when the original problem is trying to get a large set $S$.
This is less strange when one thinks
of it instead as ``safely deleting useless teams''.
Basically, if one deletes such a team,
the problem statement implies that the task must still be possible,
since $g(n,t)$ does not depend on the number of teams:
$n$ is the number of colors present, and deleting a useless team does not change this.
It turns out that this optimization is already enough to solve the problem.
\end{remark*}
\paragraph{Approach by adding colors.}
For a constructive algorithmic approach,
the idea is to greedy pick by color (rather than by team),
taking at each step the least used color.
Select the color $C_1$ with the \emph{fewest} teams using it, and a team $T_1$ using it.
Then delete all colors $T_1$ uses, and all teams which use $C_1$.
Note that
\begin{itemize}
\ii By problem condition, this deletes at most $t$ teams total.
\ii Any remaining color $C$ still has at least one user.
Indeed, if not, then $C$ had the same set of teams
as $C_1$ did (by minimality of $C$),
but then it should have deleted as a color of $T_1$.
\end{itemize}
Now repeat this algorithm with $C_2$ and $T_2$, and so on.
This operations uses at most $t$ colors each time,
so we select at least $\left\lceil n/t \right\rceil$ colors.
\begin{remark*}
A greedy approach by team \emph{does not work}.
For example, suppose we try to ``grab teams until no more can be added''.
As before, assume our league has teams, MIT and Harvard;
the colors of MIT are red/gray/black, and the colors of Harvard are red/white.
(Thus $n=4$ and $t=3$.)
If we start by selecting MIT and red, then
it is impossible to select any more teams; but $g(n,t) = 2$.
\end{remark*}
|
sols-TST-IMO-2017_2
|
Let $ABC$ be an acute scalene triangle with circumcenter $O$,
and let $T$ be on line $BC$ such that $\angle TAO = 90\dg$.
The circle with diameter $\ol{AT}$
intersects the circumcircle of $\triangle BOC$ at two points
$A_1$ and $A_2$, where $OA_1 < OA_2$.
Points $B_1$, $B_2$, $C_1$, $C_2$ are defined analogously.
\begin{enumerate}
\ii[(a)]
Prove that $\ol{AA_1}$, $\ol{BB_1}$, $\ol{CC_1}$ are concurrent.
\ii[(b)]
Prove that $\ol{AA_2}$, $\ol{BB_2}$, $\ol{CC_2}$ are concurrent
on the Euler line of triangle $ABC$.
\end{enumerate}
|
Let triangle $ABC$ have circumcircle $\Gamma$.
Let $\triangle XYZ$ be the tangential triangle of $\triangle ABC$
(hence $\Gamma$ is the incircle of $\triangle XYZ$),
and denote by $\Omega$ its circumcircle.
Suppose the symmedian $\ol{AX}$ meets $\Gamma$ again at $D$,
and let $M$ be the midpoint of $\ol{AD}$.
Finally, let $K$ be the Miquel point of quadrilateral $ZBCY$,
meaning it is the intersection of $\Omega$
and the circumcircle of $\triangle BOC$ (other than $X$).
\begin{center}
\begin{asy}
size(12cm);
pair A = dir(115);
pair B = dir(200);
pair C = dir(340);
draw(A--B--C--cycle, blue);
filldraw(circumcircle(A, B, C), opacity(0.1)+lightblue, blue);
pair O = circumcenter(A, B, C);
pair X = 2*B*C/(B+C);
pair Y = 2*C*A/(C+A);
pair Z = 2*A*B/(A+B);
filldraw(circumcircle(X, Y, Z), opacity(0.1)+lightcyan, cyan);
draw(X--Y--Z--cycle, cyan);
filldraw(circumcircle(B, O, C), opacity(0.1)+heavycyan, heavycyan);
pair T = extension(Y, Z, B, C);
draw(Z--T, lightcyan);
draw(T--B, blue+dashed);
pair D = -A+2*foot(O, A, X);
pair U = foot(A, B, C);
filldraw(circumcircle(T, A, U), opacity(0.1)+lightgreen, heavygreen);
pair V = circumcenter(X, Y, Z);
pair M = midpoint(A--D);
pair L = IP(V--(Y+Z-V), circumcircle(X, Y, Z));
pair K = extension(L, A, O, U);
draw(A--X, heavycyan);
draw(K--L, heavygreen+dashed);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$O$", O, dir(O));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
dot("$Z$", Z, dir(Z));
dot("$T$", T, dir(T));
dot("$D$", D, dir(D));
dot("$M$", M, dir(M));
dot("$L$", L, dir(L));
dot("$K$", K, dir(K));
/* TSQ Source:
!size(12cm);
A = dir 115
B = dir 200
C = dir 340
A--B--C--cycle blue
circumcircle A B C 0.1 lightblue / blue
O = circumcenter A B C
X = 2*B*C/(B+C)
Y = 2*C*A/(C+A)
Z = 2*A*B/(A+B)
circumcircle X Y Z 0.1 lightcyan / cyan
X--Y--Z--cycle cyan
circumcircle B O C 0.1 heavycyan / heavycyan
T = extension Y Z B C
Z--T lightcyan
T--B blue dashed
D = -A+2*foot O A X
U := foot A B C
circumcircle T A U 0.1 lightgreen / heavygreen
V := circumcenter X Y Z
M = midpoint A--D
L = IP V--(Y+Z-V) circumcircle X Y Z
K = extension L A O U
A--X heavycyan
K--L heavygreen dashed
*/
\end{asy}
\end{center}
We first claim that $M$ and $K$ are $A_1$ and $A_2$.
In that case $OM < OA < OK$, so $M = A_1$, $K = A_2$.
To see that $M = A_1$, note that $\angle OMX = 90\dg$,
and moreover that $\ol{TA}$, $\ol{TD}$ are tangents to $\Gamma$,
whence we also have $M = \ol{TO} \cap \ol{AD}$.
Thus $M$ lies on both $(BOC)$ and $(AT)$.
This solves part (a) of the problem:
the concurrency point is the symmedian point of $\triangle ABC$.
Now, note that since $K$ is the Miquel point,
\[ \frac{ZK}{YK} = \frac{ZB}{YC} = \frac{ZA}{YA} \]
and hence $\ol{KA}$ is an angle bisector of $\angle ZKY$.
Thus from $(TA;YZ)=-1$ we obtain $\angle TKA = 90\dg$.
It remains to show $\ol{AK}$ passes through a fixed point on the Euler line.
We claim it is the exsimilicenter of $\Gamma$ and $\Omega$.
Let $L$ be the midpoint of the arc $YZ$ of $\triangle XYZ$ not containing $X$.
Then we know that $K$, $A$, $L$ are collinear.
Now the positive homothety sending $\Gamma$ to $\Omega$ maps $A$ to $L$;
this proves the claim.
Finally, it is well-known that the line through $O$
and the circumcenter of $\triangle XYZ$
coincides with the Euler line of $\triangle ABC$;
hence done.
A second approach to (b) presented by many contestants
is to take an inversion around the circumcircle of $ABC$.
In that situation, the part reduces to the following
known lemma: if $\ol{AH_a}$, $\ol{BH_b}$, $\ol{CH_c}$
are the altitudes of a triangle,
then the circumcircles of triangles $OAH_a$, $BOH_b$, $COH_c$
are coaxial, and the radical axis coincides with the Euler line.
Indeed one simply observes that the orthocenter
has equal power to all three circles.
\paragraph{Authorship comments.}
This problem was inspired by the fact that $K$, $A$, $L$
are collinear in the figure,
which was produced by one of my students (Ryan Kim)
in a solution to a homework problem.
I realized for example that this implied that line $AK$
passed through the $X_{56}$ point of $\triangle XYZ$
(which lies on the Euler line of $\triangle ABC$).
This problem was the result of playing around with
the resulting very nice picture:
all the power comes from the ``magic'' point $K$.
|
sols-TST-IMO-2017_3
|
Let $P, Q \in \RR[x]$ be relatively prime nonconstant polynomials.
Show that there can be at most three real numbers $\lambda$
such that $P + \lambda Q$ is the square of a polynomial.
|
This is true even with $\RR$ replaced by $\CC$,
and it will be necessary to work in this generality.
\paragraph{First solution using transformations.}
We will prove the claim in the following form:
\begin{claim*}
Assume $P, Q \in \CC[x]$ are relatively prime.
If $\alpha P + \beta Q$ is a square for four different
choices of the ratio $[\alpha : \beta]$
then $P$ and $Q$ must be constant.
\end{claim*}
Call pairs $(P,Q)$ as in the claim \emph{bad};
so we wish to show the only bad pairs are pairs of constant polynomials.
Assume not, and take a bad pair with $\deg P + \deg Q$ minimal.
By a suitable M\"obius transformation,
we may transform $(P,Q)$ so that the four ratios are $[1:0]$,
$[0:1]$, $[1:-1]$ and $[1:-k]$,
so we find there are polynomials $A$ and $B$ such that
\begin{align*}
A^2 - B^2 &= C^2 \\
A^2 - k B^2 &= D^2
\end{align*}
where $A^2 = P+\lambda_1 Q$, $B^2 = P+\lambda_2 Q$, say.
Of course $\gcd(A,B) = 1$.
Consequently, we have $C^2 = (A+B)(A-B)$
and $D^2 = (A+\mu B)(A-\mu B)$ where $\mu^2 = k$.
Now $\gcd(A,B) = 1$, so $A+B$, $A-B$, $A+ \mu B$ and $A - \mu B$
are squares; id est $(A,B)$ is bad.
This is a contradiction, since $\deg A + \deg B < \deg P + \deg Q$.
\paragraph{Second solution using derivatives (by Zack Chroman).}
We will assume without loss of generality that $\deg P \neq \deg Q$;
if not, then one can replace $(P,Q)$ with $(P+cQ,Q)$
for a suitable constant $c$.
Then, there exist $\lambda_i \in \CC$ and polynomials $R_i$
for $i=1,2,3,4$ such that
\begin{align*}
P + \lambda_i Q &= R_i^2 \\
\implies P' + \lambda_i Q' &= 2 R_i R_i' \\
\implies R_i &\mid Q'(P+\lambda_i Q) - Q(P' + \lambda_i Q')
= Q'P - QP'.
\end{align*}
On the other hand by Euclidean algorithm it follows that
$R_i$ are relatively prime to each other.
Therefore
\[ R_1 R_2 R_3 R_4 \mid Q' P - Q P'. \]
However, we have
\begin{align*}
\sum_1^4 \deg R_i
&\ge \frac{3\max(\deg P, \deg Q) + \min(\deg P, \deg Q)}{2} \\
&\ge \deg P + \deg Q > \deg(Q'P - QP').
\end{align*}
This can only occur if $Q'P - QP' = 0$ or $(P/Q)' = 0$
by the quotient rule!
But $P/Q$ can't be constant, the end.
\begin{remark*}
The result is previously known; see e.g.\
Lemma 1.6 of \url{https://mathweb.ucsd.edu/~ebelmont/ec-notes.pdf}
or Exercise 6.5.L(a) of Vakil's notes.
\end{remark*}
|
sols-TST-IMO-2017_4
|
You are cheating at a trivia contest.
For each question, you can peek at each of the
$n > 1$ other contestant's guesses before writing your own.
For each question, after all guesses are submitted, the emcee announces the correct answer.
A correct guess is worth $0$ points.
An incorrect guess is worth $-2$ points for other contestants,
but only $-1$ point for you, because you hacked the scoring system.
After announcing the correct answer, the emcee proceeds to read out the next question.
Show that if you are leading by $2^{n-1}$ points at any time,
then you can surely win first place.
|
We will prove the result with $2^{n-1}$ replaced
even by $2^{n-2}+1$.
We first make the following reductions.
First, change the weights to be $+1$, $-1$, $0$ respectively
(rather than $0$, $-2$, $-1$); this clearly has no effect.
Also, WLOG that all contestants except you initially have score zero
(and that your score exceeds $2^{n-2}$).
WLOG ignore rounds in which all answers are the same.
Finally, ignore rounds in which you get the correct answer,
since that leaves you at least as well off as before --- in other words,
we'll assume your score is always fixed,
but you can pick any group of people with the same answers
and ensure they lose 1 point,
while some other group gains $1$ point.
The key observation is the following.
Consider two rounds $R_1$ and $R_2$ such that:
\begin{itemize}
\ii In round $R_1$, some set $S$ of contestants gains a point.
\ii In round $R_2$, the set $S$ of contestants all have the same answer.
\end{itemize}
Then, if we copy the answers of contestants in $S$ during $R_2$,
then the sum of the scorings in $R_1$ and $R_2$ cancel each other out.
In other words we can then ignore $R_1$ and $R_2$ forever.
We thus consider the following strategy.
We keep a list $\mathcal L$ of subsets of $\{1, \dots, n\}$, initially empty.
Now do the following strategy:
\begin{itemize}
\ii On a round, suppose there exists a set $S$ of people
with the same answer such that $S \in \mathcal L$.
$($If multiple exist, choose one arbitrarily.)
Then, copy the answer of $S$, causing them to lose a point.
Delete $S$ from $\mathcal L$.
(Importantly, we do not add any new sets to $\mathcal L$.)
\ii Otherwise, copy any set $T$ of contestants,
selecting $|T| \ge n/2$ if possible.
Let $S$ be the set of contestants who answer correctly (if any),
and add $S$ to the list $\mathcal L$.
Note that $|S| \le n/2$, since $S$ is disjoint from $T$.
\end{itemize}
By construction, $\mathcal L$ has no duplicate sets.
So the score of any contestant $c$ is bounded above by the number of times
that $c$ appears among sets in $\mathcal L$.
The number of such sets is clearly at most $\half \cdot 2^{n-1}$.
So, if you lead by $2^{n-2}+1$ then you ensure victory.
This completes the proof!
\begin{remark*}
Several remarks are in order.
First, we comment on the bound $2^{n-2} + 1$ itself.
The most natural solution using only the
list idea gives an upper bound of $(2^n-2)+1$,
which is the number of nonempty proper subsets of $\{1, \dots, n\}$.
Then, there are two optimizations one can observe:
\begin{itemize}
\ii In fact we can improve to the number of times
any particular contestant $c$ appears in some set,
rather than the total number of sets.
\ii When adding new sets $S$ to $\mathcal L$, one can ensure $|S| \le n/2$.
\end{itemize}
Either observation alone improves the bound from $2^n-1$ to $2^{n-1}$,
but both together give the bound $2^{n-2} + 1$.
Additionally, when $n$ is odd the calculation of subsets
actually gives $2^{n-2} - \half \binom{n-1}{\frac{n-1}{2}} + 1$.
This gives the best possible value at both $n=2$ and $n=3$.
It seems likely some further improvements are possible,
and the true bound is suspected to be polynomial in $n$.
Secondly, the solution is highly motivated by considering a true/false contest
in which only two distinct answers are given per question.
However, a natural mistake (which graders assessed as a two-point deduction)
is to try and prove that in fact one can ``WLOG'' we are in the two-question case.
The proof of this requires substantially more care than expected.
For instance, set $n = 3$.
If $\mathcal L = \left\{ \{1\}, \{2\}, \{3\} \right\}$
then it becomes impossible to prevent a duplicate set from appearing
in $\mathcal L$ if all contestants give distinct answers.
One might attempt to fix this by instead adding to $\mathcal L$
the \emph{complement} of the set $T$ described above.
The example $\mathcal L = \left\{ \{1,2\}, \{2,3\}, \{3,1\} \right\}$
(followed again by a round with all distinct answers)
shows that this proposed fix does not work either.
This issue affects all variations of the above approach.
Because the USA TST did not have any solution-writing process at this time,
this issue was not noticed until January 15 (less than a week before the exam).
When it was brought up by email by Evan,
every organizer who had testsolved the problem had apparently made the same error.
\end{remark*}
\begin{remark*}
Here are some motivations for the solution:
\begin{enumerate}
\ii The exponential bound $2^n$ suggests looking at subsets.
\ii The $n = 2$ case suggests the idea of ``repeated rounds''.
(I think this $n=2$ case is actually really good.)
\ii The ``two distinct answers'' case suggests looking at rounds
as partitions (even though the WLOG does not work,
at least not without further thought).
\ii There's something weird about this problem:
it's a finite bound over unbounded time.
This is a hint to \emph{not} worry excessively about
the actual scores, which turn out to be almost irrelevant.
\end{enumerate}
\end{remark*}
|
sols-TST-IMO-2017_5
|
Let $ABC$ be a triangle with altitude $\ol{AE}$.
The $A$-excircle touches $\ol{BC}$ at $D$,
and intersects the circumcircle at two points $F$ and $G$.
Prove that one can select points $V$ and $N$
on lines $DG$ and $DF$ such that quadrilateral $EVAN$ is a rhombus.
|
Let $I$ denote the incenter, $J$ the $A$-excenter,
and $L$ the midpoint of $\ol{AE}$.
Denote by $\ol{IY}$, $\ol{IZ}$ the tangents
from $I$ to the $A$-excircle.
Note that lines $\ol{BC}$, $\ol{GF}$, $\ol{YZ}$ then concur at $H$
(unless $AB=AC$, but this case is obvious),
as it's the radical center of cyclic hexagon $BICYJZ$,
the circumcircle and the $A$-excircle.
\begin{center}
\begin{asy}
size(12cm);
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
pair I = incenter(A, B, C);
pair M = dir(270);
pair J = 2*M-I;
pair D = foot(J, B, C);
pair E = foot(A, B, C);
pair F = IP(CP(J, D), unitcircle);
pair G = OP(CP(J, D), unitcircle);
filldraw(unitcircle, opacity(0.1)+lightblue, lightblue);
draw(arc(J,abs(D-J),-30,210), lightblue);
pair L = midpoint(A--E);
pair V = extension(G, D, L, L+B-C);
pair N = extension(F, D, L, L+B-C);
draw(A--B--C--cycle, lightblue);
draw(A--E, lightblue);
draw(G--V, lightblue);
draw(N--F, lightblue);
filldraw(E--V--A--N--cycle, opacity(0.1)+lightgreen, heavygreen);
draw(V--N, heavygreen);
pair H = extension(G, F, B, C);
draw(C--H--G, lightblue);
filldraw(circumcircle(B, I, C), opacity(0.1)+lightred, lightred);
pair Y = IP(circumcircle(B, I, C), circumcircle(D, F, G));
pair Z = OP(circumcircle(B, I, C), circumcircle(D, F, G));
pair T = -D+2*foot(J, D, I);
draw(T--H, blue);
draw(Z--H, blue);
draw(L--T, blue);
draw(Y--I--Z, blue);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$I$", I, dir(I));
dot("$M$", M, dir(M));
dot("$J$", J, dir(J));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$F$", F, dir(270));
dot("$G$", G, dir(270));
dot("$L$", L, dir(L));
dot("$V$", V, dir(V));
dot("$N$", N, dir(N));
dot("$H$", H, dir(H));
dot("$Y$", Y, dir(Y));
dot("$Z$", Z, dir(Z));
dot("$T$", T, dir(T));
/* TSQ Source:
!size(12cm);
A = dir 110
B = dir 210
C = dir 330
I = incenter A B C
M = dir 270
J = 2*M-I
D = foot J B C
E = foot A B C
F = IP CP J D unitcircle R270
G = OP CP J D unitcircle R270
unitcircle 0.1 lightblue / lightblue
!draw(arc(J,abs(D-J),-30,210), lightblue);
L = midpoint A--E
V = extension G D L L+B-C
N = extension F D L L+B-C
A--B--C--cycle lightblue
A--E lightblue
G--V lightblue
N--F lightblue
E--V--A--N--cycle 0.1 lightgreen / heavygreen
V--N heavygreen
H = extension G F B C
C--H--G lightblue
circumcircle B I C 0.1 lightred / lightred
Y = IP circumcircle B I C circumcircle D F G
Z = OP circumcircle B I C circumcircle D F G
T = -D+2*foot J D I
T--H blue
Z--H blue
L--T blue
Y--I--Z blue
*/
\end{asy}
\end{center}
Now let $\ol{HD}$ and $\ol{HT}$ be the tangents from $H$ to the $A$-excircle.
It follows that $\ol{DT}$ is the symmedian of $\triangle DZY$,
hence passes through $I = \ol{YY} \cap \ol{ZZ}$.
Moreover, it's well known that $\ol{DI}$ passes through $L$,
the midpoint of the $A$-altitude (for example by homothety).
Finally, $(DT;FG) = -1$, hence project through $D$ onto the line through $L$
parallel to $\ol{BC}$ to obtain $(\infty L; VN) = -1$ as desired.
\paragraph{Authorship comments.}
This is a joint proposal with Danielle Wang (mostly by her).
The formulation given was that the tangents to the $A$-excircle
at $F$ and $G$ was on line $\ol{DI}$;
I solved this formulation using the radical axis argument above.
I then got the idea to involve the point $L$,
already knowing it was on $\ol{DI}$.
Observing the harmonic quadrilateral,
I took perspectivity through $M$ onto the line through $L$ parallel to $\ol{BC}$
(before this I had tried to use the $A$-altitude with little luck).
This yields the rhombus in the problem.
|
sols-TST-IMO-2017_6
|
Prove that there are infinitely many triples $(a,b,p)$ of integers,
with $p$ prime and $0 < a \le b < p$,
for which $p^5$ divides $(a+b)^p - a^p - b^p$.
\end{enumerate}
|
The key claim is that if $p \equiv 1 \pmod 3$,
then
\[ p(x^2+xy+y^2)^2 \text{ divides } (x+y)^p - x^p - y^p \]
as polynomials in $x$ and $y$.
Since it's known that one can select $a$ and $b$ such that
$p^2 \mid a^2 + ab + b^2$, the conclusion follows.
(The theory of quadratic forms tells us we can do it with $p^2 = a^2+ab+b^2$;
Thue's lemma lets us do it by solving $x^2+x+1 \equiv 0 \pmod{p^2}$.)
To prove this, it is the same to show that
\[ (x^2+x+1)^2 \text{ divides } F(x) \coloneq (x+1)^p - x^p - 1. \]
since the binomial coefficients $\binom pk$ are clearly divisible by $p$.
Let $\zeta$ be a third root of unity.
Then $F(\zeta) = (1+\zeta)^p - \zeta^p - 1 = -\zeta^2 - \zeta - 1 = 0$.
Moreover, $F'(x) = p(x+1)^{p-1} - px^{p-1}$,
so $F'(\zeta) = p - p = 0$.
Hence $\zeta$ is a double root of $F$ as needed.
(Incidentally, $p = 2017$ works!)
\begin{remark*}
One possible motivation for this solution is the case $p = 7$.
It is nontrivial even to prove that $p^2$ can divide the expression
if we exclude the situation $a+b=p$ (which provably never achieves $p^3$).
As $p = 3, 5$ fails considering the $p = 7$ polynomial gives
\[ (x+1)^7 - x^7 - 1 = 7x(x+1) \left( x^4 + 2x^3 + 3x^2 + 2x + 1 \right). \]
The key is now to notice that the last factor is $(x^2+x+1)^2$,
which suggests the entire solution.
In fact, even if $p \equiv 2 \pmod 3$,
the polynomial $x^2+x+1$ still divides $(x+1)^p-x^p-1$.
So even the $p = 5$ case can motivate the main idea.
\end{remark*}
|
sols-TST-IMO-2018_1
|
Let $n \ge 2$ be a positive integer,
and let $\sigma(n)$ denote the sum of the positive divisors of $n$.
Prove that the $n$\textsuperscript{th} smallest positive integer
relatively prime to $n$ is at least $\sigma(n)$,
and determine for which $n$ equality holds.
|
The equality case is $n = p^e$ for
$p$ prime and a positive integer $e$.
It is easy to check that this works.
\bigskip
\paragraph{First solution.}
In what follows, by $[a,b]$ we mean $\{a,a+1,\dots,b\}$.
First, we make the following easy observation.
\begin{claim*}
If $a$ and $d$ are positive integers,
then precisely $\varphi(d)$ elements
of $[a, a + d - 1]$ are relatively prime to $d$.
\end{claim*}
Let $d_1$, $d_2$, \dots, $d_k$ denote
denote the divisors of $n$ in some order.
Consider the intervals
\begin{align*}
I_1 &= [1, d_1], \\
I_2 &= [d_1+1, d_1+d_2] \\
&\vdotswithin= \\
I_k &= [d_1+\dots+d_{k-1}+1, d_1+\dots+d_k].
\end{align*}
of length $d_1, \dots, d_k$ respectively.
The $j$th interval will have exactly $\varphi(d_j)$ elements
which are relatively prime $d_j$,
hence at most $\varphi(d_j)$ which are relatively prime to $n$.
Consequently, in $I = \bigcup_{j=1}^k I_k$
there are at most
\[ \sum_{j=1}^k \varphi(d_j) = \sum_{d \mid n} \varphi(d) = n \]
integers relatively prime to $n$.
On the other hand $I = [1,\sigma(n)]$ so this implies the inequality.
We see that the equality holds for $n = p^e$.
Assume now $p < q$ are distinct primes dividing $n$.
Reorder the divisors $d_i$ so that $d_1 = q$.
Then $p,q \in I_1$, and so $I_1$ should contain
strictly fewer than $\varphi(d_1)=q-1$ elements
relatively prime to $n$, hence the inequality is strict.
\paragraph{Second solution (Ivan Borsenco and Evan Chen).}
Let $n = p_1^{e_1} \dots p_k^{e_k}$,
where $p_1 < p_2 < \dots$.
We are going to assume $k \ge 2$,
since the $k=1$ case was resolved in the very beginning,
and prove the strict inequality.
For a general $N$, the number of relatively prime integers in $[1,N]$ is
given exactly by
\[ f(N) = N - \sum_i \left\lfloor \frac{N}{p_i} \right\rfloor
+ \sum_{i<j} \left\lfloor \frac{N}{p_i p_j} \right\rfloor - \dots \]
according to the inclusion-exclusion principle.
So, we wish to show that $f(\sigma(n)) < n$ (as $k \ge 2$).
Discarding the error terms from the floors
(noting that we get at most $1$ from the negative floors)
gives
\begin{align*}
f(N) &< 2^{k-1} + N - \sum_i \frac{N}{p_i}
+ \sum_{i<j} \frac{N}{p_i p_j} - \dots \\
&= 2^{k-1} + N \prod_i \left( 1 - p_i\inv \right) \\
&= 2^{k-1} + \prod_i \left( 1 - p_i\inv \right)
\left( 1 + p_i + p_i^2 + \dots + p_i^{e_i} \right) \\
&= 2^{k-1} + \prod_i \left( p_i^{e_i} - p_i\inv \right).
\end{align*}
The proof is now divided into two cases.
If $k=2$ we have
\begin{align*}
f(N) &< 2 + \left( p_1^{e_1} - p_1\inv \right)\left( p_2^{e_2} -
p_2\inv \right) \\
&= 2 + n - \frac{p_2^{e_2}}{p_1} - \frac{p_1^{e_1}}{p_2}
+ \frac{1}{p_1p_2} \\
&\le 2 + n - \frac{p_2}{p_1} - \frac{p_1}{p_2}
+ \frac{1}{p_1p_2} \\
&= n + \frac{1-(p_1-p_2)^2}{p_1p_2} \le n.
\end{align*}
On the other hand if $k \ge 3$ we may now write
\begin{align*}
f(N) &< 2^{k-1} + \left[ \prod_{i=2}^{k-1} \left( p_i^{e_i} \right)
\right] \left( p_1^{e_1} - p_1\inv \right) \\
&= 2^{k-1} + n - \frac{p_2^{e_2} \dots p_k^{e_k}}{p_1} \\
&\le 2^{k-1} + n - \frac{p_2 p_3 \dots p_k}{p_1}.
\end{align*}
If $p_1 = 2$, then one can show
by induction that $p_2 p_3 \dots p_k \ge 2^{k+1}-1$,
which implies the result.
If $p_1 > 2$, then one can again show by induction
$p_3 \dots p_k \ge 2^k-1$ (since $p_3 \ge 7$),
which also implies the result.
|
sols-TST-IMO-2018_2
|
Find all functions $f \colon \ZZ^2 \to [0,1]$
such that for any integers $x$ and $y$,
\[ f(x, y) = \frac{f(x-1, y) + f(x, y-1)}{2}. \]
|
We claim that the only functions $f$ are constant functions.
(It is easy to see that they work.)
\paragraph{First solution (hands-on).}
First, iterating the functional equation
relation to the $n$th level shows that
\[ f(x, y) = \frac{1}{2^n} \sum_{i=0}^n \binom{n}{i} f(x-i, y-(n-i)). \]
In particular,
\begin{align*}
|f(x, y) - f(x-1, y+1)|
&= \frac{1}{2^n} \left\lvert \sum_{i=0}^{n+1} f(x-i, y-(n-i)) \cdot
\left(\binom{n}{i} - \binom{n}{i-1} \right) \right\rvert \\
&\le \frac{1}{2^n} \sum_{i=0}^{n+1} \left\lvert
\binom{n}{i} - \binom{n}{i-1} \right\rvert \\
&= \frac{1}{2^n} \cdot 2\binom{n}{\lfloor n/2 \rfloor}
\end{align*}
where we define $\binom{n}{n+1} = \binom{n}{-1} = 0$ for convenience.
Since \[ \binom{n}{\lfloor n/2 \rfloor} = o(2^n) \]
it follows that $f$ must be constant.
\begin{remark*}
A very similar proof extends to $d$ dimensions.
\end{remark*}
\paragraph{Second solution (random walks, Mark Sellke).}
We show that if $x+y=x'+y'$ then $f(x,y)=f(x',y')$.
Let $Z_n$, $Z'_n$ be random walks starting at $(x,y)$
and $(x',y')$ and moving down/left.
Then $f(Z_n)$ is a martingale so we have
\[\mathbb E[f(Z_n)]=f(x,y), \qquad
\mathbb E[f(Z'_n)]=f(x',y') .\]
We'll take $Z_n$, $Z'_n$ to be independent until they hit each other,
after which they will stay together. Then
\[|\mathbb E[f(Z_n)-f(Z'_n)]| \leq \mathbb E[|f(Z_n)-f(Z'_n)|]
\leq p_n\]
where $p_n$ is the probability that $Z_n$, $Z'_n$ never collide.
But the distance between $Z_n$, $Z'_n$
is essentially a $1$-dimensional random walk,
so they will collide with probability $1$, meaning $\lim_{n\to\infty} p_n=0$.
Hence
\[ |f(x,y)-f(x',y')| = |\mathbb E[f(Z_n)-f(Z'_n)]| = o(1)\]
as desired.
\begin{remark*}
If the problem were in $\ZZ^d$ for large $d$,
this solution wouldn't work as written
because the independent random walks wouldn't hit each other.
However, this isn't a serious problem
because $Z_n$, $Z'_n$ don't have to be independent
before hitting each other.
Indeed, if every time $Z_n,Z'_n$ agree on a new coordinate
we force them to agree on that coordinate forever,
we can make the two walks individually have the distribution
of a coordinate-decreasing random walk but make them intersect
eventually with probability $1$.
The difference in each coordinate will be a $1$-dimensional
random walk which gets stuck at $0$.
\end{remark*}
\paragraph{Third solution (martingales).}
Imagine starting at $(x, y)$
and taking a random walk down and to the left.
This is a martingale.
As $f$ is bounded, this martingale converges with probability $1$.
Let $X_1, X_2, \dots$ each be random variables
that represent either down moves or left moves with equal probability.
Note that by the Hewitt-Savage 0-1 law,
we have that for any real numbers $a < b$,
\[ \Pr\left[ \lim_{n \to \infty} f((x, y)+X_1+X_2+\dots + X_n)
\in [a, b] \right] \in \{0,1\}. \]
Hence, there exists a single value $v$ such that with probability $1$,
\[ \Pr\left[\lim_{n \to \infty} f((x, y)+X_1+X_2+\dots + X_n)
= v\right] = 1. \]
Obviously, this value $v$ must equal $f(x, y)$.
Now, we show this value $v$ is the same for all $(x, y)$.
Note that any two starting points have a positive chance of meeting.
Therefore, we are done.
|
sols-TST-IMO-2018_4
|
Let $n$ be a positive integer and let $S \subseteq \{0,1\}^n$
be a set of binary strings of length $n$.
Given an odd number $x_1, \dots, x_{2k+1} \in S$ of binary strings
(not necessarily distinct), their \emph{majority} is defined as
the binary string $y \in \{0,1\}^n$ for which
the $i$\textsuperscript{th} bit of $y$ is the most common bit
among the $i$\textsuperscript{th} bits of $x_1$, \dots, $x_{2k+1}$.
(For example, if $n=4$ the majority of
$0000$, $0000$, $1101$, $1100$, $0101$ is $0100$.)
Suppose that for some positive integer $k$,
$S$ has the property $P_k$ that the majority of any $2k+1$
binary strings in $S$ (possibly with repetition) is also in $S$.
Prove that $S$ has the same property $P_k$ for all
positive integers $k$.
|
Let $M$ denote the majority function (of any length).
\paragraph{First solution (induction).}
We prove all $P_k$ are equivalent by induction on $n \ge 2$,
with the base case $n = 2$ being easy to check by hand.
(The case $n=1$ is also vacuous; however,
the inductive step is not able to go from $n=1$ to $n=2$.)
For the inductive step, we proceed by contradiction;
assume $S$ satisfies $P_{\ell}$, but not $P_{k}$,
so there exist $x_1, \dots, x_{2k+1} \in S$
whose majority $y = M(x_1, \dots, x_k)$ is not in $S$.
We contend that:
\begin{claim*}
Let $y_i$ be the string which differs from $y$
only in the $i$\ts{th} bit.
Then $y_i \in S$.
\end{claim*}
\begin{proof}
For a string $s \in S$
we let $\hat s$ denote the string $s$ with the $i$\ts{th} bit deleted
(hence with $n-1$ bits).
Now let \[ T = \left\{ \hat s \mid s \in S \right\}. \]
Since $S$ satisfies $P_\ell$, so does $T$;
thus by the induction hypothesis on $n$, $T$ satisfies $P_{k}$.
Consequently, $T \ni M(\hat{x}_1, \dots, \hat{x}_{2k+1}) = \hat y$.
Thus there exists $s \in S$ such that $\hat s = \hat y$.
This implies $s = y$ or $s = y_i$.
But since we assumed $y \notin S$ it follows $y_i \in S$ instead.
\end{proof}
Now take any $2\ell+1$ copies of the $y_i$, about equally often
(i.e.\ the number of times
any two $y_i$ are taken differs by at most $1$).
We see the majority of these is $y$ itself, contradiction.
\paragraph{Second solution (circuit construction).}
Note that $P_k \implies P_1$ for any $k$, since
\[ M( \underbrace{a,\dots,a}_k, \underbrace{b,\dots,b}_k, c )
= M(a,b,c) \]
for any $a$, $b$, $c$.
We will now prove $P_1 + P_k \implies P_{k+1}$ for any $k$,
which will prove the result.
Actually, we will show that the majority
of any $2k+3$ strings $x_1$, \dots, $x_{2k+3}$
can be expressed by $3$ and $(2k+1)$-majorities.
WLOG assume that $M(x_1, \dots, x_{2k+3}) = 0\dots0$,
and let $\odot$ denote binary AND.
\begin{claim*}
We have $M(x_1, x_2, M(x_3, \dots, x_{2k+3})) = x_1 \odot x_2$.
\end{claim*}
\begin{proof}
Consider any particular bit.
The result is clear if the bits are equal.
Otherwise, if they differ,
the result follows from the original hypothesis that
$M(x_1, \dots, x_{2k+3}) = 0\dots0$
(removing two differing bits does not change the majority).
\end{proof}
By analogy we can construct any $x_i \odot x_j$.
Finally, note that
\[ M(x_1 \odot x_2, x_2 \odot x_3, \dots,
x_{2k+1} \odot x_{2k+2}) = 0\dots0, \]
as desired. (Indeed, if we look at any index,
there were at most $k+1$ $1$'s in the $x_i$ strings,
and hence there will be at most $k$ $1$'s among
$x_i \odot x_{i+1}$ for $i=1,\dots,2k+1$.)
\begin{remark*}
The second solution can be interpreted in circuit language
as showing that all ``$2k+1$-majority gates'' are equivalent.
See also \url{https://cstheory.stackexchange.com/a/21399/48303},
in which Valiant gives a probabilistic construction to prove
that one can construct $(2k+1)$-majority gates from a
\emph{polynomial} number of $3$-majority gates.
No explicit construction is known for this.
\end{remark*}
|
sols-TST-IMO-2018_5
|
Let $ABCD$ be a convex cyclic quadrilateral which is not a kite,
but whose diagonals are perpendicular and meet at $H$.
Denote by $M$ and $N$ the midpoints of $\ol{BC}$ and $\ol{CD}$.
Rays $MH$ and $NH$ meet $\ol{AD}$ and $\ol{AB}$ at $S$ and $T$, respectively.
Prove there exists a point $E$, lying outside quadrilateral $ABCD$,
such that
\begin{itemize}
\ii ray $EH$ bisects both angles $\angle BES$, $\angle TED$, and
\ii $\angle BEN = \angle MED$.
\end{itemize}
|
The main claim is that $E$ is the
intersection of $(ABCD)$ with the circle with diameter $\ol{AH}$.
\begin{center}
\begin{asy}
size(10cm);
pair A = dir(100);
pair B = dir(190);
pair D = dir(-10);
pair F = -A;
pair H = foot(A, B, D);
pair E = foot(A, F, H);
pair C = -A+2*foot(origin, A, H);
filldraw(unitcircle, opacity(0.1)+lightcyan, lightblue);
draw(A--B--C--D--cycle, lightblue);
draw(A--B--F--D--cycle, lightblue);
draw(A--C, lightblue);
draw(B--D, lightblue);
pair M = midpoint(B--C);
pair N = midpoint(D--C);
pair S = foot(H, A, D);
pair T = foot(H, A, B);
draw(M--S, lightblue);
draw(N--T, lightblue);
filldraw(circumcircle(B, T, S), opacity(0.1)+yellow, orange);
// pair Q = midpoint(A--H);
draw(H--F, lightblue);
pair P = midpoint(H--F);
filldraw(B--T--S--D--cycle, opacity(0.1)+yellow, orange+1);
// draw(P--Q, blue);
draw(A--F, blue);
draw(circumcircle(E, B, M), lightgreen);
draw(circumcircle(E, D, N), lightgreen);
draw(H--E--A, orange);
draw(circumcircle(A, T, S), lightgreen);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$D$", D, dir(D));
dot("$F$", F, dir(F));
dot("$H$", H, dir(H));
dot("$E$", E, dir(E));
dot("$C$", C, dir(C));
dot("$M$", M, dir(M));
dot("$N$", N, dir(N));
dot("$S$", S, dir(S));
dot("$T$", T, dir(T));
// dot("$Q$", Q, dir(200));
dot("$P$", P, dir(P));
/* TSQ Source:
!size(10cm);
A = dir 100
B = dir 190
D = dir -10
F = -A
H = foot A B D
E = foot A F H
C = -A+2*foot origin A H
unitcircle 0.1 lightcyan / lightblue
A--B--C--D--cycle lightblue
A--B--F--D--cycle lightblue
A--C lightblue
B--D lightblue
M = midpoint B--C
N = midpoint D--C
S = foot H A D
T = foot H A B
M--S lightblue
N--T lightblue
circumcircle B T S 0.1 yellow / orange
Q = midpoint A--H R200
H--F lightblue
P = midpoint H--F
B--T--S--D--cycle 0.1 yellow / orange+1
P--Q blue
A--F blue
circumcircle E B M lightgreen
circumcircle E D N lightgreen
H--E--A orange
circumcircle A T S lightgreen
*/
\end{asy}
\end{center}
The following observation can be quickly made
without reference to $E$.
\begin{lemma*}
We have $\angle HSA = \angle HTA = 90\dg$.
Consequently, quadrilateral $BTSD$ is cyclic.
\end{lemma*}
\begin{proof}
This is direct angle chasing.
In fact, $\ol{HM}$ passes through the circumcenter of $\triangle BHC$
and $\triangle HAD \sim \triangle HCB$,
so $\ol{HS}$ ought to be the altitude of $\triangle HAD$.
\end{proof}
From here it follows that $E$ is the Miquel point of
cyclic quadrilateral $BTSD$.
Define $F$ to be the point diametrically opposite $A$,
so that $E$, $H$, $F$ are collinear, $\ol{CF} \parallel \ol{BD}$.
By now we already have
\[ \dang BEH = \dang BEF = \dang BAF = \dang CAD = \dang HAS = \dang HES \]
so $\ol{EH}$ bisects $\angle BES$, and $\angle TED$.
Hence it only remains to show $\angle BEM = \angle NED$;
we present several proofs below.
\paragraph{First proof (original solution).}
Let $P$ be the circumcenter of $BTSD$.
The properties of the Miquel point imply $P$ lies on
the common bisector $\ol{EH}$ already,
and it also lies on the perpendicular bisector of $\ol{BD}$,
hence it must be the midpoint of $\ol{HF}$.
We now contend quadrilaterals $BMPS$ and $DNPT$ are cyclic.
Obviously $\ol{MP}$ is the external angle bisector of $\angle BMS$,
and $PB = PS$, so $P$ is the arc midpoint of $(BMS)$.
The proof for $DNPT$ is analogous.
It remains to show $\angle BEN = \angle MED$,
or equivalently $\angle BEM = \angle NED$.
By properties of Miquel point we have $E \in (BMPS) \cap (TPND)$, so
\[ \dang BEM = \dang BPM = \dang PBD = \dang BDP = \dang NPD = \dang NED \]
as desired.
\paragraph{Second proof (2011 G4).}
By 2011 G4, the circumcircle of $\triangle EMN$
is tangent to the circumcircle of $ABCD$.
Hence if we extend $\ol{EM}$ and $\ol{EN}$ to meet $(ABCD)$
again at $X$ and $Y$, we get $\ol{XY} \parallel \ol{MN} \parallel \ol{BD}$.
Thus $\dang BEM = \dang BEX = \dang YED = \dang NED$.
\paragraph{Third proof (involutions, submitted by Daniel Liu).}
Let $G = \ol{BN} \cap \ol{MD}$ denote the centroid of $\triangle BCD$,
and note that it lies on $\ol{EHF}$.
Now consider the dual of Desargues involution theorem on complete
quadrilateral $BMDNCG$ at point $E$.
We get
\[ (EB,ED), \quad (EM,EN), \quad (EC,EG) \]
form an involutive pairing.
However, the bisector of $\angle BED$, say $\ell$,
is also the angle bisector of $\angle CEF$ (since $\ol{CF} \parallel \ol{BD}$).
So the involution we found must coincide with reflection across $\ell$.
This means $\angle MEN$ is bisected by $\ell$ as well, as desired.
\paragraph{Authorship comments.}
This diagram actually comes from the inverted
picture in IMO 2014/3 (which I attended).
I had heard for many years that one could
solve this problem quickly by inversion at $H$ afterwards.
But when I actually tried to do it during an OTIS class
years later, I ended up with the picture in the TST problem,
and couldn't see why it was true!
In the process of trying to reconstruct this rumored solution,
I ended up finding most of the properties
that ended up in the January TST problem
(but were overkill for the original IMO problem).
Let us make the equivalence explicit
by deducing the IMO problem from our work.
Let rays $EM$ and $EN$ meet the circumcircles of $\triangle BHC$
and $\triangle BNC$ again at $X$ and $Y$, with $EM < EX$ and $EN < EY$.
As above we concluded $EM/EX = EN/EY$
and so $\ol{MN} \parallel \ol{XY} \implies \ol{XY} \perp \ol{AHC}$.
Now consider an inversion at $H$ which swaps
$B \leftrightarrow D$ and $A \leftrightarrow C$.
The point $E$ goes to $E^\ast$ diametrically opposite $A$.
Points $X$ and $Y$ go to points on $X^\ast \in \ol{AD}$ and $Y^\ast \in \ol{AB}$.
Since the reflection of $E$ across $\ol{PX}$ is supposed to lie on $(BAE)$,
it follows that the circumcenter of $\triangle HX^\ast E^\ast$
lies on $\ol{AD}$.
Consequently $X^\ast$ plays the role of point ``$T$'' in the IMO problem.
Then $Y^\ast$ plays the role of point ``$S$'' in the IMO problem.
Now the fact that $(HX^\ast Y^\ast)$ is tangent to $\ol{BD}$
is equivalent to $\ol{XY} \perp \ol{AHC}$ which we already knew.
|
sols-TST-IMO-2018_6
|
Alice and Bob play a game.
First, Alice secretly picks a finite set $S$
of lattice points in the Cartesian plane.
Then, for every line $\ell$ in the plane
which is horizontal, vertical, or has slope $+1$ or $-1$,
she tells Bob the number of points of $S$ that lie on $\ell$.
Bob wins if he can then determine the set $S$.
Prove that if Alice picks $S$ to be of the form
\[ S = \left\{ (x,y) \in \ZZ^2 \mid m \le x^2+y^2 \le n \right\} \]
for some positive integers $m$ and $n$, then Bob can win.
(Bob does not know in advance that $S$ is of this form.)
\end{enumerate}
|
Clearly Bob can compute the number $N$ of points.
The main claim is that:
\begin{claim*}
Fix $m$ and $n$ as in the problem statement.
Among all sets $T \subseteq \ZZ^2$ with $N$ points,
the set $S$ is the \emph{unique} one which maximizes the value of
\[ F(T):=\sum_{(x,y)\in T} (x^2+y^2)(m+n-(x^2+y^2)). \]
\end{claim*}
\begin{proof}
Indeed, the different points in $T$ do not interact in this sum,
so we simply want the points $(x,y)$ with $x^2+y^2$
as close as possible to $\frac{m+n}{2}$ which is exactly what $S$ does.
\end{proof}
As a result of this observation,
it suffices to show that Bob has enough information to
compute $F(S)$ from the data given.
(There is no issue with fixing $m$ and $n$,
since Bob can find an upper bound on the magnitude
of the points and then check all pairs $(m,n)$ smaller than that.)
The idea is that he knows the full distribution of each of
$X$, $Y$, $X+Y$, $X-Y$ and hence can compute sums
over $T$ of any power of a single one of those linear functions.
By taking linear combinations we can hence compute $F(S)$.
Let us make the relations explicit.
For ease of exposition we take $Z=(X,Y)$ to be a
uniformly random point from the set $S$.
The information is precisely the individual distributions
of $X$, $Y$, $X+Y$, and $X-Y$.
Now compute
\begin{align*}
\frac{F(S)}{N} &= \mathbb E\left[ (m+n)(X^2+Y^2) - (X^2+Y^2)^2 \right] \\
&= (m+n) \left( \mathbb E[X^2] + \mathbb E[Y^2] \right)
- \mathbb E[X^4] - \mathbb E[Y^4] - 2 \mathbb E[X^2 Y^2].
\end{align*}
On the other hand,
\[ \mathbb E[X^2Y^2]
= \frac{\mathbb E[(X+Y)^4] + \mathbb E[(X-Y)^4]
- 2\mathbb E[X^4] - 2\mathbb E[Y^4]}{12}. \]
Thus we have written $F(S)$ in terms of the distributions
of $X$, $Y$, $X-Y$, $X+Y$ which completes the proof.
\begin{remark*}[Mark Sellke]
\begin{itemize}
\ii This proof would have worked just as well if we
allowed arbitrary $[0,1]$-valued weights on points
with finitely many weights non-zero.
There is an obvious continuum generalization one
can make concerning the indicator function for an annulus.
It's a simpler but fun problem to characterize
when just the vertical/horizontal directions determine the distribution.
\ii An obstruction to purely combinatorial arguments
is that if you take an octagon with points $(\pm a,\pm b)$
and $(\pm b,\pm a)$ then the two ways to pick every other point
(going around clockwise) are indistinguishable by Bob.
This at least shows that Bob's task is far from possible in general,
and hints at proving an inequality.
\ii A related and more standard fact
(among a certain type of person)
is that given a probability distribution $\mu$ on $\RR^n$,
if I tell you the distribution of \emph{all} $1$-dimensional
projections of $\mu$, that determines $\mu$ uniquely.
This works because this information gives me the Fourier transform $\hat{\mu}$,
and Fourier transforms are injective.
For the continuum version of this problem,
this connection gives a much larger family of counterexamples
to any proposed extension to arbitrary non-annular shapes.
Indeed, take a fast-decaying smooth function
$f \colon \RR^2 \to \RR$ which vanishes on the four lines
\[ x=0, \; y=0, \; x+y=0, \; x-y=0.\]
Then the Fourier transform $\hat f$ will have mean $0$
on each line $\ell$ as in the problem statement.
Hence the positive and negative parts of $\hat f$
will not be distinguishable by Bob.
\end{itemize}
\end{remark*}
|
sols-TST-IMO-2019_1
|
Let $ABC$ be a triangle and let $M$ and $N$
denote the midpoints of $\ol{AB}$ and $\ol{AC}$, respectively.
Let $X$ be a point such that $\ol{AX}$
is tangent to the circumcircle of triangle $ABC$.
Denote by $\omega_B$ the circle through $M$ and $B$ tangent to $\ol{MX}$,
and by $\omega_C$ the circle through $N$ and $C$ tangent to $\ol{NX}$.
Show that $\omega_B$ and $\omega_C$ intersect on line $BC$.
|
We present four solutions,
the second of which shows that $M$ and $N$
can be replaced by any two points on $AB$ and $AC$
satisfying $AM/AB + AN/AC = 1$.
\paragraph{First solution using symmedians (Merlijn Staps).}
Let $\ol{XY}$ be the other tangent from $X$ to $(AMN)$.
\begin{claim*}
Line $\ol{XM}$ is tangent to $(BMY)$;
hence $Y$ lies on $\omega_B$.
\end{claim*}
\begin{center}
\begin{asy}
size(11cm);
pair A = dir(130); pair B = dir(220); pair C = dir(320);
draw(B--A--C--B);
pair M = 0.5*A + 0.5*B;
pair N = 0.5*A + 0.5*C;
pair O = circumcenter(A,M,N);
pair Q = rotate(-90,A)*O;
pair X = 3*Q - 2*A;
pair Y = intersectionpoints(circle(X,abs(X-A)),circumcircle(A,M,N))[0];
pair Z = 0.5*A + 0.5*Y;
path anglemark(pair A, pair B, pair C, real t=8) {
pair M,N,P[],Q[];
path mark;
M=t*0.03*unit(A-B)+B;
N=t*0.03*unit(C-B)+B;
mark=arc(B,M,N);
mark=(mark--B--cycle);
return mark;
}
fill(anglemark(A,M,X,t=4),rgb(0,1,0.5)+opacity(0.7));
fill(anglemark(Y,M,Z,t=4),rgb(0,1,0.5)+opacity(0.7));
fill(anglemark(M,Y,B,t=4),rgb(0,1,0.5)+opacity(0.7));
filldraw(circumcircle(B,Y,M),lightcyan+opacity(0.1), blue+opacity(0.5));
filldraw(circumcircle(C,Y,N),lightcyan+opacity(0.1), blue+opacity(0.5));
filldraw(circumcircle(A,M,N),lightred+opacity(0.1), red);
draw(A--X,red);
draw(A--Y,dashed);
draw(X--Y,red);
draw(X--M--Z,gray);
draw(B--Y--M,gray);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(0));
dot("$M$", M, dir(220)); dot("$N$", N, dir(10));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(-40));
dot("$Z$", Z, dir(0));
\end{asy}
\end{center}
\begin{proof}
Let $Z$ be the midpoint of $\ol{AY}$.
Then $\ol{MX}$ is the $M$-symmedian in triangle $AMY$.
Since $\ol{MZ} \parallel \ol{BY}$,
it follows that $\dang AMX = \dang ZMY = \dang BYM$.
We conclude that $\ol{XM}$ is tangent
to the circumcircle of triangle $BMY$.
\end{proof}
Similarly, $\omega_C$ is the circumcircle of triangle $CNY$.
As $AMYN$ is cyclic too,
it follows that $\omega_B$ and $\omega_C$ intersect on $\ol{BC}$,
by Miquel's theorem.
\begin{remark*}
The converse of Miquel's theorem is true,
which means the problem is equivalent
to showing that the second intersection
of the $\omega_B$ and $\omega_C$ moves along $(AMN)$.
Thus the construction of $Y$ above is not so unnatural.
\end{remark*}
\paragraph{Second solution (Jetze Zoethout).}
Let $\omega_B$ intersect $\ol{BC}$ again at $S$
and let $\ol{MS}$ intersect $\ol{AC}$ again at $Y$.
Angle chasing gives
$\dang XMY = \dang XMS = \dang MBS = \dang ABC = \dang XAC = \dang XAY$,
so $Y$ is on the circumcircle of triangle $AMX$.
Furthermore, from $\dang XMY = \dang ABC$ and
$\dang ACB = \dang XAB = \dang XYM$ it follows that
$\triangle ABC \sim \triangle XMY$ and from $\dang XAY = \dang MBS$
and $\dang YXA = \dang YMA = \dang BMS$ it follows that
$\triangle AXY \sim \triangle BMS$.
\begin{center}
\begin{asy}
size(10cm);
pair A = dir(130); pair B = dir(220); pair C = dir(320);
draw(B--A--C--B);
pair M = 0.5*A + 0.5*B;
pair N = 0.5*A + 0.5*C;
pair O = circumcenter(A,M,N);
pair Q = rotate(-90,A)*O;
pair X = 3*Q - 2*A;
pair T = intersectionpoints(circle(X,abs(X-A)),circumcircle(A,M,N))[0];
pair S = intersectionpoints(circumcircle(B,M,T),circumcircle(C,N,T))[0];
pair Y = extension(M,S,A,C);
fill(X--M--S--cycle,green+opacity(0.1));
fill(X--A--N--cycle,green+opacity(0.1));
draw(S--Y--A);
draw(X--A,blue);
draw(X--M,red);
draw(X--N--S--X,gray);
draw(circumcircle(B,M,S),red);
draw(circumcircle(A,B,C),blue);
draw(circumcircle(X,M,A),gray+dashed);
draw(circumcircle(X,N,S),gray+dashed);
dot("$A$", A, dir(70)); dot("$B$", B, dir(B)); dot("$C$", C, dir(0));
dot("$M$", M, dir(50)); dot("$N$", N, dir(10));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
dot("$S$", S, dir(-60));
\end{asy}
\end{center}
We now find
\[
\frac{AN}{AX} = \frac{AN/BM}{AX/BM} = \frac{AC/AB}{MS/XY} = \frac{AB/AB}{MS/XM} = \frac{XM}{MS},
\]
which together with $\angle XMS = \angle XAN$ yields
$\triangle XMS \sim \triangle XAN$.
From $\dang XSY = \dang XSM = \dang XNA = \dang XNY$
we now have that $S$ is on the circumcircle of triangle $XNY$.
Finally, we have $\dang XNS = \dang XYS = \dang XYM = \dang ACB = \dang NCS$
so $\ol{XN}$ is tangent to the circle through $C$, $N$, and $S$, as desired.
\paragraph{Third solution by moving points method.}
Fix triangle $ABC$ and animate $X$ along the tangent at $A$.
We let $D$ denote the second intersection point of $\omega_C$
with line $\ol{BC}$.
\begin{claim*}
The composed map $X \mapsto D$ is a fractional linear transformation
(i.e.\ a projective map) in terms of
a real coordinate on line $\ol{AA}$, $\ol{BC}$.
\end{claim*}
\begin{proof}
Let $\ell$ denote the perpendicular bisector of $\ol{CN}$,
also equipped with a real coordinate.
We let $P$ denote the intersection of $\ol{XM}$ with $\ell$,
$S$ the circumcenter of $\triangle CMD$.
Let $T$ denote the midpoint of $\ol{BD}$.
We claim that the composed map
\begin{align*}
\ol{AA} &\to \ell \to \ell \to \ol{BC} \to \ol{BC} \\
\text{by} \quad
X &\mapsto P \mapsto S \mapsto T \mapsto D
\end{align*}
is projective, by showing each individual map is projective.
\begin{center}
\begin{asy}
size(8cm);
pair A = dir(129); pair B = dir(220); pair C = dir(320);
filldraw(A--B--C--cycle, opacity(0.1)+yellow, black);
pair M = 0.5*A + 0.5*B;
pair N = 0.5*A + 0.5*C;
pair O = circumcenter(A,M,N);
pair Q = rotate(-90,A)*O;
pair X = 4.2*Q - 3.2*A;
pair Y = intersectionpoints(circle(X,abs(X-A)),circumcircle(A,M,N))[0];
pair Z = 0.5*A + 0.5*Y;
// filldraw(circumcircle(B,Y,M),lightcyan+opacity(0.1), blue+opacity(0.5));
filldraw(circumcircle(C,Y,N),lightcyan+opacity(0.1), blue+opacity(0.5));
// filldraw(circumcircle(A,M,N),lightred+opacity(0.1), red);
draw(A--X,red);
pair S = circumcenter(C, N, Y);
pair T = foot(S, B, C);
pair D = 2*T - C;
pair P = extension(S, midpoint(N--C), X, N);
draw(P--S, deepgreen);
draw(X--P, lightblue);
draw(S--T, black);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(0));
dot("$M$", M, dir(135)); dot("$N$", N, dir(80));
dot("$X$", X, dir(X));
dot("$S$", S, dir(135));
dot("$T$", T, dir(-90));
dot("$D$", D, dir(225));
dot("$P$", P, dir(45));
dot(extension(S,P,N,C));
\end{asy}
\end{center}
\begin{itemize}
\ii The map $X \mapsto P$ is projective since it is
a perspectivity through $N$ from $\ol{AA}$ to $\ell$.
\ii The map $P \mapsto S$ is projective
since it is equivalent to a negative inversion on $\ell$
at the midpoint of $\ol{NC}$ with radius $\half NC$.
(Note $\angle PNS = 90\dg$ is fixed.)
\ii The map $S \mapsto T$ is projective
since it is a perspectivity $\ell \to \ol{BC}$
through the point at infinity perpendicular to $\ol{BC}$
(in fact, it is linear).
\ii The map $T \mapsto D$ is projective (in fact, linear)
since it is a homothety through $C$ with fixed ratio $2$.
\end{itemize}
Thus the composed map is projective as well.
\end{proof}
Similarly, if we define $D'$ so that $\ol{XM}$
is tangent to $(BMD')$, the map $X \mapsto D'$ is projective as well.
We aim to show $D = D'$, and since the maps
correspond to fractional linear transformations
in projective coordinates,
it suffices to verify it for three distinct choices of $X$.
We do so:
\begin{itemize}
\ii If $X = \ol{AA} \cap \ol{MN}$,
then $D$ and $D'$ satisfy $MB = MD'$, $NC = ND$.
This means they are the feet of the $A$-altitude on $\ol{BC}$.
\ii As $X$ approaches $A$ the points $D$ and $D'$
approach the infinity point along $\ol{BC}$.
\ii If $X$ is a point at infinity along $\ol{AA}$,
then $D$ and $D'$ coincide with the midpoint of $\ol{BC}$.
\end{itemize}
This completes the solution.
\begin{remark*}
[Anant Mudgal]
An alternative (shorter) way to show $X \mapsto D$ is projective
is to notice $\dang XND$ is a constant angle.
I left the longer ``original'' proof for instructional reasons.
\end{remark*}
\paragraph{Fourth solution by isogonal conjugates (Anant Mudgal).}
Let $Y$ be the isogonal conjugate of $X$ in $\triangle AMN$
and $Z$ be the reflection of $Y$ in $\ol{MN}$.
As $\ol{AX}$ is tangent to the circumcircle of $\triangle AMN$,
it follows that $\ol{AY} \parallel \ol{MN}$.
Thus $Z$ lies on $\ol{BC}$ since $\ol{MN}$
bisects the strip made by $\ol{AY}$ and $\ol{BC}$.
\begin{asy}
size(11cm);
pair A = dir(130); pair B = dir(220); pair C = dir(320);
draw(B--A--C--B);
pair M = 0.5*A + 0.5*B;
pair N = 0.5*A + 0.5*C;
pair O = circumcenter(A,M,N);
pair Q = rotate(-90,A)*O;
pair X = 3*Q - 2*A;
pair V = intersectionpoints(circle(X,abs(X-A)),circumcircle(A,M,N))[0];
pair Z = OP(circumcircle(B,V,M), circumcircle(C,V,N));
pair Y = -Z+2*foot(Z,M,N);
filldraw(circumcircle(B,V,M),lightcyan+opacity(0.1), blue+opacity(0.5));
filldraw(circumcircle(C,V,N),lightcyan+opacity(0.1), blue+opacity(0.5));
filldraw(circumcircle(A,M,N),lightred+opacity(0.1), red);
// draw(A--X--P);
// draw(X--B);
draw(A--X,red);
draw(B--V--M,gray);
dot(B*C/A);
filldraw(unitcircle, opacity(0.1)+yellow, black);
draw(M--Z--N, blue);
draw(N--X--M, lightblue);
draw(M--N, red);
draw(A--(B*C/A), red);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(0));
dot("$M$", M, dir(220)); dot("$N$", N, dir(10));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(-45));
dot("$Z$", Z, dir(Z));
\end{asy}
Finally, \[ \dang ZMX=\dang ZMN+\dang NMX=\dang NMY+\dang YMA
= \dang NMA=\dang ZBM \]
so $\ol{XM}$ is tangent to the circumcircle of
$\triangle ZMB$, hence $Z$ lies on $\omega_B$.
Similarly, $Z$ lies on $\omega_C$ and we're done.
|
sols-TST-IMO-2019_2
|
Let $\ZZ/n\ZZ$ denote the set of integers considered modulo $n$
(hence $\ZZ/n\ZZ$ has $n$ elements).
Find all positive integers $n$ for which there exists a bijective function
$g \colon \ZZ/n\ZZ \to \ZZ/n\ZZ$,
such that the $101$ functions
\[ g(x), \quad g(x)+x, \quad g(x)+2x, \quad \dots, \quad g(x)+100x \]
are all bijections on $\ZZ/n\ZZ$.
|
Call a function $g$ \emph{valiant} if it obeys this condition.
We claim the answer is all numbers relatively prime to $101!$.
The construction is to just let $g$ be the identity function.
Before proceeding to the converse solution,
we make a long motivational remark.
\begin{remark*}
[Motivation for both parts]
The following solution is dense,
and it is easier to think about some small cases first,
to motivate the ideas.
We consider the result
where $101$ is replaced by $2$ or $3$.
\begin{itemize}
\ii If we replaced $101$ with $2$,
you can show $2 \nmid n$ easily: write
\[ \sum_x x \equiv \sum_x g(x) \equiv \sum_x (g(x) + x ) \pmod n \]
which implies
\[ 0 \equiv \sum_x x = \half n (n+1) \pmod n \]
which means $\half n(n+1) \equiv 0 \pmod n$, hence $n$ odd.
\ii If we replaced $101$ with $3$,
then you can try a similar approach using squares, since
\begin{align*}
0 &\equiv \sum_x \left[ \left( g(x)+2x \right)^2
- 2\left( g(x)+x \right)^2
+ g(x)^2 \right] \pmod n\\
&= \sum_x 2x^2 = 2 \cdot \frac{n(n+1)(2n+1)}{6}
\end{align*}
which is enough to force $3 \nmid n$.
\end{itemize}
\end{remark*}
We now present several different proofs of the converse,
all of which generalize the ideas contained here.
In everything that follows we assume $n > 1$ for convenience.
\paragraph{First solution (original one).}
The proof is split into two essentially orthogonal claims,
which we state as lemmas.
\begin{lemma*}
[Lemma I: elimination of $g$]
Assume valiant $g \colon \ZZ/n\ZZ \to \ZZ/n\ZZ$ exists.
Then \[ k! \sum_{x \in \ZZ/n\ZZ} x^k \equiv 0 \pmod{n} \]
for $k = 0, 1, \dots, 100$.
\end{lemma*}
\begin{proof}
Define $g_x(T) = g(x) + Tx$ for any integer $T$.
If we view $g_x(T)^k$
as a polynomial in $\ZZ[T]$ of degree $k$
with leading coefficient $x^k$,
then taking the $k$th finite difference implies that,
for any $x$,
\[ k! x^k = \binom k0 g_x(k)^k
- \binom k1 g_x(k-1)^k + \binom k2 g_x(k-2)^k - \dots
+ (-1)^k \binom kk g_x(0)^k. \]
On the other hand, for any $1 \le k \le 100$
we should have
\begin{align*}
\sum_x g_x(0)^k \equiv \sum_x g_x(1)^k
&\equiv \dots \equiv \sum_x g_x(k)^k \\
&\equiv S_k \coloneq 0^k + \dots + (n-1)^k \pmod n
\end{align*}
by the hypothesis.
Thus we find
\[ k! \sum_x x^k
\equiv \left[ \binom k0 - \binom k1 + \binom k2 - \dotsb \right] S_k
\equiv 0 \pmod n \]
for any $1 \le k \le 100$, but also obviously for $k = 0$.
\end{proof}
We now prove the following self-contained lemma.
\begin{lemma*}
[Lemma II: power sum calculation]
Let $p$ be a prime, and let $n$, $M$ be positive integers such that
\[ M \quad\text{ divides }\quad 1^k + 2^k + \dots + n^k \]
for $k = 0, 1, \dots, p-1$.
If $p \mid n$ then $\nu_p(M) < \nu_p(n)$.
\end{lemma*}
\begin{proof}
The hypothesis means that
that any polynomial $f(T) \in \ZZ[T]$
with $\deg f \le p-1$ will have $\sum_{x=1}^n f(x) \equiv 0 \pmod{M}$.
In particular, we have
\begin{align*}
0 &\equiv \sum_{x=1}^n (x-1)(x-2) \dotsm (x-(p-1)) \\
&= (p-1)! \sum_{x=1}^{n} \binom{x-1}{p-1}
= (p-1)! \binom np \pmod{M}.
\end{align*}
But now $\nu_p(M) \le \nu_p(\binom np) = \nu_p(n) - 1$.
\end{proof}
Now assume for contradiction that valiant
$g \colon \ZZ/n\ZZ \to \ZZ/n\ZZ$ exists,
and $p \le 101$ is the \emph{smallest} prime dividing $n$.
Lemma I implies that $k! \sum_x x^k \equiv 0 \pmod n$
for $k = 1, \dots, p-1$
and hence $\sum_x x^k \equiv 0 \pmod n$ too.
Thus $M = n$ holds in the previous lemma, impossible.
\paragraph{A second solution.}
Both lemmas above admit variations
where we focus on working modulo $p^e$ rather than working modulo $n$.
\begin{lemma*}
[Lemma I']
Assume valiant $g \colon \ZZ/n\ZZ \to \ZZ/n\ZZ$ exists.
Let $p \le 101$ be a prime, and $e = \nu_p(n)$.
Then \[ \sum_{x \in \ZZ/n\ZZ} x^k \equiv 0 \pmod{p^e} \]
for $k = 0, 1, \dots, p-1$.
\end{lemma*}
\begin{proof}
This is weaker than Lemma I,
but we give an independent specialized proof.
Begin by writing
\[ \sum_x \left( g(x) + Tx \right)^k \equiv \sum_x x^k \pmod{p^e}. \]
Both sides are integer polynomials in $T$,
which vanish at $T = 0, 1, \dots, p-1$ by hypothesis
(since $p-1 \le 100$).
We now prove the following more general fact:
if $f(T) \in \ZZ[T]$ is an integer polynomial with $\deg f \le p-1$,
such that $f(0) \equiv \dots \equiv f(p-1) \equiv 0 \pmod{p^e}$,
then all coefficients of $f$ are divisible by $p^e$.
The proof is by induction on $e \ge 1$.
When $e = 1$, this is just the assertion that
the polynomial has at most $\deg f$ roots modulo $p$.
When $e \ge 2$, we note that the previous result
implies all coefficients are divisible by $p$,
and then we divide all coefficients by $p$.
Applied here, we have that all coefficients of
\[ f(T) \coloneq \sum_x \left( g(x) + Tx \right)^k - \sum_x x^k \]
are divisible by $p^e$.
The leading $T^k$ coefficient is $\sum_k x^k$ as desired.
\end{proof}
\begin{lemma*}
[Lemma II']
If $e \ge 1$ is an integer,
and $p$ is a prime, then
\[ \nu_p\left( 1^{p-1} + 2^{p-1} + \dots + (p^e-1)^{p-1} \right)
= e-1. \]
\end{lemma*}
\begin{proof}
First, note that the cases where $p = 2$ or $e = 1$
are easy; since if $p = 2$ we have
$\sum_{x = 0}^{2^e - 1} x \equiv 2^{e - 1}(2^e - 1)
\equiv -2^{e - 1}\pmod{2^e}$,
while if $e = 1$ we have $1^{p-1} + \dots + (p-1)^{p-1} \equiv -1 \pmod p$.
Henceforth assume that $p > 2$, $e > 1$.
Let $g$ be an integer which is a primitive root modulo $p^e$.
Then, we can sum the terms which are relatively prime to $p$ as
\[ S_0 \coloneq \sum_{\gcd(x,p) = 1} x^{p-1}
\equiv \sum_{i=1}^{\varphi(p^e)} g^{(p-1) \cdot i}
\equiv \frac{g^{p^{e-1}(p-1)^2} - 1}{g^{p-1}-1}
\pmod{p^e}
\]
which implies $\nu_p(S_0) = e-1$, by lifting the exponent.
More generally, for $r \ge 1$ we may set
\[ S_r \coloneq \sum_{\nu_p(x) = r} x^{p-1}
\equiv (p^r)^{p-1} \sum_{i=1}^{\varphi(p^{e-r})}
g_r^{(p-1) \cdot i}
\pmod{p^e}
\]
where $g_r$ is a primitive root modulo $p^{e-r}$.
Repeating the exponent-lifting calculation
shows that $\nu_p(S_r) = r(p-1) + \left( (e-r)-1 \right) > e$,
as needed.
\end{proof}
Assume to the contrary that $p \le 101$ is a prime dividing
$n$, and a valiant $g \colon \ZZ / n \ZZ \to \ZZ / n \ZZ$ exists.
Take $k = p-1$ in Lemma I' to contradict Lemma II'
\paragraph{A third remixed solution.}
We use Lemma I and Lemma II' from before.
As before, assume $g \colon \ZZ / n \ZZ \to \ZZ / n \ZZ$ is valiant,
and $n$ has a prime divisor $p \le 101$.
Also, let $e = \nu_p(n)$.
Then $(p-1)! \sum_x x^{p-1} \equiv 0 \pmod{n}$ by Lemma I,
and now
\begin{align*}
0 & \equiv \sum_x x^{p-1} \pmod{p^e} \\
&\equiv \frac{n}{p^e} \sum_{x=1}^{p^e-1} x^{p-1} \not\equiv 0 \pmod{p^e}
\end{align*}
by Lemma II', contradiction.
\paragraph{A fourth remixed solution.}
We also can combine Lemma I' and Lemma II.
As before, assume $g \colon \ZZ / n \ZZ \to \ZZ / n \ZZ$ is valiant,
and let $p$ be the smallest prime divisor of $n$.
Assume for contradiction $p \le 101$.
By Lemma I' we have
\[ \sum_x x^k \equiv 0 \pmod{p^e} \]
for $k = 0, \dots, p-1$.
This directly contradicts Lemma II with $M = p^e$.
|
sols-TST-IMO-2019_3
|
A \emph{snake of length $k$} is an animal
which occupies an ordered $k$-tuple
$(s_1, \dots, s_k)$ of cells in an $n \times n$ grid of square unit cells.
These cells must be pairwise distinct,
and $s_i$ and $s_{i+1}$ must share a side for $i=1,\dots,k-1$.
If the snake is currently occupying $(s_1, \dots, s_k)$
and $s$ is an unoccupied cell sharing a side with $s_1$,
the snake can \emph{move} to occupy $(s, s_1, \dots, s_{k-1})$ instead.
The snake has \emph{turned around} if it occupied $(s_1, s_2, \dots, s_k)$
at the beginning, but after a finite number of moves
occupies $(s_k, s_{k-1}, \dots, s_1)$ instead.
Determine whether there exists an integer $n > 1$
such that one can place some snake of length at least
$0.9n^2$ in an $n \times n$ grid which can turn around.
|
The answer is yes (and $0.9$ is arbitrary).
\paragraph{First grid-based solution.}
The following solution is due to Brian Lawrence.
For illustration reasons, we give below a figure
of a snake of length $89$ turning around in an $11 \times 11$ square
(which generalizes readily to odd $n$).
We will see that a snake of length $(n-1)(n-2)-1$ can turn
around in an $n \times n$ square,
so this certainly implies the problem.
\begin{center}
\begin{asy}
unitsize(0.20cm);
defaultpen(fontsize(8pt));
int n = 0;
picture snake(path p) {
picture pic;
draw(pic, shift(-0.5,-0.5)*scale(11)*unitsquare, black+1);
for (int i=0; i<=10; ++i) {
for (int j=0; j<=10; ++j) {
if ( (i+j)#2 == (i+j)/2 )
fill(pic, shift(i-0.5,j-0.5)*unitsquare, opacity(0.3)+gray);
}
}
path q = subpath(p, arctime(p, arclength(p)-88), arctime(p, arclength(p)));
draw(pic, q, deepgreen+1.3, EndArrow(TeXHead,1.3));
++n;
label(pic, "Figure " + (string) n, (5,0), 2*dir(-90));
return pic;
}
real M = 13;
add(snake( (0,1)--(0,10)--
(8,10)--(8,9)--(1,9)--(1,8)--(8,8)--(8,7)--(1,7)--(1,6)--
(8,6)-- (8,5)--(1,5)--(1,4)--(8,4)--(8,3)--(1,3)--(1,2)--
(8,2)--(8,1)--(1,1)));
add(shift(M,0)*snake((7,9)--
(1,9)--(1,8)--(8,8)--(8,7)--(1,7)--(1,6)--(8,6)--(8,5)--
(1,5)--(1,4)--(8,4)--(8,3)--(1,3)--(1,2)--(8,2)--(8,1)--
(1,1)--(1,0)--(10,0)--(10,9)));
add(shift(2*M,0)*snake(shift(-1,1)*((7,9)--
(1,9)--(1,8)--(8,8)--(8,7)--(1,7)--(1,6)--(8,6)--(8,5)--
(1,5)--(1,4)--(8,4)--(8,3)--(1,3)--(1,2)--(8,2)--(8,1)--
(1,1)--(1,0)--(10,0)--(10,9))));
add(shift(3*M,0)*snake(
(0,9)--(7,9)--(7,8)--(0,8)--(0,7)--(7,7)--(7,6)--
(0,6)--(0,5)--(7,5)--(7,4)--(0,4)--(0,3)--(7,3)--(7,2)--
(0,2)--(0,1)--(9,1)--(9,10)--(2,10)));
add(shift(0,-M)*snake(
(0,9)--(7,9)--(7,8)--(0,8)--(0,7)--(7,7)--(7,6)--
(0,6)--(0,5)--(7,5)--(7,4)--(0,4)--(0,3)--(7,3)--(7,2)--
(0,2)--(0,1)--(9,1)--
(9,2)--(8,2)--(8,3)--(9,3)--
(9,10)--(4,10)));
add(shift(M,-M)*snake(
(0,9)--(7,9)--(7,8)--(0,8)--(0,7)--(7,7)--(7,6)--
(0,6)--(0,5)--(7,5)--(7,4)--(0,4)--(0,3)--(6,3)--(6,2)--
(0,2)--(0,1)--(9,1)--
(9,2)--(8,2)--(8,3)--(9,3)--
(9,10)--(2,10)));
add(shift(2*M,-M)*snake(
(0,9)--(5,9)--(5,8)--(0,8)--(0,7)--(3,7)--(3,6)--
(0,6)--(0,5)--(1,5)--(1,4)--(0,4)--(0,3)--(0,3)--(0,2)--
(0,2)--(0,1)--(9,1)--(9,2)--
(2,2)--(2,3)--(9,3)--(9,4)--(3,4)--(3,5)--(9,5)--(9,6)--
(5,6)--(5,7)--(9,7)--(9,8)--(7,8)--(7,9)--(9,9)--(9,10)--(2,10)
));
add(shift(3*M,-M)*snake((0,9)--
(0,2)--(0,1)--(9,1)--(9,2)--
(2,2)--(2,3)--(9,3)--(9,4)--(2,4)--(2,5)--(9,5)--(9,6)--
(2,6)--(2,7)--(9,7)--(9,8)--(2,8)--(2,9)--(9,9)--(9,10)--(2,10)
));
add(shift(0,-2*M)*snake((0,7)--
(0,2)--(0,1)--(9,1)--(9,2)--
(1,2)--(1,3)--(9,3)--(9,4)--(2,4)--(2,5)--(9,5)--(9,6)--
(2,6)--(2,7)--(9,7)--(9,8)--(2,8)--(2,9)--(9,9)--(9,10)--(2,10)
));
add(shift(M,-2*M)*snake(
(0,1)--(9,1)--(9,2)--
(1,2)--(1,3)--(9,3)--(9,4)--(1,4)--(1,5)--(9,5)--(9,6)--
(1,6)--(1,7)--(9,7)--(9,8)--(1,8)--(1,9)--(9,9)--(9,10)--(2,10)
));
add(shift(2*M,-2*M)*snake((0,1)--(8,1)--(8,2)--
(1,2)--(1,3)--(8,3)--(8,4)--(1,4)--(1,5)--(9,5)--(9,6)--
(1,6)--(1,7)--(9,7)--(9,8)--(1,8)--(1,9)--(9,9)--(9,10)--
(0,10)--(0,8)
));
add(shift(3*M,-2*M)*snake((0,1)--(8,1)--(8,2)--
(1,2)--(1,3)--(8,3)--(8,4)--(1,4)--(1,5)--(8,5)--(8,6)--
(1,6)--(1,7)--(8,7)--(8,8)--(1,8)--(1,9)--(8,9)--(8,10)--
(0,10)--(0,2)
));
\end{asy}
\end{center}
Use the obvious coordinate system with $(1,1)$ in the bottom left.
Start with the snake as shown in Figure 1,
then have it move to $(2,1)$, $(2,n)$, $(n,n-1)$
as in Figure 2.
Then, have the snake shift to the position in Figure 3;
this is possible since the snake can just walk to $(n,n)$,
then start walking to the left and then follow the route;
by the time it reaches the $i$th row
from the top its tail will have vacated by then.
Once it achieves Figure 3, move the head of the snake
to $(3,n)$ to achieve Figure 4.
In Figure 5 and 6, the snake begins to ``deform'' its loop continuously.
In general, this deformation by two squares
is possible in the following way.
The snake walks first to $(1,n)$ then retraces the steps
left by its tail,
except when it reaches $(n-1,3)$ it makes a brief detour to
$(n-2,3)$, $(n-2,4)$, $(n-1,4)$ and continues along its way;
this gives the position in Figure 5.
Then it retraces the entire loop again,
except that when it reaches $(n-4,4)$ it turns directly
down, and continues retracing its path;
thus at the end of this second revolution, we arrive at Figure 6.
By repeatedly doing perturbations of two cells,
we can move move all the ``bumps'' in the path gradually
to protrude from the right; Figure 7 shows a partial application of the
procedure, with the final state as shown in Figure 8.
In Figure 9, we stretch the bottom-most bump by two more cells;
this shortens the ``tail'' by two units, which is fine.
Doing this for all $(n-3)/2$ bumps arrives at the situation in
Figure 10, with the snake's head at $(3,n)$.
We then begin deforming the turns on the bottom-right
by two steps each as in Figure 11,
which visually will increase the length of the head.
Doing this arrives finally at the situation in Figure 12.
Thus the snake has turned around.
\paragraph{Second solution phrased using graph theory (Nikolai Beluhov).}
Let $G$ be any undirected graph. Consider a snake of length $k$ lying within $G$, with each segment of the snake occupying one vertex, consecutive segments occupying adjacent vertices, and no two segments occupying the same vertex. One move of the snake consists of the snake's head advancing to an adjacent empty vertex and segment $i$ advancing to the vertex of segment $i - 1$ for $i = 2$, 3, \dots, $k$.
The solution proceeds in two stages. First we construct a planar graph $G$ such that it is possible for a snake that occupies nearly all of $G$ to turn around inside $G$. Then we construct a subgraph $H$ of a grid adjacency graph such that $H$ is isomorphic to $G$ and $H$ occupies nearly all of the grid.
For the first stage of the solution, we construct $G$ as follows.
Let $r$ and $\ell$ be positive integers. Start with $r$ disjoint \emph{main} paths $p_1$, $p_2$, \dots, $p_r$, each of length at least $\ell$, with $p_i$ leading from $A_i$ to $B_i$ for $i = 1$, 2, \dots, $r$. Add to those $r$ \emph{linking} paths, one leading from $B_i$ to $A_{i + 1}$ for each $i = 1$, 2, \dots, $r - 1$, and one leading from $B_r$ to $A_1$. Finally, add to those two families of \emph{transit} paths, with one family containing one transit path joining $A_1$ to each of $A_2$, $A_3$, \dots, $A_r$ and the other containing one path joining $B_r$ to each of $B_1$, $B_2$, \dots, $B_{r - 1}$. We require that all paths specified in the construction have no interior vertices in common, with the exception of transit paths in the same family.
We claim that a snake of length $(r - 1)\ell$ can turn around inside $G$.
To this end, let the concatenation $A_1B_1A_2B_2\dots A_r B_r$ of all main and linking paths be the \emph{great cycle}. We refer to $A_1B_1A_2B_2\dots A_r B_r$ as the counterclockwise orientation of the great cycle, and to $B_r A_r B_{r - 1}A_{r - 1}\dots B_1A_1$ as its clockwise orientation.
Place the snake so that its tail is at $A_1$ and its body extends counterclockwise along the great cycle. Then let the snake manoeuvre as follows. (We track only the snake's head, as its movement uniquely determines the movement of the complete body of the snake.)
At phase 1, advance counterclockwise along the great cycle to $B_{r - 1}$, take a detour along a transit path to $B_r$, and advance clockwise along the great cycle to $A_r$.
For $i = 2$, 3, \dots, $r - 1$, at phase $i$, take a detour along a transit path to $A_1$, advance counterclockwise along the great cycle to $B_{r - i}$, take a detour along a transit path to $B_r$, and advance clockwise along the great cycle to $A_{r - i + 1}$.
At phase $r$, simply advance clockwise along the great cycle to $A_1$.
For the second stage of the solution, let $n$ be a sufficiently large positive integer. Consider an $n \times n$ grid $S$. Number the columns of $S$ from 1 to $n$ from left to right, and its rows from 1 to $n$ from bottom to top.
Let $a_1$, $a_2$, \dots, $a_{r + 1}$ be cells of $S$ such that all of $a_1$, $a_2$, \dots, $a_{r + 1}$ lie in column 2, $a_1$ lies in row 2, $a_{r + 1}$ lies in row $n - 1$, and $a_1$, $a_2$, \dots, $a_{r + 1}$ are approximately equally spaced. Let $b_1$, $b_2$, \dots, $b_r$ be cells of $S$ such that all of $b_1$, $b_2$, \dots, $b_r$ lie in column $n - 2$ and $b_i$ lies in the row of $a_{i + 1}$ for $i = 1$, 2, \dots, $r$.
Construct $H$ as follows. For $i = 1$, 2, \dots, $r$, let the main path from $a_i$ to $b_i$ fill up the rectangle bounded by the rows and columns of $a_i$ and $b_i$ nearly completely. Then every main path is of length approximately $\frac{1}{r}n^2$.
For $i = 1$, 2, \dots, $r - 1$, let the linking path that leads from $b_i$ to $a_{i + 1}$ lie inside the row of $b_i$ and $a_{i + 1}$ and let the linking path that leads from $b_r$ to $a_1$ lie inside row $n$, column $n$, and row $1$.
Lastly, let the union of the first family of transit paths be column 1 and let the union of the second family of transit paths be column $n - 1$, with the exception of their bottommost and topmost squares.
As in the first stage of the solution, by this construction a snake of length $k$ approximately equal to $\frac{r - 1}{r}n^2$ can turn around inside an $n \times n$ grid $S$. When $r$ is fixed and $n$ tends to infinity, $\frac{k}{n^2}$ tends to $\frac{r - 1}{r}$. Furthermore, when $r$ tends to infinity, $\frac{r - 1}{r}$ tends to $1$. This gives the answer.
|
sols-TST-IMO-2019_4
|
We say a function $f \colon \ZZ_{\ge 0} \times \ZZ_{\ge 0} \to \ZZ$
is \emph{great} if for any nonnegative integers $m$ and $n$,
\[ f(m+1, n+1) f(m,n) - f(m+1,n) f(m,n+1) = 1. \]
If $A = (a_0, a_1, \dots)$ and $B = (b_0, b_1, \dots)$
are two sequences of integers,
we write $A \sim B$ if there exists a great function $f$
satisfying $f(n,0) = a_n$ and $f(0,n) = b_n$
for every nonnegative integer $n$ (in particular, $a_0=b_0$).
Prove that if $A$, $B$, $C$, and $D$ are four sequences of integers
satisfying $A \sim B$, $B \sim C$, and $C \sim D$, then $D \sim A$.
|
We present two solutions.
In what follows, we say $(A, B)$ form a great pair if $A \sim B$.
\paragraph{First solution (Nikolai Beluhov).}
Let $k = a_0 = b_0 = c_0 = d_0$.
We let $f$, $g$, $h$ be great functions for $(A,B)$, $(B,C)$, $(C,D)$
and write the following infinite array:
\[
\begin{bmatrix}
& \vdots & \vdots & {\color{blue} b_3} & \vdots & \vdots & \\[5pt]
\dotsb & g(2,2) & g(2,1) & {\color{blue}b_2} & f(1,2) & f(2,2) & \dotsb \\[5pt]
\dotsb & g(1,2) & g(1,1) & {\color{blue}b_1} & f(1,1) & f(2,1) & \dotsb \\[5pt]
{\color{blue}c_3} & {\color{blue}c_2} & {\color{blue}c_1} &
{\color{blue}k} &
{\color{blue}a_1} & {\color{blue}a_2} & {\color{blue}a_3} \\[5pt]
\dotsb & h(2,1) & h(1,1) & {\color{blue}d_1} & & & \\[5pt]
\dotsb & h(2,2) & h(1,2) & {\color{blue}d_2} & & & \\[5pt]
& \vdots & \vdots & {\color{blue}d_3} & & & \ddots \\[5pt]
\end{bmatrix}
\]
The greatness condition is then equivalent to saying
that any $2 \times 2$ sub-grid has determinant $\pm1$
(the sign is $+1$ in two quadrants and $-1$ in the other two),
and we wish to fill in the lower-right quadrant.
To this end, it suffices to prove the following.
\begin{lemma*}
Suppose we have a $3 \times 3$ sub-grid
\[
\begin{bmatrix}
a & b & c \\
x & y & z \\
p & q &
\end{bmatrix}
\]
satisfying the determinant conditions.
Then we can fill in the ninth entry in the lower right
with an integer while retaining greatness.
\end{lemma*}
\begin{proof}
We consider only the case where the $3 \times 3$
is completely contained inside the bottom-right quadrant,
since the other cases are exactly the same
(or even by flipping the signs of the top row
or left column appropriately).
If $y = 0$ we have $-1 = bz = bx = xq$, hence $qz = -1$,
and we can fill in the entry arbitrarily.
Otherwise, we have $bx \equiv xq \equiv bz \equiv -1 \pmod{y}$.
This is enough to imply $qz \equiv -1 \pmod y$,
and so we can fill in the integer $\frac{qz+1}{y}$.
\end{proof}
\begin{remark*}
In this case (of all $+1$ determinants),
I think it turns out the bottom entry is
exactly equal to $qza - cyp - c - p$,
which is obviously an integer.
\end{remark*}
\paragraph{Second solution (Ankan Bhattacharya).}
We will give an explicit classification of great sequences:
\begin{lemma*}
The pair $(A,B)$ is great if and only if
$a_0 = b_0$, $a_0 \mid a_1b_1 + 1$, and
$a_n \mid a_{n-1} + a_{n+1}$ and $b_n \mid b_{n-1} + b_{n+1}$ for all $n$.
\end{lemma*}
\begin{proof}[Proof of necessity]
It is clear that $a_0 = b_0$.
Then $a_0f(1, 1) - a_1b_1 = 1$, i.e. $a_0 \mid a_1b_1 + 1$.
Now, focus on six entries $f(x, y)$
with $x \in \{n-1, n, n+1\}$ and $y \in \{0, 1\}$.
Let $f(n-1, 1) = u$, $f(n, 1) = v$, and $f(n+1, 1) = w$, so
\begin{align*}
v a_{n-1} - u a_n & = 1,\\
w a_n - v a_{n+1} & = 1.
\end{align*}
Then
\[u + w = \frac{v(a_{n-1} + a_{n+1})}{a_n}\]
and from above $\gcd(v, a_n) = 1$,
so $a_n \mid a_{n-1} + a_{n+1}$;
similarly for $b_n$.
(If $a_n = 0$,
we have $va_{n-1} = 1$ and $va_{n+1} = -1$,
so this is OK.)
\end{proof}
\begin{proof}[Proof of sufficiency]
Now consider two sequences $a_0, a_1, \dots$
and $b_0, b_1, \dots$ satisfying our criteria.
We build a great function $f$ by induction on $(x, y)$.
More strongly, we will assume as part of the inductive hypothesis
that any two adjacent entries of $f$ are relatively prime
and that for any three consecutive entries horizontally or vertically,
the middle one divides the sum of the other two.
First we set $f(1, 1)$ so that $a_0 f(1, 1) = a_1 b_1 + 1$,
which is possible.
Consider an uninitialized $f(s, t)$;
without loss of generality suppose $s \ge 2$.
Then we know five values of $f$
and wish to set a sixth one $z$,
as in the matrix below:
\[
\begin{matrix}
u & x\\
v & y\\
w & {\color{gray} z} \\
\end{matrix}
\]
(We imagine $a$-indices to increase southwards and $b$-indices to increase eastwards.)
If $v \neq 0$,
then the choice $y \cdot \frac{u + w}{v} - x$ works
as $uy - vx = 1$.
If $v = 0$,
it easily follows that $\{u, w\} = \{1, -1\}$
and $y = w$ as $yw = 1$.
Then we set the uninitialized entry to anything.
Now we verify that this is compatible with the inductive hypothesis.
From the determinant 1 condition,
it easily follows that $\gcd(w, z) = \gcd(v, z) = 1$.
The proof that $y \mid x + z$
is almost identical to a step performed in the ``necessary'' part
of the lemma and we do not repeat it here.
By induction, a desired great function $f$ exists.
\end{proof}
We complete the solution.
Let $A$, $B$, $C$, and $D$ be integer sequences
for which $(A, B)$, $(B, C)$, and $(C, D)$ are great.
Then $a_0 = b_0 = c_0 = d_0$,
and each term in each sequence (after the zeroth term)
divides the sum of its neighbors.
Since $a_0$ divides all three of $a_1b_1 + 1$, $b_1c_1 + 1$, and $c_1d_1 + 1$,
it follows $a_0$ divides $d_1a_1 + 1$,
and thus $(D, A)$ is great as desired.
\begin{remark*}
To simplify the problem, we may restrict the codomain of great functions
and elements in great pairs of sequences to $\ZZ_{> 0}$.
This allows the parts of the solution dealing with zero entries to be ignored.
\end{remark*}
\begin{remark*}
Of course, this solution also shows that any odd path
(in the graph induced by the great relation on sequences)
completes to an even cycle.
If we stipulate that great functions must have $f(0, 0) = \pm 1$,
then even paths complete to cycles as well.
Alternatively, we could change the great functional equation to
\[f(x + 1, y + 1) f(x, y) - f(x + 1, y) f(x, y + 1) = -1.\]
A quick counterexample to transitivity of $\sim$ as is
without the condition $f(0,0)=1$, for concreteness:
let $a_n = c_n = 3+n$ and $b_n = 3+2n$ for $n \ge 0$.
\end{remark*}
|
sols-TST-IMO-2019_5
|
Let $n$ be a positive integer.
Tasty and Stacy are given a circular necklace
with $3n$ sapphire beads and $3n$ turquoise beads,
such that no three consecutive beads have the same color.
They play a cooperative game where they alternate turns
removing three consecutive beads, subject to the following conditions:
\begin{itemize}
\ii Tasty must remove three consecutive beads
which are turquoise, sapphire, and turquoise, in that order,
on each of his turns.
\ii Stacy must remove three consecutive beads
which are sapphire, turquoise, and sapphire, in that order,
on each of her turns.
\end{itemize}
They win if all the beads are removed in $2n$ turns.
Prove that if they can win with Tasty going first,
they can also win with Stacy going first.
|
In the necklace, we draw a \emph{divider}
between any two beads of the same color.
Unless there are no dividers,
this divides the necklace into several \emph{zigzags}
in which the beads in each zigzag alternate.
Each zigzag has two \emph{endpoints}
(adjacent to dividers).
Observe that the condition about not having
three consecutive matching beads is equivalent
to saying there are no zigzags of lengths $1$.
\begin{center}
\begin{asy}
size(7cm);
string[] s = {"T", "S", "S", "T", "T", "S", "T", "T", "S", "T", "T", "S", "S", "T", "S", "T", "S", "S", "T", "S", "S", "T", "S", "T"};
pen[] colors = {white, red, deepgreen, blue, heavygreen, gray,
purple, orange, deepcyan};
int j = 0;
for (int i=0; i<24; ++i) {
if ((i == 0) || (s[i] == s[i-1])) {
++j;
draw(0.9*dir(187.5-15*i)--1.1*dir(187.5-15*i), black+1.5);
}
label(s[i], dir(180-15*i), dir(180-15*i), colors[j]);
}
draw(unitcircle);
\end{asy}
\end{center}
The main claim is that
the game is winnable (for either player going first)
if and only if there are at most $2n$ dividers.
We prove this in two parts,
the first part not using the hypothesis about three consecutive letters.
\begin{claim*}
The game cannot be won with Tasty going first
if there are more than $2n$ dividers.
\end{claim*}
\begin{proof}
We claim each move removes at most one divider,
which proves the result.
Consider removing a $TST$ in some zigzag
(necessarily of length at least $3$).
We illustrate the three possibilities in the following table,
with Tasty's move shown in red.
\begin{center}
\begin{tabular}{ccl}
Before & After & Change \\ \hline
% ------------
$\dots \mathtt{ST} \mid \mathtt{\color{red} TST}
\mid \mathtt{TS}\dots$
& $\dots \mathtt{ST} \mid \mathtt{TS} \dots$
& One less divider; two zigzags merge \\
$\dots \mathtt{ST} \mid \mathtt{\color{red} TST}
\mathtt{ST}\dots$
& $\dots \mathtt{STST} \dots$
& One less divider; two zigzags merge \\
$\dots \mathtt{S} \mathtt{\color{red} TST} \mathtt{S} \dots$
& $\dots \mathtt{S} \mid \mathtt{S} \dots$
& One more divider; a zigzag splits in two
\end{tabular}
\end{center}
The analysis for Stacy's move is identical.
\end{proof}
\begin{claim*}
If there are at most $2n$ dividers
and there are no zigzags of length $1$
then the game can be won (with either player going first).
\end{claim*}
\begin{proof}
By symmetry it is enough to prove Tasty wins going first.
At any point if there are no dividers at all,
then the necklace alternates $TSTST\dots$
and the game can be won.
So we will prove that on each of Tasty's turns,
if there exists at least one divider,
then Tasty and Stacy can each make a move
at an endpoint of some zigzag
(i.e.\ the first two cases above).
As we saw in the previous proof, such moves will
(a) decrease the number of dividers by exactly one,
(b) not introduce any singleton zigzags
(because the old zigzags merge, rather than split).
Since there are fewer than $2n$ dividers,
our duo can eliminate all dividers and then win.
Note that as the number of $S$ and $T$'s are equal,
there must be an equal number of
\begin{itemize}
\ii zigzags of odd length ($\ge 3$) with $T$ at the endpoints
(i.e.\ one more $T$ than $S$), and
\ii zigzags of odd length ($\ge 3$) with $S$ at the endpoints
(i.e.\ one more $S$ than $T$).
\end{itemize}
Now iff there is at least one of each,
then Tasty removes a $TST$ from the end of such a zigzag
while Stacy removes an $STS$ from the end of such a zigzag.
Otherwise suppose all zigzags have even size.
Then Tasty finds any zigzag of length $\ge 4$
(which must exist since the \emph{average} zigzag length is $3$)
and removes $TST$ from the end containing $T$.
The resulting merged zigzag is odd and hence $S$ endpoints,
hence Stacy can move as well.
\end{proof}
\begin{remark*}
There are many equivalent ways to phrase the solution.
For example, the number of dividers is equal to the
number of pairs of two consecutive letters
(rather than singleton letters).
So the win condition can also be phrased in terms
of the number of adjacent pairs of letters being at least $2n$,
or equivalently the number of differing pairs being at least $4n$.
If one thinks about the game as a process,
this is a natural ``monovariant'' to consider anyways,
so the solution is not so unmotivated.
\end{remark*}
\begin{remark*}
The constraint of no three consecutive identical beads is actually
needed: a counterexample without this constraint is
$\mathtt{TTSTSTSTTSSS}$.
(They win if Tasty goes first and lose if Stacy goes first.)
\end{remark*}
\begin{remark*}
[Why induction is unlikely to work]
Many contestants attempted induction.
However, in doing so they often implicitly proved
a different problem: ``prove that if they can win
with Tasty going first \emph{without ever creating a triplet},
they can also win in such a way with Stacy going first''.
This essentially means nearly all induction attempts fail.
Amusingly, even the modified problem
(which is much more amenable to induction)
sill seems difficult without \emph{some} sort of global argument.
Consider a position in which Tasty wins going first,
with the sequence of winning moves being Tasty's first move in red
below and Stacy's second move in blue below:
\[
\dots \mathtt{TTSSTT}
\underbrace{\mathtt{\color{blue}S}
\overbrace{\mathtt{\color{red}TST}}^{\text{Tasty}}
\mathtt{\color{blue}TS}}_{\text{Stacy}}
\mathtt{STTSST} \dots.
\]
There is no ``nearby'' $STS$ that Stacy can remove
instead on her first turn,
without introducing a triple-$T$ and also
preventing Tasty from taking a $TST$.
So it does not seem possible to easily change
a Tasty-first winning sequence to a Stacy-first one,
even in the modified version.
\end{remark*}
|
sols-TST-IMO-2019_6
|
Let $ABC$ be a triangle with incenter $I$,
and let $D$ be a point on line $BC$ satisfying $\angle AID = 90\dg$.
Let the excircle of triangle $ABC$ opposite the vertex $A$
be tangent to $\ol{BC}$ at point $A_1$.
Define points $B_1$ on $\ol{CA}$ and $C_1$ on $\ol{AB}$ analogously,
using the excircles opposite $B$ and $C$, respectively.
Prove that if quadrilateral $AB_1A_1C_1$ is cyclic,
then $\ol{AD}$ is tangent to the circumcircle of $\triangle DB_1C_1$.
\end{enumerate}
|
We present two solutions.
\paragraph{First solution using spiral similarity (Ankan Bhattacharya).}
First, we prove the part of the problem
which does not depend on the condition $A B_1 A_1 C_1$ is cyclic.
\begin{lemma*}
Let $ABC$ be a triangle and define $I$, $D$, $B_1$, $C_1$
as in the problem.
Moreover, let $M$ denote the midpoint of $\ol{AD}$.
Then $\ol{AD}$ is tangent to $(AB_1C_1)$,
and moreover $\ol{B_1 C_1} \parallel \ol{IM}$.
\end{lemma*}
\begin{proof}
Let $E$ and $F$ be the tangency points of the incircle.
Denote by $Z$ the Miquel point of $BFEC$,
i.e.\ the second intersection
of the circle with diameter $\ol{AI}$ and the circumcircle.
Note that $A$, $Z$, $D$ are collinear,
by radical axis on $(ABC)$, $(AFIE)$, $(BIC)$.
\begin{center}
\begin{asy}
size(11cm);
pair A = dir(130);
pair B = dir(210);
pair C = dir(330);
pair I = incenter(A, B, C);
pair E = foot(I, C, A);
pair B_1 = A+C-E;
pair F = foot(I, A, B);
pair C_1 = A+B-F;
pair T = foot(I, B, C);
pair Z = foot(A, I, foot(T, E, F));
filldraw(A--B--C--cycle, opacity(0.1)+lightcyan, blue);
draw(circumcircle(A, E, F), orange);
pair D = extension(A, Z, B, C);
draw(arc(dir(270), C, B), orange);
draw(D--B, blue);
filldraw(A--B_1--C_1--cycle, opacity(0.1)+deepgreen, blue);
pair M = midpoint(A--D);
filldraw(A--D--I--cycle, opacity(0.1)+lightred, red);
draw(I--M, red);
filldraw(Z--E--F--cycle, opacity(0.1)+deepgreen, deepgreen);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
draw(incircle(A, B, C), dotted+blue);
draw(circumcircle(A, B_1, C_1), dashed+red);
dot("$A$", A, dir(A));
dot("$B$", B, dir(225));
dot("$C$", C, dir(315));
dot("$I$", I, dir(270));
dot("$E$", E, dir(30));
dot("$B_1$", B_1, dir(B_1));
dot("$F$", F, dir(F));
dot("$C_1$", C_1, dir(C_1));
dot("$Z$", Z, dir(Z));
dot("$D$", D, dir(D));
dot("$M$", M, dir(M));
/* TSQ Source:
!size(11cm);
A = dir 130
B = dir 210 R225
C = dir 330 R315
I = incenter A B C R270
E = foot I C A R30
B_1 = A+C-E
F = foot I A B
C_1 = A+B-F
T := foot I B C
Z = foot A I foot T E F
A--B--C--cycle 0.1 lightcyan / blue
circumcircle A E F orange
D = extension A Z B C
!draw(arc(dir(270), C, B), orange);
D--B blue
A--B_1--C_1--cycle 0.1 deepgreen / blue
M = midpoint A--D
A--D--I--cycle 0.1 lightred / red
I--M red
Z--E--F--cycle 0.1 deepgreen / deepgreen
unitcircle 0.1 lightcyan / blue
incircle A B C dotted blue
circumcircle A B_1 C_1 dashed red
*/
\end{asy}
\end{center}
Then the spiral similarity gives us
\[ \frac{ZF}{ZE} = \frac{BF}{CE} = \frac{AC_1}{AB_1} \]
which together with $\dang FZE = \dang FAE = \dang BAC$
implies that $\triangle ZFE$
and $\triangle AC_1B_1$ are (directly) similar.
(See IMO Shortlist 2006 G9 for a similar application
of spiral similarity.)
Now the remainder of the proof is just angle chasing.
First, since
\[ \dang DAC_1 = \dang ZAF = \dang ZEF = \dang AB_1C_1 \]
we have $\ol{AD}$ is tangent to $(AB_1C_1)$.
Moreover, to see that $\ol{IM} \parallel \ol{B_1C_1}$, write
\begin{align*}
\dang (\ol{AI}, \ol{B_1C_1})
&= \dang IAC + \dang AB_1C_1 = \dang BAI + \dang ZEF
= \dang FAI + \dang ZAF \\
&= \dang ZAI = \dang MAI = \dang AIM
\end{align*}
the last step since $\triangle AID$ is right with hypotenuse
$\ol{AD}$, and median $\ol{IM}$.
\end{proof}
Now we return to the present problem
with the additional condition.
\begin{center}
\begin{asy}
unitsize(100);
pair A, B, C, I, A1, B1, C1, D, E, F, J, K, Z, M;
B = dir(193); C = reflect((0, 1), (0, -1)) * B;
I = intersectionpoints(circle(dir(270), abs(dir(270)-B)),
-B--(-C))[1];
A = 2 * foot(origin, I, dir(270)) - dir(270);
D = extension(B, C, I, rotate(90, I) * A);
E = foot(I, C, A);
F = foot(I, A, B);
A1 = B + C - foot(I, B, C);
B1 = C + A - E;
C1 = A + B - F;
M = (A + D)/2;
Z = foot(I, A, D);
K = 2 * foot(circumcenter(A, B1, C1), A, I) - A;
//draw(unitcircle);
filldraw(circumcircle(A, B1, C1), pink+opacity(0.5), red);
filldraw(circumcircle(A, E, F), lightcyan+opacity(0.5), heavycyan+dashed);
draw(incircle(A, B, C), heavycyan+dotted);
filldraw(circumcircle(A, B, C), lightcyan+opacity(0.2), cyan);
filldraw(circumcircle(B, I, C), lightcyan+opacity(0.3), heavycyan+dashed);
draw(arc(circumcenter(D, B1, C1), B1, D), heavyred);
draw(A--D, heavyred);
draw(A--B--C--cycle, heavymagenta);
draw(D--B, lightmagenta);
draw(A--I--D, heavycyan);
// draw(I--K, heavyred);
draw(M--B1, dashed+heavyred);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(10));
dot("$I$", I, dir(-I));
dot("$A_1 = V$", A1, dir(A1));
dot("$B_1$", B1, dir(10));
dot("$C_1$", C1, dir(130));
dot("$D$", D, dir(D));
dot("$E$", E, dir(40));
dot("$F$", F, dir(170));
dot("$M$", M, dir(130));
dot("$Z$", Z, dir(Z));
// dot("$K$", K, dir(K-circumcenter(A, B1, C1)));
clip(box((-3, -0.4), (1.2, 1.25)));
\end{asy}
\end{center}
\begin{claim*}
Given the condition, we actually have
$\angle AB_1A_1 = \angle AC_1A_1 = 90\dg$.
\end{claim*}
\begin{proof}
Let $I_A$, $I_B$ and $I_C$ be the excenters of $\triangle ABC$.
Then the perpendiculars to $\ol{BC}$, $\ol{CA}$, $\ol{AB}$
from $A_1$, $B_1$, $C_1$ respectively meet at
the so-called \emph{Bevan point} $V$
(which is the circumcenter of $\triangle I_A I_B I_C$).
Now $\triangle AB_1C_1$ has circumdiameter $\ol{AV}$.
We are given $A_1$ lies on this circle,
so if $V \neq A_1$ then $\ol{AA_1} \perp \ol{A_1V}$.
But $\ol{A_1V} \perp \ol{BC}$ by definition,
which would imply $\ol{AA_1} \parallel \ol{BC}$, which is absurd.
\end{proof}
\begin{claim*}
Given the condition the points $B_1$, $I$, $C_1$ are collinear
(hence with $M$).
\end{claim*}
\begin{proof}
By Pappus theorem on $\ol{I_B A I_C}$ and $\ol{BA_1C}$
after the previous claim.
\end{proof}
To finish, since $\ol{DMA}$ was tangent to the circumcircle of
$\triangle AB_1C_1$, we have $MD^2 = MA^2 = MC_1 \cdot MB_1$,
implying the required tangency.
\begin{remark*}
The triangles satisfying the problem hypothesis
are exactly the ones satisfying $r_A = 2R$,
where $R$ and $r_A$ denote the circumradius and $A$-exradius.
\end{remark*}
\begin{remark*}
If $P$ is the foot of the $A$-altitude
then this should also imply $AB_1PC_1$ is harmonic.
\end{remark*}
\paragraph{Second solution by inversion and mixtilinears (Anant Mudgal).}
As in the end of the preceding solution, we have
$\angle AB_1A_1=\angle AC_1A_1=90^{\circ}
\quad\text{and}\quad I \in \ol{B_1 C_1}$.
Let $M$ be the midpoint of minor arc $BC$
and $N$ be the midpoint of arc $\widehat{BAC}$.
Let $L$ be the intouch point on $\ol{BC}$.
Let $O$ be the circumcenter of $\triangle ABC$.
Let $K=\ol{AI} \cap \ol{BC}$.
\begin{center}
\begin{asy}
/* Geogebra to Asymptote conversion, documentation at aops.com/Wiki, go to User:Azjps/geogebra */
import graph; size(12cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -4., xmax = 6., ymin = -3., ymax = 4.; /* image dimensions */
/* draw figures */
draw(circle((2.7,0.6), 2.7658633371878665), linewidth(0.4));
draw((1.892579421398536,1.2)--(3.5074205786014643,0.), linewidth(0.4) + dotted);
draw((1.445226775603362,3.064861893765504)--(-3.109814760356735,0.), linewidth(0.4));
draw((-3.109814760356735,0.)--(2.7,-2.1658633371878664), linewidth(0.4));
draw((0.8894305730917524,2.690894150918487)--(3.9547732243966385,-1.864861893765504), linewidth(0.4) + linetype("2 2"));
draw((1.445226775603362,3.064861893765504)--(0.,0.), linewidth(0.4));
draw((1.445226775603362,3.064861893765504)--(5.4,0.), linewidth(0.4));
draw((5.4,0.)--(-3.109814760356735,0.), linewidth(0.4));
draw((-3.5571674061519136,1.864861893765503)--(2.7,3.3658633371878666), linewidth(0.4));
draw((1.445226775603362,3.064861893765504)--(2.7,-2.1658633371878664), linewidth(0.4));
draw((-3.5571674061519136,1.864861893765503)--(-3.109814760356735,0.), linewidth(0.4));
draw((-3.109814760356735,0.)--(1.892579421398536,1.2), linewidth(0.4));
draw((1.892579421398536,1.2)--(1.892579421398536,0.), linewidth(0.4) + dotted);
draw((2.7,3.3658633371878666)--(2.7,-2.1658633371878664), linewidth(0.4));
draw((0.8894305730917524,2.690894150918487)--(0.8894305730917507,-1.4908941509184854), linewidth(0.4) + linetype("2 2"));
draw((2.7,3.3658633371878666)--(0.8894305730917507,-1.4908941509184854), linewidth(0.4) + linetype("2 2"));
draw((-3.5571674061519136,1.864861893765503)--(4.217574445111032,0.9163536869921585), linewidth(0.4));
draw((1.445226775603362,3.064861893765504)--(0.8894305730917507,-1.4908941509184854), linewidth(0.4) + linetype("2 2"));
draw((3.5074205786014643,0.)--(4.217574445111032,0.9163536869921585), linewidth(0.4) + dotted);
draw((0.6380284948350587,1.353053551156462)--(3.5074205786014643,0.), linewidth(0.4) + dotted);
draw((0.8894305730917524,2.690894150918487)--(2.7,-2.1658633371878664), linewidth(0.4) + linetype("2 2"));
/* dots and labels */
dot((0.,0.),linewidth(4.pt) + dotstyle);
label("$B$", (-0.32,-0.42), NE * labelscalefactor);
dot((2.7,0.6),linewidth(4.pt) + dotstyle);
label("$O$", (2.78,0.76), NE * labelscalefactor);
dot((2.7,-2.1658633371878664),linewidth(4.pt) + dotstyle);
label("$M$", (2.78,-2.), NE * labelscalefactor);
dot((1.892579421398536,1.2),linewidth(4.pt) + dotstyle);
label("$I$", (1.5,1.32), NE * labelscalefactor);
dot((1.445226775603362,3.064861893765504),linewidth(4.pt) + dotstyle);
label("$A$", (1.52,3.22), NE * labelscalefactor);
dot((-3.109814760356735,0.),linewidth(4.pt) + dotstyle);
label("$D$", (-3.22,-0.46), NE * labelscalefactor);
dot((1.892579421398536,0.),linewidth(4.pt) + dotstyle);
label("$L$", (1.72,-0.42), NE * labelscalefactor);
dot((4.217574445111032,0.9163536869921585),linewidth(4.pt) + dotstyle);
label("$B_1$", (4.3,1.08), NE * labelscalefactor);
dot((0.6380284948350587,1.353053551156462),linewidth(4.pt) + dotstyle);
label("$C_1$", (0.2,1.02), NE * labelscalefactor);
dot((-3.5571674061519136,1.864861893765503),linewidth(4.pt) + dotstyle);
label("$G$", (-3.48,2.02), NE * labelscalefactor);
dot((3.5074205786014643,0.),linewidth(4.pt) + dotstyle);
label("$V$", (3.38,-0.4), NE * labelscalefactor);
dot((2.7,3.3658633371878666),linewidth(4.pt) + dotstyle);
label("$N$", (2.78,3.52), NE * labelscalefactor);
dot((0.8894305730917524,2.690894150918487),linewidth(4.pt) + dotstyle);
label("$Z$", (0.36,2.5), NE * labelscalefactor);
dot((0.8894305730917507,-1.4908941509184854),linewidth(4.pt) + dotstyle);
label("$T$", (0.46,-1.78), NE * labelscalefactor);
dot((3.9547732243966385,-1.864861893765504),linewidth(4.pt) + dotstyle);
label("$A'$", (4.04,-1.7), NE * labelscalefactor);
dot((-0.8322939923766871,1.5324309468827513),linewidth(4.pt) + dotstyle);
label("$X$", (-0.98,1.74), NE * labelscalefactor);
dot((5.4,0.),linewidth(4.pt) + dotstyle);
label("$C$", (5.38,-0.36), NE * labelscalefactor);
dot((2.1804415825316923,0.),linewidth(4.pt) + dotstyle);
label("$K$", (2.2,-0.4), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
\end{asy}
\end{center}
\begin{claim*}
We have $\angle (\ol{AI}, \ol{B_1C_1})=\angle IAD$.
\end{claim*}
\begin{proof} Let $Z$ lie on $(ABC)$ with $\angle AZI=90^{\circ}$. By radical axis theorem on $(AIZ), (BIC),$ and $(ABC)$, we conclude that $D$ lies on $\ol{AZ}$. Let $\ol{NI}$ meet $(ABC)$ again at $T \neq N$.
Inversion in $(BIC)$ maps $\ol{AI}$ to $\ol{KI}$ and $(ABC)$ to $\ol{BC}$. Thus, $Z$ maps to $L$, so $Z, L, M$ are collinear. Since $BL=CV$ and $OI=OV$, we see that $MLIN$ is a trapezoid with $\ol{IL} \parallel \ol{MN}$. Thus, $\ol{ZT} \parallel \ol{MN}$.
It is known that $\ol{AT}$ and $\ol{AA_1}$ are isogonal in angle $BAC$. Since $\ol{AV}$ is a circumdiameter in $(AB_1C_1)$, so $\ol{AT} \perp \ol{B_1C_1}$. So $\dang ZAI=\dang NMT=90^{\circ}-\dang TAI=\dang (\ol{AI}, \ol{B_1C_1})$.
\end{proof}
Let $X$ be the midpoint of $\ol{AD}$ and $G$ be the reflection of $I$ in $X$. Since $AIDG$ is a rectangle, we have $\dang AIG=\dang ZAI=\dang (\ol{AI}, \ol{B_1C_1})$, by the previous claim. So $\ol{IG}$ coincides with $\ol{B_1C_1}$. Now $\ol{AI}$ bisects $\angle B_1AC_1$ and $\angle IAG=90^{\circ}$, so $(\ol{IG}; \ol{B_1C_1})=-1$.
Since $\angle IDG=90^{\circ}$, we see that $\ol{DI}$ and $\ol{DG}$ are bisectors of angle $B_1DC_1$. Now $\angle XDI=\angle XID \implies \angle XDC_1=\angle XID-\angle IDB_1=\angle DB_1C_1$, so $\ol{XD}$ is tangent to $(DB_1C_1)$.
|
sols-TST-IMO-2020_1
|
Choose positive integers $b_1$, $b_2$, \dots\ satisfying
\[ 1 = \frac{b_1}{1^2} > \frac{b_2}{2^2} > \frac{b_3}{3^2} > \frac{b_4}{4^2} > \dotsb \]
and let $r$ denote the largest real number
satisfying $\frac{b_n}{n^2} \ge r$ for all positive integers $n$.
What are the possible values of $r$ across all
possible choices of the sequence $(b_n)$?
|
The answer is $0 \le r \le 1/2$. Obviously $r \ge 0$.
In one direction, we show that
\begin{claim*}
[Greedy bound]
For all integers $n$, we have
\[ \frac{b_n}{n^2} \le \half + \frac{1}{2n}. \]
\end{claim*}
\begin{proof}
This is by induction on $n$. For $n=1$ it is given.
For the inductive step we have
\begin{align*}
b_n &< n^2 \frac{b_{n-1}}{(n-1)^2}
\le n^2 \left( \frac12 + \frac{1}{2(n-1)} \right)
= \frac{n^3}{2(n-1)} \\
&= \half \left[ n^2+n+1 + \frac{1}{n-1} \right] \\
&= \frac{n(n+1)}{2} + \half \left[ 1 + \frac{1}{n-1} \right] \\
&\le \frac{n(n+1)}{2} + 1
\end{align*}
So $b_n < \frac{n(n+1)}{2} + 1$
and since $b_n$ is an integer, $b_n \le \frac{n(n+1)}{2}$.
This implies the result.
\end{proof}
We now give a construction.
For $r=1/2$ we take $b_n = \half n(n+1)$ for $r=0$ we take $b_n = 1$.
\begin{claim*}
[Explicit construction, given by Nikolai Beluhov]
Fix $0 < r < 1/2$.
Let $N$ be large enough that $\left\lceil rn^2+n \right\rceil < \half n(n+1)$ for all $n \ge N$.
Then the following sequence works:
\[ b_n = \begin{cases}
\left\lceil rn^2 + n \right\rceil & n \ge N \\
\frac{n^2+n}{2} & n < N.
\end{cases} \]
\end{claim*}
\begin{proof}
We certainly have
\[ \frac{b_n}{n^2} = \frac{rn^2 + n + O(1)}{n^2} \xrightarrow{n \to \infty} r. \]
Mainly, we contend $b_n n^{-2}$ is strictly decreasing.
We need only check this for $n \ge N$; in fact
\[ \frac{b_n}{n^2} \ge \frac{rn^2+n}{n^2}
> \frac{[r(n+1)^2+(n+1)]+1}{(n+1)^2}
> \frac{b_{n+1}}{(n+1)^2} \]
where the middle inequality is true
since it rearranges to $\frac1n > \frac{n+2}{(n+1)^2}$.
\end{proof}
|
sols-TST-IMO-2020_2
|
Two circles $\Gamma_1$ and $\Gamma_2$
have common external tangents $\ell_1$ and $\ell_2$ meeting at $T$.
Suppose $\ell_1$ touches $\Gamma_1$ at $A$ and $\ell_2$ touches $\Gamma_2$ at $B$.
A circle $\Omega$ through $A$ and $B$ intersects $\Gamma_1$
again at $C$ and $\Gamma_2$ again at $D$, such that quadrilateral $ABCD$ is convex.
Suppose lines $AC$ and $BD$ meet at point $X$,
while lines $AD$ and $BC$ meet at point $Y$.
Show that $T$, $X$, $Y$ are collinear.
|
We present four solutions.
\paragraph{First solution, elementary (original).}
We have $\triangle YAC \sim \triangle YBD$, from which it follows
\[ \frac{d(Y,AC)}{d(Y,BD)} = \frac{AC}{BD}. \]
Moreover, if we denote by $r_1$ and $r_2$ the radii of $\Gamma_1$ and $\Gamma_2$, then
\[
\frac{d(T,AC)}{d(T,BD)}
= \frac{TA \sin \angle(AC,\ell_1)}{TB \sin \angle (BD,\ell_2)}
= \frac{2r_1 \sin \angle(AC,\ell_1)}{2r_2 \sin \angle(BD,\ell_2)}
= \frac{AC}{BD}
\]
the last step by the law of sines.
\begin{center}
\begin{asy}
size(12.5cm);
pair A,E,O_A,O_B,T;
T = (0,0);
A = (14.4,-6);
O_A = (16.9,0);
E = 1.6 * A;
O_B = 1.6 * O_A;
draw(circle(O_A,abs(O_A-A)),gray);
draw(circle(O_B,abs(O_B-E)),gray);
pair B = intersectionpoint(circumcircle(T,E,O_B),circle(O_B,abs(O_B-E)));
draw(T--1.2*E);
draw(T--1.2*B);
pair G = (17,-3);
// draw(circumcircle(A,B,G),red);
pair C = intersectionpoints(circle(O_A,abs(O_A-A)),circumcircle(A,B,G))[0];
pair D = intersectionpoints(circle(O_B,abs(O_B-B)),circumcircle(A,B,G))[1];
pair X = extension(A,C,B,D);
pair Y = extension(A,D,B,C);
draw(A--C,red);
draw(B--D,red);
draw(-0.2*X--1.6*X,dashed);
draw(A--Y--B, orange);
dot("$A$", A, dir(-90));
dot("$B$",B,dir(150));
dot("$C$",C,dir(-20));
dot("$D$",D,dir(D));
dot("$X$",X,dir(135));
dot("$Y$",Y,dir(-45));
dot("$T$",T,dir(-90));
label("$\ell_1$",8*dir(A),dir(-90));
label("$\ell_2$",8*dir(B),dir(90));
label("$\Gamma_1$",O_A+dir(170)*abs(O_A-A),dir(170),gray);
label("$\Gamma_2$",O_B+dir(70)*abs(O_B-B),dir(70),gray);
\end{asy}
\end{center}
This solves the problem up to configuration issues:
we claim that $Y$ and $T$ both lie
inside $\angle AXB \equiv \angle CXD$.
WLOG $TA < TB$.
\begin{itemize}
\ii The former is since $Y$ lies outside segments $BC$ and $AD$,
since we assumed $ABCD$ was convex.
\ii For the latter, we note that $X$ lies inside both $\Gamma_1$ and $\Gamma_2$
in fact on the radical axis of the two circles
(since $X$ was an interior point of both chords $AC$ and $BD$).
In particular, $X$ is contained inside $\angle ATB$, and moreover $\angle ATB < 90\dg$,
and this is enough to imply the result.
\end{itemize}
\paragraph{Second solution, inversive.}
This is based on the solution posted by \texttt{kapilpavase} on AoPS.
Consider the inversion at $T$ swapping $\Gamma_1$ and $\Gamma_2$;
we let it send $A$ to $E$, $B$ to $F$, $C$ to $V$, $D$ to $W$, as shown.
Draw circles $ADWE$ and $BCVF$.
\begin{center}
\begin{asy}
size(12.5cm);
pair A,E,O_A,O_B,T;
T = (0,0);
A = (14.4,-6);
O_A = (16.9,0);
E = 1.6 * A;
O_B = 1.6 * O_A;
draw(circle(O_A,abs(O_A-A)),gray);
draw(circle(O_B,abs(O_B-E)),gray);
pair B = intersectionpoint(circumcircle(T,E,O_B),circle(O_B,abs(O_B-E)));
draw(T--1.2*E);
draw(T--1.2*B);
pair G = (17,-3);
// draw(circumcircle(A,B,G),red);
pair C = intersectionpoints(circle(O_A,abs(O_A-A)),circumcircle(A,B,G))[0];
pair D = intersectionpoints(circle(O_B,abs(O_B-B)),circumcircle(A,B,G))[1];
pair X = extension(A,C,B,D);
pair Y = extension(A,D,B,C);
pair F = extension(T,B,A,A+B-E);
filldraw(circumcircle(A,D,E), opacity(0.1)+yellow, red);
filldraw(circumcircle(B,C,F), opacity(0.1)+yellow, red);
pair V = extension(T,C,X,E);
pair W = extension(T,D,X,F);
draw(A--C,red);
draw(B--D,red);
// draw(-0.2*X--1.6*X,dashed);
draw(A--Y--B, gray+dotted);
draw(W--T--C, lightblue);
draw(V--E, lightblue+dashed);
draw(W--F, lightblue+dashed);
dot("$A$", A, dir(225));
dot("$B$",B,dir(80));
dot("$C$",C,dir(-10));
dot("$D$",D,dir(-45));
dot("$X$",X,dir(175));
dot("$Y$",Y,dir(-45));
dot("$T$",T,dir(-90));
label("$\ell_1$",8*dir(A),dir(-90));
label("$\ell_2$",8*dir(B),dir(90));
label("$\Gamma_1$",O_A+dir(170)*abs(O_A-A),dir(170),gray);
label("$\Gamma_2$",O_B+dir(70)*abs(O_B-B),dir(70),gray);
dot("$E$", E, dir(-90));
dot("$F$", F, dir(160));
dot("$V$", V, dir(110));
dot("$W$", W, dir(-10));
\end{asy}
\end{center}
\begin{claim*}
Points $T$ and $Y$ lie on the radical axis of $(ADE)$ and $(BCF)$.
\end{claim*}
\begin{proof}
Because $TF \cdot TB = TA \cdot TE$ and $YA \cdot YD = YC \cdot YB$.
\end{proof}
\begin{claim*}
Point $X$ has equal power to $(ADE)$ and $(BCF)$.
\end{claim*}
\begin{proof}
Since $TV \cdot TC = TA \cdot TE$, quadrilateral $VCEA$ is cyclic too,
so by radical axis with $\Gamma_1$ and $\Gamma_2$ we find $X$ lies on $VE$.
Similarly, $X$ lies on $FW$.
Thus, $X$ is the center of negative inversion between $(ADE)$ and $(BCF)$.
But since $AE = BF$ and moreover
\begin{align*}
\dang BCF + \dang ADE
&= (\dang BCA + \dang ACF) + (\dang ADB + \dang BDE) \\
&= (\dang BCA + \dang ADB) + (\dang ACF + \dang BDE) = 0 + 0 = 0
\end{align*}
we conclude that $(ADE)$ and $(BCF)$ are \emph{congruent}.
As $X$ was the center of negative inversion between them, we're done.
\end{proof}
\paragraph{Third solution, projective (Nikolai Beluhov).}
We start with some definitions.
Let $\ell_1$ touch $\Gamma_2$ at $E$, $\ell_2$ touch $\Gamma_1$ at $F$,
$K = \ell_1 \cap \ol{BD}$, $L = \ell_2 \cap \ol{AC}$,
line $FX$ meet $\Gamma_1$ again at $M$,
line $EX$ meet $\Gamma_2$ again at $N$,
and lines $AB$, $AD$, and $BC$ meet line $TX$ at $Z$, $Y_1$, and $Y_2$.
Thus the desired statement is equivalent to $Y_1 = Y_2$.
\begin{claim*}
$(EB; ND)_{\Gamma_2} = (FA; MC)_{\Gamma_1}$.
\end{claim*}
\begin{proof}
Note that $AX \cdot XC = BX \cdot XD = EX \cdot XN$, so $AECN$ is cyclic.
Likewise $BFDM$ is cyclic.
Consider the inversion with center $T$ which swaps $\Gamma_1$ and $\Gamma_2$;
it also swaps the pairs $\{A, E\}$ and $\{B, F\}$.
Since $AECN$ is cyclic, $C$ is on $\Gamma_1$, and $N$ is on $\Gamma_2$,
it also swaps $\{C, N\}$; similarly it swaps $\{D, M\}$.
Thus $(EB; ND)_{\Gamma_2} = (AF; CM)_{\Gamma_1} = (FA; MC)_{\Gamma_1}$ as desired.
\end{proof}
With this claim, the remainder of the proof is chasing cross-ratios:
\[
(TZ; XY_1)
\stackrel{A}{=} (KB; XD)
\stackrel{E}{=} (EB; ND)_{\Gamma_2}
= (FA; MC)_{\Gamma_1}
\stackrel{F}{=} (LA; XC)
\stackrel{B}{=} (TZ; XY_2)
\]
implies $Y_1 = Y_2$ as desired.
\paragraph{Fourth solution by untethered moving points.}
Fix $\ell_1$, $\ell_2$, $T$, $\Gamma_1$ and $\Gamma_2$,
and let $\Gamma_1$ and $\Gamma_2$ meet at $U$ and $V$.
By the radical axis theorem, $X$ lies on $UV$.
Thus we instead treat $X$ as a variable point on line $UV$
and let $C = AX \cap \Gamma_1$, $D = BX \cap \Gamma_2$.
By definition, $X$ has degree $1$ and $T$ has degree $0$.
We apply \textbf{Zack's lemma} to untethered point $Y$.
Note that $C$ and $D$ move projectively on conics,
and therefore have degree $2$.
Then, lines $AD$ and $BC$ each have degree at most $\deg(A)+\deg(D)=0+2=2$,
and so their intersection $Y$ has degree at most $2+2=4$.
But when $X \in AB$, the lines $AD$ and $BC$ are the same,
so Zack's lemma implies that
\[ \deg Y \le 4 - 1 = 3. \]
Thus the assertion that $T$, $X$, $Y$ are collinear
(which for example can be seen as a certain vanishing determinant)
is a statement of degree at most $0+1+3=4$.
Thus it suffices to find $5$ values of $X$
(other than $X \in \ol{AB}$, which we used already).
This is remarkably easy:
\begin{enumerate}
\item When $X=U$ or $X=V$, then $X=C=D=Y$ and the statement is obvious.
\item When $X \in \ell_1$, say, then $A=C$ and so $Y$ lies on $AC=\ell_1$ as well.
The case $X \in \ell_2$ is symmetric.
\item Finally, take $X$ at infinity along $UV$.
Then $C$ and $D$ are the other tangency points
of the circles with $\ell_1$, $\ell_2$, and so $AC=\ell_1$, $BD=\ell_2$, so $Y=T$.
\end{enumerate}
This finishes the problem.
|
sols-TST-IMO-2020_3
|
Let $\alpha \ge 1$ be a real number.
Hephaestus and Poseidon play a turn-based game
on an infinite grid of unit squares.
Before the game starts, Poseidon chooses a finite
number of cells to be \emph{flooded}.
Hephaestus is building a \emph{levee},
which is a subset of unit edges of the grid, called \emph{walls},
forming a connected, non-self-intersecting path or loop.
The game then begins with Hephaestus moving first.
On each of Hephaestus's turns, he adds one or more walls
to the levee, as long as the total length of the levee
is at most $\alpha n$ after his $n$th turn.
On each of Poseidon's turns,
every cell which is adjacent to an already flooded cell
and with no wall between them becomes flooded as well.
Hephaestus wins if the levee forms a closed loop
such that all flooded cells are
contained in the interior of the loop ---
hence stopping the flood and saving the world.
For which $\alpha$ can Hephaestus guarantee victory
in a finite number of turns
no matter how Poseidon chooses the initial cells to flood?
|
We show that if $\alpha > 2$
then Hephaestus wins,
but when $\alpha = 2$ (and hence $\alpha \le 2$)
Hephaestus cannot contain even a single-cell flood initially.
\bigskip
\textbf{Strategy for $\alpha > 2$}:
Impose $\ZZ^2$ coordinates on the cells.
Adding more flooded cells does not make our task easier,
so let us assume that initially
the cells $(x,y)$ with $|x|+|y| \le d$ are flooded
for some $d \ge 2$;
thus on Hephaestus's $k$th turn,
the water is contained in $|x|+|y| \le d+k-1$.
Our goal is to contain the flood with a large rectangle.
We pick large integers $N_1$ and $N_2$ such that
\begin{align*}
\alpha N_1 &> 2N_1 + (2d + 3) \\
\alpha (N_1+N_2) &> 2N_2 + (6N_1 + 8d + 4).
\end{align*}
Mark the points $X_i$, $Y_i$ as shown
in the figure for $1 \le i \le 6$.
The red figures indicate the distance
between the marked points on the rectangle.
\begin{center}
\begin{asy}
size(10cm);
transform t = shift(-0.5,-0.5);
path c(real x, real y) {
return t * shift(x,y) * unitsquare ;
}
for (int x=-7; x<=7; ++x) {
for (int y=-16; y<=2; ++y) {
if (abs(x)+abs(y) <= 2) {
filldraw(c(x,y), blue, gray);
}
else if (abs(x)+abs(y) <= 7) {
filldraw(c(x,y), mediumcyan, gray);
}
else if (abs(x)+abs(y) <= 16) {
filldraw(c(x,y), paleblue, gray);
}
else {
filldraw(c(x,y), white, gray);
}
}
}
dotfactor *= 1.5;
dot("$X_1$", (-0.5, 2.5), dir(115));
dot("$Y_1$", ( 0.5, 2.5), dir(65));
dot("$X_2$", (-5.5, 2.5), dir(90));
dot("$Y_2$", ( 5.5, 2.5), dir(90));
dot("$X_3$", (-7.5, 2.5), dir(135));
dot("$Y_3$", ( 7.5, 2.5), dir(45));
dot("$X_4$", (-7.5,-0.5), dir(180));
dot("$Y_4$", ( 7.5,-0.5), dir(0));
dot("$X_5$", (-7.5, -9.5), dir(180));
dot("$Y_5$", ( 7.5, -9.5), dir(0));
dot("$X_6$", (-7.5,-16.5), dir(180));
dot("$Y_6$", ( 7.5,-16.5), dir(0));
draw(box( (-7.5,-16.5), (7.5,2.5) ), black+1.4 );
label("$1$", (0, 2.5), dir(90), red);
label("$N_1$", (-3, 2.5), dir(90), red);
label("$N_1$", ( 3, 2.5), dir(90), red);
label("$d$", (-6.5, 2.5), dir(90), red);
label("$d$", ( 6.5, 2.5), dir(90), red);
label("$d+1$", (-7.5, 1), dir(180), red);
label("$d+1$", ( 7.5, 1), dir( 0), red);
label("$N_2$", (-7.5, -5), dir(180), red);
label("$N_2$", ( 7.5, -5), dir( 0), red);
label("$N_1+d$", (-7.5, -13), dir(180), red);
label("$N_1+d$", ( 7.5, -13), dir( 0), red);
\end{asy}
\end{center}
We follow the following plan.
\begin{itemize}
\ii Turn $1$: place wall $X_1 Y_1$.
This cuts off the flood to the north.
\ii Turns $2$ through $N_1+1$: extend
the levee to segment $X_2 Y_2$.
This prevents further flooding to the north.
\ii Turn $N_1+2$:
add in broken lines $X_4 X_3 X_2$ and $Y_4 Y_3 Y_2$ all at once.
This cuts off the flood west and east.
\ii Turns $N_1+2$ to $N_1+N_2+1$:
extend the levee along segments $X_4 X_5$ and $Y_4 Y_5$.
This prevents further flooding west and east.
\ii Turn $N_1 + N_2 + 2$:
add in the broken line $X_5 X_6 Y_6 Y_5$ all at once and win.
\end{itemize}
\bigskip
\textbf{Proof for $\alpha = 2$}:
Suppose Hephaestus contains the flood
on his $(n+1)$st turn.
We prove that $\alpha > 2$ by showing
that in fact at least $2n+4$ walls have been constructed.
Let $c_0$, $c_1$, \dots, $c_n$ be a path of cells such that
$c_0$ is the initial cell flooded,
and in general $c_i$ is flooded on
Poseidon's $i$th turn from $c_{i-1}$.
The levee now forms a closed loop enclosing all $c_i$.
\begin{claim*}
If $c_i$ and $c_j$ are adjacent
then $|i-j|=1$.
\end{claim*}
\begin{proof}
Assume $c_i$ and $c_j$ are adjacent but $|i-j|>1$.
Then the two cells must be separated by a wall.
But the levee forms a closed loop,
and now $c_i$ and $c_j$ are on opposite sides.
\end{proof}
Thus the $c_i$ actually form a path.
We color \emph{green} any edge of the unit grid (wall or not)
which is an edge of exactly one $c_i$
(i.e.\ the boundary of the polyomino).
It is easy to see there are exactly $2n+4$ green edges.
Now, from the center of each cell $c_i$,
shine a laser towards each green edge of $c_i$
(hence a total of $2n+4$ lasers are emitted).
An example below is shown for $n = 6$,
with the levee marked in brown.
\begin{center}
\begin{asy}
for (int i=0; i<=4; ++i) {
for (int j=0; j<=4; ++j) {
draw(shift(i,j)*unitsquare, lightgray);
}
}
draw( (0,0)--(0,2)--(2,2)--(2,3)--(0,3)--(0,5)
--(5,5)--(5,2)--(3,2)--(3,0)--cycle, brown+1.5 );
filldraw( (0.1,1.1)--(2.9,1.1)--(2.9,4.9)--(1.1,4.9)
--(1.1,4.1)--(2.1,4.1)--(2.1,1.9)--(0.1,1.9)--cycle,
palecyan, green+1.5);
label("$c_0$", (1.5,4.5));
label("$c_1$", (2.5,4.5));
label("$c_2$", (2.5,3.5));
label("$c_3$", (2.5,2.5));
label("$c_4$", (2.5,1.5));
label("$c_5$", (1.5,1.5));
label("$c_6$", (0.5,1.5));
// Lasers from c_6
draw( (0.5,1.5)--(0,1.5), red, EndArrow(TeXHead), Margin(2,1) );
draw( (0.5,1.5)--(0.5,2), red, EndArrow(TeXHead), Margin(2,1) );
draw( (0.5,1.5)--(0.5,0), red, EndArrow(TeXHead), Margin(2,1) );
// Lasers from c_5
draw( (1.5,1.5)--(1.5,2), red, EndArrow(TeXHead), Margin(2,1) );
draw( (1.5,1.5)--(1.5,0), red, EndArrow(TeXHead), Margin(2,1) );
// Lasers from c_4
draw( (2.5,1.5)--(3,1.5), red, EndArrow(TeXHead), Margin(2,1) );
draw( (2.5,1.5)--(2.5,0), red, EndArrow(TeXHead), Margin(2,1) );
// Lasers from c_3
draw( (2.5,2.5)--(2,2.5), red, EndArrow(TeXHead), Margin(2,1) );
draw( (2.5,2.5)--(5,2.5), red, EndArrow(TeXHead), Margin(2,1) );
// Lasers from c_2
draw( (2.5,3.5)--(0,3.5), red, EndArrow(TeXHead), Margin(2,1) );
draw( (2.5,3.5)--(5,3.5), red, EndArrow(TeXHead), Margin(2,1) );
// Lasers from c_1
draw( (2.5,4.5)--(2.5,5), red, EndArrow(TeXHead), Margin(2,1) );
draw( (2.5,4.5)--(5,4.5), red, EndArrow(TeXHead), Margin(2,1) );
// Lasers from c_0
draw( (1.5,4.5)--(1.5,5), red, EndArrow(TeXHead), Margin(2,1) );
draw( (1.5,4.5)--(0,4.5), red, EndArrow(TeXHead), Margin(2,1) );
draw( (1.5,4.5)--(1.5,3), red, EndArrow(TeXHead), Margin(2,1) );
\end{asy}
\end{center}
\begin{claim*}
No wall is hit by more than one laser.
\end{claim*}
\begin{proof}
Assume for contradiction that a wall $w$ is hit
by lasers from $c_i$ and $c_j$.
WLOG that laser is vertical, so
$c_i$ and $c_j$ are in the same column
(e.g.\ $(i,j) = (0,5)$ in figure).
We consider two cases on the position of $w$.
\begin{itemize}
\ii If $w$ is between $c_i$ and $c_j$,
then we have found a segment intersecting
the levee exactly once.
But the endpoints of the segment lie inside the levee.
This contradicts the assumption that the levee is a closed loop.
\ii Suppose $w$ lies above both $c_i$ and $c_j$
and assume WLOG $i < j$.
Then we have found that there is no levee at all
between $c_i$ and $c_j$.
Let $\rho \ge 1$ be the distance between
the centers of $c_i$ and $c_j$.
Then $c_j$ is flooded in a straight line from $c_i$
within $\rho$ turns, and this is the unique
shortest possible path.
So this situation can only occur if $j = i+\rho$
and $c_i$, \dots, $c_j$ form a column.
But then no vertical lasers from $c_i$ and $c_j$
may point in the same direction, contradiction.
\end{itemize}
Since neither case is possible, the proof ends here.
\end{proof}
This implies the levee has at least $2n+4$ walls
(the number of lasers) on Hephaestus's $(n+1)$st turn.
So $\alpha \ge \frac{2n+4}{n+1} > 2$.
\begin{remark*}
[Author comments]
The author provides the following remarks.
\begin{itemize}
\item Even though the flood can be stopped when $\alpha = 2 + \eps$, it takes a very long time to do that. Starting from a single flooded cell, the strategy I have outlined requires $\Theta(1 / \eps^2)$ days. Starting from several flooded cells contained within an area of diameter $D$, it takes $\Theta( D / \eps^2)$ days. I do not know any strategies that require fewer days than that.
\item There is a gaping chasm between $\alpha \le 2$ and $\alpha > 2$. Since $\alpha \le 2$ does not suffice even when only one cell is flooded in the beginning, there are in fact no initial configurations at all for which it is sufficient. On the other hand, $\alpha > 2$ works for all initial configurations.
\item The second half of the solution essentially estimates the perimeter of a polyomino in terms of its diameter (where diameter is measured entirely within the polyomino).
It appears that this has not been done before, or at least I was unable to find any reference for it. I did find tons of references where the perimeter of a polyomino is estimated in terms of its area, but nothing concerning the diameter.
My argument is a formalisation of the intuition that if $P$ is any shortest path within some weirdly-shaped polyomino, then the boundary of that polyomino must hug $P$ rather closely so that $P$ cannot be shortened.
\end{itemize}
\end{remark*}
|
sols-TST-IMO-2020_4
|
For a finite simple graph $G$, we define $G'$ to be the graph
on the same vertex set as $G$, where for any two vertices $u \neq v$,
the pair $\{u,v\}$ is an edge of $G'$ if and only if
$u$ and $v$ have a common neighbor in $G$.
Prove that if $G$ is a finite simple graph which is isomorphic to $(G')'$,
then $G$ is also isomorphic to $G'$.
|
We say a vertex of a graph is \emph{fatal} if it has degree at least $3$,
and some two of its neighbors are not adjacent.
\begin{claim*}
The graph $G'$ has at least as many triangles as $G$,
and has strictly more if $G$ has any fatal vertices.
\end{claim*}
\begin{proof}
Obviously any triangle in $G$ persists in $G'$.
Moreover, suppose $v$ is a fatal vertex of $G$.
Then the neighbors of $G$ will form a clique in $G'$
which was not there already, so there are more triangles.
\end{proof}
Thus we only need to consider graphs $G$ with no fatal vertices.
Looking at the connected components, the only possibilities are cliques
(including single vertices), cycles, and paths.
So in what follows we restrict our attention to graphs $G$
only consisting of such components.
\begin{remark*}
[Warning]
Beware: assuming $G$ is connected loses generality.
For example, it could be that $G = G_1 \sqcup G_2$,
where $G_1' \cong G_2$ and $G_2' \cong G_1$.
\end{remark*}
First, note that the following are stable under the operation:
\begin{itemize}
\ii an isolated vertex,
\ii a cycle of odd length, or
\ii a clique with at least three vertices.
\end{itemize}
In particular, $G \cong G''$ holds for such graphs.
On the other hand, cycles of even length or paths of nonzero
length will break into more connected components.
For this reason, a graph $G$ with any of these components will not satisfy $G \cong G''$
because $G'$ will have strictly more connected components than $G$,
and $G''$ will have at least as many as $G'$.
Therefore $G \cong G''$ if and only if $G$
is a disjoint union of the three types of connected components named earlier.
Since $G \cong G'$ holds for such graphs as well, the problem statement follows right away.
\begin{remark*}
Note that the same proof works equally well
for an arbitrary number of iterations $G^{''\dots''} \cong G$,
rather than just $G'' \cong G$.
\end{remark*}
\begin{remark*}
The proposers included a variant of the problem where given any graph $G$,
the operation stabilized after at most $O(\log n)$ operations,
where $n$ was the number of vertices of $G$.
\end{remark*}
|
sols-TST-IMO-2020_5
|
Find all integers $n \ge 2$ for which there exists an integer $m$ and
a polynomial $P(x)$ with integer coefficients satisfying the following three conditions:
\begin{itemize}
\item $m > 1$ and $\gcd(m,n) = 1$;
\item the numbers $P(0)$, $P^2(0)$, \dots, $P^{m-1}(0)$ are not divisible by $n$; and
\item $P^m(0)$ is divisible by $n$.
\end{itemize}
Here $P^k$ means $P$ applied $k$ times, so $P^1(0) = P(0)$, $P^2(0) = P(P(0))$, etc.
|
The answer is that this is possible if and only if
there exists primes $p' < p$ such that $p \mid n$ and $p' \nmid n$.
(Equivalently, the radical $\opname{rad}(n)$
must not be the product of the first several primes.)
For a polynomial $P$, and an integer $N$, we introduce the notation
\[ \mathbf{zord}(P \bmod N)
\coloneq \min \left\{ e > 0 \mid P^e(0) \equiv 0 \bmod N \right\} \]
where we set $\min\varnothing=0$ by convention.
Note that in general we have
\[ \mathbf{zord}(P \bmod N) = \opname*{lcm}_{q \mid N}
\left( \mathbf{zord}(P \bmod q) \right) \qquad (\dagger) \]
where the index runs over all prime powers
$q \mid N$ (by Chinese remainder theorem).
This will be used in both directions.
\textbf{Construction}:
First, we begin by giving a construction.
The idea is to first use the following prime power case.
\begin{claim*}
Let $p^e$ be a prime power, and $1 \le k < p$.
Then
\[ f(X) = X+1 - k \cdot \frac{X(X-1)(X-2) \dots (X-(k-2))}{(k-1)!} \]
viewed as a polynomial in $(\ZZ/p^e)[X]$ satisfies $\mathbf{zord}(f \bmod{p^e}) = k$.
\end{claim*}
\begin{proof}
Note $f(0) = 1$, $f(1) = 2$, \dots, $f(k-2) = k-1$, $f(k-1) = 0$ as needed.
\end{proof}
This gives us a way to do the construction now.
For the prime power $p^e \mid n$, we choose $1 \le p' < p$ and require
$\mathbf{zord}(P \bmod{p^e}) = p'$ and
$\mathbf{zord}(P \bmod{q}) = 1$ for every other prime power $q$ dividing $n$.
Then by $(\dagger)$ we are done.
\begin{remark*}
The claim can be viewed as a special case of Lagrange interpolation
adapted to work over $\ZZ/p^e$ rather than $\ZZ/p$.
Thus the construction of the polynomial $f$ above is quite natural.
\end{remark*}
\textbf{Necessity}: by $(\dagger)$ again,
it will be sufficient to prove the following claim.
\begin{claim*}
For any prime power $q = p^e$, and any polynomial $f(x) \in \ZZ[x]$,
if the quantity $\mathbf{zord}(f \bmod q)$ is nonzero
then it has all prime factors at most $p$.
\end{claim*}
\begin{proof}
This is by induction on $e \ge 1$.
For $e=1$, the pigeonhole principle immediately
implies that $\mathbf{zord}(P \bmod p) \le p$.
Now assume $e \ge 2$.
Let us define
\[ k \coloneq \mathbf{zord}(P \bmod{p^{e-1}}),
\qquad Q \coloneq P^k. \]
Since being periodic modulo $p^e$ requires periodic modulo $p^{e-1}$,
it follows that
\[ \mathbf{zord}(P \bmod{p^e}) = k \cdot \mathbf{zord}(Q \bmod{p^e}). \]
However since $Q(0) \equiv 0\bmod p^{e-1}$, it follows $\{Q(0), Q^2(0), \dots\}$
are actually all multiples of $p^{e-1}$.
There are only $p$ residues modulo $p^e$ which are also multiples of $p^{e-1}$,
so $\mathbf{zord}(Q \bmod{p^e}) \le p$, as needed.
\end{proof}
\begin{remark*}
One reviewer submitted the following test-solving comments:
This is one of these problems where you can make many useful natural observations,
and if you make enough of them eventually they cohere into a solution.
For example, here are some things I noticed while solving:
\begin{itemize}
\ii The polynomial $1-x$ shows that $m=2$ works for any odd $n$.
\ii In general, if $\zeta$ is a primitive $m$th root of unity
modulo $n$, then $\zeta(x+1)-1$ has the desired property (assuming $\gcd(m,n) = 1$).
We can extend this using the Chinese remainder theorem
to find that if $p \mid n$, $m \mid p-1$, and $\gcd(m,n)=1$, then $n$ works.
So by this point I already have something
about the prime factors of $n$ being sort-of closed downwards.
\ii By iterating $P$ we see it is enough to consider $m$ prime.
\ii In the case where $n=2^k$,
it is not too difficult to show that no odd prime $m$ works.
\end{itemize}
\end{remark*}
|
sols-TST-IMO-2020_6
|
Let $P_1P_2 \dots P_{100}$ be a cyclic $100$-gon,
and let $P_i = P_{i+100}$ for all $i$.
Define $Q_i$ as the intersection of diagonals
$\ol{P_{i-2}P_{i+1}}$ and $\ol{P_{i-1}P_{i+2}}$ for all integers $i$.
Suppose there exists a point $P$
satisfying $\ol{PP_i} \perp \ol{P_{i-1}P_{i+1}}$
for all integers $i$.
Prove that the points $Q_1$, $Q_2$, \dots, $Q_{100}$
are concyclic.
\end{enumerate}
|
We show two solutions.
In addition, Luke Robitaille has a reasonable complex numbers solution
posted at \url{https://aops.com/community/p26795631}.
\paragraph{Solution to proposed problem.}
We let $\ol{PP_2}$ and $\ol{P_1P_3}$
intersect (perpendicularly) at point $K_2$,
and define $K_\bullet$ cyclically.
\begin{center}
\begin{asy}
size(16cm);
pair P_1 = dir(178);
pair P_2 = dir(136);
pair P_3 = dir(82);
pair P_4 = dir(32);
pair K_2 = foot(P_2, P_1, P_3);
pair K_3 = foot(P_3, P_2, P_4);
pair P = extension(P_2, K_2, P_3, K_3);
pair K_4 = foot(P_3, P, P_4);
pair P_5 = -P_3+2*foot(origin, P_3, K_4);
pair K_5 = foot(P_4, P, P_5);
pair P_6 = -P_4+2*foot(origin, P_4, K_5);
pair K_1 = foot(P_2, P, P_1);
pair P_0 = -P_2+2*foot(origin, P_2, K_1);
fill(P--P_1--P_2--cycle, opacity(0.1)+paleblue);
fill(P--P_2--P_3--cycle, opacity(0.1)+lightgreen);
fill(P--P_3--P_4--cycle, opacity(0.1)+paleblue);
fill(P--P_4--P_5--cycle, opacity(0.1)+lightgreen);
draw(unitcircle, blue);
draw(P_0--P_1--P_2--P_3--P_4--P_5--P_6, blue);
draw(P--P_1, red);
draw(P--P_2, red);
draw(P--P_3, red);
draw(P--P_4, red);
draw(P--P_5, red);
draw(P_1--P_3--P_5, deepgreen);
draw(P_0--P_2--P_4--P_6, deepgreen);
draw(P_0--P_3, orange);
draw(P_1--P_4, orange);
draw(P_2--P_5, orange);
draw(P_3--P_6, orange);
pair H_1 = orthocenter(P, P_1, P_2);
pair H_2 = orthocenter(P, P_2, P_3);
pair H_3 = orthocenter(P, P_3, P_4);
pair H_4 = orthocenter(P, P_4, P_5);
pair Q_2 = extension(P_0, P_3, P_1, P_4);
pair Q_3 = extension(P_1, P_4, P_2, P_5);
pair Q_4 = extension(P_2, P_5, P_3, P_6);
pair N = circumcenter(K_1, K_2, K_3);
draw(CP(N, K_1), deepgreen);
pair E = 2*N-P;
pair E_2 = foot(E, P_1, P_3);
pair E_3 = foot(E, P_2, P_4);
pair E_4 = foot(E, P_3, P_5);
draw(E_2--E_3--E_4, lightblue+1.5);
draw(K_2--K_3--K_4, orange+1.5);
draw(P--H_1--E, deepcyan);
draw(P--H_2--E, deepcyan);
draw(P--H_3--E, deepcyan);
draw(P--E);
draw(E_2--E, brown);
draw(E_3--E, brown);
draw(E_4--E, brown);
pair F_1 = extension(E, H_1, P_0, P_3);
pair F_2 = extension(E, H_2, P_1, P_4);
pair F_3 = extension(E, H_3, P_2, P_5);
draw(E_2--E_4, deepgreen+1.5);
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("$K_2$", K_2, dir(K_2));
dot("$K_3$", K_3, dir(K_3));
dot("$P$", P, dir(P));
dot("$K_4$", K_4, dir(K_4));
dot("$P_5$", P_5, dir(P_5));
dot("$K_5$", K_5, dir(K_5));
dot(P_6);
dot("$K_1$", K_1, dir(K_1));
dot(P_0);
dot("$H_1$", H_1, dir(H_1));
dot("$H_2$", H_2, dir(H_2));
dot("$H_3$", H_3, dir(H_3));
dot("$H_4$", H_4, dir(H_4));
dot("$Q_2$", Q_2, dir(Q_2));
dot("$Q_3$", Q_3, dir(Q_3));
dot("$Q_4$", Q_4, dir(Q_4));
dot(N);
dot("$E$", E, dir(E));
dot("$E_2$", E_2, dir(E_2));
dot("$E_3$", E_3, dir(E_3));
dot("$E_4$", E_4, dir(E_4));
dot(F_1);
dot(F_2);
dot(F_3);
/* TSQ Source:
!size(16cm);
P_1 = dir 178
P_2 = dir 136
P_3 = dir 82
P_4 = dir 32
K_2 = foot P_2 P_1 P_3
K_3 = foot P_3 P_2 P_4
P = extension P_2 K_2 P_3 K_3
K_4 = foot P_3 P P_4
P_5 = -P_3+2*foot(origin, P_3, K_4)
K_5 = foot P_4 P P_5
P_6 .= -P_4+2*foot(origin, P_4, K_5)
K_1 = foot P_2 P P_1
P_0 .= -P_2+2*foot(origin, P_2, K_1)
!fill(P--P_1--P_2--cycle, opacity(0.1)+paleblue);
!fill(P--P_2--P_3--cycle, opacity(0.1)+lightgreen);
!fill(P--P_3--P_4--cycle, opacity(0.1)+paleblue);
!fill(P--P_4--P_5--cycle, opacity(0.1)+lightgreen);
unitcircle blue
P_0--P_1--P_2--P_3--P_4--P_5--P_6 blue
P--P_1 red
P--P_2 red
P--P_3 red
P--P_4 red
P--P_5 red
P_1--P_3--P_5 deepgreen
P_0--P_2--P_4--P_6 deepgreen
P_0--P_3 orange
P_1--P_4 orange
P_2--P_5 orange
P_3--P_6 orange
H_1 = orthocenter P P_1 P_2
H_2 = orthocenter P P_2 P_3
H_3 = orthocenter P P_3 P_4
H_4 = orthocenter P P_4 P_5
Q_2 = extension P_0 P_3 P_1 P_4
Q_3 = extension P_1 P_4 P_2 P_5
Q_4 = extension P_2 P_5 P_3 P_6
N .= circumcenter K_1 K_2 K_3
CP N K_1 deepgreen
E = 2*N-P
E_2 = foot E P_1 P_3
E_3 = foot E P_2 P_4
E_4 = foot E P_3 P_5
E_2--E_3--E_4 lightblue+1.5
K_2--K_3--K_4 orange+1.5
P--H_1--E deepcyan
P--H_2--E deepcyan
P--H_3--E deepcyan
P--E
E_2--E brown
E_3--E brown
E_4--E brown
F_1 .= extension E H_1 P_0 P_3
F_2 .= extension E H_2 P_1 P_4
F_3 .= extension E H_3 P_2 P_5
E_2--E_4 deepgreen+1.5
*/
\end{asy}
\end{center}
\begin{claim*}
The points $K_\bullet$ are concyclic
say with circumcircle $\gamma$.
\end{claim*}
\begin{proof}
Note that $PP_1 \times PK_1 = PP_2 \times PK_2 = \dots$
so the result follows by inversion at $P$.
\end{proof}
Let $E_i$ be the second intersection
of line $\ol{P_{i-1} K_i P_{i+1}}$ with $\gamma$;
then it follows that the perpendiculars to $\ol{P_{i-1} P_{i+1}}$
at $E_i$ all concur at a point $E$,
which is the reflection of $P$ across the center of $\gamma$.
We let $H_2 = \ol{P_1P_3} \cap \ol{P_2P_4}$
denote the orthocenter of $\triangle PP_2P_3$
and define $H_\bullet$ cyclically.
\begin{claim*}
We have
\[ \ol{EH_2} \perp \ol{P_1 P_4} \parallel \ol{K_2 K_3}
\quad\text{and}\quad
\ol{PH_2} \perp \ol{E_2 E_3} \parallel \ol{P_2 P_3}. \]
\end{claim*}
\begin{proof}
Both parallelisms follow by Reim's theorem through
$\angle E_2 H_2 E_3 = \angle K_2 H_2 K_3$,
So we need to show the perpendicularities.
Note that $\ol{H_2 P}$ and $\ol{H_2 E}$
are respectively circum-diameters of
$\triangle H_2K_2K_3$ and $\triangle H_2E_2E_3$.
As $\ol{K_2 K_3}$ and $\ol{E_2 E_3}$ are anti-parallel,
it follows $\ol{H_2P}$ and $\ol{H_2E}$ are isogonal
and we derive both perpendicularities.
\end{proof}
\begin{claim*}
The points $E$, $Q_3$, $E_3$ are collinear.
\end{claim*}
\begin{proof}
We use the previous claim.
The parallelisms imply that
\[ \frac{E_3H_2}{E_3P_2} = \frac{E_2H_2}{E_2P_3}
= \frac{E_4H_3}{E_4P_3} = \frac{E_3H_3}{E_3P_4}. \]
Now consider a homothety centered at $E_3$ sending $H_2$ to $P_2$
and $H_3$ to $P_4$.
Then it should send the orthocenter of $\triangle EH_2H_3$ to $Q_3$,
proving the result.
\end{proof}
From all this it follows that $\triangle EQ_2Q_3 \sim \triangle PK_2K_3$
as the opposite sides are all parallel.
Repeating this we actually find a homothety of $100$-gons
\[ Q_1 Q_2 Q_3 \dots \sim K_1 K_2 K_3 \dots \]
and that concludes the proof.
\begin{remark*}
The proposer remarks that in fact,
if one lets $s$ be an integer
and instead defines $R_i = P_i P_{i+s} \cap P_{i+1} P_{i+s+1}$,
then the $R_\bullet$ are concyclic.
The present problem is the case $s=3$.
We comment on a few special cases:
\begin{itemize}
\ii There is nothing to prove for $s=1$.
\ii If $s=0$, this amounts to proving
that poles of $\ol{P_i P_{i+1}}$ are concyclic;
by inversion this is equivalent to showing the
midpoints of the sides are concyclic.
This is an interesting problem but not as difficult.
\ii The problem for $s=2$ is to show that our $H_\bullet$
are concyclic, which uses the $s=0$ case as a lemma.
\end{itemize}
\end{remark*}
\paragraph{Solution to generalization (Nikolai Beluhov).}
We are going to need some well-known lemmas.
\begin{lemma*}
Suppose that $ABCD$ is inscribed in a circle $\Gamma$.
Let the tangents to $\Gamma$ at $A$ and $B$ meet at $E$,
let the tangents to $\Gamma$ at $C$ and $D$ meet at $F$,
and let diagonals $AC$ and $BD$ meet at $P$.
Then points $E$, $F$, and $P$ are collinear.
\end{lemma*}
\begin{proof}
Let the circle of center $E$ and radius $EA = EB$
meet lines $AC$ and $BD$ for the second time at points $U$ and $V$.
By a simple angle chase, triangles $EUV$ and $FCD$ are homothetic.
\end{proof}
\begin{lemma*}
Suppose that points $X$ and $Y$ are isogonal conjugates
in polygon $\mathcal{A} = A_1A_2 \dots A_n$.
(This means that lines $A_iX$ and $A_iY$
are symmetric with respect to the interior angle bisector of
$\angle A_{i - 1}A_iA_{i + 1}$ for all $i$,
where $A_{n + j} \equiv A_j$ for all $j$.)
Then the $2n$ projections of $X$ and $Y$ on the sides of $\mathcal{A}$ are concyclic.
\end{lemma*}
\begin{proof}
By a simple angle chase, for all $i$ we have that
the four projections on sides
$A_{i - 1}A_i$ and $A_iA_{i + 1}$ are concyclic.
Say that they lie on circle $\Gamma_i$.
Consider the midpoint $M$ of segment $XY$.
For every side $s$ of $\mathcal{A}$,
we have that $M$ is equidistant from the projections of $X$ and $Y$ on $s$.
Therefore, $M$ is the center of $\Gamma_i$ for all $i$,
and thus all of the $\Gamma_i$ coincide.
\end{proof}
\begin{lemma*}
Let $\Gamma'$ and $\Gamma''$ be two circles
and let $r$ be some fixed real number.
Then the locus of points $X$ satisfying
$\opname{Pow}(X, \Gamma') \colon \opname{Pow}(X, \Gamma'') = r$
is a circle.
\end{lemma*}
\begin{proof}
This is a classical result in circle geometry.
A full proof is given, for example,
in item 115 of Roger Johnson's \emph{Advanced Euclidean Geometry}.
\end{proof}
We are ready to solve the problem.
Let $\mathcal{P}$ be our polygon,
let $O$ be its the circumcenter,
and let $\Gamma$ be its circumcircle.
Fix any index $i$. In triangle $P_{i - 1}P_iP_{i + 1}$,
we have that line $P_iP$ contains the altitude through $P_i$
and line $P_iO$ contains the circumradius through $P_i$.
Therefore, these two lines are symmetric with respect
to the interior angle bisector of $\angle P_{i - 1}P_iP_{i + 1}$.
Thus points $P$ and $O$ are isogonal conjugates in $\mathcal{P}$.
By Lemma 2, it follows that the projections of $O$
onto the sides of $\mathcal{P}$ are concyclic.
In other words, the midpoints of the sides of
$\mathcal{P}$ lie on some circle $\omega$.
Let $M_i$ be the midpoint of segment $P_iP_{i + 1}$
and let the tangents to $\Gamma$ at points $P_i$ and $P_{i + 1}$ meet at $T_i$.
Since inversion with respect to $\Gamma$ swaps $M_i$ and $T_i$ for all $i$,
and also since all of the $M_i$ lie on the same circle $\omega$,
we obtain that all of the $T_i$ lie on the same circle $\Omega$.
\begin{center}
\begin{asy}
import geometry;
pair O = (0, 0);
real R = 135;
circle Gamma = circle((point) O, R);
pair P = (15, 10);
pair[] PP;
PP.push(rotate(95, O)*(R, 0));
PP.push(rotate(125, O)*(R, 0));
int n = 6;
pair get_next(pair A, pair B) {
pair F = projection(line(P, B))*A;
pair C = 2*(projection(line(A, F))*O) - A;
return C;
}
for (int i = 2; i <= n - 1; i = i + 1) {
PP.push(get_next(PP[i - 2], PP[i - 1]));
}
pair[] QQ;
QQ.push((0, 0));
QQ.push((0, 0));
for (int i = 2; i <= n - 3; i = i + 1) {
QQ.push(extension(PP[i - 2], PP[i + 1], PP[i - 1], PP[i + 2]));
}
pair[] TT;
for (int i = 0; i <= n - 2; i = i + 1) {
TT.push(extension(PP[i], rotate(90, PP[i])*O, PP[i + 1], rotate(90, PP[i + 1])*O));
}
circle Omega = circle(TT[2], TT[3], TT[4]);
draw(arc(Omega.C, Omega.r, 85, 330), 1 + green);
draw(arc(O, R, 85, 330), 1 + blue);
for (int i = 0; i <= n - 2; i = i + 1) {
draw(PP[i]--PP[i + 1], 1 + blue);
draw(PP[i]--TT[i]--PP[i + 1], 1 + green);
}
for (int i = 0; i <= n - 4; i = i + 1) {
draw(PP[i]--PP[i + 3], 1 + red);
}
for (int i = 0; i <= n - 5; i = i + 1) {
draw(TT[i]--TT[i + 3], 1 + yellow);
}
for (int i = 0; i <= n - 1; i = i + 1) {
dot(PP[i], UnFill);
}
for (int i = 2; i <= n - 3; i = i + 1) {
dot(QQ[i], UnFill);
}
for (int i = 0; i <= n - 2; i = i + 1) {
dot(TT[i], UnFill);
}
label("$P_{i - 2}$", PP[0], unit((1, -1)));
label("$P_{i - 1}$", PP[1], unit((1, -0.5)));
label("$P_i$", PP[2], unit((1, 0.2)));
label("$P_{i + 1}$", PP[3], unit((1, 0.5)));
label("$P_{i + 2}$", PP[4], 2*unit((0.2, 1)));
label("$P_{i + 3}$", PP[5], 2*unit((-0.2, 1)));
label("$Q_i$", QQ[2], unit(-QQ[2]));
label("$Q_{i + 1}$", QQ[3], unit(-QQ[3]));
label("$T_{i - 2}$", TT[0], unit(TT[0]));
label("$T_{i - 1}$", TT[1], unit(TT[1]));
label("$T_i$", TT[2], unit(TT[2]));
label("$T_{i + 1}$", TT[3], unit(TT[3]));
label("$T_{i + 2}$", TT[4], unit(TT[4]));
\end{asy}
\end{center}
Again, fix any index $i$.
By Lemma 1 applied to cyclic quadrilateral $P_{i - 2}P_{i - 1}P_{i + 1}P_{i + 2}$,
we have that point $Q_i$ lies on line $T_{i - 2}T_{i + 1}$.
Similarly, point $Q_{i + 1}$ lies on line $T_{i - 1}T_{i + 2}$.
Define \[f_i = \frac{\opname{Pow}(Q_i, \Gamma)}{\opname{Pow}(Q_i, \Omega)}.\]
\begin{claim*}
We have $f_i = f_{i + 1}$ for all $i$.
\end{claim*}
\begin{proof}
Note that
\begin{align*}
\opname{Pow}(Q_i, \Gamma)
&= Q_iP_{i - 1} \cdot Q_iP_{i + 2} \\
\opname{Pow}(Q_{i + 1}, \Gamma)
&= Q_{i + 1}P_{i - 1} \cdot Q_{i + 1}P_{i + 2}. \\
\opname{Pow}(Q_i, \Omega)
&= Q_iT_{i - 2} \cdot Q_iT_{i + 1} \\
\opname{Pow}(Q_{i + 1}, \Omega)
&= Q_{i + 1}T_{i - 1} \cdot Q_{i + 1}T_{i + 2}.
\end{align*}
Consider cyclic quadrilateral $T_{i - 2}T_{i - 1}T_{i + 1}T_{i + 2}$.
Since $\Gamma$ touches its opposite sides $T_{i - 2}T_{i - 1}$
and $T_{i + 1}T_{i + 2}$ at points $P_{i - 1}$ and $P_{i + 2}$,
we have that line $P_{i - 1}P_{i + 2}$
makes equal angles with these opposite sides.
From here, a simple angle chase shows
that triangles $P_{i - 1}Q_iT_{i - 2}$ and $P_{i + 2}Q_{i + 1}T_{i + 2}$
are similar.
Thus
\[\frac{Q_iP_{i - 1}}{Q_iT_{i - 2}} = \frac{Q_{i + 1}P_{i + 2}}{Q_{i + 1}T_{i + 2}}.\]
Similarly,
\[\frac{Q_iP_{i + 2}}{Q_iT_{i + 1}} = \frac{Q_{i + 1}P_{i - 1}}{Q_{i + 1}T_{i - 1}}.\]
From these, the desired identity $f_i = f_{i + 1}$ follows.
\end{proof}
Therefore, the power ratio $f_i$ is the same for all $i$.
By Lemma 3 for circles $\Gamma$ and $\Omega$, the solution is complete.
\begin{remark*}
This solution applies to the full generalization (from $3$ to $s$)
mentioned in the end of the previous solution,
essentially with no change.
\end{remark*}
\begin{remark*}
Polygon $T_1T_2 \dots T_{100}$ is both
circumscribed about a circle and inscribed inside a circle.
Polygons like that are known as \emph{Poncelet polygons}.
See, for example, \url{https://en.wikipedia.org/wiki/Poncelet's_closure_theorem}.
This solution borrows a lot from the discussion of
Poncelet's closure theorem in \emph{Advanced Euclidean Geometry},
referenced above for Lemma 3.
\end{remark*}
|
sols-TST-IMO-2021_1
|
Determine all integers $s \ge 4$
for which there exist positive integers $a$, $b$, $c$, $d$
such that $s = a+b+c+d$
and $s$ divides $abc+abd+acd+bcd$.
|
The answer is $s$ composite.
\paragraph{Composite construction.}
Write $s = (w+x)(y+z)$,
where $w$, $x$, $y$, $z$ are positive integers.
Let $a=wy$, $b=wz$, $c=xy$, $d=xz$.
Then
\[ abc+abd+acd+bcd = wxyz(w+x)(y+z) \]
so this works.
\paragraph{Prime proof.}
Choose suitable $a$, $b$, $c$, $d$. Then
\[
(a+b)(a+c)(a+d)
= (abc+abd+acd+bcd) + a^2(a+b+c+d)
\equiv 0 \pmod s.
\]
Hence $s$ divides a product of positive integers less than $s$,
so $s$ is composite.
\begin{remark*}
Here is another proof that $s$ is composite.
Suppose that $s$ is prime.
Then the polynomial $(x-a)(x-b)(x-c)(x-d) \in \FF_s[x]$
is even, so the roots come in two opposite pairs in $\FF_s$.
Thus the sum of each pair is at least $s$,
so the sum of all four is at least $2s > s$, contradiction.
\end{remark*}
|
sols-TST-IMO-2021_2
|
Points $A$, $V_1$, $V_2$, $B$, $U_2$, $U_1$
lie fixed on a circle $\Gamma$, in that order,
and such that $BU_2 > AU_1 > BV_2 > AV_1$.
Let $X$ be a variable point on the arc $V_1 V_2$ of $\Gamma$
not containing $A$ or $B$.
Line $XA$ meets line $U_1 V_1$ at $C$,
while line $XB$ meets line $U_2 V_2$ at $D$.
Prove there exists a fixed point $K$, independent of $X$,
such that the power of $K$ to the circumcircle
of $\triangle XCD$ is constant.
|
For brevity, we let $\ell_i$ denote line $U_iV_i$ for $i=1,2$.
We first give an explicit description of the fixed point $K$.
Let $E$ and $F$ be points on $\Gamma$ such that $\ol{AE} \parallel \ell_1$
and $\ol{BF} \parallel \ell_2$.
The problem conditions imply that $E$ lies between $U_1$ and $A$
while $F$ lies between $U_2$ and $B$.
Then we let \[ K = \ol{AF} \cap \ol{BE}. \]
This point exists because $AEFB$ are the vertices
of a convex quadrilateral.
\begin{remark*}
[How to identify the fixed point]
If we drop the condition that $X$ lies on the arc,
then the choice above is motivated by choosing $X \in \{E,F\}$.
Essentially, when one chooses $X \to E$,
the point $C$ approaches an infinity point.
So in this degenerate case, the only points whose
power is finite to $(XCD)$ are bounded are those on line $BE$.
The same logic shows that $K$ must lie on line $AF$.
Therefore, if the problem is going to work,
the fixed point must be exactly $\ol{AF} \cap \ol{BE}$.
\end{remark*}
We give two possible approaches for proving the power
of $K$ with respect to $(XCD)$ is fixed.
\paragraph{First approach by Vincent Huang.}
We need the following claim:
\begin{claim*}
Suppose distinct lines $AC$ and $BD$ meet at $X$.
Then for any point $K$
\[ \opname{pow}(K, XAB) + \opname{pow}(K, XCD)
= \opname{pow}(K, XAD) + \opname{pow}(K, XBC). \]
\end{claim*}
\begin{proof}
The difference between the left-hand side and right-hand
side is a linear function in $K$,
which vanishes at all of $A$, $B$, $C$, $D$.
\end{proof}
Construct the points $P = \ell_1 \cap \ol{BE}$
and $Q = \ell_2 \cap \ol{AF}$, which do not depend on $X$.
\begin{claim*}
Quadrilaterals $BPCX$ and $AQDX$ are cyclic.
\end{claim*}
\begin{proof}
By Reim's theorem: $\dang CPB = \dang AEB = \dang AXB = \dang CXB$, etc.
\end{proof}
\begin{center}
\begin{asy}
pair A = dir(190);
pair B = -conj(A);
pair U_1 = dir(125);
pair V_1 = dir(220);
pair U_2 = dir(80);
pair V_2 = dir(310);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
draw(A--B, blue);
draw(U_1--V_1, deepgreen);
draw(U_2--V_2, deepgreen);
pair E = V_1*U_1/A;
pair F = V_2*U_2/B;
pair X = dir(265);
pair C = extension(A, X, U_1, V_1);
pair D = extension(B, X, U_2, V_2);
draw(A--X--B, lightred);
draw(A--E, deepcyan);
draw(B--F, deepcyan);
pair K = extension(A, F, B, E);
draw(A--F, blue);
draw(B--E, blue);
pair P = extension(B, E, U_1, V_1);
pair Q = extension(A, F, U_2, V_2);
filldraw(circumcircle(X, C, B), opacity(0.05)+yellow, dotted+orange);
filldraw(circumcircle(X, A, D), opacity(0.05)+yellow, dotted+orange);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$U_1$", U_1, dir(U_1));
dot("$V_1$", V_1, dir(V_1));
dot("$U_2$", U_2, dir(U_2));
dot("$V_2$", V_2, dir(V_2));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$X$", X, dir(X));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$K$", K, dir(270));
dot("$P$", P, dir(290));
dot("$Q$", Q, dir(240));
/* TSQ Source:
A = dir 190
B = -conj(A)
U_1 = dir 125
V_1 = dir 220
U_2 = dir 80
V_2 = dir 310
unitcircle 0.1 lightcyan / blue
A--B blue
U_1--V_1 deepgreen
U_2--V_2 deepgreen
E = V_1*U_1/A
F = V_2*U_2/B
X = dir 265
C = extension A X U_1 V_1
D = extension B X U_2 V_2
A--X--B lightred
A--E deepcyan
B--F deepcyan
K = extension A F B E R270
A--F blue
B--E blue
P = extension B E U_1 V_1 R290
Q = extension A F U_2 V_2 R240
circumcircle X C B 0.05 yellow / dotted orange
circumcircle X A D 0.05 yellow / dotted orange
*/
\end{asy}
\end{center}
Now, for the particular $K$ we choose, we have
\begin{align*}
\opname{pow}(K, XCD) &=
\opname{pow}(K, XAD) + \opname{pow}(K, XBC) - \opname{pow}(K, XAB) \\
&= KA \cdot KQ + KB \cdot KP - \opname{pow}(K, \Gamma).
\end{align*}
This is fixed, so the proof is completed.
\paragraph{Second approach by authors.}
Let $Y$ be the second intersection of $(XCD)$ with $\Gamma$.
Let $S = \ol{EY} \cap \ell_1$ and $T = \ol{FY} \cap \ell_2$.
\begin{claim*}
Points $S$ and $T$ lies on $(XCD)$ as well.
\end{claim*}
\begin{proof}
By Reim's theorem: $\dang CSY = \dang AEY = \dang AXY = \dang CXY$, etc.
\end{proof}
Now let $X'$ be any other choice of $X$,
and define $C'$ and $D'$ in the obvious way.
We are going to show that $K$ lies on the radical axis
of $(XCD)$ and $(X'C'D')$.
\begin{center}
\begin{asy}
pair A = dir(190);
pair B = -conj(A);
pair U_1 = dir(125);
pair V_1 = dir(220);
pair U_2 = dir(80);
pair V_2 = dir(310);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
draw(A--B, blue);
draw(U_1--V_1, deepgreen);
draw(U_2--V_2, deepgreen);
pair E = V_1*U_1/A;
pair F = V_2*U_2/B;
pair X = dir(265);
pair C = extension(A, X, U_1, V_1);
pair D = extension(B, X, U_2, V_2);
draw(A--X--B, lightred);
draw(A--E, deepcyan);
draw(B--F, deepcyan);
pair K = extension(A, F, B, E);
draw(A--F, blue);
draw(B--E, blue);
pair Y = -X+2*foot(X, origin, circumcenter(X, C, D));
draw(circumcircle(X, C, D), red);
pair S = extension(Y, E, U_1, V_1);
pair T = extension(Y, F, U_2, V_2);
draw(E--Y--F, brown);
pair Xp = dir(300);
pair Cp = extension(Xp, A, U_1, V_1);
pair Dp = extension(Xp, B, U_2, V_2);
draw(A--Xp--B, orange);
pair L = extension(S, Y, Cp, Xp);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$U_1$", U_1, dir(U_1));
dot("$V_1$", V_1, dir(V_1));
dot("$U_2$", U_2, dir(U_2));
dot("$V_2$", V_2, dir(V_2));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$X$", X, dir(X));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$K$", K, dir(270));
dot("$Y$", Y, dir(Y));
dot("$S$", S, dir(S));
dot("$T$", T, dir(T));
dot("$X'$", Xp, dir(Xp));
dot("$C'$", Cp, dir(170));
dot("$D'$", Dp, dir(10));
dot("$L$", L, dir(L));
/* TSQ Source:
A = dir 190
B = -conj(A)
U_1 = dir 125
V_1 = dir 220
U_2 = dir 80
V_2 = dir 310
unitcircle 0.1 lightcyan / blue
A--B blue
U_1--V_1 deepgreen
U_2--V_2 deepgreen
E = V_1*U_1/A
F = V_2*U_2/B
X = dir 265
C = extension A X U_1 V_1
D = extension B X U_2 V_2
A--X--B lightred
A--E deepcyan
B--F deepcyan
K = extension A F B E R270
A--F blue
B--E blue
Y = -X+2*foot X origin circumcenter X C D
circumcircle X C D red
S = extension Y E U_1 V_1
T = extension Y F U_2 V_2
E--Y--F brown
X' = dir 300
C' = extension Xp A U_1 V_1 R170
D' = extension Xp B U_2 V_2 R10
A--Xp--B orange
L = extension S Y Cp Xp
*/
\end{asy}
\end{center}
The main idea is as follows:
\begin{claim*}
The point $L = \ol{EY} \cap \ol{AX'}$ lies on the radical axis.
By symmetry, so does the point $M = \ol{FY} \cap \ol{BX'}$ (not pictured).
\end{claim*}
\begin{proof}
Again by Reim's theorem, $SC'YX'$ is cyclic.
Hence we have
\[ \opname{pow}(L, X'C'D') = LC' \cdot LX'
= LS \cdot LY = \opname{pow}(L, XCD). \qedhere \]
\end{proof}
To conclude, note that by Pascal theorem on
\[ EYFAX'B \]
it follows $K$, $L$, $M$ are collinear,
as needed.
\begin{remark*}
All the conditions about $U_1$, $V_1$, $U_2$, $V_2$
at the beginning are there to eliminate configuration issues,
making the problem less obnoxious to the contestant.
In particular, without the various assumptions,
there exist configurations in which the point $K$ is at infinity.
In these cases, the center of $XCD$ moves along a fixed line.
\end{remark*}
|
sols-TST-IMO-2021_3
|
Find all functions $f \colon \RR \to \RR$ that satisfy the inequality
\[
f(y) -
\left(\frac{z-y}{z-x} f(x) + \frac{y-x}{z-x}f(z)\right) \leq
f\left(\frac{x+z}{2}\right) - \frac{f(x)+f(z)}{2}
\]
for all real numbers $x < y < z$.
\end{enumerate}
|
Answer: all functions of the form $f(y) = a y^2 + by + c$, where
$a, b, c$ are constants with $a \leq 0$.
If $I = (x,z)$ is an interval,
we say that a real number $\alpha$ is a
\emph{supergradient} of $f$ at $y \in I$
if we always have
\[ f(t) \le f(y) + \alpha(t-y) \]
for every $t \in I$.
(This inequality may be familiar as the so-called ``tangent line trick''.
A cartoon of this situation is drawn below for intuition.)
We will also say $\alpha$ is a supergradient of $f$ at $y$,
without reference to the interval,
if $\alpha$ is a supergradient of \emph{some} open interval containing $y$.
\begin{center}
\begin{asy}
size(4cm);
pair X = (0,0);
pair Z = (7,3);
pair Y = (3,4);
draw(X..Y..Z, blue);
pair t = 3*dir(X..Y..Z, 1);
draw( (Y-t)--(Y+t), red, Arrows);
label(rotate(15)*"slope $\alpha$", Y+0.45*t, dir(90), red );
dot("$x$", X, dir(-90), blue);
dot("$z$", Z, dir(-90), blue);
dot("$y$", Y, dir(90), blue);
\end{asy}
\qquad
\begin{asy}
size(5cm);
pair X = (-3,-9/3);
pair Z = (2,-4/3);
draw( X..(-2,-4/3)..(-1,-1/3)..(0,0)..(1,-1/3)..Z, blue );
pair Y = (-0.5,-0.25/3);
draw(X--Z, deepgreen);
draw((Y-1.5*dir(Z-X))--(Y+1.5*dir(Z-X)), red, Arrows );
label(rotate(18)*"slope $\frac{f(z)-f(x)}{z-x}$",
midpoint(X--Z), dir(110), deepgreen);
dot("$x$", X, dir(-90), blue);
dot("$y = \frac{x+z}{2}$", Y, 1.5*dir(-60), blue);
dot("$z$", Z, dir(-90), blue);
\end{asy}
\end{center}
\begin{claim*}
The problem condition is equivalent to asserting
that $\frac{f(z) - f(x)}{z-x}$ is a supergradient of $f$
at $\frac{x+z}{2}$ for the interval $(x,z)$, for every $x < z$.
\end{claim*}
\begin{proof}
This is just manipulation.
\end{proof}
At this point, we may readily verify that all claimed
quadratic functions $f(x) = ax^2+bx+c$ work --- these functions are concave,
so the graphs lie below the tangent line at any point.
Given $x < z$, the tangent at $\frac{x+z}{2}$ has slope
given by the derivative $f'(x)=2ax+b$, that is
\[ f'\left(\frac{x+z}{2}\right) = 2a \cdot \frac{x+z}{2} + b
= \frac{f(z)-f(x)}{z-x} \]
as claimed.
(Of course, it is also easy to verify the condition directly
by elementary means, since it is just a polynomial inequality.)
Now suppose $f$ satisfies the required condition; we prove that it has
the above form.
\begin{claim*}
The function $f$ is concave.
\end{claim*}
\begin{proof}
Choose any $\Delta > \max\{z-y,y-x\}$.
Since $f$ has a supergradient $\alpha$ at $y$ over the interval
$(y-\Delta,y+\Delta)$, and this interval includes $x$ and $z$, we have
\begin{align*}
\frac{z-y}{z-x}f(x) + \frac{y-x}{z-x}f(z) &\leq
\frac{z-y}{z-x}(f(y) + \alpha(x-y)) + \frac{y-x}{z-x}(f(y) +
\alpha(z-y)) \\
&= f(y).
\end{align*}
That is, $f$ is a concave function.
Continuity follows from the fact that any concave
function on $\RR$ is automatically continuous.
\end{proof}
\begin{lemma*}
[see e.g.\ {\url{https://math.stackexchange.com/a/615161}} for picture]
Any concave function $f$ on $\RR$ is continuous.
\end{lemma*}
\begin{proof}
Suppose we wish to prove continuity at $p \in \RR$.
Choose any real numbers $a$ and $b$ with $a < p < b$.
For any $0 < \eps < \max(b-p,p-a)$ we always have
\[ f(p) + \frac{f(b)-f(p)}{b-p} \eps \le f(p+\eps) \le f(p) + \frac{f(p)-f(a)}{p-a} \eps \]
which implies right continuity; the proof for left continuity is the same.
\end{proof}
\begin{claim*}
The function $f$ cannot have more than one supergradient
at any given point.
\end{claim*}
\begin{proof}
Fix $y \in \RR$.
For $t > 0$, let's define the function
\[ g(t) = \frac{f(y)-f(y-t)}{t} - \frac{f(y+t)-f(y)}{t}. \]
We contend that $g(\eps) \le \frac35g(3\eps)$
for any $\eps > 0$.
Indeed by the problem condition,
\noindent
\begin{minipage}{0.5\textwidth}
\begin{align*}
f(y) &\le f(y-\eps) + \frac{f(y+\eps)-f(y-3\eps)}{4} \\
f(y) &\le f(y+\eps) - \frac{f(y+3\eps)-f(y-\eps)}{4}.
\end{align*}
Summing gives the desired conclusion.
\end{minipage}
\begin{minipage}{0.4\textwidth}
\begin{asy}
size(6cm);
real f(real x) { return -x*x/6; }
pair A = (-3, f(3));
pair B = (-2, f(2));
pair C = (-1, f(1));
pair D = (-0, f(0));
pair E = (1, f(1));
pair F = (2, f(2));
pair G = (3, f(3));
draw(A..B..C..D..E..F..G, blue);
draw(A--E, lightred);
draw(C--G, deepgreen);
draw( (C-2*dir(E-A))--(C+2*dir(E-A)), lightred, Arrows );
draw( (E-2*dir(G-C))--(E+2*dir(G-C)), deepgreen, Arrows );
dot("$y-3\varepsilon$", A, dir(-90), blue);
dot("$y-\varepsilon$", C, dir(-90), blue);
dot("$y$", D, dir(-90), blue);
dot("$y+\varepsilon$", E, dir(-90), blue);
dot("$y+3\varepsilon$", G, dir(-90), blue);
\end{asy}
\end{minipage}
Now suppose that $f$ has two supergradients $\alpha < \alpha'$ at point $y$.
For small enough $\eps$, we should have
we have $f(y-\eps) \leq f(y) - \alpha'\eps$
and $f(y+\eps) \leq f(y) + \alpha\eps$, hence
\[ g(\eps) = \frac{f(y)-f(y-\eps)}{\eps} - \frac{f(y+\eps)-f(y)}{\eps}
\geq \alpha' - \alpha. \]
This is impossible since $g(\eps)$ may be arbitrarily small.
\end{proof}
\begin{claim*}
The function $f$ is quadratic on the rational numbers.
\end{claim*}
\begin{proof}
Consider any four-term arithmetic progression $x, x+d, x+2d, x+3d$.
Because $(f(x+2d)-f(x+d))/d$ and $(f(x+3d)-f(x))/3d$ are both
supergradients of $f$ at the point $x+3d/2$, they must be equal, hence
\begin{equation} \label{quadratic}
f(x+3d) - 3f(x+2d) + 3f(x+d) - f(x) = 0.
\end{equation}
If we fix $d = 1/n$, it follows inductively
that $f$ agrees with a quadratic function $\wt f_n$ on the set $\frac1n \ZZ$.
On the other hand, for any $m \neq n$,
we apparently have $\wt f_n = \wt f_{mn} = \wt f_m$,
so the quadratic functions on each ``layer'' are all equal.
\end{proof}
Since $f$ is continuous, it follows $f$ is quadratic, as needed.
\begin{remark*}
[Alternate finish using differentiability due to Michael Ren]
In the proof of the main claim (about uniqueness of supergradients),
we can actually notice the two terms $\frac{f(y)-f(y-t)}{t}$
and $\frac{f(y+t)-f(y)}{t}$ in the definition of $g(t)$ are both monotonic
(by concavity).
Since we supplied a proof that $\lim_{t \to 0} g(t) = 0$,
we find $f$ is differentiable.
Now, if the derivative at some point exists,
it must coincide with all the supergradients;
(informally, this is why ``tangent line trick'' always has the slope
as the derivative, and formally, we use the mean value theorem).
In other words, we must have
\[ f(x+y) - f(x-y) = 2f'(x) \cdot y \]
holds for all real numbers $x$ and $y$.
By choosing $y=1$ we obtain that $f'(x) = f(x+1) - f(x-1)$
which means $f'$ is also continuous.
Finally differentiating both sides with respect to $y$ gives
\[ f'(x+y) - f'(x-y) = 2f'(x) \]
which means $f'$ obeys Jensen's functional equation.
Since $f'$ is continuous, this means $f'$ is linear.
Thus $f$ is quadratic, as needed.
\end{remark*}
|
sols-TST-IMO-2023_1
|
There are $2022$ equally spaced points
on a circular track $\gamma$ of circumference $2022$.
The points are labeled $A_1$, $A_2$, \dots, $A_{2022}$ in some order, each label used once.
Initially, Bunbun the Bunny begins at $A_1$.
She hops along $\gamma$ from $A_1$ to $A_2$, then from $A_2$ to $A_3$,
until she reaches $A_{2022}$, after which she hops back to $A_1$.
When hopping from $P$ to $Q$, she always hops along the
shorter of the two arcs $\arc{PQ}$ of $\gamma$;
if $\ol{PQ}$ is a diameter of $\gamma$, she moves along either semicircle.
Determine the maximal possible sum of the lengths of the $2022$ arcs
which Bunbun traveled, over all possible labellings of the $2022$ points.
|
Replacing $2022$ with $2n$, the answer is $2n^2 - 2n + 2$.
(When $n=1011$, the number is $2042222$.)
\begin{center}
\begin{asy}
unitsize(48);
int n = 5;
int m = 2*n;
int[] a = {};
a.push(0);
for (int i = 1; i < m; ++i)
a.push(a[i-1] + (i % 2 > 0 ? n : n-1));
int[] b = new int[m];
pair[] B = new pair[m];
b.cyclic = true;
B.cyclic = true;
for (int i = 0; i < m; ++i) {
b[a[i]] = i+1;
B[i] = dir(a[i]*360/m);
}
draw(unitcircle);
for (int i = 0; i < m; ++i) {
draw(B[i]--B[i+1], gray(0.7));
}
for (int i = 0; i < m; ++i) {
pair P = dir(i*360/m);
dot((string) b[i], P, P);
}
\end{asy}
\hspace{3em}
\begin{asy}
unitsize(48);
int n = 5;
int m = 2*n;
draw(unitcircle);
for (int i = 0; i < m; ++i)
dot(dir(i*360/m));
for (int i = 0; i < n; ++i) {
pair P = dir(i*360/m + 180/m);
draw(scale(1.15) * (P--(-P)), red);
}
\end{asy}
\end{center}
\paragraph{Construction.}
The construction for $n = 5$ shown on the left half of the figure
easily generalizes for all $n$.
\begin{remark*}
The validity of this construction
can also be seen from the below proof.
\end{remark*}
\paragraph{First proof of bound.}
Let $d_i$ be the shorter distance from $A_{2i-1}$ to $A_{2i+1}$.
\begin{claim*}
The distance of the leg of the journey $A_{2i-1} \to A_{2i} \to A_{2i+1}$
is at most $2n - d_i$.
\end{claim*}
\begin{proof}
Of the two arcs from $A_{2i-1}$ to $A_{2i+1}$,
Bunbun will travel either $d_i$ or $2n-d_i$.
One of those arcs contains $A_{2i}$ along the way.
So we get a bound of $\max(d_i, 2n-d_i) = 2n - d_i$.
\end{proof}
That means the total distance is at most
\[ \sum_{i=1}^n \left( 2n - d_i \right)
= 2n^2 - (d_1 + d_2 + \dots + d_n). \]
\begin{claim*}
We have \[ d_1 + d_2 + \dots + d_n \ge 2n-2. \]
\end{claim*}
\begin{proof}
The left-hand side is the sum of the walk
$A_1 \to A_3 \to \dots \to A_{2n-1} \to A_1$.
Among the $n$ points here, two of them must have distance at least $n-1$ apart;
the other $d_i$'s contribute at least $1$ each.
So the bound is $(n-1) + (n-1) \cdot 1 = 2n-2$.
\end{proof}
\paragraph{Second proof of bound.}
Draw the $n$ diameters through the $2n$ arc midpoints,
as shown on the right half of the figure for $n = 5$ in red.
\begin{claim*}[Interpretation of distances]
The distance between any two points equals the number of diameters crossed to travel between the points.
\end{claim*}
\begin{proof}
Clear.
\end{proof}
With this in mind, call a diameter \emph{critical} if it is crossed by all $2n$ arcs.
\begin{claim*}
At most one diameter is critical.
\end{claim*}
\begin{proof}
Suppose there were two critical diameters; these divide the circle into four arcs.
Then all $2n$ arcs cross both diameters, and so travel between opposite arcs.
But this means that points in two of the four arcs are never accessed --- contradiction.
\end{proof}
\begin{claim*}
Every diameter is crossed an even number of times.
\end{claim*}
\begin{proof}
Clear: the diameter needs to be crossed an even number of times for the loop to return to its origin.
\end{proof}
This immediately implies that the maximum possible total distance
is achieved when one diameter is crossed all $2n$ times,
and every other diameter is crossed $2n-2$ times,
for a total distance of at most
\[ n \cdot (2n-2) + 2 = 2n^2 - 2n + 2. \]
|
sols-TST-IMO-2023_2
|
Let $ABC$ be an acute triangle.
Let $M$ be the midpoint of side $BC$,
and let $E$ and $F$ be the feet of the altitudes from $B$ and $C$, respectively.
Suppose that the common external tangents
to the circumcircles of triangles $BME$ and $CMF$ intersect at a point $K$,
and that $K$ lies on the circumcircle of $ABC$.
Prove that line $AK$ is perpendicular to line $BC$.
|
We present several distinct approaches.
\paragraph{Inversion solution submitted by Ankan Bhattacharya and Nikolai Beluhov.}
Let $H$ be the orthocenter of $\triangle ABC$.
We use inversion in the circle with diameter $\ol{BC}$.
We identify a few images:
\begin{itemize}
\ii The circumcircles of $\triangle BME$ and $\triangle CMF$ are mapped to lines $BE$ and $CF$.
\ii The common external tangents are mapped to the two circles through $M$
which are tangent to lines $BE$ and $CF$.
\ii The image of $K$, denoted $K^\ast$, is the second intersection of these circles.
\ii The assertion that $K$ lies on $(ABC)$ is equivalent to $K^\ast$ lying on $(BHC)$.
\end{itemize}
However, now $K^\ast$ is simple to identify directly:
it's just the reflection of $M$ in the bisector of $\angle BHC$.
\begin{center}
\begin{asy}
size(12cm);
pair B = dir(202.82);
pair C = dir(337.18);
pair A = dir(136);
pair K = -B*C/A;
pair E = foot(B, C, A);
pair F = foot(C, A, B);
draw(B--E, lightred);
draw(C--F, lightred);
draw(A--K, lightred);
pair M = midpoint(B--C);
filldraw(unitcircle, opacity(0.1)+yellow, red);
filldraw(A--B--C--cycle, opacity(0.1)+yellow, red);
filldraw(circumcircle(B, M, E), opacity(0.1)+cyan, gray);
filldraw(circumcircle(C, M, F), opacity(0.1)+cyan, gray);
pair O_2 = circumcenter(C, M, F);
pair K_u = IP(CP(midpoint(K--O_2), K), CP(O_2, M));
pair K_v = OP(CP(midpoint(K--O_2), K), CP(O_2, M));
draw(K_u--K--K_v, gray);
pair H = orthocenter(A, B, C);
pair Q = foot(A, H, M);
filldraw(circumcircle(B, H, C), opacity(0.1)+green, deepgreen);
pair K_asterisk = ((2*foot(Q, B, C))-Q);
draw(circumcircle(K_asterisk, B, M), dotted);
draw(circumcircle(K_asterisk, C, M), dotted);
clip(box((-2,-2),(2,2)));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$A$", A, dir(A));
dot("$K$", K, dir(K));
dot("$E$", E, dir(70));
dot("$F$", F, dir(55));
dot("$M$", M, dir(285));
dot("$H$", H, dir(315));
dot("$K^{\ast}$", K_asterisk, dir(K_asterisk));
/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
!size(12cm);
B = dir 202.82
C = dir 337.18
A = dir 136
K = -B*C/A
E 70= foot B C A
F 55 = foot C A B
B--E lightred
C--F lightred
A--K lightred
M 285 = midpoint B--C
unitcircle / 0.1 yellow / red
A--B--C--cycle / 0.1 yellow / red
circumcircle B M E / 0.1 cyan / gray
circumcircle C M F / 0.1 cyan / gray
O_2 := circumcenter C M F
K_u := IP (CP (midpoint K--O_2) K) (CP O_2 M)
K_v := OP (CP (midpoint K--O_2) K) (CP O_2 M)
K_u--K--K_v / gray
H 315 = orthocenter A B C
Q := foot A H M
circumcircle B H C / 0.1 green / deepgreen
K& = (minus (mult 2 (foot Q B C)) Q)
circumcircle K& B M / dotted
circumcircle K& C M / dotted
!clip(box((-2,-2),(2,2)));
*/
\end{asy}
\end{center}
In particular, $\ol{HK^\ast}$ is a symmedian of $\triangle BHC$.
However, since $K^\ast$ lies on $(BHC)$,
this means $(HK^\ast; BC) = -1$.
Then, we obtain that $\ol{BC}$ bisects $\angle HMK^\ast \equiv \angle HMK$.
However, $K$ also lies on $(ABC)$,
which forces $K$ to be the reflection of $H$ in $\ol{BC}$.
Thus $\ol{AK} \perp \ol{BC}$, as wanted.
\paragraph{Solution with coaxial circles (Pitchayut Saengrungkongka).}
Let $H$ be the orthocenter of $\triangle ABC$. Let $Q$ be the second intersection of $\odot(BME)$ and $\odot(CMF)$.
We first prove the following well-known properties of $Q$.
\begin{claim*}
$Q$ is the Miquel point of $BCEF$. In particular, $Q$ lies on both $\odot(AEF)$ and $\odot(ABC)$.
\end{claim*}
\begin{proof}
Follows since $BCEF$ is cyclic with $M$ being the circumcenter.
\end{proof}
\begin{claim*}
$A(Q,H;B, C) = -1$.
\end{claim*}
\begin{proof}
By the radical center theorem on $\odot(AEF)$, $\odot(ABC)$, and $\odot(BCEF)$,
we get that $AQ$, $EF$, and $BC$ are concurrent. Now, the result follows from a
well-known harmonic property.
\end{proof}
Now, we get to the meat of the solution. Let the circumcircle of
$\odot(QMK)$ meet $BC$ again at $T\neq M$. The key claim is the following.
\begin{claim*}
$QT$ is tangent to $\odot(BQC)$.
\end{claim*}
\begin{proof}
We use the ``forgotten coaxiality lemma".
\begin{align*}
\frac{BT}{TC} &= \frac{TB\cdot TM}{TC\cdot TM} \\
&= \frac{\operatorname{pow}(T, \odot(BME))}
{\operatorname{pow}(T, \odot(CMF))} \\
&= \frac{\operatorname{pow}(K, \odot(BME))}
{\operatorname{pow}(K, \odot(CMF))} \\
&= \left(\frac{r_{\odot(BME)}}{r_{\odot(CMF)}}\right)^2 \\
&= \left(\frac{BQ/\sin\angle QMB}{CQ/\sin\angle QMC}\right)^2 \\
&= \frac{BQ^2}{CQ^2},
\end{align*}
implying the result.
\end{proof}
To finish, let $O$ be the center of $\odot(ABC)$. Then, from the claim, $\angle OQT = 90^\circ = \angle OMT$, so $O$ also lies on $\odot(QMTK)$.
Thus, $\angle OKT=90^\circ$, so $KT$ is also tangent to $\odot(ABC)$ as well. This implies that $QBKC$ is harmonic quadrilateral, and the result follows from the second claim.
\paragraph{Solution by Luke Robitaille.}
Let $Q$ be the second intersection of $\odot(BME)$ and
$\odot(CMF)$. We use the first two claims of the previous solution.
In particular, $Q\in\odot(ABC)$. We have the following claim.
\begin{claim*}[Also appeared in ISL 2017 G7]
We have $\measuredangle QKM = \measuredangle QBM + \measuredangle QCM$.
\end{claim*}
\begin{proof}
Let $KQ$ and $KM$ meet $\odot(BME)$ again at $Q'$ and $M'$. Then, by homothety, $\measuredangle Q'QM' = \measuredangle QCM$, so
\begin{align*}
\measuredangle QKM
&= \measuredangle Q'QM' + \measuredangle QM'M \\
&= \measuredangle QCM + \measuredangle QBM,
\end{align*}
as desired.
\end{proof}
Now, we extend $KM$ to meet $\odot(ABC)$ again at $Q_1$. We have
\begin{align*}
\measuredangle Q_1QB
= \measuredangle Q_1KB
&= \measuredangle Q_1KQ + \measuredangle QCB \\
&= \measuredangle MKQ + \measuredangle QKB \\
&= (\measuredangle MBQ + \measuredangle MCQ)
+ \measuredangle QCB \\
&= \measuredangle CBQ,
\end{align*}
implying that $QQ_1\parallel BC$. This implies that $QBKC$ is harmonic quadrilateral, so we are done.
\paragraph{Synthetic solution due to Andrew Gu (Harvard 2026).}
Define $O_1$ and $O_2$ as the circumcenters of $(BME)$ and $(CMF)$.
Let $T$ be the point on $(ABC)$ such that $\ol{AT} \perp \ol{BC}$.
Denote by $L$ the midpoint of minor arc $\arc{BC}$.
We are going to ignore the condition that $K$ lies on the circumcircle of $ABC$,
and prove the following unconditional result:
\begin{proposition*}
The points $T$, $L$, $K$ are collinear.
\end{proposition*}
This will solve the problem because if $K$ is on the circumcircle of $ABC$,
it follows $K = T$ or $K = L$; but $K = L$ can never occur
since $O_1$ and $O_2$ are obviously on different sides of line $LM$
so line $LM$ must meet $O_1 O_2$ inside segment $O_1 O_2$,
and $K$ lies outside this segment.
\begin{center}
\begin{asy}
size(14cm);
pair B = dir(200);
pair C = dir(340);
pair A = dir(113);
pair L = dir(270);
pair T = -B*C/A;
pair E = foot(B, C, A);
pair F = foot(C, A, B);
draw(B--E, lightred);
draw(C--F, lightred);
draw(A--T, lightred);
pair M = midpoint(B--C);
filldraw(unitcircle, opacity(0.1)+yellow, red);
filldraw(A--B--C--cycle, opacity(0.1)+yellow, red);
pair Q = foot(A, M, -A);
filldraw(circumcircle(B, M, Q), opacity(0.1)+cyan, gray);
filldraw(circumcircle(C, M, Q), opacity(0.1)+cyan, gray);
pair O_2 = circumcenter(C, M, F);
pair O_1 = circumcenter(B, M, E);
pair P_2 = 2*O_2-M;
pair P_1 = 2*O_1-M;
draw(P_2--M--P_1, lightcyan);
pair O = circumcenter(A, B, C);
pair Q_2 = extension(A, B, M, O);
pair Q_1 = extension(A, C, M, O);
draw(L--Q_2--A, lightred);
draw(B--Q_1, lightcyan);
draw(C--Q_2, lightcyan);
draw(B--P_1--Q_1, blue);
draw(C--P_2--Q_2, blue);
pair K = extension(L, T, O_1, O_2);
pair K_u = IP(CP(midpoint(K--O_2), K), CP(O_2, M));
pair K_v = OP(CP(midpoint(K--O_2), K), CP(O_2, M));
draw(K_u--K--K_v, gray);
draw(K--L, deepgreen);
draw(K--O_2, gray);
pair N = dir(90);
draw(K--M, deepgreen);
draw(A--N, deepgreen);
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$A$", A, dir(A));
dot("$L$", L, dir(L));
dot("$T$", T, dir(T));
dot("$E$", E, dir(80));
dot("$F$", F, dir(35));
dot("$M$", M, dir(285));
dot("$Q$", Q, dir(150));
dot("$O_2$", O_2, dir(350));
dot("$O_1$", O_1, dir(315));
dot("$P_2$", P_2, dir(P_2));
dot("$P_1$", P_1, dir(130));
dot("$Q_2$", Q_2, dir(Q_2));
dot("$Q_1$", Q_1, dir(45));
dot("$K$", K, dir(K));
dot("$N$", N, dir(45));
/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
!size(16cm);
B = dir 200
C = dir 340
A = dir 113
L = dir 270
T = -B*C/A
E 80 = foot B C A
F 35 = foot C A B
B--E lightred
C--F lightred
A--T lightred
M 285 = midpoint B--C
unitcircle / 0.1 yellow / red
A--B--C--cycle / 0.1 yellow / red
Q 150 = foot A M -A
circumcircle B M Q / 0.1 cyan / gray
circumcircle C M Q / 0.1 cyan / gray
O_2 350 = circumcenter C M F
O_1 315 = circumcenter B M E
P_2 = 2*O_2-M
P_1 130 = 2*O_1-M
P_2--M--P_1 / lightcyan
O := circumcenter A B C
Q_2 = extension A B M O
Q_1 45 = extension A C M O
L--Q_2--A / lightred
B--Q_1 / lightcyan
C--Q_2 / lightcyan
B--P_1--Q_1 / blue
C--P_2--Q_2 / blue
K = extension L T O_1 O_2
K_u := IP (CP (midpoint K--O_2) K) (CP O_2 M)
K_v := OP (CP (midpoint K--O_2) K) (CP O_2 M)
K_u--K--K_v / gray
K--L deepgreen
K--O_2 / gray
N 45 = dir 90
K--M deepgreen
A--N deepgreen
*/
\end{asy}
\end{center}
We now turn to the proof of the main lemma.
Let $P_1$ and $P_2$ be the antipodes of $M$ on these circles.
\begin{claim*}
Lines $AC$ and $LM$ meet at the antipode $Q_1$ of $B$ on $(BME)$,
so that $BP_1Q_1M$ is a rectangle.
Similarly,
lines $AB$ and $LM$ meet at the antipode $Q_2$ of $C$ on $(CMF)$,
so that $CP_2Q_2M$ is a rectangle.
\end{claim*}
\begin{proof}
Let $Q_1' = \omega_1 \cap AC \neq E$.
Then $\dang BMQ_1' = \dang BEQ_1' = 90^\circ$ hence $Q_1' \in LM$.
The other half of the lemma follows similarly.
\end{proof}
From this, it follows that $P_1Q_1 = BM = \half BC = MC = P_2Q_2$.
Letting $r_1$ denote the radius of $\omega_1$
(and similarly for $\omega_2$), we deduce that $CQ_1 = BQ_1 = 2r_1$.
\begin{claim*}
$KM = KL$.
\end{claim*}
\begin{proof}
I first claim that $\ol{CL}$ is the external bisector of $\angle Q_1 C Q_2$;
this follows from
\[ \dang Q_1 C L = \dang ACL = \dang ABL = \dang Q_2 B L = \dang Q_2 C L. \]
The external angle bisector theorem then gives an equality of directed ratios
\[ \frac{LQ_1}{LQ_2} = \frac{|CQ_1|}{|CQ_2|} = \frac{|BQ_1|}{|CQ_2|}
= \frac{2r_1}{2r_2} = \frac{r_1}{r_2} \]
Let the reflection of $M$ over $K$ be $P$; then $P$ lies on $\ol{P_1P_2}$ and
\[ \frac{PP_1}{PP_2} = \frac{2KO_1}{2KO_2} = \frac{KO_1}{KO_2}
= \frac{r_1}{r_2} = \frac{LQ_1}{LQ_2} \]
where again the ratios are directed.
Projecting everything onto line $LM$, so that $P_1$ lands at $Q_1$
and $P_2$ lands at $Q_2$,
we find that the projection of $P$ must land exactly at $L$.
\end{proof}
\begin{claim*}
Line $KM$ is an external angle bisector of $\angle O_1MO_2$.
\end{claim*}
\begin{proof}
Because $\frac{KO_1}{KO_2} = \frac{r_1}{r_2} = \frac{MO_1}{MO_2}$.
\end{proof}
To finish, note that we know that
$\ol{MP_1} \parallel \ol{CQ_1} \equiv \ol{AC}$
and $\ol{MP_2} \parallel \ol{BQ_2} \equiv \ol{AB}$,
meaning the angles $\angle O_1 M O_2$ and $\angle CAB$ have parallel legs.
Hence, if $N$ is the antipode of $L$,
it follows that $\ol{MK} \parallel \ol{AN}$.
Now from $MK = KL$ and the fact that $ANLT$ is an isosceles trapezoid,
we deduce that $\ol{LT}$ and $\ol{LK}$ are lines in the same direction
(namely, the reflection of $MK \parallel AN$ across $\ol{BC}$), as needed.
\paragraph{Complex numbers approach with Apollonian circles, by Carl Schildkraut.}
We use complex numbers.
As in the first approach, we will ignore the hypothesis that $K$ lies on $(ABC)$.
Let $Q \coloneq (AH) \cap (ABC) \cap (AEF) \neq A$
be the Miquel point of $BFEC$ again.
Construct the point $T$ on $(ABC)$ for which $AT\perp BC$;
note that $T=-\frac{bc}a$.
This time the unconditional result is:
\begin{proposition*}
We have $Q$, $M$, $T$, $K$ are concyclic (or collinear)
on an Apollonian circle of $\ol{O_1O_2}$.
\end{proposition*}
This will solve the original problem since once $K$ lies on $(ABC)$
it must be either $Q$ or $T$.
But since $K$ is not on $(BME)$, $K\neq Q$, it will have to be $T$.
We now prove the proposition.
Suppose $(ABC)$ is the unit circle and let $A=a$, $B=b$, $C=c$.
Let $H=a+b+c$ be the orthocenter of $\triangle ABC$.
By the usual formulas,
\[ E \coloneq \half\left(a+b+c-\frac{bc}{a}\right). \]
Let $O_1$ be the center of $(BME)$ and $O_2$ be the center of $(CMF)$.
\begin{claim*}
[Calculation of the Miquel point]
We have $Q = \frac{2a+b+c}{a\left( \frac 1a + \frac 1b + \frac 1c \right)+1}$.
\end{claim*}
\begin{proof}
We now compute that $Q=q$ satisfies $\ol q=1/q$
(since $Q$ is on the unit circle)
and $\frac{q-h}{q-a}\in i\RR$ (since $AQ\perp QH$),
which expands to
\[0=\frac{q-h}{q-a}+\frac{1/q-\ol h}{1/q-1/a}
=\frac{q-h}{q-a}-\frac{a(1-q\ol h)}{q-a}.\]
This solves to $q=\frac{h+a}{a\ol h+1}=\frac{2a+b+c}{a\ol h+1}$.
\end{proof}
\begin{claim*}
[Calculation of $O_1$ and $O_2$]
We have $O_1 = \frac{b(2a+b+c)}{2(a+b)}$
and $O_2 = \frac{c(2a+b+c)}{2(a+c)}$.
\end{claim*}
\begin{proof}
We now compute $O_1$ and $O_2$.
For $x,y,z\in\CC$, let $\opname{Circum}(x,y,z)$
denote the circumcenter of the triangle
defined by vertices $x$, $y$, and $z$ in $\CC$. We have
\begin{align*}
O_1&=\opname{Circum}(B,M,E)\\
&=b+\frac12\opname{Circum}\left(0,c-b,\frac{(a-b)(b-c)}b\right)\\
&=b-\frac{b-c}{2b}\opname{Circum}\left(0,b,b-a\right)\\
&=b-\frac{b-c}{2b}\left(b-\opname{Circum}\left(0,b,a\right)\right)\\
&=b-\frac{b-c}{2b}\left(b-\frac{ab}{a+b}\right)=b-\frac{b(b-c)}{2(a+b)}=\frac{b(2a+b+c)}{2(a+b)}.
\end{align*}
Similarly, $O_2=\frac{c(2a+b+c)}{2(a+c)}$.
\end{proof}
We are now going to prove the following:
\begin{claim*}
We have
\[\frac{TO_1}{TO_2}=\frac{MO_1}{MO_2}=\frac{QO_1}{QO_2}.\]
\end{claim*}
\begin{proof}
We now compute
\[MO_1=BO_1=\left|b-\frac{b(2a+b+c)}{2(a+b)}\right|
=\left|\frac{b(b-c)}{2(a+b)}\right|=\frac12\left|\frac{b-c}{a+b}\right|\]
and
\[QO_1=\left|r-\frac{b(2a+b+c)}{2(a+b)}\right|
=\left|1-\frac{b(a+h)}{2(a+b)r}\right|=\left|1-\frac{b(a\ol h+1)}{2(a+b)}\right|
=\left|\frac{a-\frac{ab}c}{2(a+b)}\right|=\frac12\left|\frac{b-c}{a+b}\right|.\]
This implies both (by symmetry) that
$\frac{MO_1}{MO_2}=\frac{QO_1}{QO_2}=\big|\frac{a+c}{a+b}\big|$
and that $Q$ is on $(BME)$ and $(CMF)$.
Also,
\[\frac{TO_1}{TO_2}
=\frac{\left|\frac{b(2a+b+c)}{2(a+b)}+\frac{bc}a\right|}%
{\left|\frac{c(2a+b+c)}{2(a+c)}+\frac{bc}a\right|}
=\left|\frac{\frac{b(2a^2+ab+ac+2ac+2bc)}{2a(a+b)}}{\frac{c(2a^2+ab+ac+2ab+2bc)}{2a(a+c)}}\right|
=\left|\frac{a+c}{a+b}\right|\cdot\left|\frac{2a^2+2bc+ab+3ac}{2a^2+2bc+3ab+ac}\right|;\]
if $z=2a^2+2bc+ab+3ac$, then $a^2bc\ol z=2a^2+2bc+3ab+ac$, so the second
term has magnitude $1$.
This means $\frac{TO_1}{TO_2}=\frac{MO_1}{MO_2}=\frac{QO_1}{QO_2}$, as desired.
\end{proof}
To finish, note that this common ratio is the ratio between the radii
of these two circles, so it is also $\frac{KO_1}{KO_2}$.
By Apollonian circles the points $\{Q,M,T,K\}$ lie on a circle or a line.
|
sols-TST-IMO-2023_3
|
Consider pairs $(f,g)$ of functions
from the set of nonnegative integers to itself such that
\begin{itemize}
\ii $f(0) \ge f(1) \ge f(2) \ge \dots \ge f(300) \ge 0$;
\ii $f(0) + f(1) + f(2) + \dots + f(300) \leq 300$;
\ii for any $20$ nonnegative integers $n_1$, $n_2$, \dots, $n_{20}$,
not necessarily distinct, we have
\[ g(n_1 + n_2 + \dots + n_{20}) \leq f(n_1) + f(n_2) + \dots + f(n_{20}). \]
\end{itemize}
Determine the maximum possible value of $g(0) + g(1) + \dots + g(6000)$
over all such pairs of functions.
|
Replace $300 = \frac{24 \cdot 25}{2}$ with $\frac{s(s+1)}{2}$ where $s=24$, and
$20$ with $k$. The answer is $115440 = \frac{ks(ks+1)}{2}$. Equality is achieved
at $f(n) = \max (s-n, 0)$ and $g(n) = \max (ks-n,0)$. To prove
\[g(n_1+\dotsb+n_k)\leq f(n_1)+\dotsb+f(n_k),\]
write it as
\[\max(x_1+\dotsb+x_k, 0)\leq \max(x_1, 0)+\dotsb+\max(x_k, 0)\]
with $x_i=s-n_i$. This can be proven from the $k=2$ case and induction.
\medskip
It remains to show the upper bound. For this problem, define a \emph{partition}
to be a nonincreasing function $p \colon \ZZ_{\geq 0}\to \ZZ_{\geq 0}$
such that $p(n)=0$ for some $n$. The sum of $p$ is defined to be
$\sum_{n=0}^{\infty}p(n)$, which is finite under the previous assumption. Let
$L=\ZZ_{\geq 0}^2$. The
\emph{Young diagram} of the partition is the set of points
\[\mathcal{P} := \{(x, y)\in L: y < p(x)\}.\]
The number of points in $\mathcal{P}$ is equal to the sum of $p$.
The \emph{conjugate} of a partition defined as
\[p_*(n) = \text{the number of $i$ for which $p(i) > n$}.\]
This is a partition with the same sum as $p$. Geometrically, the Young diagrams
of $p$ and $p_*$ are reflections about $x=y$.
Since each $g(n)$ is independent, we may
maximize each one separately for all $n$ and assume that
\[
g(n) = \min_{n_1+\dotsb+n_k=n}(f(n_1)+\dotsb+f(n_k)) \tag{*}.
\]
The conditions of the problem statement imply that $f\big(\tfrac{s(s+1)}{2}\big)=0$.
Then, for any $n\leq k\frac{s(s+1)}{2}$, there exists an optimal combination
$(n_1, \dots, n_k)$ in (*) where all $n_i$ are at most $\frac{s(s+1)}{2}$, by
replacing any term in an optimum greater than $\frac{s(s+1)}{2}$ by
$\frac{s(s+1)}{2}$ and shifting the excess to smaller terms (because $f$ is
nonincreasing). Therefore we may extend $f$ to a partition by letting $f(n) = 0$
for $n > \frac{s(s+1)}{2}$ without affecting the relevant values of $g$. Then
(*) implies that $g$ is a partition as well.
The problem can be restated as follows: $f$ is a partition with sum
$\frac{s(s+1)}{2}$, and $g$ is a partition defined by (*). Find the maximum
possible sum of $g$. The key claim is that the problem is the
same under conjugation.
\begin{claim*}
Under these conditions, we have
\[
g_*(n) = \min_{n_1+\dotsb+n_k=n}(f_*(n_1)+\dotsb+f_*(n_k)).
\]
\end{claim*}
\begin{proof}
Let $\mathcal{F}$ and $\mathcal{G}$ be the Young diagrams of $f$ and $g$
respectively, and $\ol{\mathcal{F}} = L\setminus \mathcal{F}$ and
$\ol{\mathcal{G}} = L\setminus \mathcal{G}$ be their complements. The
lower boundary of $\ol{\mathcal{F}}$ is formed by the points $(n, f(n))$
for $i\in \ZZ_{\geq 0}$. By the definition of $g$, the lower boundary
of $\ol{\mathcal{G}}$
consists of points $(n, g(n))$ which are formed by adding $k$ points of
$\ol{\mathcal{F}}$. This means
\[
\ol{\mathcal{G}} = \underbrace{\ol{\mathcal{F}} + \dots +
\ol{\mathcal{F}}}_{k \ \ol{\mathcal{F}}\text{'s}}
\]
where $+$ denotes set addition. This definition remains invariant under
reflection about $x=y$, which swaps $f$ and $g$ with their conjugates.
\end{proof}
Let $A$ be the sum of $g$. We now derive different bounds on $A$. First, by
Hermite's identity
\[n = \sum_{i=0}^{k-1}\left\lfloor\tfrac{n+i}{k}\right\rfloor\]
we have
\begin{align*}
A &= \sum_{n=0}^{\infty} g(n) \\
& \leq \sum_{n=0}^{\infty} \sum_{i=0}^{k-1} f\left( \left\lfloor
\tfrac{n+i}{k} \right\rfloor \right) \\
&= k^2 \sum_{n=0}^{\infty}f(n) - \frac{k(k-1)}{2}f(0) \\
&= k^2\frac{s(s+1)}{2}-\frac{k(k-1)}{2}f(0).
\end{align*}
By the claim, we also get the second bound $A \leq k^2 \frac{s(s+1)}{2} -
\frac{k(k-1)}{2}f_*(0)$.
For the third bound, note that $f(f_*(0)) = 0$ and thus $g(k f_*(0)) = 0$. Moreover,
\[g(q f_*(0) + r) \le q \cdot f(f_*(0)) + (k-q-1) f(0) + f(r) = (k-q-1)f(0) +
f(r),\]
so we have
\begin{align*}
A &= \sum_{\substack{0 \leq q < k \\ 0 \leq r < f_*(0)}} g(qf_*(0) + r) \\
& \le \frac{k(k-1)}{2}f_*(0)f(0) + k \sum_{0 \leq r < f_*(0)} f(r) \\
&= \frac{k(k-1)}{2} f_*(0)f(0) + k\frac{s(s+1)}{2}.
\end{align*}
Now we have three cases:
\begin{itemize}
\item If $f(0)\geq s$ then
\[
A\leq k^2 \frac{s(s+1)}{2} - \frac{k(k-1)}{2}f(0) \leq \frac{ks(ks+1)}{2}.
\]
\item If $f_*(0)\geq s$ then
\[
A\leq k^2 \frac{s(s+1)}{2} - \frac{k(k-1)}{2}f_*(0) \leq \frac{ks(ks+1)}{2}.
\]
\item Otherwise, $f(0)f_*(0) \leq s^2$ and
\[
A\leq \frac{k(k-1)}{2} f_*(0)f(0) + k\frac{s(s+1)}{2} \leq \frac{ks(ks+1)}{2}.
\]
\end{itemize}
\medskip
In all cases, $A \leq \frac{ks(ks+1)}{2}$, as desired.
\begin{remark*}
One can estimate the answer to be around $k^2 \frac{s(s+1)}{2}$ by
observing the set addition operation ``dilates'' $\mathcal{F}$ by a factor of
$k$, but significant care is needed to sharpen the bound.
\end{remark*}
|
sols-TST-IMO-2023_4
|
For nonnegative integers $a$ and $b$,
denote their \emph{bitwise xor} by $a \oplus b$.
(For example, $9 \oplus 10 = 1001_2 \oplus 1010_2 = 0011_2 = 3$.)
Find all positive integers $a$ such that
for any integers $x > y \ge 0$, we have
\[ x \oplus ax \neq y \oplus ay. \]
|
Answer: the function $x \mapsto x \oplus ax$ is injective if and only if
$a$ is an even integer.
\paragraph{Even case.}
First, assume $\nu_2(a) = k > 0$.
We wish to recover $x$ from $c \coloneq x \oplus ax$.
Notice that:
\begin{itemize}
\ii The last $k$ bits of $c$ coincide with the last $k$ bits of $x$.
\ii Now the last $k$ bits of $x$ give us also the last $2k$ bits of $ax$,
so we may recover the last $2k$ bits of $x$ as well.
\ii Then the last $2k$ bits of $x$ give us also the last $3k$ bits of $ax$,
so we may recover the last $3k$ bits of $x$ as well.
\ii \dots and so on.
\end{itemize}
\paragraph{Odd case.}
Conversely, suppose $a$ is odd.
To produce the desired collision:
\begin{claim*}
Let $n$ be any integer such that $2^n > a$, and define
\[ x = \underbrace{1\dots1}_n = 2^n-1,
\qquad y = 1\underbrace{0\dots0}_n1 = 2^n+1. \]
Then $x \oplus ax = y \oplus ay$.
\end{claim*}
\begin{proof}
Let $P$ be the binary string for $a$, zero-padded to length $n$,
and let $Q$ be the binary string for $a - 1$, zero-padded to length $n$,
Then let $R$ be the bitwise complement of $Q$.
(Hence all three of are binary strings of length $n$.)
Then
\begin{align*}
ax = \ol{QR} &\implies x \oplus ax = \ol{QQ} \\
ay = \ol{PP} &\implies y \oplus ay = \ol{QQ}.
\end{align*}
We're done.
\end{proof}
|
sols-TST-IMO-2023_5
|
Let $m$ and $n$ be fixed positive integers.
Tsvety and Freyja play a game on an infinite grid of unit square cells.
Tsvety has secretly written a real number inside of each cell so that
the sum of the numbers within every rectangle of size
either $m \times n$ or $n \times m$ is zero.
Freyja wants to learn all of these numbers.
One by one, Freyja asks Tsvety about some cell in the grid,
and Tsvety truthfully reveals what number is written in it.
Freyja wins if, at any point, Freyja can simultaneously deduce
the number written in every cell of the entire infinite grid.
(If this never occurs, Freyja has lost the game and Tsvety wins.)
In terms of $m$ and $n$, find the smallest number of
questions that Freyja must ask to win,
or show that no finite number of questions can suffice.
|
The answer is the following:
\begin{itemize}
\ii If $\gcd(m, n) > 1$, then Freyja cannot win.
\ii If $\gcd(m, n) = 1$, then Freyja
can win in a minimum of $(m-1)^2 + (n-1)^2$ questions.
\end{itemize}
First, we dispose of the case where $\gcd(m, n) > 1$.
Write $d = \gcd(m, n)$.
The idea is that any labeling where each $1 \times d$ rectangle has sum zero is valid.
Thus, to learn the labeling, Freyja must ask at least one question in every row,
which is clearly not possible in a finite number of questions.
Now suppose $\gcd(m, n) = 1$.
We split the proof into two halves.
\paragraph{Lower bound.}
Clearly, any labeling where each $m \times 1$ and $1 \times m$ rectangle has sum zero is valid.
These labelings form a vector space with dimension $(m-1)^2$, by inspection.
(Set the values in an $(m-1)\times (m-1)$ square arbitrarily and every other
value is uniquely determined.)
Similarly, labelings where each $n \times 1$ and $1 \times n$ rectangle have sum
zero are also valid, and have dimension $(n-1)^2$.
It is also easy to see that no labeling other than the all-zero labeling
belongs to both categories;
labelings in the first space are periodic in both directions with period $m$,
while labelings in the second space are periodic in both directions with period $n$;
and hence any labeling in both categories must be constant, ergo all-zero.
Taking sums of these labelings gives a space of valid labelings
of dimension $(m-1)^2 + (n-1)^2$.
Thus, Freyja needs at least $(m-1)^2 + (n-1)^2$ questions to win.
\paragraph{Proof of upper bound using generating functions, by Ankan Bhattacharya.}
We prove:
\begin{claim*}[Periodicity]
Any valid labeling is doubly periodic with period $mn$.
\end{claim*}
\begin{proof}
By Chicken McNugget, there exists $N$ such that $N$ and $N+1$
are both nonnegative integer linear combinations of $m$ and $n$.
Then both $mn \times N$ and $mn \times (N+1)$ rectangles have zero sum,
so $mn \times 1$ rectangles have zero sum.
This implies that any two cells with a vertical displacement of $mn$ are equal;
similarly for horizontal displacements.
\end{proof}
With that in mind, consider a valid labeling.
It naturally corresponds to a generating function
\[ f(x, y) = \sum_{a=0}^{mn-1} \sum_{b=0}^{mn-1} c_{a, b} x^a y^b \]
where $c_{a, b}$ is the number in $(a, b)$.
The generating function corresponding to sums over $n \times m$ rectangles is
\[
f(x, y) (1 + x + \dots + x^{m-1}) (1 + y + \dots + y^{n-1})
= f(x, y) \cdot \frac{x^m - 1}{x - 1} \cdot \frac{y^n - 1}{y - 1}.
\]
Similarly, the one for $m \times n$ rectangles is
\[ f(x, y) \cdot \frac{x^n - 1}{x - 1} \cdot \frac{y^m - 1}{y - 1}. \]
Thus, the constraints for $f$ to be valid are equivalent to
\[
f(x, y) \cdot \frac{x^m - 1}{x - 1} \cdot \frac{y^n - 1}{y - 1}
\quad \text{and} \quad
f(x, y) \cdot \frac{x^n - 1}{x - 1} \cdot \frac{y^m - 1}{y - 1}
\]
being zero when reduced modulo $x^{mn} - 1$ and $y^{mn} - 1$,
or, letting $\omega = \exp(2\pi i/mn)$,
both terms being zero when powers of $\omega$ are plugged in.
To restate the constraints one final time, we need
\[
f(\omega^a, \omega^b) \cdot \frac{\omega^{am} - 1}{\omega^a - 1} \cdot \frac{\omega^{bn} - 1}{\omega^b - 1}
= f(\omega^a, \omega^b) \cdot \frac{\omega^{an} - 1}{\omega^a - 1} \cdot \frac{\omega^{bm} - 1}{\omega^b - 1} = 0
\]
for all $a, b \in \{0, \dots, mn-1\}$.
\begin{claim*}
This implies that $f(\omega^a, \omega^b) = 0$
for all but at most $(m-1)^2 + (n-1)^2$ values of $(a,b) \in \{0, \dots, mn-1\}^2$.
\end{claim*}
\begin{proof}
Consider a pair $(a,b)$ such that $f(\omega^a, \omega^b) \neq 0$.
Then we need
\[
\frac{\omega^{am} - 1}{\omega^a - 1} \cdot \frac{\omega^{bn} - 1}{\omega^b - 1}
= \frac{\omega^{an} - 1}{\omega^a - 1} \cdot \frac{\omega^{bm} - 1}{\omega^b - 1} = 0.
\]
This happens when (at least) one fraction in either product is zero.
\begin{itemize}
\ii If the first fraction is zero,
then either $n \mid a$ and $a > 0$, or $m \mid b$ and $b > 0$.
\ii If the second fraction is zero,
then either $m \mid a$ and $a > 0$, or $n \mid b$ and $b > 0$.
\end{itemize}
If the first condition holds in both cases, then $mn \mid a$, but $0 < a < mn$,
a contradiction. Thus if $n \mid a$, then we must have $n \mid b$, and similarly
if $m \mid a$ then $m \mid b$.
The former case happens $(m-1)^2$ times, and the latter case happens $(n-1)^2$ times.
Thus, at most $(m-1)^2 + (n-1)^2$ values of $f(\omega^a, \omega^b)$ are nonzero.
\end{proof}
\begin{claim*}
The $(mn)^2$ equations $f(\omega^a, \omega^b) = 0$ are linearly independent
when viewed as linear equations in $(mn)^2$ variables $c_{a,b}$.
Hence, any subset of these equations is also linearly independent.
\end{claim*}
\begin{proof}
In general, the equation $f(x,y) = 0$ is a polynomial relation with
$\deg_x f(x,y) = \deg_y f(x,y) < mn$.
However, if we let $S = \{\omega^0, \omega^1, \dots, \omega^{mn-1}\}$, then
$|S| = mn$ and we are given $f(s,s') = 0$ for all $s,s' \in S$.
This can only happen if $f$ is the zero polynomial,
that is, $c_{a,b} = 0$ for all $a$ and $b$.
\end{proof}
It follows that the dimension of the space of valid labelings is at
most $(m-1)^2 + (n-1)^2$, as desired.
\paragraph{Explicit version of winning algorithm by Freyja, from author.}
Suppose that $\gcd(m, n) = 1$ and $m \le n$. Let $[a, b]$ denote the set of integers
between $a$ and $b$ inclusive.
Let Freyja ask about all cells $(x, y)$ in the two squares
\begin{align*}
S_\text{1} &= [1, m - 1] \times [1, m - 1] \\
S_\text{2} &= [m, m + n - 2] \times [1, n - 1].
\end{align*}
In the beginning, one by one,
Freyja determines all values inside of the rectangle
$Q \coloneq [1, m - 1] \times [m, n - 1]$.
To that end, on each step she considers some
rectangle with $m$ rows and $n$ columns such that its top left corner is in $Q$
and all of the other values in it have been determined already.
In this way, Freyja uncovers all of $Q$, starting with its lower right corner and
then proceeding upwards and to the left.
Thus Freyja can learn all numbers inside of the rectangle
\[ R \coloneq [1, m + n - 2]
\times [1, n - 1] = Q \cup S_\text{1} \cup S_\text{2}. \]
See the figure below for an illustration for $(m,n) = (5,8)$.
The first cell of $Q$ is uncovered using the dotted green rectangle.
\begin{center}
\begin{asy}
defaultpen(fontsize(18pt));
size(9cm);
for (int i=-1; i<=9; ++i) {
draw( (-1,i)--(12,i), mediumgray );
}
for (int i=-1; i<=12; ++i) {
draw( (i,-1)--(i,9), mediumgray );
}
filldraw(box((0,0),(4,4)), opacity(0.1)+lightred, black+1.5);
filldraw(box((4,0),(11,7)), opacity(0.1)+yellow, black+1.5);
filldraw(box((0,4),(4,7)), opacity(0.1)+cyan, black+1.5);
filldraw(box((3,0),(11,5)), opacity(0.1)+green, deepgreen+dotted+1.1);
filldraw(box((0,7),(11,8)), opacity(0.1)+purple, purple+dashed+1.1);
label("$S_1$", (2,2), red);
label("$S_2$", (7.5,3.5), brown);
label("$Q$", (2,5.5), blue);
label("$T$", (5.5,7.5), purple);
label("$(m,n)=(5,8)$", (5.5,-1), dir(-90));
\end{asy}
\end{center}
We need one lemma:
\begin{lemma*}
Let $m$ and $n$ be positive integers with $\gcd(m, n) = 1$.
Consider an unknown sequence of real numbers $z_1$, $z_2$, $\dots$, $z_s$
with $s \ge m + n - 2$. Suppose that we know the sums of all contiguous blocks
of size either $m$ or $n$ in this sequence. Then we can determine all
individual entries in the sequence as well.
\end{lemma*}
\begin{proof}
By induction on $m + n$. Suppose, without loss of generality, that $m \le n$.
Our base case is $m = 1$, which is clear. For the induction step, set $\ell =
n - m$. Each contiguous block of size $\ell$ within $z_1$, $z_2$, $\dots$,
$z_{s - n}$ is the difference of two contiguous blocks of sizes $m$ and $n$
within the original sequence. By the induction hypothesis for $\ell$ and $m$,
it follows that we can determine all of $z_1$, $z_2$, $\dots$, $z_{s - n}$.
Then we determine the remaining $z_i$ as well, one by one, in order from left
to right, by examining on each step an appropriate contiguous block of size $m$.
\end{proof}
Let $T$ be the rectangle $[1, m + n - 2] \times \{n\}$. By looking at
appropriate rectangles of sizes $m \times n$ and $n \times m$ such that
their top row is contained within $T$ and all of their other rows are
contained within $R$, Freyja can learn the sums of all contiguous blocks of
values of sizes $m$ and $n$ within $T$. By the Lemma, it follows that Freyja
can uncover all of $T$.
In this way, with the help of the Lemma, Freyja can extend her rectangular area
of knowledge both upwards and downwards. Once its height reaches $m + n -
2$, by the same method she will be able to extend it to the left and right as
well. This allows Freyja to determine all values in the grid. Therefore, $(m -
1)^2 + (n - 1)^2$ questions are indeed sufficient.
\begin{remark*}
The ideas in the solution also yield a proof of the following result:
\begin{quote}
Let $m$ and $n$ be relatively prime positive integers.
Consider an infinite grid of unit square cells coloured in
such a way that every rectangle of size either $m \times n$ or $n \times m$
contains the same multiset of colours.
Then the colouring is either doubly periodic with period length $m$
or doubly periodic with period length $n$.
\end{quote}
(Here, ``doubly periodic with period length $s$''
means ``both horizontally and vertically periodic with period length $s$''.)
Here is a quick sketch of the proof. Given two positive integers $i$ and $j$
with $1 \le i, j \le m - 1$, we define
\[
f_{ij}(x, y) \coloneq
\begin{cases}
+1 & \text{ when } (x, y) \equiv (0, 0) \text{ or } (i, j) \pmod{m}; \\
-1 & \text{ when } (x, y) \equiv (0, j) \text{ or } (i, 0) \pmod{m};
\text{ and } \\
\phantom{+}0 & \text{ otherwise.}
\end{cases}
\]
Define $g_{i, j}$ similarly, but with $1 \le i, j \le n - 1$ and ``mod $m$''
everywhere replaced by ``mod $n$''. First we show that if a linear combination
$h := \sum \alpha_{i, j}f_{i, j}+\sum\beta_{i, j}g_{i, j}$ of the $f_{i, j}$ and
$g_{i, j}$ contains only two distinct values, then either all of the $\alpha_{i,
j}$ vanish or all of the $\beta_{i, j}$ do. It follows that each colour,
considered in isolation, is either doubly periodic with period length $m$ or
doubly periodic with period length $n$. Finally, we check that different period
lengths cannot mix.
On the other hand, if $m$ and $n$ are not relatively prime,
then there exist infinitely many non-isomorphic valid colourings.
Furthermore, when $\gcd(m, n) = 2$, there exist valid colourings
which are not horizontally periodic; and, when $\gcd(m, n) \ge 3$,
there exist valid colourings which are neither horizontally nor vertically periodic.
\end{remark*}
|
sols-TST-IMO-2023_6
|
Fix a function $f \colon \NN \to \NN$ and for any $m,n \in \NN$ define
\[
\Delta(m,n) = \underbrace{f(f(\dots f}_{f(n)\text{ times}} (m)\dots)) %chktex 9
- \underbrace{f(f(\dots f}_{f(m)\text{ times}} (n)\dots)). %chktex 9
\]
Suppose $\Delta(m,n) \neq 0$ for any distinct $m,n \in \NN$.
Show that $\Delta$ is unbounded, meaning that for any constant $C$
there exist $m,n \in \NN$ with $\left\lvert \Delta(m,n) \right\rvert > C$.
\end{enumerate}
|
Suppose for the sake of contradiction that $|\Delta(m,n)|\le N$ for all $m$, $n$.
Note that $f$ is injective, as
\[ f(m)=f(n) \implies \Delta(m,n) = 0 \implies m=n, \]
as desired.
Let $G$ be the ``arrow graph'' of $f$,
which is the directed graph with vertex set $\NN$ and edges $n\to f(n)$.
The first step in the solution is to classify the structure of $G$.
Injectivity implies that $G$ is a disjoint collection of chains
(infinite and half-infinite) and cycles.
We have the following sequence of claims that further refine the structure.
\begin{claim*}
The graph $G$ has no cycles.
\end{claim*}
\begin{proof}
Suppose for the sake of contradiction that $f^k(n)=n$
for some $k\ge 2$ and $n\in\NN$.
As $m$ varies over $\NN$, we have $|\Delta(m,n)|\le N$,
so $f^{f(n)}(m)$ can only take on some finite set of values.
In particular, this means that
\[f^{f(n)}(m_1)=f^{f(n)}(m_2)\]
for some $m_1\ne m_2$, which contradicts injectivity.
\end{proof}
\begin{claim*}
The graph $G$ has at most $2N+1$ chains.
\end{claim*}
\begin{proof}
Suppose we have numbers $m_1$, \dots, $m_k$ in distinct chains.
Select a positive integer $B > \max\{f(m_1), \dots,f(m_k)\}$.
Now,
\[ \left|\Delta\left(m_i,f^{B-f(m_i)}(1)\right)\right|\le N
\implies \left|f^B(1)-f^{f^{B-f(m_i)+1}(1)}(m_i)\right|\le N.\]
Since the $m_i$s are in different chains,
we have that $f^{f^{B-f(m_i)+1}(1)}(m_i)$ are distinct for each $i$,
which implies that $k\le 2N+1$, as desired.
\end{proof}
\begin{claim*}
The graph $G$ consists of exactly one half-infinite chain.
\end{claim*}
\begin{proof}
Fix some $c\in\NN$.
Call an element of $\NN$ \textit{bad} if
it is not of the form $f^k(c)$ for some $k\ge 0$.
It suffices to show that there are only finitely many bad numbers.
Since there are only finitely many chains, $f^{f(c)}(n)$ achieves all
sufficiently large positive integers, say all positive integers at least $M$.
Fix $A$ and $B$ such that $B>A\ge M$.
If $f^{f(c)}(n)\in[A,B]$, then $f^{f(n)}(c)\in[A-N,B+N]$,
and distinct $n$ generate distinct $f^{f(n)}(c)$ due to the structure of $G$.
Therefore, we have at least $B-A+1$ good numbers in $[A-N,B+N]$,
so there are at most $2N$ bad numbers in $[A-N,B+N]$.
Varying $B$, this shows there are at most $2N$ bad numbers at least $A-N$.
\end{proof}
Let $c$ be the starting point of the chain,
so every integer is of the form $f^k(c)$, where $k\geq 0$.
Define a function $g \colon \ZZ_{\ge 0}\to\NN$
by \[ g(k) \coloneq f^k(c). \]
Due to the structure of $G$, $g$ is a bijection.
Define
\[\delta(a,b) \coloneq \Delta(f^a(c),f^b(c)) = g(g(b+1)+a)-g(g(a+1)+b),\]
so the conditions are equivalent to $|\delta(a,b)|\le N$
for all $a,b\in\ZZ_{\ge 0}$ and $\delta(a,b)\ne 0$ for $a\ne b$,
which is equivalent to $g(a+1)-a\ne g(b+1)-b$ for $a\ne b$.
This tells us that $g(x)-x$ is injective for $x\ge 1$.
\begin{lemma*}
For all $M$, there exists a nonnegative integer $x$ with $g(x)\leq x-M$.
\end{lemma*}
\begin{proof}
Assume for the sake of contradiction that $g(x)-x$ is bounded below.
Fix some large positive $K$.
Since $g(x)-x$ is injective,
there exists $B$ such that $g(x)-x\geq K$ for all $x\geq B$.
Then $\min \{g(B+1),g(B+2),\dots\} \ge B+K$,
while $\{g(0),\dots,g(B)\}$ only achieve $B+1$ values.
Thus, at least $K-1$ values are not achieved by $g$, which is a contradiction.
\end{proof}
Now pick $B$ such that $g(B)+N\le B$ and $g(B) > N$.
Note that infinitely many such $B$ exist,
since we can take $M$ to be arbitrarily small in the above lemma.
Let
\[ t = \max \{g^{-1}(g(B)-N), g^{-1}(g(B)-N+1), \dots, g^{-1}(g(B)+N)\}. \]
Note that $g(t)\le g(B)+N\le B$, so we have
\[ |\delta(t-1,B-g(t))| = |g(B)-g(t-1+g(B+1-g(t)))| \le N, \]
so
\[ t-1+g(B+1-g(t)) \in
\{g^{-1}(g(B)-N),g^{-1}(g(B)-N+1),\dots,g^{-1}(g(B)+N)\},\]
so by the maximality of $t$, we must have $g(B+1-g(t))=1$,
so $B+1-g(t)=g^{-1}(1)$.
We have $|g(t)-g(B)|\le N$, so
\[|(B-g(B))+1-g^{-1}(1)|\le N.\]
This is true for infinitely many values of $B$,
so infinitely many values of $B-g(B)$ (by injectivity of $g(x)-x$),
which is a contradiction.
This completes the proof.
|
sols-TST-IMO-2024_1
|
Find the smallest constant $C > 1$ such that the following statement holds:
for every integer $n \ge 2$ and sequence of non-integer
positive real numbers $a_1$, $a_2$, \dots, $a_n$ satisfying
\[ \frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_n} = 1, \]
it's possible to choose positive integers $b_i$ such that
\begin{enumerate}
\ii[(i)] for each $i=1,2,\dots,n$,
either $b_i = \left\lfloor a_i \right\rfloor$ or $b_i = \left\lfloor a_i \right\rfloor + 1$; and
\ii[(ii)] we have
\[ 1 < \frac1{b_1} + \frac1{b_2} + \dots + \frac{1}{b_n} \le C. \]
\end{enumerate}
|
\paragraph{Answer.} The answer is $C=\frac{3}{2}$.
\paragraph{Lower bound.}
Note that if $a_1 = \frac{4n-3}{2n-1}$ and $a_i = \frac{4n-3}{2}$ for $i > 1$,
then we must have $b_1 \in \{1,2\}$ and $b_i \in \{2n-2,2n-1\}$ for $i > 1$. If
we take $b_1 = 2$ then we obtain
\[
\frac{1}{b_1} + \frac{1}{b_2} + \dots + \frac{1}{b_n} \le \frac{1}{2} + (n-1)
\cdot \frac{1}{2n-2} = 1,
\]
whereas if we take $b_1 = 1$ then we obtain
\[
\frac{1}{b_1} + \frac{1}{b_2} + \dots + \frac{1}{b_n} \ge 1 + (n-1) \cdot
\frac{1}{2n-1} = \frac{3n-2}{2n-1}.
\]
This shows that $C \ge \frac{3n-2}{2n-1}$, and as $n\to \infty$ this shows that
$C\geq \frac{3}{2}$.
\paragraph{Upper bound.}
For $0\leq k\leq n$, define
\[
c_i = \sum_{i=1}^{k}\frac{1}{\lfloor a_i\rfloor}
+ \sum_{i=k+1}^{n}\frac{1}{\lfloor a_i\rfloor+1}.
\]
Note that $c_0 < c_1 < \dots < c_n$ and
\[c_0 < \frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_n} = 1 < c_n.\]
This means there exists a unique value of $k$ for which $c_{k-1} < 1 < c_{k}$. For
this $k$ we have
\[
1 < c_{k} = c_{k-1} + \frac{1}{(\lfloor a_k\rfloor)(\lfloor a_k\rfloor+1)} < 1 +
\frac{1}{1\cdot 2} = \frac{3}{2}.
\]
Therefore we may choose $b_i = \lfloor a_i\rfloor$ for $i\leq k$ and $b_i =
\lfloor a_i\rfloor+1$ for $i > k$.
\begin{remark*}
The solution can be phrased in the following ``motion-based'' way.
Imagine starting with all floors (corresponding to $c_0$),
then changing each floor to a ceiling one by one
until after $n$ steps every floor is a ceiling (arriving at $c_n$).
As we saw, $c_0 < 1 < c_n$, but $c_0 < \dots < c_n$.
Moreover, each discrete step increases the sum by at most
\[ \frac{1}{\left\lfloor a_i \right\rfloor}
- \frac{1}{\left\lfloor a_i \right\rfloor + 1} \leq \frac 12 \]
and so the changing sum must be in the interval $[1, 3/2]$ at some point.
\end{remark*}
\paragraph{Upper bound (alternate).}
First suppose $a_i < 2$ for some $i$. Assume without loss of generality $i=1$
here. Let $b_1=1$ and $b_i=\lfloor a_i\rfloor+1$ for all other $i$. Then
\begin{align*}
1 &< \frac{1}{b_1} + \dots + \frac{1}{b_n} = 1 + \frac{1}{\lfloor a_2\rfloor +
1} + \dotsb + \frac{1}{\lfloor a_n\rfloor + 1} \\
&< \left(\frac{1}{2} +
\frac{1}{a_1}\right) + \frac{1}{a_2} + \dotsb + \frac{1}{a_n} = \frac{3}{2}.
\end{align*}
Now suppose $a_i > 2$ always. Then $\frac{a_i}{\lfloor a_i\rfloor} <
\frac{3}{2}$, so
\[
1 = \frac{1}{a_1} + \dotsb + \frac{1}{a_n} < \frac{1}{\lfloor a_1\rfloor} +
\dots + \frac{1}{\lfloor a_n\rfloor} < \frac{3}{2}\left(\frac{1}{a_1} + \dotsb
+ \frac{1}{a_n}\right) = \frac{3}{2}.
\]
Therefore we may let $b_i=\lfloor a_i\rfloor$ for all $i$.
\begin{remark*}
The original proposal asked to find the optimal $C$ for a fixed $n$. The
answer is $\frac{3n-2}{2n-1}$, i.e. the lower bound construction in the
solution is optimal.
\end{remark*}
|
sols-TST-IMO-2024_2
|
Let $ABC$ be a triangle with incenter $I$.
Let segment $AI$ intersect the incircle of triangle $ABC$ at point $D$.
Suppose that line $BD$ is perpendicular to line $AC$.
Let $P$ be a point such that $\angle BPA = \angle PAI = 90^\circ$.
Point $Q$ lies on segment $BD$ such that the
circumcircle of triangle $ABQ$ is tangent to line $BI$.
Point $X$ lies on line $PQ$ such that $\angle IAX = \angle XAC$.
Prove that $\angle AXP = 45^\circ$.
|
We show several approaches.
\paragraph{First solution, by author.}
\begin{center}
\begin{asy}
unitsize(1.0inches);
real eps = 1.15104332322704; /* trololol */
real v = 52 + eps;
real w = 150 + eps;
pair V = dir(v);
pair W = dir(w);
pair A = 2*V*W/(V+W);
pair D = dir((v+w)/2);
pair B = (D/(V*V)-1/D+2/W)/(1/(V*V)+1/(W*W));
pair I = 0;
pair U = 2*foot(W,I,B)-W;
pair C = 2*U*V/(U+V);
pair P = foot(B,A,A+A*dir(90));
pair Q = B + dir(D-B)*(P-B)/dir(P-B);
pair X = extension(P,Q,A,incenter(A,I,V));
draw(circumcircle(V,-V,W));
draw(A--B--C--cycle);
draw(A--I);
draw(I--U,linewidth(0.3));
draw(B--P--A);
draw(B--Q--D);
draw(B--I,linewidth(0.3));
draw(P--Q--X);
draw(A--X,linewidth(0.3));
pair O = circumcenter(A,Q,B);
pair AA = (A-O)*dir(20)+O;
pair BB = (B-O)*dir(-20)+O;
draw(arc(O,BB,AA),linewidth(0.4)+dashed);
dot("$A$",A,dir(45));
dot("$D$",D,dir(-30));
dot("$B$",B,dir(260));
dot("$I$",I,dir(0));
dot("$E$",U,dir(U));
dot("$C$",C,dir(C));
dot("$P$",P,dir(P));
dot("$Q$",Q,dir(-55));
dot("$X$",X,dir(270));
\end{asy}
\end{center}
\begin{claim*}
We have $BP=BQ$.
\end{claim*}
\begin{proof}
For readability, we split the proof into three unconditional parts.
\begin{itemize}
\ii We translate the condition $\ol{BD} \perp \ol{AC}$.
It gives $\angle DBA = 90\dg - A$, so that
\begin{align*}
\angle DBI &= \left| \frac B2 - (90\dg - A) \right| = \frac{|A-C|}{2} \\
\angle BDI &= \angle DBA + \angle BAD = (90\dg-A) + \frac A2 = 90\dg - \frac A2.
\end{align*}
Hence, letting $r$ denote the inradius, we can translate $\ol{BD} \perp \ol{AC}$
into the following trig condition:
\[ \sin \frac B2 = \frac{r}{BI} = \frac{DI}{BI}
= \frac{\sin \angle DBI}{\sin \angle BDI}
= \frac{\sin\frac{|A-C|}{2}}{\sin\left(90\dg-\frac A2\right)}. \]
\ii The length of $BP$ is given from right triangle $APB$ as
\[ BP = BA \cdot \sin \angle PAB = BA \cdot \sin\left( 90\dg-\frac A2 \right). \]
\ii The length of $BQ$ is given from the law of sines on triangle $ABQ$.
The tangency gives $\angle BAQ = \angle DBI$ and
$\angle BQA = 180\dg - \angle ABI = 180\dg - \angle IBE$ and thus
\[ BQ = BA \cdot \frac{\sin \angle BAQ}{\sin \angle AQB}
= BA \cdot \frac{\sin \angle DBI}{\sin \angle ABI}
= BA \cdot \frac{\sin \frac{|A-C|}{2}}{\sin \frac B2}. \]
\end{itemize}
The first bullet implies the expressions in the second and third bullet
for $BP$ and $BQ$ are equal, as needed.
\end{proof}
\begin{remark*}
In the above proof, one dos not actually need to compute $\angle DBI = \frac{|A-C|}{2}$.
The proof works equally leaving that expression intact as $\sin \angle DBI$
in place of $\sin \frac{|A-C|}{2}$.
\end{remark*}
Now we can finish by angle chasing. We have
\[ \angle PBQ = \angle PBA + \angle ABD = \frac{A}{2} + 90\dg - A =
90\dg - \frac{A}{2}.
\]
Then
\[
\angle BPQ = \frac{180\dg - \angle PBQ}{2} = 45\dg + \frac{A}{4},
\]
so $\angle APQ = 90\dg - \angle BPQ = 45\dg - \frac{A}{4}$.
Also, if we let $J$ be the incenter of $IAC$,
then $\angle PAJ = 90 \dg + \frac{A}{4}$, and clearly $X$ lies on line $AJ$.
Then $\angle APQ + \angle PAJ = 135 \dg < 180 \dg$,
so $X$ lies on the same side of $AP$ as $Q$ and $J$ (by the parallel postulate).
Therefore $\angle AXP = 180 \dg - 135 \dg = 45 \dg$, as desired.
\begin{remark*}
The problem was basically written backwards by starting from the $BD \perp AC$
condition, turning that into trig, and then contriving $P$ and $Q$ so that the
$BD \perp AC$ condition implied $BP=BQ$.
\end{remark*}
\paragraph{Second solution, by Jeffrey Kwan.}
We prove the following restatement:
\begin{quote}
Consider isosceles triangle $AEF$ with $AE = AF$ and incenter $D$.
Let $B$ be the point on ray $AE$ such that $BD\perp AF$,
and let $P$ be the projection of $B$ onto the line through $A$ parallel to $EF$.
Let $I$ be the point diametrically opposite $A$ in the circumcircle of $AEF$,
and let $Q$ be the point on line $BD$ such that $BI$ is tangent to the circumcircle of
$AQB$. Then $\angle APQ = 45^\circ - \angle A / 4$.
\end{quote}
First note that $\angle DFE = 45^\circ - \angle A / 4$, so it suffices to show that $\ol{PQ}\parallel \ol{DF}$.
Let $U = \ol{BD} \cap \ol{EF}$, and let $V = BI\cap (AEF)$. Observe that:
\begin{itemize}
\item $P$ and $V$ both lie on the circle with diameter $AB$, so $\angle BVP = \angle PAB = 90^\circ - \angle A / 2$.
\item We have $\angle EVB = \angle EVI = \angle A / 2 = \angle DUF = \angle BUE$.
Hence $BEUV$ is cyclic.
\end{itemize}
Now $\angle BVU = \angle AEU = 90^\circ - \angle A / 2 = \angle BVP$, so $\ol{PUV}$ are collinear.
\begin{center}
\begin{asy}
unitsize(0.9inches);
size(600);
pair A, E, F, D, B, I, P, Q, U, V;
A = dir(90);
E = dir(191);
F = dir(349);
D = incenter(A, E, F);
I = dir(270);
B = extension(A, E, D, foot(D, A, F));
P = foot(B, A, A + E - F);
Q = extension(B, D, P, P + F - D);
U = extension(B, D, E, F);
V = extension(B, I, P, U);
draw(A--E--F--cycle, orange);
draw(circumcircle(A, E, F), red);
draw(incircle(A, E, F), heavyred);
draw(E--B--foot(D, A, F), orange);
draw(A--P--B, heavycyan);
draw(A--Q--B^^P--Q, lightblue);
draw(P--V, dotted+magenta);
draw(B--V, dotted+magenta);
draw(circumcircle(A, P, B), magenta);
draw(circumcircle(A, P, Q), purple + dashed);
dot("$A$", A, dir(60));
dot("$B$", B, dir(B));
dot("$E$", E, dir(160));
dot("$F$", F, dir(F));
dot("$D$", D, dir(330));
dot("$I$", I, dir(270));
dot("$P$", P, dir(135));
dot("$Q$", Q, dir(300));
dot("$U$", U, dir(270));
dot("$V$", V, dir(300));
dot(foot(D, A, F));
\end{asy}
\end{center}
From the tangency condition, we have that $\angle AQB = 180^\circ - \angle ABI$, which implies that
\[\angle AQU + \angle APU = \angle AQB + \angle APV = (180^\circ - \angle ABI) + \angle ABI = 180^\circ,\]
and so $APUQ$ is cyclic. Finally, note that $D$ is the orthocenter of $\triangle AUF$, which implies that
\[\angle APQ = \angle AUQ = \angle AUD = \angle AFD = \angle DFE.\]
This forces $\ol{PQ}\parallel \ol{DF}$, as desired.
\paragraph{Third solution by Pitchayut Saengrungkongka and Maxim Li.}
We provide yet another proof that $BP = BQ$.
\begin{center}
\begin{asy}
size(8cm);
import geometry;
real eps = 1.15104332322704; /* trololol */
real v = 52 + eps;
real v = 52 + eps;
real w = 150 + eps;
real w = 150 + eps;
pair V = dir(v);
pair W = dir(w);
pair A = 2*V*W/(V+W);
pair D = dir((v+w)/2);
pair B = (D/(V*V)-1/D+2/W)/(1/(V*V)+1/(W*W));
pair I = 0;
pair U = 2*foot(W,I,B)-W;
pair C = 2*U*V/(U+V);
pair P = foot(B,A,A+A*dir(90));
pair Q = B + dir(D-B)*(P-B)/dir(P-B);
pair X = extension(P,Q,A,incenter(A,I,V));
pair T = extension(U,W,D,D+A-P);
fill(P--A--Q--B--cycle, red+opacity(0.2));
fill(T--D--I--W--cycle, blue+opacity(0.2));
draw(A--B--C--cycle, linewidth(1.0));
draw(incircle(A,B,C), linewidth(0.8));
draw(A--I, linewidth(0.7));
draw(P--A--Q--B--cycle, red+linewidth(0.8));
draw(T--D--I--W--cycle, blue+linewidth(0.8));
draw(P--X, linewidth(0.7));
draw(T--U, blue+linewidth(0.8));
draw(Q--D, red+linewidth(0.7));
dot("$V$", V, dir(53));
dot("$W$", W, dir(180));
dot("$A$", A, dir(77));
dot("$D$", D, dir(54));
dot("$B$", B, dir(-130));
dot("$I$", I, dir(-54));
dot("$U$", U, dir(-90));
dot("$C$", C, dir(-19));
dot("$P$", P, dir(146));
dot("$Q$", Q, dir(-70));
dot("$X$", X, dir(-12));
dot("$T$", T, dir(-141));
\end{asy}
\end{center}
Let the incircle be $\omega$ and touch $BC$ and $AB$ at point $U$ and $W$.
Let the tangent to $\omega$ at $D$ meet $UW$ at $T$.
Notice that $T$ is the pole of $BD$ with respect to $\omega$, so $IT\perp BD$.
Now, we make the following critical claim, which in particular implies $BP = BQ$.
\begin{claim*}
Quadrilaterals $DIWT$ and $PBQA$ are inversely similar.
\end{claim*}
\begin{proof}
This follows from four angle relations.
\begin{itemize}
\item $\dang IDT = \dang BPA = 90^\circ$.
\item $\dang TIW = \dang ABQ$.
\item $\dang DIT = \dang IAC = \dang BAI = \dang ABP$.
\item $\dang ITW = \dang QBI = \dang QAB$. \qedhere
\end{itemize}
\end{proof}
With $BP = BQ$ obtained,
one finishes with the same angle chasing as in the first solution.
|
sols-TST-IMO-2024_3
|
Let $n > k \ge 1$ be integers and let $p$ be a prime dividing $\tbinom nk$.
Prove that the $k$-element subsets of $\{1, \dots, n\}$ can be
split into $p$ classes of equal size, such that
any two subsets with the same sum of elements belong to the same class.
|
Let $\sigma(S)$ denote the sum of the elements of $S$, so that
\[ P(x) \coloneq \sum_{\substack{S\subseteq \{1, \dots, n\} \\ |S|=k}} x^{\sigma(S)} \]
is the generating function for the sums of $k$-element subsets of $\{1, \dots, n\}$.
By Legendre's formula,
\[ \nu_p\left(\binom{n}{k}\right) = \sum_{r=1}^{\infty}
\left(\left\lfloor\frac{n}{p^r}\right\rfloor - \left\lfloor\frac{k}{p^r}\right\rfloor
- \left\lfloor\frac{n-k}{p^r}\right\rfloor\right) \]
so there exists a positive integer $r$ with
\[ \left\lfloor\frac{n}{p^r}\right\rfloor - \left\lfloor\frac{k}{p^r}\right\rfloor
- \left\lfloor\frac{n-k}{p^r}\right\rfloor > 0. \]
The main claim is the following:
\begin{claim*}
$P(x)$ is divisible by
\[
\Phi_{p^r}(x) = x^{(p-1)p^{r-1}} + \dots + x^{p^{r-1}} + 1.
\]
\end{claim*}
Before proving this claim, we will show how it solves the problem. It implies
that there exists a polynomial $Q$ with integer coefficients satisfying
\begin{align*}
P(x) & = \Phi_{p^r}(x) Q(x)\\
& = (x^{(p-1)p^{r-1}} + \dots + x^{p^{r-1}} + 1) Q(x).
\end{align*}
Let $c_0,\ c_1,\ \dots$ denote the coefficients of $P$, and define
\[ s_i = \sum_{j \equiv i \pmod{p^r}} c_j. \]
Then it's easy to see that
\begin{alignat*}{4}
& s_0 && = s_{p^{r-1}} && = \cdots && = s_{(p-1)p^{r-1}}\\
& s_1 && = s_{p^{r-1}+1} && = \cdots && = s_{(p-1)p^{r-1}+1}\\
& && && \vdots &&\\
& s_{p^{r-1}-1} && = s_{2p^{r-1}-1} && = \cdots && = s_{p^r-1}.
\end{alignat*}
This means we can construct the $p$ classes by placing a set with sum $z$ in
class $\left\lfloor\frac{z\mod{p^r}}{p^{r-1}}\right\rfloor$.
Now we present two ways to prove the claim.
\paragraph{First proof of claim.}
There's a natural bijection between $k$-element subsets of $\{1, \dots, n\}$
and binary strings consisting of $k$ zeroes and $\ell$ ones:
the set $\{a_1, \dots, a_k\}$ corresponds to the string
which has zeroes at positions $a_1,\ \dots,\ a_k$.
Moreover, the inversion count of this string is simply $(a_1 + \dots + a_k) - \tfrac12k(k+1)$,
so we only deal with these inversion counts (equivalently, we are factoring
$x^{\frac{k(k+1)}{2}}$ out of $P$).
Recall that the generating function for these inversion counts is given by the
$q$-binomial coefficient
\[
P(x) = \frac{(x-1) \cdots (x^{k+\ell}-1)}
{\big[(x-1)\cdots(x^k-1)\big] \times \big[(x-1)\cdots(x^\ell-1)\big]}.
\]
By choice of $r$, the numerator of $P(x)$ has more factors of $\Phi_{p^r}(x)$
than the denominator, so $\Phi_{p^r}(x)$ divides $P(x)$.
\begin{remark*}
Here is a proof that $P(x)$ is divisible by $\Phi_{p^r}(x)$ for some $r$ using
the $q$-binomial formula, without explicitly identifying $r$. We know that
$P(x)$ is the product of several cyclotomic polynomials, and that $P(1)$ is a
multiple of $p$. Thus there is a factor $\Phi_q(x)$ for which $\Phi_q(1)$ is a
multiple of $p$, which is equivalent to $q$ being a power of $p$.
\end{remark*}
\paragraph{Second proof of claim.}
Note that $P(x)$ is the coefficient of $y^k$ in the polynomial
\[ Q(x, y) \coloneq (1+xy)(1+x^2y)\dotsm(1+x^ny). \]
Let $a$ be the remainder when $n$ is divided by $p^r$, and let $b$ be the
remainder when $k$ is divided by $p^r$; then we have $a<b$ by the choice of $r$.
Let $q = \lfloor n/p^r\rfloor$ so $n=p^rq+a$.
Consider taking $x$ to be a primitive $p^r$th root of unity, say $\omega$. Then
\[
Q(\omega, y) = \left[(1+\omega y)(1+\omega^2y)\cdots (1+\omega^{p^r}y)\right]^q
(1+\omega y)(1+\omega^2y)\dots (1+\omega^a y).
\]
Now $\omega$, $\omega^2$, \dots, $\omega^{p^r}$ are all the $p^r$th roots of unity,
each exactly once; then we can see that
\begin{align*}
& (1+\omega y)(1+\omega^2y)\cdots (1+\omega^{p^r}y) \\
&= (1-\omega(-y))(1-\omega^2(-y))\cdots (1-\omega^{p^r}(-y)) \\
&= 1-(-y)^{p^r},
\end{align*}
so
\[ Q(\omega,y) = (1-(-y)^{p^r})^q (1+\omega y)(1+\omega^2y)\dots (1+\omega^ay). \]
In particular, for any $m$, if the coefficient of $y^m$ in $Q(w,x)$ is nonzero,
then $m$ must be congruent to one of $0,1, \dots, a \pmod{p^r}$.
Therefore the coefficient of $y^k$ in $Q(\omega, y)$ is zero.
This means that $P(\omega)=0$ whenever $\omega$ is a primitive $p^r$th root
of unity, which proves the claim.
|
sols-TST-IMO-2024_4
|
Find all integers $n \ge 2$ for which there exists
a sequence of $2n$ pairwise distinct points $(P_1, \dots, P_n, Q_1, \dots, Q_n)$
in the plane satisfying the following four conditions:
\begin{enumerate}[label={(\roman*)}]
\ii no three of the $2n$ points are collinear;
\ii $P_iP_{i+1} \ge 1$ for all $i=1,2,\dots,n$, where $P_{n+1} = P_1$;
\ii $Q_iQ_{i+1} \ge 1$ for all $i=1,2,\dots,n$, where $Q_{n+1} = Q_1$; and
\ii $P_i Q_j \leq 1$ for all $i=1,2,\dots,n$ and $j=1,2,\dots,n$.
\end{enumerate}
|
\paragraph{Answer.}
Even integers only.
\paragraph{Proof that even $n$ work.}
If we ignore the conditions that the points are pairwise distinct and form no
collinear triples, we may take
\[
P_{2i+1}=(0.51, 0), \quad P_{2i}=(-0.51, 0), \quad
Q_{2i+1}=(0, 0.51), \quad Q_{2i}=(0,-0.51).
\]
The distances $P_iP_{i+1}$ and $Q_iQ_{i+1}$ are $1.02 > 1$, while the distances
$P_iQ_j$ are $0.51\sqrt{2} < 1$. We may then perturb each point by a small
amount to ensure that the distance inequalities still hold and have the points
in general position.
\paragraph{Proof that odd $n$ do not work.}
The main claim is the following.
\begin{claim*}
For $1\leq i\leq n$, points $Q_i$ and $Q_{i+1}$ must lie on opposite sides of
line $P_1P_2$.
\end{claim*}
To isolate the geometry component of the problem,
we rewrite the claim in the following contrapositive form,
without referencing the points $Q_i$ and $Q_{i+1}$:
\begin{lemma*}
Suppose $A$ and $B$ are two points such that $\max(P_1A, P_1B, P_2A, P_2B) \le 1$,
and moreover $A$ and $B$ lie on the same side of line $P_1 P_2$.
Further assume no three of $\{P_1, P_2, A, B\}$ are collinear.
Then $AB < 1$.
\end{lemma*}
\begin{center}
\begin{asy}
size(13cm);
pair P_1 = (0,0);
pair P_2 = (1,0);
pair T = (0.5, 3^0.5/2);
pair A = P_1 + dir(20);
pair B = P_2 + 0.96*dir(145);
pair X = extension(P_1, A, P_2, B);
picture lp, rp;
filldraw(lp, arc(P_1, P_2, T)--P_1--cycle, opacity(0.1)+yellow, blue+1.2);
filldraw(lp, arc(P_2, T, P_1)--P_2--cycle, opacity(0.1)+yellow, blue+1.2);
draw(lp, A--B, red);
draw(lp, P_1--A, deepgreen);
draw(lp, P_2--B, deepgreen);
dot(lp, T);
dot(lp, "$X$", X, dir(-90));
dot(lp, "$P_1$", P_1, dir(-90));
dot(lp, "$P_2$", P_2, dir(-90));
dot(lp, "$A$", A, dir(A-P_1));
dot(lp, "$B$", B, dir(245));
pair A = P_1 + 0.9*dir(55);
pair B = (0.85,0.03);
filldraw(rp, arc(P_1, P_2, T)--P_1--cycle, opacity(0.1)+yellow, blue+1.2);
filldraw(rp, arc(P_2, T, P_1)--P_2--cycle, opacity(0.1)+yellow, blue+1.2);
draw(rp, A--B, red);
draw(rp, P_1--A--P_2, deepgreen);
draw(rp, P_1--B--P_2, dotted);
dot(rp, "$P_1$", P_1, dir(-90));
dot(rp, "$P_2$", P_2, dir(-90));
dot(rp, "$A$", A, dir(260));
dot(rp, "$B$", B, dir(65));
add(lp);
add(shift(1.2,0)*rp);
\end{asy}
\end{center}
\begin{proof}[Proof of lemma.]
Suppose for the sake of contradiction
that $A$ and $B$ lie on the same side of $P_1P_2$.
The convex hull of these four points is either a quadrilateral or a triangle.
\begin{itemize}
\ii If the convex hull is a quadrilateral,
assume WLOG that the vertices are $P_1P_2AB$ in order.
Let $X$ denote the intersection of segments $P_1A$ and $P_2B$.
Then \[ 1 + AB = P_1P_2 + AB < P_1X + XP_2 + AX + XB = P_1A + P_2B \leq 2. \]
\ii Otherwise, assume WLOG that $B$ is in the interior of triangle $P_1P_2A$.
Since $\angle P_1BA + \angle P_2BA = 360\dg-\angle P_1BP_2 > 180\dg$,
at least one of $\angle P_1BA$ and $\angle P_2BA$ is obtuse.
Assume WLOG the former angle is obtuse; then $AB < P_1A \leq 1$.
\qedhere
\end{itemize}
\end{proof}
\begin{remark*}
Another proof of the lemma can be found by replacing segment $AB$ with
the intersection of this line on the boundary of the blue region above,
which does not decrease the distance.
In other words, one can assume WLOG that $A$ and $B$ lie on either segment $AB$
or one of the two circular arcs.
One then proves that $AB \le 1$, and that for equality to occur,
one of $A$ and $B$ must lie on segment $P_1 P_2$
However, this approach seems to involve a fair bit more calculation.
Yet another clever approach uses the trivia-fact that a
\href{https://w.wiki/7VaD}{Reuleaux triangle} happens to have constant width.
In any case, it's important to realize that this claim is \emph{not trivial};
while it looks like it is easy to prove, it is not,
owing to the two near-equality cases.
\end{remark*}
It follows from the claim that $Q_i$ is on the same side of line $P_1P_2$
as $Q_1$ if $i$ is odd, and on the opposite side if $i$ is even.
Since $Q_1 = Q_{n+1}$, this means the construction is not possible when $n$ is odd.
\begin{remark*}
The fact that $n$ cannot be odd follows from Theorem 3 of
\href{https://arxiv.org/pdf/1803.01822.pdf}{EPTAS for Max Clique on Disks and
Unit Balls}. In the language of that paper, if $G$ is a unit ball graph, then
the \emph{induced odd cycle parking number} of $\bar G$ is at most 1.
In earlier versions of the proposed problem,
the points were not necessarily distinct to make the even $n$ case nicer,
but this resulted in annoying boundary conditions for the odd $n$ case.
\end{remark*}
|
sols-TST-IMO-2024_5
|
Suppose $a_1 < a_2 < \dots < a_{2024}$ is an arithmetic sequence of positive integers,
and $b_1 < b_2 < \dots < b_{2024}$ is a geometric sequence of positive integers.
Find the maximum possible number of integers that could appear in both sequences,
over all possible choices of the two sequences.
|
\paragraph{Answer.} $11$ terms.
\paragraph{Construction.} Let $a_i = i$ and $b_i = 2^{i-1}$.
\paragraph{Bound.}
We show a $\nu_p$-based approach communicated by Derek Liu,
which seems to be the shortest one.
At first, we completely ignore the geometric sequence $b_i$
and focus only on the arithmetic sequence.
\begin{claim*}
Let $p$ be any prime, and consider the sequence
\[ \nu_p(a_1), \; \nu_p(a_2), \; \dots, \; \nu_p(a_{2024}). \]
Set $C \coloneq \left\lfloor \log_p(2023) \right\rfloor$.
Then there are at most $C+2$ different values in this sequence.
\end{claim*}
\begin{proof}
By scaling, assume the $a_i$ do not have a common factor,
so that $a_i = a + di$ where $\gcd(a,d) = 1$.
\begin{itemize}
\ii If $p \mid d$, then $p \nmid a$ and $\nu_p(a_i)$ is constant.
\ii Otherwise, assume $p \nmid d$.
We will in fact prove that every term in the sequence
is contained in $\{0, 1, \dots, C\}$ with at most one exception.
Define $M \coloneq \max_i \nu_p(a_i)$.
If $M \le C$, there is nothing to prove.
Otherwise, fix some index $m$ such that $\nu_p(a_m) = M$.
We know $\nu_p(i-m) \le C$ since $|i-m| \le 2023$.
But now for any other index $i \neq m$;
\begin{align*}
\nu_p(d(i-m)) &= \nu_p(i-m) \le C < M = \nu_p(a_m) \\
\implies \nu_p(a_i) &= \nu_p(a_m + d(i-m)) = \nu_p(i-m) \le C.
\end{align*}
so $\nu_p(a_m)$ is the unique exceptional term of the sequence exceeding $C$.
\qedhere
\end{itemize}
\end{proof}
\begin{remark*}
The bound in the claim is best possible by taking
$a_i = p^M + (i-1)$ for any $M > C$.
Then indeed, the sequence $\nu_p(a_i)$ takes on values in $\{0,1,\dots,C\}$ for $i>1$
while $\nu_p(a_1) = M$.
\end{remark*}
Back to the original problem with $(b_i)$.
Consider the common ratio $r \in \QQ$ of the geometric sequence $(b_i)$.
If $p$ is any prime with $\nu_p(r) \neq 0$,
then every term of $(b_i)$ has a different $\nu_p$.
So there are two cases to think about.
\begin{itemize}
\ii Suppose any $p \ge 3$ has $\nu_p(r) \neq 0$.
Then there are at most
\[ 2 + \log_p(2023) = 2 + \log_3(2023) \approx 8.929 < 11 \]
overlapping terms, as needed.
\ii Otherwise, suppose $r$ is a power of $2$ (in particular, $r \ge 2$ is an integer).
We already have an upper bound of $12$; we need to improve it to $11$.
As in the proof before, we may assume WLOG by scaling down by a power of $2$
that the common difference of $a_i$ is odd.
(This may cause some $b_i$ to become rational non-integers, but that's OK.
However, unless $\nu_2(a_i)$ is constant, the $a_i$ will still be integers.)
Then in order for the bound of $12$ to be achieved,
the sequence $\nu_2(a_i)$ must be contained in $\{0,1,\dots,10,M\}$ for some $M \ge 11$.
In particular, we only need to work with $r = 2$.
Denote by $b$ the unique odd-integer term in the geometric sequence,
which must appear among $(a_i)$.
Then $2b$ appears too, so the common difference of $a_i$ is at most $b$.
But if $(a_i)$ is an arithmetic progression of integers that contains $b$
and has common difference at most $b$,
then no term of the sequence can ever exceed $b + 2023 \cdot b = 2024b$.
Hence $2^M b$ cannot appear for any $M \ge 11$. This completes the proof.
\end{itemize}
\begin{remark*}
There are several other approaches to the problem, but most take some time to execute.
The primary issue is that the common difference of the $a_i$'s could
share prime factors with the common ratio in the $b_i$'s,
which means that merely trying to write out a lot of modular arithmetic equations
leads to a lot of potential technical traps that are not pleasant to defuse.
One unusual thing is that many solutions end up proving a bound of $12$
(in the case the common ratio is $2$) and then having to adjust it to $11$ later.
\end{remark*}
|
sols-TST-IMO-2024_6
|
Solve over $\RR$ the functional equation
\[ f(xf(y)) + f(y)=f(x+y)+f(xy). \]
\end{enumerate}
|
In addition to all constant functions, $f(x) \equiv x + 1$ clearly works too.
We prove these are the only solutions.
The solution that follows is by the original proposer.
Let $P(x,y)$ denote the given assertion.
\begin{claim}\label{periodic}
If $f$ is periodic, then $f$ is constant.
\end{claim}
\begin{proof}
Let $f$ have period $d \neq 0$.
From $P(x,y+d)$, we have
\[ f(x(y+d))=f(x+y+d)-f(y+d)-f(xf(y+d))=f(x+y)-f(y)-f(xf(y)). \]
Applying $P(x,y)$ gives
\[ f(x(y+d))=f(xy). \]
In particular, taking $y=0$ yields that $f(dx)=f(0)$.
Thus $f$ is constant, as $d\neq 0$.
\end{proof}
\begin{claim}\label{symmetry}
For all real numbers $x$ and $y$, we have $f(f(x)+y)=f(f(y)+x)$.
\end{claim}
\begin{proof}
Applying $P(f(x),y)$ and then $P(y,x)$ gives us
\begin{align*}
f(f(x)f(y)) &= f(f(x)+y)+f(f(x)y)-f(y) \\
&= f(f(x)+y)+f(x+y)+f(xy)-f(x)-f(y).
\end{align*}
Now swapping $x$ and $y$ gives us $f(f(x)+y)=f(f(y)+x)$.
\end{proof}
\begin{claim}\label{distribute}
If $f$ is nonconstant, then $f(f(x)+y)=f(x)+f(y)$ for all reals $x,y$.
\end{claim}
\begin{proof}
Let $x, y \in \RR$ and let $d \coloneq f(f(x)+y)-f(x)-f(y)$.
Let $z$ be an arbitrary real number.
By repeatedly applying \Cref{symmetry}, we have
\begin{align*}
f(z+f(f(x)+y)) &= f(f(z)+f(x)+y) \\
&= f(f(x)+f(z)+y) \\
&= f(x+f(f(z)+y)) \\
&= f(x+f(f(y)+z)) \\
&= f(f(x)+f(y)+z) \\
&= f(z+f(x)+f(y)).
\end{align*}
If $d \neq 0$, then $f$ is periodic with period $d$, contradicting \Cref{periodic}.
\end{proof}
\begin{claim}\label{zero}
If $f$ is nonconstant, then $f(0)=1$ and $f(x+1)=f(x)+1$.
\end{claim}
\begin{proof}
From $P(z,0)$ we have $f(zf(0))=f(z)$ for all real $z$.
Then $P(xf(0),y)$ and $P(x,y)$ give us
\begin{align*}
f(xf(0)+y) &= f(y)+f(xf(0)f(y))-f(xf(0)y) \\
&=f(y)+f(xf(y))-f(xy)=f(x+y).
\end{align*}
If $f(0) \neq 1$, then $xf(0)-x$ is a period of $f$ for all $x$, violating \Cref{periodic}.
So we must have $f(0) = 1$.
Now putting $x=0$ in \Cref{distribute} gives $f(x+1)=f(x)+1$.
\end{proof}
\begin{claim}\label{additive}
If $f$ is nonconstant, then $f(x)+f(y)=f(x+y)+1$.
\end{claim}
\begin{proof}
From $P(x+1,y)$ and \Cref{zero}, we have
\[ f((x+1)f(y))=f(x+y+1)+f(xy+y)-f(y)=f(x+y)+f(xy+y)-f(y)+1. \]
Also, from \Cref{distribute} and $P(x,y)$, we have
\[ f((x+1)f(y))=f(xf(y))+f(y)=f(x+y)+f(xy). \]
Thus $f(xy)=f(xy+y)-f(y)+1$.
Replacing $x$ with $\frac{x}{y}$ gives the claim for all $y\ne 0$
(whereas $y=0$ follows from \Cref{zero}).
\end{proof}
\begin{claim}
If $f$ is nonconstant, then $f(x)\equiv x+1$.
\end{claim}
\begin{proof}
We apply \Cref{distribute}, \Cref{additive}, and \Cref{zero}:
\[ f(f(x)+y)=f(x)+f(y)=f(x+y)+1=f(x+y+1). \]
If $f(x) \neq x+1$ for some $x$, then \Cref{periodic} again gives a contradiction.
\end{proof}
|
sols-TSTST-2011_1
|
Find all real-valued functions $f$ defined on pairs of real numbers,
having the following property: for all real numbers $a, b, c$,
the median of $f(a,b), f(b,c), f(c,a)$ equals the median of $a, b, c$.
(The \emph{median} of three real numbers, not necessarily distinct,
is the number that is in the middle when the three numbers
are arranged in nondecreasing order.)
|
The following solution is joint with Andrew He.
We prove the following main claim,
from which repeated applications can deduce the problem.
\begin{claim*}
Let $a < b < c$ be arbitrary. On $\{a,b,c\}^2$,
$f$ takes one of the following two forms,
where the column indicates the $x$-value
and the row indicates the $y$-value.
\[
\begin{array}{c|rrr}
f & a & b & c \\ \hline
a & a & b & \ge c \\
b & \le a & b & \ge c \\
c & \le a & b & c
\end{array}
\qquad
\text{or}
\qquad
\begin{array}{c|rrr}
f & a & b & c \\ \hline
a & a & \le a & \le a \\
b & b & b & b \\
c & \ge c & \ge c & c
\end{array}
\]
\end{claim*}
\begin{proof}
First, we of course have $f(x,x) = x$ for all $x$.
Now:
\begin{itemize}
\ii By considering the assertion for $(a,a,c)$ and $(a,c,c)$
we see that one of $f(a,c)$ and $f(c,a)$ is $\ge c$
and the other is $\le a$.
\ii Hence, by considering $(a,b,c)$ we find that
one of $f(a,b)$ and $f(b,c)$ must be $b$,
and similarly for $f(b,a)$ and $f(c,b)$.
\ii Now, WLOG $f(b,a) = b$; we prove we get the first case.
\ii By considering $(a,a,b)$ we deduce $f(a,b) \le a$,
so $f(b,c) = b$ and then $f(c,b) \ge c$.
\ii Finally, considering $(c,b,a)$ once again
in conjunction with the first bullet, we arrive at the conclusion
that $f(a,c) \le a$; similarly $f(c,a) \ge c$.
\qedhere
\end{itemize}
\end{proof}
From this it's easy to obtain that $f(x,y) \equiv x$
or $f(x,y) \equiv y$ are the only solutions.
|
sols-TSTST-2011_2
|
Two circles $\omega_1$ and $\omega_2$ intersect at points $A$ and $B$.
Line $\ell$ is tangent to $\omega_1$ at $P$
and to $\omega_2$ at $Q$ so that $A$ is closer to $\ell$ than $B$.
Let $X$ and $Y$ be points on major arcs $\arc{PA}$
(on $\omega_1$) and $\arc{AQ}$ (on $\omega_2$), respectively,
such that $AX/PX = AY/QY = c$.
Extend segments $PA$ and $QA$ through $A$ to $R$ and $S$,
respectively, such that $AR = AS = c\cdot PQ$.
Given that the circumcenter of triangle $ARS$ lies on line $XY$,
prove that $\angle XPA = \angle AQY$.
|
We begin as follows:
\begin{claim*}
There is a spiral similarity centered at $X$ mapping $AR$ to $PQ$.
Similarly there is a spiral similarity centered at $Y$ mapping $SA$ to $PQ$.
\end{claim*}
\begin{proof}
Since $\dang XAR = \dang XAP = \dang XPQ$,
and $AR/AX = PQ/PX$ is given.
\end{proof}
Now the composition of the two spiral similarities
\[ AR \xmapsto{X} PQ \xmapsto{Y} SA \]
must be a rotation, since $AR = AS$.
The center of this rotation must coincide with the circumcenter $O$ of $\triangle ARS$,
which is known to lie on line $XY$.
\begin{center}
\begin{asy}
import graph; size(10cm);
pen uququq = rgb(0.25098,0.25098,0.25098); pen zzttqq = rgb(0.6,0.2,0.);
draw((13.97096,-7.13004)--(20.23013,-5.53182)--(27.19232,-12.23004)--cycle, linewidth(0.6) + zzttqq);
draw(circle((25.,-3.), 9.48683), linewidth(0.6) + uququq);
draw(circle((13.,-3.), 4.24264), linewidth(0.6) + uququq);
draw((13.97096,-7.13004)--(16.,0.), linewidth(0.6) + red);
draw((13.97096,-7.13004)--(11.14589,0.81605), linewidth(0.6) + red);
draw((11.14589,0.81605)--(25.35640,-1.57297), linewidth(0.6) + blue);
draw((27.19232,-12.23004)--(20.85410,5.53296), linewidth(0.6) + red);
draw((27.19232,-12.23004)--(16.,0.), linewidth(0.6) + red);
draw((20.85410,5.53296)--(9.74299,-7.13207), linewidth(0.6) + blue);
draw((13.97096,-7.13004)--(20.23013,-5.53182), linewidth(0.6) + zzttqq);
draw((20.23013,-5.53182)--(27.19232,-12.23004), linewidth(0.6) + zzttqq);
draw((27.19232,-12.23004)--(13.97096,-7.13004), linewidth(0.6) + zzttqq);
draw((20.23013,-5.53182)--(19.26755,-9.17733), linewidth(0.6));
draw((11.14589,0.81605)--(20.85410,5.53296), linewidth(0.6));
dot("$A$", (16.,0.), dir(100));
dot("$Q$", (20.85410,5.53296), dir(120));
dot("$P$", (11.14589,0.81605), dir(120));
dot("$X$", (13.97096,-7.13004), dir(280));
dot("$Y$", (27.19232,-12.23004), dir(280));
dot("$R$", (25.35640,-1.57297), dir((8.043, 16.110)));
dot("$S$", (9.74299,-7.13207), dir(225));
dot("$O$", (19.26755,-9.17733), dir(45));
dot("$O'$", (20.23013,-5.53182), dir(135));
\end{asy}
\end{center}
As $O$ is a fixed-point of the composed map above,
we may let $O'$ be the image of $O$ under the rotation at $X$,
so that
\[ \triangle XPA \overset{+}{\sim} \triangle XO'O, \qquad
\triangle YQA \overset{+}{\sim} \triangle YO'O. \]
Because
\[ \frac{XO}{XO'} = \frac{XA}{XP} = c \frac{YQ}{YA} = \frac{YO}{YO'} \]
it follows $\ol{O'O}$ bisects $\angle XO'Y$.
Finally, we have
\[ \dang XPA = \dang XO'O = \dang OO'Y = \dang AQY. \]
\begin{remark*}
Indeed, this also shows $\ol{XP} \parallel \ol{YQ}$; so the positive homothety
from $\omega_1$ to $\omega_2$ maps $P$ to $Q$ and $X$ to $Y$.
\end{remark*}
|
sols-TSTST-2011_3
|
Prove that there exists a real constant $c$
such that for any pair $(x,y)$ of real numbers,
there exist relatively prime integers $m$ and $n$
satisfying the relation
\[ \sqrt{(x-m)^2 + (y-n)^2} < c \log(x^2+y^2+2). \]
|
This is actually the same problem as USAMO 2014/6. Surprise!
|
sols-TSTST-2011_4
|
Acute triangle $ABC$ is inscribed in circle $\omega$.
Let $H$ and $O$ denote its orthocenter and circumcenter, respectively.
Let $M$ and $N$ be the midpoints of sides $AB$ and $AC$, respectively.
Rays $MH$ and $NH$ meet $\omega$ at $P$ and $Q$, respectively.
Lines $MN$ and $PQ$ meet at $R$.
Prove that $\ol{OA} \perp \ol{RA}$.
|
Let $MH$ and $NH$ meet the nine-point circle again at $P'$ and $Q'$, respectively.
Recall that $H$ is the center of the homothety between
the circumcircle and the nine-point circle.
From this we can see that $P$ and $Q$ are the images of this homothety, meaning that
\[ HQ = 2HQ' \quad\text{and}\quad HP = 2HP'. \]
Since $M$, $P'$, $Q'$, $N$ are cyclic, Power of a Point gives us
\[ MH \cdot HP' = HN \cdot HQ'. \]
Multiplying both sides by two, we thus derive
\[ HM \cdot HP = HN \cdot HQ. \]
It follows that the points $M$, $N$, $P$, $Q$ are concyclic.
\begin{center}
\begin{asy}
size(11cm);
defaultpen(fontsize(8pt));
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
pair H = A+B+C;
pair O = circumcenter(A, B, C);
pair M = midpoint(A--B);
pair N = midpoint(A--C);
pair U = -C;
pair V = -B;
pair P = -U+2*foot(O, U, H);
pair Q = -V+2*foot(O, V, H);
pair R = extension(M, N, P, Q);
draw(A--B--C--cycle, blue);
draw(unitcircle, blue);
draw(M--P);
draw(N--Q);
draw(circumcircle(M, N, midpoint(B--C)), green);
draw(circumcircle(P, Q, M), red+dashed);
draw(P--R--A);
draw(R--N, dotted);
draw(circumcircle(A, M, N), dashed);
pair Qp = midpoint(H--Q);
pair Pp = midpoint(H--P);
dot("$A$", A, dir(A));
dot("$B$", B, dir(250));
dot("$C$", C, dir(C));
dot("$H$", H, dir(-90));
dot("$O$", O, dir(315));
dot("$M$", M, 1.4*dir(150));
dot("$N$", N, dir(20));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
dot("$R$", R, dir(R));
dot("$Q'$", Qp, dir(255));
dot("$P'$", Pp, dir(Pp));
/* Source generated by TSQ */
\end{asy}
\end{center}
Let $\omega_1$, $\omega_2$, $\omega_3$ denote the circumcircles of
$MNPQ$, $AMN$, and $ABC$, respectively.
The radical axis of $\omega_1$ and $\omega_2$ is line $MN$, while the
radical axis of $\omega_1$ and $\omega_3$ is line $PQ$.
Hence the line $R$ lies on the radical axis of $\omega_2$ and $\omega_3$.
But we claim that $\omega_2$ and $\omega_3$ are internally tangent at $A$.
This follows by noting the homothety at $A$ with ratio $2$ sends
$M$ to $B$ and $N$ to $C$.
Hence the radical axis of $\omega_2$ and $\omega_3$ is a line tangent to
both circles at $A$.
Hence $\ol{RA}$ is tangent to $\omega_3$.
Therefore, $\ol{RA} \perp \ol{OA}$.
|
sols-TSTST-2011_5
|
At a certain orphanage, every pair of orphans are either friends or enemies.
For every three of an orphan's friends,
an even number of pairs of them are enemies.
Prove that it's possible to assign each orphan two parents
such that every pair of friends shares exactly one parent,
no pair of enemies shares any parents,
and no three parents are in a love triangle (where each pair of them has a child).
|
Of course, we consider the graph with vertices as children and edges as friendships.
Consider all the maximal cliques in the graph.
\begin{claim*}
Every vertex is in at most two maximal cliques.
\end{claim*}
\begin{proof}
Indeed, consider a vertex $v$ adjacent to $w_1$ and $w_2$,
but with $w_1$ not adjacent to $w_2$.
Then by condition, any third vertex $u$
must be adjacent to exactly one of $w_1$ and $w_2$.
Moreover, given vertices $u$ and $u'$ adjacent to $w_1$,
vertices $u$ and $u'$ are adjacent too.
This proves the claim.
\end{proof}
Now, for every maximal clique we assign a particular parent to all vertices in that clique,
adding in additional distinct parents if there are any deficient children.
This satisfies the friendship/enemy condition.
Moreover, one can readily check that there are no love triangles:
given children $a$, $b$, $c$ such that $a$ and $b$ share a parent
while $a$ and $c$ share another parent,
according to the claim $b$ and $c$ can't share a third parent.
This completes the problem.
\begin{remark*}
This solution is highly motivated for the following reason:
by experimenting with small cases,
one quickly finds that given some vertices which form a clique,
one \emph{must} assign some particular parent to all vertices in that clique.
That is, the requirements of the problem are sufficiently rigid
that there is no room for freedom on our part,
so we know \emph{a priori} that an assignment based on cliques
(as above) must work.
From there we know exactly what to prove,
and everything else follows through.
Ironically, the condition that there be no love triangle
actually makes the problem easier, because it tells us exactly what to do!
\end{remark*}
|
sols-TSTST-2011_6
|
Let $a,b,c$ be real numbers in the interval $[0,1]$
with $a+b,b+c,c+a \ge 1$. Prove that
\[ 1 \le (1-a)^2 + (1-b)^2 + (1-c)^2
+ \frac{2\sqrt2 abc}{\sqrt{a^2+b^2+c^2}}. \]
|
The following approach is due to Ashwin Sah.
We will prove the inequality for any $a$, $b$, $c$
the sides of a possibly degenerate triangle
(which is implied by the condition),
ignoring the particular constant $1$.
Homogenizing, we instead prove the problem
in the following form:
\begin{claim*}
We have
\[ k^2 \le (k-a)^2 + (k-b)^2 + (k-c)^2
+ \frac{2\sqrt2 abc}{\sqrt{a^2+b^2+c^2}} \]
for any $a$, $b$, $c$, $k$
with $(a,b,c)$ the sides of a
possibly degenerate triangle.
\end{claim*}
\begin{proof}
For any particular $(a,b,c)$ this is a quadratic in $k$
of the form $2k^2 - 2(a+b+c)k + C \ge 0$;
thus we will verify it holds for $k = \half(a+b+c)$.
Letting $x = \half(b+c-a)$ as is usual,
the problem rearranges to
In that case, the problem amounts to
\[ (x+y+z)^2 \le x^2+y^2+z^2
+ \frac{2(x+y)(y+z)(z+x)}{\sqrt{x^2+y^2+z^2+xy+yz+zx}} \]
or equivalently
\[ x^2+y^2+z^2+xy+yz+zx
\le \left( \frac{(x+y)(y+z)(z+x)}{xy+yz+zx} \right)^2. \]
To show this, one may let $t = xy+yz+zx$,
then using $(x+y)(x+z) = x^2+B$ this becomes
\[ t^2(x^2+y^2+z^2+t)
\le (x^2+t)(y^2+t)(z^2+t) \]
which is obvious upon expansion.
\end{proof}
\begin{remark*}
The inequality holds actually for
all real numbers $a$, $b$, $c$,
with very disgusting proofs.
\end{remark*}
|
sols-TSTST-2011_7
|
Let $ABC$ be a triangle.
Its excircles touch sides $BC$, $CA$, $AB$ at $D$, $E$, $F$.
Prove that the perimeter of triangle $ABC$ is
at most twice that of triangle $DEF$.
|
Solution by August Chen:
It turns out that it is enough to take the
orthogonal projection of $EF$ onto side $BC$
(which has length $a-(s-a)(\cos B + \cos C)$)
and sum cyclically:
\begin{align*}
-s + \sum_{\text{cyc}} EF &\ge
-s +
\sum_{\text{cyc}}
\left[ a - (s-a)\left( \cos B + \cos C \right) \right] \\
&= s - \sum_{\text{cyc}} a \cos A
= \sum_{\text{cyc}} a \left( \half - \cos A \right) \\
&= R \sum_{\text{cyc}}
\sin A (1 - 2 \cos A) \\
&= R \sum_{\text{cyc}} \left( \sin A - \sin2A \right).
\end{align*}
Thus we're done upon noting that
\[ \frac{\sin2B + \sin2C}{2} = \sin(B+C) \cos(B-C)
= \sin A \cos(B-C) \le \sin A. \]
(Alternatively, one can avoid trigonometry by substituting
$\cos A = \frac{b^2+c^2-a^2}{2bc}$
and doing some routine but long calculation.)
|
sols-TSTST-2011_8
|
Let $x_0$, $x_1$, \dots, $x_{n_0-1}$ be integers,
and let $d_1$, $d_2$, \dots, $d_k$ be positive integers
with $n_0 = d_1 > d_2 > \dotsb > d_k$ and
$\gcd(d_1, d_2, \dots, d_k) = 1$.
For every integer $n \geq n_0$, define
\[ x_n = \left\lfloor \frac{x_{n-d_1} + x_{n-d_2}
+ \dots + x_{n-d_k}}{k} \right\rfloor. \]
Show that the sequence $(x_n)$ is eventually constant.
|
Note that if the initial terms are contained
in some interval $[A,B]$ then they will remain in that interval.
Thus the sequence is eventually periodic.
Discard initial terms and let the period be $T$;
we will consider all indices modulo $T$ from now on.
Let $M$ be the maximal term in the sequence
(which makes sense since the sequence is periodic).
Note that if $x_n = M$, we must have
$x_{n-d_i} = M$ for all $i$ as well.
By taking a linear combination $\sum c_i d_i \equiv 1 \pmod T$
(possibly be Bezout's theorem, since $\gcd_i(d_i)=1$),
we conclude $x_{n-1} = M$, as desired.
|
sols-TSTST-2011_9
|
Let $n$ be a positive integer.
Suppose we are given $2^n+1$ distinct sets,
each containing finitely many objects.
Place each set into one of two categories, the red sets and the blue sets,
so that there is at least one set in each category.
We define the \textit{symmetric difference} of two sets as
the set of objects belonging to exactly one of the two sets.
Prove that there are at least $2^n$ different sets which
can be obtained as the symmetric difference of a red set and a blue set.
\end{enumerate}
|
We can interpret the problem as working with
binary strings of length $\ell \ge n+1$,
with $\ell$ the number of elements across all sets.
Let $F$ be a field of cardinality $2^\ell$,
hence $F \cong \FF_2^{\oplus \ell}$.
Then, we can think of red/blue as elements of $F$,
so we have some $B \subseteq F$, and an $R \subseteq F$.
We wish to prove that $|B+R| \ge 2^n$.
Want $|B + R| \ge 2^n$.
Equivalently, any element of a set $X$ with $|X| = 2^n - 1$
should omit some element of $|B+R|$.
To prove this: we know $|B| + |R| = 2^n+1$, and define
\[ P(b,r) = \prod_{x \in X} (b+r-x). \]
Consider $b^{|B|-1} r^{|R|-1}$.
The coefficient of is $\binom{2^n-1}{|B|-1}$,
which is odd (say by Lucas theorem),
so the nullstellensatz applies.
|
sols-TSTST-2012_1
|
Determine all infinite strings of letters with the following properties:
\begin{enumerate}
\item[(a)] Each letter is either $T$ or $S$,
\item[(b)] If position $i$ and $j$ both have the letter $T$,
then position $i+j$ has the letter $S$,
\item[(c)] There are infinitely many integers $k$ such that position $2k-1$ has the $k$th $T$.
\end{enumerate}
|
We wish to find all infinite sequences $a_1, a_2, \dots$
of positive integers satisfying the following properties:
\begin{enumerate}[(a)]
\item $a_1 < a_2 < a_3 < \dotsb$,
\item there are no positive integers $i$, $j$, $k$, not necessarily distinct, such that $a_i+a_j=a_k$,
\item there are infinitely many $k$ such that $a_k = 2k-1$.
\end{enumerate}
If $a_k=2k-1$ for some $k>1$, let $A_k=\{a_1,a_2,\dots,a_k\}$.
By (b) and symmetry, we have
\[ 2k-1
\ge \frac{|A_k-A_k|-1}{2}+|A_k|
\ge \frac{2|A_k|-2}{2}+|A_k| = 2k-1. \]
But in order for $|A_k-A_k|=2|A_k|-1$,
we must have $A_k$ an arithmetic progression,
whence $a_n=2n-1$ for all $n$ by taking $k$ arbitrarily large.
|
sols-TSTST-2012_2
|
Let $ABCD$ be a quadrilateral with $AC = BD$.
Diagonals $AC$ and $BD$ meet at $P$.
Let $\omega_1$ and $O_1$ denote the circumcircle and circumcenter of triangle $ABP$.
Let $\omega_2$ and $O_2$ denote the circumcircle and circumcenter of triangle $CDP$.
Segment $BC$ meets $\omega_1$ and $\omega_2$ again at $S$ and $T$
(other than $B$ and $C$), respectively.
Let $M$ and $N$ be the midpoints of minor arcs $\arc{SP}$ (not including $B$)
and $\arc{TP}$ (not including $C$).
Prove that $\ol{MN} \parallel \ol{O_1O_2}$.
|
Let $Q$ be the second intersection point of $\omega_1$, $\omega_2$.
Suffice to show $\ol{QP} \perp \ol{MN}$.
Now $Q$ is the center of a spiral \emph{congruence}
which sends $\ol{AC} \mapsto \ol{BD}$.
So $\triangle QAB$ and $\triangle QCD$ are similar isosceles triangles.
Now, \[ \dang QPA = \dang QBA = \dang DCQ = \dang DPQ \]
and so $\ol{QP}$ is bisects $\angle BPC$.
\begin{center}
\begin{asy}
size(12cm);
pair Q = dir(80);
pair B = dir(190);
pair C = dir(350);
pair A = shift(Q)*rotate(-40)*shift(-Q)*B;
pair D = shift(Q)*rotate(40)*shift(-Q)*C;
pair P = extension(A, C, B, D);
draw(circumcircle(A, B, P));
draw(circumcircle(C, D, P));
filldraw(A--B--C--D--cycle, opacity(0.1)+lightcyan, lightcyan);
filldraw(Q--A--C--cycle, opacity(0.1)+lightgreen, lightblue);
filldraw(Q--B--D--cycle, opacity(0.1)+lightgreen, lightblue);
pair O_1 = circumcenter(A, B, P);
pair O_2 = circumcenter(C, D, P);
pair S = -B+2*foot(O_1, B, C);
pair T = -C+2*foot(O_2, B, C);
pair M = circumcenter(S, P, incenter(B, S, P));
pair N = circumcenter(T, P, incenter(C, T, P));
draw(M--N, blue);
draw(O_1--O_2, blue);
pair I = incenter(P, B, C);
draw(Q--I, blue);
draw(B--I--C, blue);
dot("$Q$", Q, dir(Q));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$A$", A, dir(A));
dot("$D$", D, dir(D));
dot("$P$", P, dir(60));
dot("$O_1$", O_1, dir(150));
dot("$O_2$", O_2, dir(10));
dot("$S$", S, dir(-90));
dot("$T$", T, dir(-90));
dot("$M$", M, dir(100));
dot("$N$", N, dir(35));
dot("$I$", I, dir(45));
/* TSQ Source:
Q = dir 80
B = dir 190
C = dir 350
A = shift(Q)*rotate(-40)*shift(-Q)*B
D = shift(Q)*rotate(40)*shift(-Q)*C
P = extension A C B D
circumcircle A B P
circumcircle C D P
A--B--C--D--cycle 0.1 lightcyan / lightcyan
Q--A--C--cycle 0.1 lightgreen / lightblue
Q--B--D--cycle 0.1 lightgreen / lightblue
O_1 = circumcenter A B P
O_2 = circumcenter C D P
S = -B+2*foot O_1 B C
T = -C+2*foot O_2 B C
M = circumcenter S P incenter B S P
N = circumcenter T P incenter C T P
M--N blue
O_1--O_2 blue
I = incenter P B C
Q--I blue
B--I--C blue
*/
\end{asy}
\end{center}
Now, let $I = \ol{BM} \cap \ol{CN} \cap \ol{PQ}$
be the incenter of $\triangle PBC$.
Then $IM \cdot IB = IP \cdot IQ = IN \cdot IC$, so $BMNC$ is cyclic,
meaning $\ol{MN}$ is antiparallel to $\ol{BC}$ through $\angle BIC$.
Since $\ol{QPI}$ passes through the circumcenter of $\triangle BIC$,
it follows now $\ol{QPI} \perp \ol{MN}$ as desired.
|
sols-TSTST-2012_3
|
Let $\NN$ be the set of positive integers.
Let $f \colon \NN \to \NN$ be a function satisfying the following two conditions:
\begin{enumerate}[(a)]
\ii $f(m)$ and $f(n)$ are relatively prime whenever $m$ and $n$ are relatively prime.
\ii $n \le f(n) \le n+2012$ for all $n$.
\end{enumerate}
Prove that for any natural number $n$ and any prime $p$,
if $p$ divides $f(n)$ then $p$ divides $n$.
|
\paragraph{First short solution, by Jeffrey Kwan.}
Let $p_0$, $p_1$, $p_2$, \dots\ denote the
sequence of all prime numbers, in any order.
Pick \emph{any} primes $q_i$ such that
\[ q_0 \mid f(p_0), \quad q_1 \mid f(p_1), \quad
q_2 \mid f(p_2), \; \text{ etc}. \]
This is possible since each $f$ value above exceeds $1$.
Also, since by hypothesis the $f(p_i)$ are pairwise coprime,
the primes $q_i$ are all pairwise distinct.
\begin{claim*}
We must have $q_i = p_i$ for each $i$.
(Therefore, $f(p_i)$ is a power of $p_i$ for each $i$.)
\end{claim*}
\begin{proof}
Assume to the contrary that $q_0 \neq p_0$.
By changing labels if necessary, assume
$\min(p_1, p_2, \dots, p_{2012}) > 2012$.
Then by Chinese remainder theorem we can choose
an integer $m$ such that
\begin{align*}
m+i &\equiv 0 \pmod{q_i} \\
m &\not\equiv 0 \pmod{p_i}
\end{align*}
for $0 \le i \le 2012$.
But now $f(m)$ should be coprime to all $f(p_i)$,
ergo coprime to $q_0 q_1 \dots q_{2012}$,
violating $m \le f(m) \le m+2012$.
\end{proof}
All that is left to do is note that whenever $p \nmid n$,
we have $\gcd(f(p), f(n)) = 1$, hence $p \nmid f(n)$.
This is the contrapositive of the problem statement.
\paragraph{Second solution with a grid.}
Fix $n$ and $p$, and assume for contradiction $p \nmid n$.
\begin{claim*}
There exists a large integer $N$ with $f(N) = N$,
that also satisfies $N \equiv 1 \pmod n$ and $N \equiv 0 \pmod p$.
\end{claim*}
\begin{proof}
We'll need to pick both $N$ and an ancillary integer $M$.
Here is how: pick $2012 \cdot 2013$ distinct primes $q_{i,j} > n+p+2013$
for every $i=1,\dots,2012$ and $j=0,\dots,2012$,
and use it to fill in the following table:
\[
\begin{array}{c|cccc}
& N+1 & N+2 & \dots & N+2012 \\ \hline
M & q_{0,1} & q_{0,2} & \dots & q_{0,2012} \\
M+1 & q_{1,1} & q_{1,2} & \dots & q_{1,2012} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
M+2012 & q_{2012,1} & q_{2012,2} & \dots & q_{2012,2012}
\end{array}.
\]
By the Chinese Remainder Theorem, we can construct $N$
such that $N+1 \equiv 0 \pmod{q_{i,1}}$ for every $i$,
and similarly for $N+2$, and so on.
Moreover, we can also tack on the extra conditions
$N \equiv 0 \pmod p$ and $N \equiv 1 \pmod n$ we wanted.
Notice that $N$ cannot be divisible by any of the $q_{i,j}$'s,
since the $q_{i,j}$'s are greater than $2012$.
After we've chosen $N$, we can pick $M$ such that $M \equiv 0 \pmod{q_{0,j}}$ for every $j$,
and similarly $M+1 \equiv 0 \pmod{q_{1,j}}$, et cetera.
Moreover, we can tack on the condition $M \equiv 1 \pmod N$,
which ensures $\gcd(M,N) = 1$.
What does this do?
We claim that $f(N) = N$ now.
Indeed $f(M)$ and $f(N)$ are relatively prime; but look at the table!
The table tells us that $f(M)$ must have a common factor
with each of $N+1, \dots, N+2012$.
So the only possibility is that $f(N) = N$.
\end{proof}
Now we're basically done.
Since $N \equiv 1 \pmod n$, we have $\gcd(N,n) = 1$
and hence $1 = \gcd(f(N), f(n)) = \gcd(N, f(n))$.
But $p \mid N$ and $p \mid f(n)$, contradiction.
|
sols-TSTST-2012_4
|
In scalene triangle $ABC$, let the feet of the perpendiculars
from $A$ to $\ol{BC}$, $B$ to $\ol{CA}$, $C$ to $\ol{AB}$
be $A_1, B_1, C_1$, respectively.
Denote by $A_2$ the intersection of lines $BC$ and $B_1C_1$.
Define $B_2$ and $C_2$ analogously.
Let $D$, $E$, $F$ be the respective midpoints
of sides $\ol{BC}$, $\ol{CA}$, $\ol{AB}$.
Show that the perpendiculars from $D$ to $\ol{AA_2}$,
$E$ to $\ol{BB_2}$ and $F$ to $\ol{CC_2}$ are concurrent.
|
We claim that they pass through the orthocenter $H$.
Indeed, consider the circle with diameter $\ol{BC}$,
which circumscribes quadrilateral $BCB_1C_1$ and has center $D$.
Then by Brocard theorem, $\ol{AA_2}$ is the polar of line $H$.
Thus $\ol{DH} \perp \ol{AA_2}$.
|
sols-TSTST-2012_5
|
A rational number $x$ is given.
Prove that there exists a sequence
$x_0, x_1, x_2, \dots$ of rational numbers
with the following properties:
\begin{enumerate}[(a)]
\item $x_0=x$;
\item for every $n\ge1$, either $x_n = 2x_{n-1}$ or $x_n = 2x_{n-1} + \frac{1}{n}$;
\item $x_n$ is an integer for some $n$.
\end{enumerate}
|
Think of the sequence as a process over time.
We'll show that:
\begin{claim*}
At any given time $t$,
if the denominator of $x_t$ has some odd prime power $q = p^e$,
then we can delete a factor of $p$ from the denominator,
while only adding powers of two to the denominator.
\end{claim*}
(Thus we can just delete off all the odd primes one by one
and then double appropriately many times.)
\begin{proof}
The idea is to add only fractions of the form $(2^k q)\inv$.
Indeed, let $n$ be large,
and suppose $t < 2^{r+1} q < 2^{r+2} q < \dots < 2^{r+m} q < n$.
For some binary variables $\eps_i \in \{0,1\}$ we can have
\[ x_n = 2^{n-t} x_t
+ c_1 \cdot \frac{\eps_1}{q}
+ c_2 \cdot \frac{\eps_2}{q}
\dots
+ c_s \cdot \frac{\eps_m}{q}
\]
where $c_i$ is some power of $2$
(to be exact, $c_i = \frac{2^{n-2^{r+i}q}}{2^{r+1}}$,
but the exact value doesn't matter).
If $m$ is large enough the set $\{0,c_1\} + \{0,c_2\} + \dots + \{0, c_m\}$
spans everything modulo $p$.
(Actually, Cauchy-Davenport implies $m=p$ is enough,
but one can also just use Pigeonhole to notice some residue appears
more than $p$ times, for $m = O(p^2)$.)
Thus we can eliminate one factor of $p$ from the denominator, as desired.
\end{proof}
|
sols-TSTST-2012_6
|
Positive real numbers $x, y, z$ satisfy $xyz+xy+yz+zx = x+y+z+1$.
Prove that
\[ \frac{1}{3} \left( \sqrt{\frac{1+x^2}{1+x}}
+ \sqrt{\frac{1+y^2}{1+y}}
+ \sqrt{\frac{1+z^2}{1+z}} \right)
\le \left( \frac{x+y+z}{3} \right)^{5/8} . \]
|
The key is the identity
\begin{align*}
\frac{x^2+1}{x+1} &= \frac{(x^2+1)(y+1)(z+1)}{(x+1)(y+1)(z+1)} \\
&= \frac{x(xyz+xy+xz)+x^2+yz+y+z+1}{2(1+x+y+z)} \\
&= \frac{x(x+y+z+1-yz) + x^2+yz+y+z+1}{2(1+x+y+z)} \\
&= \frac{(x+y)(x+z) + x^2 + \left( x-xyz+y+z+1 \right)}{2(1+x+y+z)} \\
&= \frac{2(x+y)(x+z)}{2(1+x+y+z)} \\
&= \frac{(x+y)(x+z)}{1+x+y+z}.
\end{align*}
\begin{remark*}
The ``trick'' can be rephrased as
$(x^2+1)(y+1)(z+1) = 2(x+y)(x+z)$.
\end{remark*}
After this, straight Cauchy in the obvious way will do it
(reducing everything to an inequality in $s = x+y+z$).
One writes
\begin{align*}
\left( \sum_{\text{cyc}} \frac{\sqrt{(x+y)(x+z)}}{\sqrt{1+s}} \right)^2
&\le \frac{\left( \sum_{\text{cyc}} x+y \right)
\left( \sum_{\text{cyc}} x+z \right)}{1+s} \\
&= \frac{4s^2}{1+s}
\end{align*}
and so it suffices to check that $\frac{4s^2}{1+s} \le 9(s/3)^{5/4}$,
which is true because
\[ (s/3)^5 \cdot 9^4 \cdot (1+s)^4 - (4s^2)^4
= s^5 (s-3)^2 (27s^2 + 14s + 3) \ge 0. \]
|
sols-TSTST-2012_7
|
Triangle $ABC$ is inscribed in circle $\Omega$.
The interior angle bisector of angle $A$ intersects side $BC$
and $\Omega$ at $D$ and $L$ (other than $A$), respectively.
Let $M$ be the midpoint of side $BC$.
The circumcircle of triangle $ADM$ intersects sides
$AB$ and $AC$ again at $Q$ and $P$ (other than $A$), respectively.
Let $N$ be the midpoint of segment $PQ$,
and let $H$ be the foot of the perpendicular from $L$ to line $ND$.
Prove that line $ML$ is tangent to the circumcircle of triangle $HMN$.
|
By angle chasing, equivalent to show $\ol{MN} \parallel \ol{AD}$,
so discard the point $H$.
We now present a three solutions.
\paragraph{First solution using vectors.}
We first contend that:
\begin{claim*}
We have $QB = PC$.
\end{claim*}
\begin{proof}
Power of a Point gives $BM \cdot BD = AB \cdot QB$.
Then use the angle bisector theorem.
\end{proof}
Now notice that the vector
\[ \overrightarrow{MN}
= \half \left( \overrightarrow{BQ} + \overrightarrow{CP} \right) \]
which must be parallel to the angle bisector
since $\overrightarrow{BQ}$ and $\overrightarrow{CP}$
have the same magnitude.
\paragraph{Second solution using spiral similarity.}
let $X$ be the arc midpoint of $BAC$.
Then $ADMX$ is cyclic with diameter $\ol{XD}$,
and hence $X$ is the Miquel point of $QBPC$ and thus the midpoint of arc $BAC$.
Moreover $\ol{XND}$ collinear (as $XP=XQ$, $DP=DQ$) on $(APQ)$.
\begin{center}
\begin{asy}
pair A = dir(130);
pair B = dir(210);
pair C = dir(330);
draw(A--B--C--cycle, lightblue);
filldraw(unitcircle, opacity(0.1)+lightblue, blue);
pair L = dir(-90);
pair D = extension(A, L, B, C);
pair M = midpoint(B--C);
path w = circumcircle(A,D,M);
draw(w, dashed+orange);
pair Q = OP(A--B, w);
pair P = IP(A--C, w);
pair N = midpoint(P--Q);
pair H = foot(L, N, D);
pair X = dir(90);
draw(A--L, blue);
draw(circumcircle(H, D, M), lightblue+dotted);
draw(P--Q, blue);
draw(Q--X--C, blue);
draw(N--H, lightblue+dotted);
draw(M--L, lightblue+dotted);
filldraw(X--N--M--cycle, opacity(0.1)+lightgreen, heavygreen);
filldraw(X--P--C--cycle, opacity(0.1)+lightgreen, heavygreen);
filldraw(X--Q--B--cycle, opacity(0.1)+lightgreen, heavygreen);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$L$", L, dir(L));
dot("$D$", D, dir(D));
dot("$M$", M, dir(-90));
dot("$Q$", Q, dir(Q));
dot("$P$", P, dir(10));
dot("$N$", N, dir(130));
dot("$H$", H, dir(H));
dot("$X$", X, dir(X));
/* TSQ Source:
A = dir 130
B = dir 210
C = dir 330
A--B--C--cycle lightblue
unitcircle 0.1 lightblue / blue
L = dir -90
D = extension A L B C
M = midpoint B--C R-90
!path w = circumcircle(A,D,M);
w dashed orange
Q = OP A--B w
P = IP A--C w R10
N = midpoint P--Q R130
H = foot L N D
X = dir 90
A--L blue
circumcircle H D M lightblue dotted
P--Q blue
Q--X--C blue
N--H lightblue dotted
M--L lightblue dotted
X--N--M--cycle 0.1 lightgreen / heavygreen
X--P--C--cycle 0.1 lightgreen / heavygreen
X--Q--B--cycle 0.1 lightgreen / heavygreen
*/
\end{asy}
\end{center}
Then $\triangle XNM \sim \triangle XPC$ spirally, and
\[ \dang XMN = \dang XCP = \dang XCA = \dang XLA \]
thus done.
\paragraph{Third solution using barycentrics (mine).}
Once reduced to $\ol{MN} \parallel \ol{AB}$,
straight bary will also work.
By power of a point one obtains
\begin{align*}
P &= \left( a^2 : 0 : 2b(b+c)-a^2 \right) \\
Q &= \left( a^2 : 2c(b+c)-a^2 : 0 \right) \\
\implies N &= \left( a^2(b+c) : 2bc(b+c) - ba^2 : 2bc(b+c)-ca^2 \right).
\end{align*}
Now the point at infinity along $\ol{AD}$ is $(-(b+c):b:c)$
and so we need only verify
\[
\det \begin{bmatrix}
a^2(b+c) & 2bc(b+c) - ba^2 & 2bc(b+c)-ca^2 \\
0 & 1 & 1 \\
-(b+c) & b & c
\end{bmatrix}
= 0
\]
which follows since the first row is $-a^2$
times the third row plus $2bc(b+c)$ times the second row.
|
sols-TSTST-2012_8
|
Let $n$ be a positive integer.
Consider a triangular array of nonnegative integers as follows:
\begin{center}
\begin{asy}
size(8cm);
defaultpen(fontsize(10pt));
label(scale(0.8)*"Row $1$:", (-6, 2.9));
label(scale(0.8)*"Row $2$:", (-6, 2.2));
label(scale(0.8)*"$\vdots$", (-6, 1.5));
label(scale(0.8)*"Row $n-1$:", (-6, 0.8));
label(scale(0.8)*"Row $n$:", (-6, 0.1));
label("$a_{0,1}$", (0,2.8));
label("$a_{0,2}$", (-1,2.1));
label("$a_{1,2}$", ( 1,2.1));
label(rotate(90)*"$\ddots$", (-2,1.5));
label("$\vdots$", ( 0,1.5));
label("$\ddots$", ( 2,1.5));
label("$a_{0,n-1}$", (-3,0.7));
label("$a_{1,n-1}$", (-1,0.7));
label("$\dots$", (1,0.7));
label("$a_{n-2,n-1}$", (3,0.7));
label("$a_{0,n}$", (-4,0));
label("$a_{1,n}$", (-2,0));
label("$a_{2,n}$", (-0,0));
label("$\dots$", (2,0));
label("$a_{n-1,n}$", (4,0));
\end{asy}
\end{center}
Call such a triangular array \textit{stable} if for every $0 \le i < j < k \le n$ we have
\[ a_{i,j} + a_{j,k} \le a_{i,k} \le a_{i,j} + a_{j,k} + 1. \]
For $s_1$, \dots, $s_n$ any nondecreasing sequence of nonnegative integers,
prove that there exists a unique stable triangular array such that
the sum of all of the entries in row $k$ is equal to $s_k$.
|
Firstly, here are illustrative examples showing the arrays
for $(s_1, s_2, s_3, s_4) = (2, 5, 9, x)$ where $9 \le x \le 14$.
(The array has been left justified.)
\[
\begin{bmatrix}
2 & \swarrow \\
4 & 1 & \swarrow \\
5 & 3 & 1 & \swarrow \\
5 & 3 & 1 & 0
\end{bmatrix}
\begin{bmatrix}
2 & \swarrow \\
4 & 1 & \swarrow \\
5 & 3 & 1 & \swarrow \\
\mathbf{\color{red} 6} & 3 & 1 & 0
\end{bmatrix}
\begin{bmatrix}
2 & \swarrow \\
4 & 1 & \swarrow \\
5 & 3 & 1 & \swarrow \\
6 & 3 & \mathbf{\color{red} 2} & 0
\end{bmatrix}
\]
\[
\begin{bmatrix}
2 & \swarrow \\
4 & 1 & \swarrow \\
5 & 3 & 1 & \swarrow \\
6 & \mathbf{\color{red} 4} & 2 & 0
\end{bmatrix}
\begin{bmatrix}
2 & \swarrow \\
4 & 1 & \swarrow \\
5 & 3 & 1 & \swarrow \\
6 & 4 & 2 & \mathbf{\color{red} 1}
\end{bmatrix}
\begin{bmatrix}
2 & \swarrow \\
4 & 1 & \swarrow \\
5 & 3 & 1 & \swarrow \\
\mathbf{\color{red} 7} & 4 & 2 & 1
\end{bmatrix}
\]
Now we outline the proof.
By induction on $n$, we may assume the first $n-1$ rows are fixed.
Now, let $N = s_n$ vary.
Now, we prove our result by (another) induction on $N \ge s_{n-1}$.
The base case $N = s_{n-1}$ is done by copying
the $n-1$st row and adding a zero at the end.
This is also unique,
since $a_{i,n} \ge a_{i-1,n} + a_{n-1,n}$ for all $i=0, \dots, n-2$,
whence $\sum a_{i,n} \ge s_{n-1}$ follows.
Now the inductive step is based on the following lemma,
which illustrates the idea of a ``unique increasable entry''.
\begin{lemma*}
Fix a stable array
Construct a tournament on the $n$ entries
of the last row as follows: for $i < j$,
\begin{itemize}
\ii $a_{i,n} \to a_{j,n}$
if $a_{i,n} = a_{i,j} + a_{j,n}$, and
\ii $a_{j,n} \to a_{i,n}$
if $a_{i,n} = a_{i,j} + a_{j,n} + 1$.
\end{itemize}
Then this tournament is transitive.
Also, except for $N = s_{n-1}$, a $0$ entry is never a source.
\end{lemma*}
Intuitively, $a_{i,n} \to a_{j,n}$
if $a_{i,n}$ blocks $a_{j,n}$ from increasing.
For instance, in the example
\[
\begin{bmatrix}
2 & \swarrow \\
4 & 1 & \swarrow \\
5 & 3 & 1 & \swarrow \\
\mathbf 6 & 3 & 1 & 0
\end{bmatrix}
\]
the tournament is $1 \to 3 \to 0 \to 6$.
\begin{proof}
[Proof of lemma]
Let $0 \le i < j < k < n$ be indices.
Let $x = a_{i,n}$, $y = a_{j,n}$, $z = a_{k,n}$,
$p = a_{i,j}$, $s = a_{i,k}$, $q = a_{j,k}$. Picture:
\[
\begin{bmatrix}
p & \swarrow \\
s & q & \swarrow \\
x & y & z
\end{bmatrix}
\]
If $x \to y \to z \to x$ happens, that means
$x = y+p$, $y = q+z$, $x = s+z+1$,
which gives $s=p+q-1$, contradiction.
Similarly if $x \leftarrow y \leftarrow z \leftarrow x$ then
$x = y+p+1$, $y = q+z+1$, $x = s+z$,
which gives $s=p+q+2$, also contradiction.
\end{proof}
Now this allows us to perform our induction.
Indeed, to show existence from $N$ to $N+1$
we take a source of the tournament above and increase it.
Conversely, to show uniqueness for $N$,
note that we can take the (nonzero) sink of the tournament and decrement it,
which gives $N-1$; our uniqueness inductive hypothesis now finishes.
\begin{remark*}
Colin Tang found a nice proof of uniqueness:
\[
s_k + \sum_{i=1}^{k-1} a_{0,i} \le ka_{0,k}
\le s_k + \sum_{i=1} \left( a_{0,i} + 1 \right)
\]
and similarly for other entries.
\end{remark*}
|
sols-TSTST-2012_9
|
Given a set $S$ of $n$ variables, a binary operation $\times$ on $S$ is called
\textit{simple} if it satisfies $(x \times y) \times z = x \times (y \times z)$
for all $x,y,z \in S$ and $x \times y \in \{x,y\}$ for all $x,y \in S$.
Given a simple operation $\times$ on $S$, any string of elements in $S$
can be reduced to a single element, such as $xyz \to x \times (y \times z)$.
A string of variables in $S$ is called \textit{full}
if it contains each variable in $S$ at least once,
and two strings are \textit{equivalent} if they evaluate to the
same variable regardless of which simple $\times$ is chosen.
For example $xxx$, $xx$, and $x$ are equivalent, but these are only full if $n=1$.
Suppose $T$ is a set of full strings such that any full string is
equivalent to exactly one element of $T$. Determine the number of elements of $T$.
\end{enumerate}
|
The answer is $(n!)^2$.
In fact it is possible to essentially find all $\times$:
one assigns a real number to each variable in $S$.
Then $x \times y$ takes the larger of $\{x,y\}$,
and in the event of a tie picks either ``left'' or ``right'',
where the choice of side is fixed among elements of each size.
\paragraph{First solution (Steven Hao).}
The main trick is the two lemmas, which are not hard to show
(and are motivated by our conjecture).
\begin{align*}
xx &= x \\
xyxzx &= xyzx.
\end{align*}
Consequently, define a \textbf{double rainbow}
to be the concatenation of two full strings of length $n$,
of which there are $(n!)^2$.
We claim that these form equivalence classes for $T$.
To see that any string $s$ is equivalent to a double rainbow,
note that $s = ss$, and hence using the second identity above
repeatedly lets us reduce $ss$ to a double rainbow.
To see two distinct double rainbows $R_1$ and $R_2$ aren't equivalent,
one can use the construction mentioned in the beginning.
Specifically, take two variables $a$ and $b$ which do not
appear in the same order in $R_1$ and $R_2$.
Then it's not hard to see that $abab$, $abba$, $baab$, $baba$
are pairwise non-equivalent by choosing ``left'' or ``right'' appropriately.
Now construct $\times$ on the whole set
by having $a$ and $b$ be the largest variables,
so the rest of the variables don't matter in the evaluation of the string.
\paragraph{Second solution outline (Ankan Bhattacharya).}
We outline a proof of the characterization claimed earlier,
which will also give the answer $(n!)^2$.
We say $a \sim b$ if $ab \neq ba$.
Also, say $a > b$ if $ab = ba = a$.
The following are proved by finite casework,
using the fact that $\{ab, bc, ca\}$ always has exactly
two distinct elements for any different $a$, $b$, $c$.
\begin{itemize}
\ii If $a > b$ and $b > c$ then $a > c$.
\ii If $a \sim b$ and $b \sim c$
then $ab = a$ if and only if $bc = b$.
\ii If $a \sim b$ and $b \sim c$ then $a \sim c$.
\ii If $a \sim b$ and $a > c$ then $b > c$.
\ii If $a \sim b$ and $c > a$ then $c > b$.
\end{itemize}
This gives us the total ordering on the elements
and the equivalence classes by $\sim$.
In this we way can check the claimed operations are the only ones.
We can then (as in the first solution) verify that every full string
is equivalent to a unique double rainbow --- but this
time we prove it by simply considering all possible $\times$,
because we have classified them all.
|
sols-TSTST-2013_1
|
Let $ABC$ be a triangle and $D$, $E$, $F$ be the midpoints of arcs $BC$, $CA$, $AB$ on the circumcircle.
Line $\ell_a$ passes through the feet of the perpendiculars
from $A$ to $\ol{DB}$ and $\ol{DC}$.
Line $m_a$ passes through the feet of the perpendiculars from $D$ to $\ol{AB}$ and $\ol{AC}$.
Let $A_1$ denote the intersection of lines $\ell_a$ and $m_a$.
Define points $B_1$ and $C_1$ similarly.
Prove that triangles $DEF$ and $A_1B_1C_1$ are similar to each other.
|
In fact, it is true for any points $D$, $E$, $F$ on the circumcircle.
More strongly we contend:
\begin{claim*}
Point $A_1$ is the midpoint of $\ol{HD}$.
\end{claim*}
\begin{proof}
Lines $m_a$ and $\ell_a$ are Simson lines,
so they both pass through the point $(a+b+c+d)/2$
in complex coordinates.
\end{proof}
\begin{center}
\begin{asy}
pair A = dir(130);
pair B = dir(200);
pair C = dir(-20);
pair D = dir(-90);
pair P1 = foot(A,B,D);
pair P2 = foot(A,C,D);
pair Q1 = foot(D,A,B);
pair Q2 = foot(D,A,C);
pair H = A+B+C;
draw(H--D, deepgreen);
draw(unitcircle);
filldraw(A--B--C--cycle, lightcyan + opacity(0.1), blue);
draw(B--D--C, blue);
draw(P1--A--P2, dashed+red);
draw(Q1--D--Q2, dashed+orange);
draw(P1--P2, red);
draw(Q1--Q2, orange);
draw(P1--B, blue);
draw(Q1--B, blue);
dot(Q1, orange);
dot(Q2, orange);
dot(P1, red);
dot(P2, red);
dot("$A_1$", (A+B+C+D)/2, dir(-90));
dot("$A$", A, A);
dot("$B$", B, B);
dot("$C$", C, C);
dot("$D$", D, D);
dot("$H$", H, dir(90));
\end{asy}
\end{center}
Hence $A_1 B_1 C_1$ is similar to $DEF$ through a homothety at $H$ with
ratio $\half$.
|
sols-TSTST-2013_2
|
A finite sequence of integers $a_1, a_2, \dots, a_n$ is called
\emph{regular} if there exists a real number $x$ satisfying
\[ \left\lfloor kx \right\rfloor = a_k \quad \text{for } 1 \le k \le n. \]
Given a regular sequence $a_1, a_2, \dots, a_n$, for $1 \le k \le n$ we say that
the term $a_k$ is \emph{forced} if the following condition is satisfied:
the sequence \[ a_1, \; a_2, \; \dots, \; a_{k-1}, \; b \]
is regular if and only if $b = a_k$.
Find the maximum possible number of forced terms
in a regular sequence with $1000$ terms.
|
The answer is $985$. WLOG, by shifting $a_1 = 0$ (clearly $a_1$ isn't forced).
Now, we construct regular sequences inductively using the following procedure.
Start with the inequality \[ \tfrac01 \le x < \tfrac11. \]
Then for each $k = 2, 3, \dots, 1000$ we perform the following procedure.
If there is no fraction of the form $F = \frac mk$ in the interval $A \le x < B$,
then $a_k$ is forced,
and the interval of possible $x$ values does not change.
Otherwise, $a_k$ is not forced,
and we pick a value of $a_k$ and update the interval accordingly.
The theory of \href{https://en.wikipedia.org/wiki/Farey_sequence}{Farey sequences}
tells us that when we have a stage $\tfrac ab \le x < \tfrac cd$
then the next time we will find a fraction
in that interval is exactly $\tfrac{a+c}{b+d}$ (at time $k=b+d$),
and it will be the only such fraction.
So essentially, starting with $\tfrac 01 \le x < \tfrac 11$
we repeatedly replace one of the endpoints of the intervals with the mediant,
until one of the denominators exceeds $1000$;
we are trying to minimize the number of non-forced terms,
which is the number of denominators that appear in this process.
It is not hard to see that this optimum occurs by always replacing the smaller of the denominators,
so that the sequence is
\begin{align*}
\tfrac{0}{1} &\le x < \tfrac{1}{1} \\
\tfrac{0}{1} &\le x < \tfrac{1}{2} \\
\tfrac{1}{3} &\le x < \tfrac{1}{2} \\
\tfrac{1}{3} &\le x < \tfrac{2}{5} \\
\tfrac{3}{8} &\le x < \tfrac{2}{5} \\
\tfrac{3}{8} &\le x < \tfrac{5}{13}
\end{align*}
and so on; we see that the non-forced terms in this optimal configuration
are exactly the Fibonacci numbers.
There are $15$ Fibonacci numbers less than $1000$,
hence the answer $1000 - 15 = 985$.
|
sols-TSTST-2013_3
|
Divide the plane into an infinite square grid by drawing
all the lines $x=m$ and $y=n$ for $m,n \in \ZZ$.
Next, if a square's upper-right corner has both coordinates even, color it black;
otherwise, color it white (in this way, exactly $1/4$ of the squares are black
and no two black squares are adjacent).
Let $r$ and $s$ be odd integers,
and let $(x,y)$ be a point in the interior of any white square such that $rx-sy$ is irrational.
Shoot a laser out of this point with slope $r/s$;
lasers pass through white squares and reflect off black squares.
Prove that the path of this laser will from a closed loop.
|
Here is Sammy Luo's solution.
Fix the speed of light at $\sqrt{r^2+s^2}$ units per second.
We prove periodicity every six seconds.
We re-color the white squares as red, blue, or green
according as to whether they have a black square directly
to the left/right, above/below, or neither, as shown below.
Finally, we fix time zero to be a moment just before the laser passes
a horizontal (WLOG) lattice line (not necessarily a wall).
Shown below is an example for $(r,s) = (3,5)$.
\begin{center}
\begin{asy}
size(11cm);
int Mx = 3;
int My = 6;
for (int i=0; i<=Mx; ++i) {
for (int j=0; j<=My; ++j) {
fill(shift(2*i, 2*j) * unitsquare, opacity(0.1)+lightgreen);
if (i != Mx) { fill(shift(2*i+1, 2*j) * unitsquare, opacity(0.1)+lightred); }
if (j != My) { fill(shift(2*i, 2*j+1) * unitsquare, opacity(0.1)+lightblue); }
if ((i != Mx) && (j != My)) { fill(shift(2*i+1, 2*j+1) * unitsquare, black); }
}
}
pair ne_until_y(pair O, real y) {
return extension(O, O+(3,5), (0,y), (1,y) );
}
pair ne_until_x(pair O, real x) {
return extension(O, O+(3,5), (x,0), (x,1) );
}
pair se_until_y(pair O, real y) {
return extension(O, O+(-3,5), (0,y), (1,y) );
}
pair se_until_x(pair O, real x) {
return extension(O, O+(-3,5), (x,0), (x,1) );
}
pair[] P = new pair[10];
P[0] = (2.0, 0.84);
P[1] = ne_until_y(P[0], 3);
P[2] = se_until_y(P[1], 2);
P[3] = ne_until_x(P[2], 5);
P[4] = se_until_x(P[3], 4);
P[5] = ne_until_x(P[4], 5);
P[6] = se_until_y(P[5], 9);
P[7] = ne_until_y(P[6], 8);
P[8] = se_until_y(P[7], 11);
P[9] = ne_until_x(P[8], 1);
for (int i=0; i<=8; ++i) { draw(P[i]--P[i+1], orange); }
transform t = yscale(-1)*rotate(-90);
pair[] Q = new pair[8];
Q[0] = t*P[0];
Q[1] = t*ne_until_y(P[0], 1);
Q[2] = t*ne_until_y(P[0], 2);
Q[3] = t*ne_until_x(P[0], 3);
Q[4] = t*ne_until_y(P[0], 3); // = P[1]
Q[5] = t*se_until_y(P[1], 2);
Q[6] = t*ne_until_x(P[2], 4);
Q[7] = t*ne_until_y(P[2], 3);
// for (int i=0; i<=7; ++i) { dot(Q[i], lightblue); }
picture rotated = t*currentpicture;
currentpicture.erase();
add(rotated);
dotfactor *= 0.5;
defaultpen(fontsize(6pt));
dot("$h$", Q[0], dir(-90), red);
dot("$v$", Q[1], dir(90), orange);
dot("$v$", Q[2], dir(-45), orange);
dot("$h$", Q[3], dir(-45), orange);
dot("$v$", Q[4], dir(180)*2, orange);
dot("$v$", Q[5], dir(105)*2, orange);
dot("$h$", Q[6], dir(90), orange);
dot("$v$", Q[7], dir(90), orange);
dot(t*P[3], red);
dot(t*se_until_x(P[5], 4), red);
dot(t*P[9], red);
\end{asy}
\end{center}
The main idea is to keep track of every time the laser passes
a lattice line (again, not necessarily a wall).
There are four possible types of events:
\begin{itemize}
\ii A horizontal $h$ event where the laser switches from red to green (or vice-versa);
\ii A horizontal $h$ event where the laser rebounds off a wall,
remaining in a blue square, but flips the $x$-component of its velocity;
\ii A vertical $v$ event where the laser switches from blue to green (or vice-versa)
\ii A vertical $v$ event where the laser rebounds off a wall,
remaining in a red square, but flips the $y$-component of its velocity.
\end{itemize}
The first key observation is that:
\begin{claim*}
In the first second, the laser will encounter exactly
$r$ horizontal events and $s$ vertical events.
In every second after that, the same sequence of $r+s$ events occurs.
\end{claim*}
\begin{proof}
Bouncing off a wall doesn't change this
as opposed to if the laser had passed through the wall.
\end{proof}
We let the \emph{key-word} be the sequence $w$ of $r+s$ letters corresponding to the sequence.
For example, the picture above denotes an example with keyword $w = hvvhvvhv$;
so no matter what, every second, the laser will encounter eight lattice lines,
which are horizontal and vertical in that order.
\begin{claim*}
Color is periodic every $3$ seconds.
\end{claim*}
\begin{proof}
The free group generated by $h$ and $v$
acts on the set $\{R,G,B\}$ of colors in an obvious way; consider this right action.
First we consider the color of the square after each second.
Note that with respect to color, each letter is an involution;
so as far as color changes are concerned,
it's enough to work with the reduced word $w'$ obtained by modding out by $h^2 = 1$ and $v^2 = 1$.
(For example, $w' = hv$ in our example.)
In general, $w' = (hv)^k$ or $w' = (vh)^k$,
for some odd integer $k$ (since $k \equiv r \equiv s \equiv 1 \pmod 2$).
Now we see that the action of $hv$
on the set of colors is
$\text{red} \mapsto \text{blue} \mapsto \text{green} \mapsto \text{red}$,
and similarly for $vh$ (being the inverse).
This implies that the color is periodic every three seconds.
\end{proof}
Now in a 3-second period,
consider the $3r$ horizontal events and $3s$ vertical events (both are odd).
In order for the color to remain the same
(as the only color changes are $R \leftrightarrow G$ for $h$ and $B \leftrightarrow G$ for $v$)
there must have been an even number of color swaps for each orientation.
Therefore there was an odd number of wall collisions of each orientation.
So, the laser is pointing in the opposite direction at the end of 3 seconds.
Finally, let $x_t$ be the fractional part of the $x$
coordinate after $t$ seconds
(the $y$-coordinate is always zero by our setup at these moments).
Note that
\[ x_{t+1} =
\begin{cases}
x_t & \text{even number of vertical wall collisions} \\
1 - x_t & \text{odd numbers of vertical wall collisions}
\end{cases}
\]
Since over the three seconds there were an odd number of vertical collisions;
it follows $x_3 = 1 - x_0$.
Thus at the end of three seconds, the laser is in a symmetric position from the start;
and in $6$ seconds it will form a closed loop.
|
sols-TSTST-2013_4
|
Circle $\omega$, centered at $X$, is internally tangent to circle $\Omega$,
centered at $Y$, at $T$.
Let $P$ and $S$ be variable points on $\Omega$ and $\omega$,
respectively, such that line $PS$ is tangent to $\omega$ (at $S$).
Determine the locus of $O$ --- the circumcenter of triangle $PST$.
|
The answer is a circle centered at $Y$
with radius $\sqrt{YX \cdot YT}$,
minus the two points on line $XY$ itself.
We let $PS$ meet $\Omega$ again at $P'$,
and let $O'$ be the circumcenter of $\triangle TPS'$.
Note that $O'$, $X$, $O$ are collinear on the perpendicular
bisector of line $\ol{TS}$
Finally, we let $M$ denote the arc midpoint of $PP'$
which lies on line $TS$ (by homothety).
\begin{center}
\begin{asy}
import graph; size(9cm);
pen qqffff = rgb(0.,1.,1.); pen qqwuqq = rgb(0.,0.39215,0.); pen ffdxqq = rgb(1.,0.84313,0.); pen yqqqqq = rgb(0.50196,0.,0.);
pair X = (-2.,0.), Y = (2.,0.), T = (-9.50185,0.), P = (7.49229,10.10580), O = (5.46686,-5.82994), M = (-0.78960,11.15843);
pair S = (-3.81946,7.27786);
pair Pp = (-7.60181,6.33228);
pair Op = (-4.49671,1.94937);
fill(T--S--P--cycle, palegreen);
fill(T--Op--Y--cycle, palegreen);
fill(T--S--Pp--cycle, palecyan);
fill(T--O--Y--cycle, palecyan);
draw(circle(X, 7.50185), linewidth(0.6) + red);
draw(circle(Y, 11.50185), linewidth(0.6) + red);
draw(circle(Y, 6.78287), linewidth(0.6) + qqwuqq);
draw(T--Op--Y--O--T--Y, gray);
draw(T--Pp--M--P--T--M, black);
draw(Pp--P, black);
draw(Op--O, gray);
dot("$X$", X, dir((8.005, 21.422)));
dot("$Y$", Y, dir((8.687, 21.422)));
dot("$T$", T, dir(T-Y));
dot("$S$", S, dir(135));
dot("$P$", P, dir(P-Y));
dot("$O$", O, dir(-45));
dot("$P'$", Pp, dir(Pp-Y));
dot("$O'$", Op, dir(40));
dot("$M$", M, dir((8.912, 22.013)));
\end{asy}
\end{center}
By three applications of Salmon theorem, we have the following spiral
similarities all centered at $T$:
\begin{align*}
\triangle TSP &\overset{+}{\sim} \triangle TO'Y \\
\triangle TP'S &\overset{+}{\sim} \triangle TYO \\
\triangle TP'P &\overset{+}{\sim} \triangle TO'O.
\end{align*}
However, the shooting lemma also gives us two similarities:
\begin{align*}
\triangle TP'M &\overset{+}{\sim} \triangle TSP \\
\triangle TMP &\overset{+}{\sim} \triangle TP'S.
\end{align*}
Putting everything together, we find that
\[ TP'MP \overset{+}{\sim} TO'YO. \]
Then by shooting lemma, $YO'^2 = YX \cdot YT$, so $O$
indeed lies on the claimed circle.
As the line $\ol{O'O}$ may be any line through $X$ other than line $XY$
(one takes $S$ to be the reflection of $T$ across this line)
one concludes the only two non-achievable points are
the diametrically opposite ones on line $XY$ of this circle
(because this leads to the only degenerate situation where $S=T$).
|
sols-TSTST-2013_5
|
Let $p$ be a prime.
Prove that in a complete graph with $1000p$ vertices
whose edges are labelled with integers,
one can find a cycle whose sum of labels is divisible by $p$.
|
Select $p-1$ disjoint triangles arbitrarily. If any of these triangles have $0$ sum modulo $p$ we are done. Otherwise, we may label the vertices $u_i$, $x_i$, and $v_i$ (where $1 \le i \le p-1$) in such a way that $u_ix_i + x_iv_i \neq u_iv_i$.
Let $A_i = \left\{ u_ix_i+x_iv_i, u_iv_i \right\}$. We can show that $\left\lvert A_1 + A_2 + \dots + A_t \right\rvert \ge \min \left\{ p,t+1 \right\}$ for each $1 \le t \le p-1$, by using induction on $t$ alongside Cauchy-Davenport. So, $A_1 + A_2 + \dots + A_{p-1}$ spans all of $\ZZ_p$. All that's left to do is join the triangles together to form a cycle, and then delete either $u_ix_i,x_iv_i$ or $u_iv_i$ from each triangle in such a way that the final sum is $0$ mod $p$.
|
sols-TSTST-2013_6
|
Let $\NN$ be the set of positive integers.
Find all functions $f \colon \NN \to \NN$ that satisfy the equation
\[ f^{abc-a}(abc) + f^{abc-b}(abc) + f^{abc-c}(abc) = a + b + c \]
for all $a,b,c \ge 2$. (Here $f^k$ means $f$ applied $k$ times.)
|
The answer is $f(n) = n-1$ for $n \ge 3$ with $f(1)$ and $f(2)$ arbitrary;
check these work.
\begin{lemma*}
We have $f^{t^2-t}(t^2) = t$ for all $t$.
\end{lemma*}
\begin{proof}
We say $1 \le k \le 8$ is good if $f^{t^9-t^k}(t^9) = t^k$ for all $t$.
First, we observe that
\[ f^{t^9-t^3}(t^9) = t^3 \quad\text{and}\quad f^{t^3-t}(t^3) = t
\implies f^{t^9-t}(t^9) = t. \]
so $k=1$ and $k=3$ are good.
Then taking $(a,b,c) = (t,t^4,t^4)$,
$(a,b,c) = (t^2, t^3, t^4)$
gives that $k = 4$ and $k = 2$ are good, respectively.
% $(a,b,c) = (t,t^3,t^5)$, gives $5$ good
% Similarly, $(a,b,c) = (t,t^2,t^6)$ and $(a,b,c) = (t,t,t^7)$
% gives $6$ and $7$ are good.
The lemma follows from this $k=1$ and $k=2$ being good.
\end{proof}
Now, letting $t = abc$ we combine
\begin{align*}
f^{t-a}(t) + f^{t-b}(t) + f^{t-c}(t) &= a+b+c \\
f^{t^2-ab} (t^2) + f^{t^2-t} (t^2) + f^{t^2-c} (t^2) &= ab + t + c \\
\implies \left[ f^{t-a}(t) - a \right]
+ \left[ f^{t-b}(t) - b \right]
&=
\left[ f^{t-ab}(t) - ab \right]
\end{align*}
by subtracting and applying the lemma repeatedly.
In other words, we have proven the second lemma:
\begin{lemma*}
Let $t$ be fixed, and define $g_t(n) = f^{t-n}(t) - n$ for $n < t$.
If $a,b \ge 2$ and $ab \mid t$, $ab < t$, then $g_t(a) + g_t(b) = g_t(ab)$.
\end{lemma*}
Now let $a,b \ge 2$ be arbitrary, and let $p > q > \max\{a,b\}$ be primes.
Suppose $s = a^p b^q$ and $t = s^2$; then
\[ p g_t(a) + q g_t(b)
= g_t\left( a^p b^q \right) = g_t(s)
= f^{s^2-s}(s^2) - s = 0. \]
Now \[ q \mid g_t(a) > -a
\quad\text{and}\quad p \mid g_t(b) > -b
\implies g_t(a) = g_t(b) = 0. \]
and so we conclude $f^{t-a}(t) = a$ and $f^{t-b}(t) = b$ for $a,b \ge 2$.
In particular, if $a = n$ and $b = n+1$
then we deduce $f(n+1) = n$ for all $n \ge 2$, as desired.
\begin{remark*}
If you let $c = (ab)^2$ after the first lemma, you recover the $2$-variable version!
\end{remark*}
|
sols-TSTST-2013_7
|
A country has $n$ cities, labelled $1,2,3,\dots,n$.
It wants to build exactly $n-1$ roads between certain pairs of cities
so that every city is reachable from every other city via some sequence of roads.
However, it is not permitted to put roads between pairs of cities
that have labels differing by exactly $1$,
and it is also not permitted to put a road between cities $1$ and $n$.
Let $T_n$ be the total number of possible ways to build these roads.
\begin{enumerate}[(a)]
\ii For all odd $n$, prove that $T_n$ is divisible by $n$.
\ii For all even $n$, prove that $T_n$ is divisible by $n/2$.
\end{enumerate}
|
You can just spin the tree!
Fixing $n$, the group $G = \ZZ/n\ZZ$ acts on the set of trees
by rotation (where we imagine placing $1,2,\dots,n$ along a circle).
\begin{claim*}
For odd $n$, all trees have trivial stabilizer.
\end{claim*}
\begin{proof}
One way to see this is to look at the degree sequence.
Suppose $g^e$ fixes a tree $T$.
Then so does $g^k$, for $k = \gcd(e,n)$.
Then it follows that $n/k$ divides $\sum_v \deg v = 2n-2$.
Since $\gcd(2n-2, n) = 1$ we must then have $k = n$.
\end{proof}
The proof for even $n$ is identical except that $\gcd(2n-2, n) = 2$
and hence each tree either has stabilizer with size $\le 2$.
There is also a proof using linear algebra,
using Kirchoff's tree formula. (Overkill.)
|
sols-TSTST-2013_8
|
Define a function $f \colon \NN \to \NN$ by $f(1) = 1$,
$f(n+1) = f(n) + 2^{f(n)}$ for every positive integer $n$.
Prove that $f(1)$, $f(2)$, \dots, $f(3^{2013})$
leave distinct remainders when divided by $3^{2013}$.
|
I'll prove by induction on $k \ge 1$ that any $3^k$
consecutive values of $f$ produce distinct residues modulo $3^k$.
The base case $k=1$ is easily checked
($f$ is always odd, hence $f$ cycles $1$, $0$, $2$ mod $3$).
For the inductive step, assume it's true for $k$.
Note that $2^\bullet \pmod{3^{k+1}}$ cycles every $2 \cdot 3^k$
so we look at the outputs of $f$ modulo $2 \cdot 3^k$.
\begin{claim*}
We have
\[ \{ f(n), f(n+1), \dots, f(n+3^k-1) \} \pmod{2 \cdot 3^k} \]
is a permutation
\[ \{ 1, 3, \dots, 2 \cdot 3^k-1 \} \pmod{2 \cdot 3^k}. \]
\end{claim*}
\begin{proof}
The $f$-values were given to be distinct modulo $3^k$, and $f$ is always odd.
\end{proof}
Thus, we conclude that
\[ \{ 2^{f(n)}, 2^{f(n+1)}, \dots, 2^{f(n+3^k-1)}\} \pmod{3^{k+1}} \]
is a permutation of
\[ \{ 2^{1}, 2^{3}, \dots, 2^{2 \cdot 3^k - 1}\} \pmod{3^{k+1}} \]
and so
\begin{align*}
f(n+3^k) - f(n)
&= 2^{f(n)} + 2^{f(n+1)} + \dots + 2^{f(n+3^k-1)} \pmod{3^{k+1}} \\
&\equiv 2^1 + 2^3 + \dots + 2^{2 \cdot 3^k-1} \pmod{3^{k+1}} \\
&= 2 \cdot \frac{4^{3^k}-1}{4-1}.
\end{align*}
Hence
\[
f(n+3^k)-f(n) \equiv C \pmod{3^{k+1}}
\qquad\text{where} \qquad C = 2 \cdot \frac{4^{3^k}-1}{4-1}
\]
noting that $C$ does not depend on $n$. Exponent lifting gives $\nu_3(C) = k$
hence $f(n)$, $f(n+3^k)$, $f(n+2\cdot3^k)$ differ mod $3^{k+1}$ for all $n$,
and the inductive hypothesis now solves the problem.
|
sols-TSTST-2013_9
|
Let $r$ be a rational number in the interval $[-1,1]$
and let $\theta = \cos^{-1} r$.
Call a subset $S$ of the plane good if $S$ is unchanged
upon rotation by $\theta$ around any point of $S$
(in both clockwise and counterclockwise directions).
Determine all values of $r$ satisfying the following property:
The midpoint of any two points in a good set also lies in the set.
\end{enumerate}
|
The answer is that $r$ has this property
if and only if $r = \frac{4n-1}{4n}$ for some integer $n$.
Throughout the solution, we will let $r = \frac ab$ with $b > 0$ and
$\gcd(a,b) = 1$. We also let
\[ \omega = e^{i\theta} = \frac ab \pm \frac{\sqrt{b^2-a^2}}{b} i. \]
This means we may work with complex multiplication in the usual way;
the rotation of $z$ through center $c$ is given by $z \mapsto \omega(z-c)+c$.
For most of our proof, we start by constructing a good set as follows.
\begin{itemize}
\ii Start by letting $S_0 = \{0, 1\}$.
\ii Let $S_i$ consist of $S_{i-1}$ plus
all points that can be obtained by rotating a point of $S_{i-1}$
through a different point of $S_{i-1}$ (with scale factor $\omega$).
\ii Let $S_\infty = \bigcup_{i \ge 0} S_i$.
\end{itemize}
The set $S_\infty$ is the (minimal, by inclusion) good set containing
$0$ and $1$.
We are going to show that for most values of $r$,
we have $\half \notin S_\infty$.
\begin{claim*}
If $b$ is odd, then $\half \notin S_\infty$.
\end{claim*}
\begin{proof}
Idea: denominators that appear are always odd.
Consider the ring
\[ A = \ZZ_{\{b\}} = \left\{ \frac{s}{t} \mid s,t \in \ZZ, t \mid b^\infty \right\} \]
which consists of all rational numbers
whose denominators divide $b^\infty$.
Then, $0, 1, \omega \in A[\sqrt{b^2-a^2}]$
and hence $S_\infty \subseteq A[\sqrt{b^2-a^2}]$ too.
(This works even if $\sqrt{b^2-a^2} \in \ZZ$,
in which case $S_\infty \subseteq A = A[\sqrt{b^2-a^2}]$.)
But $\half \notin A[\sqrt{b^2-a^2}]$.
\end{proof}
\begin{claim*}
If $b$ is even and $|b-a| \neq 1$, then $\half \notin S_\infty$.
\end{claim*}
\begin{proof}
Idea: take modulo a prime dividing $b-a$.
Let $D = b^2-a^2 \equiv 3 \pmod 4$.
Let $p$ be a prime divisor of $b-a$.
Because $\gcd(a,b) = 1$, we have $p \neq 2$ and $p \nmid b$.
Consider the ring
\[ A = \ZZ_{(p)} = \left\{ \frac{s}{t} \mid s,t \in \ZZ, p \perp t \right\} \]
which consists of all rational numbers whose denominators are coprime to $p$.
Then, $0, 1, \omega \in A[\sqrt{-D}]$ and hence $S_\infty \subseteq A[\sqrt{-D}]$ too.
Now, there is a well-defined ``mod-$p$'' ring homomorphism
\[ \Psi \colon A[\sqrt{-D}] \to \FF_p \quad\text{by}\quad
x+y\sqrt{-D} \mapsto x \bmod p \]
which commutes with addition and multiplication (as $p \mid D$).
Under this map,
\[ \omega \mapsto \frac ab \bmod p = 1. \]
Consequently, the rotation $z \mapsto \omega(z-c)+c$
is just the identity map modulo $p$.
In other words, the pre-image of any point in $S_\infty$
under $\Psi$ must be either $\Psi(0) = 0$ or $\Psi(1) = 1$.
However, $\Psi(1/2) = 1/2$ is neither of these. So this point cannot be achieved.
\end{proof}
\begin{claim*}
Suppose $a = 2n-1$ and $b=2n$ for $n$ an odd integer.
Then $\half \notin S_\infty$
\end{claim*}
\begin{proof}
Idea: $\omega$ is ``algebraic integer'' sans odd denominators.
This time, we define the ring
\[ B = \ZZ_{(2)} = \left\{ \frac{s}{t} \mid s,t \in \ZZ, t \text{ odd} \right\} \]
of rational numbers with odd denominator.
We carefully consider the ring $B[\omega]$ where
\[ \omega = \frac{2n-1 \pm \sqrt{1-4n}}{2n}. \]
So $S_\infty \subseteq B[\omega]$ as $0, 1, \omega \in B[\omega]$.
I claim that $B[\omega]$ is an integral extension of $B$;
equivalently that $\omega$ is integral over $B$.
Indeed, $\omega$ is the root of the monic polynomial
\[ (T-1)^2 + \frac1n (T-1) - \frac 1n = 0 \]
where $\frac1n \in B$ makes sense as $n$ is odd.
On the other hand, $\half$ is not integral over $B$
so it is not an element of $B[\omega]$.
\end{proof}
It remains to show that if $r = \frac{4n-1}{4n}$, then goods sets
satisfy the midpoint property.
Again starting from the points $z_0 = 0$, $z_1 = 1$ construct the sequence
\begin{align*}
z_2 &= \omega(z_1-z_0) + z_0 \\
z_3 &= \omega\inv(z_0-z_2) + z_2 \\
z_4 &= \omega\inv(z_2-z_3) + z_3 \\
z_5 &= \omega(z_3-z_4) + z_4
\end{align*}
as shown in the diagram below.
\begin{center}
\begin{asy}
size(8cm);
pair z0 = (0,0);
pair z1 = (1,0);
real theta = 28.96;
pair z2 = rotate(theta, z0) * z1;
pair z3 = rotate(-theta, z2) * z0;
pair z4 = rotate(-theta, z3) * z2;
pair z5 = rotate(theta, z4) * z3;
draw(arc(z0, z1, z2), lightgreen);
draw(arc(z3, z4, z2), lightgreen);
draw(arc(z2, z3, z0), lightgreen);
draw(arc(z4, z3, z5), lightgreen);
draw(z1--z0--z2--z3--z4--z5, deepcyan);
dot("$z_0=0$", z0, dir(-45), red);
dot("$z_1=1$", z1, dir(-45), red);
dot("$z_2$", z2, dir(90));
dot("$z_3$", z3, dir(90));
dot("$z_4=2r-1$", z4, dir(-90));
dot("$z_5=2r-2$", z5, dir(-90));
\end{asy}
\end{center}
This construction shows that if we have the length-one segment
$\{0,1\}$ then we can construct the length-one segment $\{2r-2, 2r-1\}$.
In other words, we can shift the segment to the left by
\[ 1-(2r-1) = 2(1-r) = \frac{1}{2n}. \]
Repeating this construction $n$ times gives the desired midpoint $\half$.
|
sols-TSTST-2014_1
|
Let $\leftarrow$ denote the left arrow key on a standard keyboard. If
one opens a text editor and types the keys
``ab$\leftarrow$ cd $\leftarrow \leftarrow$ e $\leftarrow \leftarrow$ f'',
the result is ``faecdb''.
We say that a string $B$ is \emph{reachable} from a string $A$ if
it is possible to insert some amount of $\leftarrow$'s in $A$,
such that typing the resulting characters produces $B$.
So, our example shows that ``faecdb'' is reachable from ``abcdef''.
Prove that for any two strings $A$ and $B$,
$A$ is reachable from $B$ if and only if $B$ is reachable from $A$.
|
Obviously $A$ and $B$ should have the
same multiset of characters, and we focus only on that situation.
\begin{claim*}
If $A = 123 \dots n$
and $B = \sigma(1) \sigma(2) \dots \sigma(n)$
is a permutation of $A$,
then $B$ is reachable if and only if
it is \textbf{213-avoiding},
i.e.\ there are no indices $i < j < k$
such that $\sigma(j) < \sigma(i) < \sigma(k)$.
\end{claim*}
\begin{proof}
This is clearly necessary.
To see its sufficient, one can just type $B$ inductively:
after typing $k$, the only way to get stuck is if
$k+1$ is to the right of $k$ and there is some character in the way;
this gives a 213 pattern.
\end{proof}
\begin{claim*}
A permutation $\sigma$ on $\{1, \dots, n\}$
is 213-avoiding if and only if the inverse $\sigma\inv$ is.
\end{claim*}
\begin{proof}
Suppose $i < j < k$ and $\sigma(j) < \sigma(i) < \sigma(k)$.
Let $i' = \sigma(j)$, $j' = \sigma(i)$, $k' = \sigma(k)$;
then $i' < j' < k'$ and $\sigma\inv(j') < \sigma\inv(i') < \sigma\inv(k')$.
\end{proof}
This essentially finishes the problem.
Suppose $B$ is reachable from $A$.
By using the typing pattern,
we get some permutation $\sigma \colon \{1, \dots, n\}$
such that the $i$th character of $A$
is the $\sigma(i)$th character of $B$,
and which is 213-avoiding by the claim.
(The permutation is unique if $A$ has all distinct characters,
but there could be multiple if $A$ has repeated ones.)
Then $\sigma\inv$ is 213-avoiding too
and gives us a way to change $B$ into $A$.
|
sols-TSTST-2014_2
|
Consider a convex pentagon circumscribed about a circle.
We name the lines that connect vertices of the pentagon with
the opposite points of tangency with the circle \emph{gergonnians}.
\begin{enumerate}
\ii[(a)] Prove that if four gergonnians are concurrent,
then all five of them are concurrent.
\ii[(b)] Prove that if there is a triple of gergonnians
that are concurrent, then there is another triple
of gergonnians that are concurrent.
\end{enumerate}
|
This problem is insta-killed by taking a homography
sending the concurrency point (in either part)
to the center of the circle while fixing the incircle.
Alternatively, one may send any four of the tangency points to a rectangle.
Here are the details.
Let $ABCDE$ be a pentagon with gergonnians
$\ol{AV}$, $\ol{BW}$, $\ol{CX}$, $\ol{DY}$, $\ol{EZ}$.
We prove the following lemma, which
(up to a suitable permutation of point names)
solves both parts (a) and (b).
\begin{lemma*}
The gergonnians $\ol{AV}$, $\ol{CX}$, $\ol{DY}$
are concurrent if and only if the gergonnians
$\ol{AV}$, $\ol{BW}$, $\ol{EZ}$ are concurrent.
\end{lemma*}
\begin{proof}
We prove the first set implies the second
(the converse direction being identical).
Suppose $\ol{AV}$, $\ol{CX}$, $\ol{DY}$ intersect at $P$
and take a homography fixing the circle
and moving $P$ to its center.
\begin{center}
\begin{asy}
size(7cm);
pair Y = dir(115);
pair X = -conj(Y);
pair V = dir(270);
pair Z = X*X/V;
pair W = Y*Y/V;
pair A = 2*X*Y/(X+Y);
pair B = 2*Y*Z/(Y+Z);
pair C = 2*Z*V/(Z+V);
pair D = 2*V*W/(V+W);
pair E = 2*W*X/(W+X);
filldraw(A--B--C--D--E--cycle, opacity(0.1)+lightblue, blue);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
pair P = origin;
draw(C--X, red);
draw(D--Y, red);
draw(A--V, red);
draw(B--W, heavygreen+dashed);
draw(E--Z, heavygreen+dashed);
dot("$Y$", Y, dir(Y));
dot("$X$", X, dir(X));
dot("$V$", V, dir(V));
dot("$Z$", Z, dir(Z));
dot("$W$", W, dir(W));
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("$P$", P, dir(0));
/* Source generated by TSQ */
\end{asy}
\end{center}
Then $X$ and $Y$ are symmetric around $\ol{APV}$
by hypothesis.
Since $D = \ol{VV} \cap \ol{PY}$, $C = \ol{VV} \cap \ol{PX}$,
it follows that $C$ and $D$,
and hence $Z$ and $W$,
are also symmetric around $\ol{APV}$.
Consequently $B$ and $E$ are symmetric too.
So $\ol{BW}$ and $\ol{EZ}$ meet on $\ol{AV}$.
\end{proof}
|
sols-TSTST-2014_3
|
Find all polynomials $P(x)$ with real coefficients that satisfy
\[ P(x \sqrt 2) = P(x + \sqrt{1-x^2}) \]
for all real numbers $x$ with $\lvert x \rvert \le 1$.
|
The answer is any polynomial of the form
$P(x) = f(U(x/\sqrt2))$, where $f \in \RR[x]$
and $U$ is the unique polynomial satisfying $U(\cos\theta) = \cos(8\theta)$.
Let $Q(x) = P(x \sqrt 2)$; then the condition reads
\[
Q(\cos\theta) = Q\left( \frac{1}{\sqrt2}(\cos\theta+\sin\theta) \right)
= Q(\cos(\theta-45\dg)) \qquad \forall \; 0 \le \theta \le 180\dg.
\]
We call a polynomial \emph{good} if it satisfies this functional equation.
It's easy to see $U$ itself is good, and hence so is any polynomial of the form
$g(U(x))$ for $g \in \RR[x]$.
Our goal is to show that every good $Q$ is of this form.
The proof is by induction on $\deg Q$, where the case $\deg Q = 0$ is vacuous.
Note that
\begin{align*}
Q(\cos136\dg) &= Q(\cos91\dg)
= Q(\cos46\dg) = Q(\cos1\dg) = Q(\cos-44\dg) \\
&= Q(\cos44\dg) = Q(\cos89\dg) = Q(\cos134\dg) = Q(\cos179\dg).
\end{align*}
Hence $Q$ is equal at eight distinct values
(not nine since $\cos -44\dg = \cos44\dg$ is repeated).
So it follows that the quotient
\[ \wt Q(x) \coloneq \frac{Q(x) - Q(\cos1\dg)}{U(x) - U(\cos1\dg)} \]
is also a good polynomial, because the polynomial
$U(x) - U(\cos1\dg)$ is the degree-eight polynomial with those distinct values as roots.
In particular, $\deg Q \ge 8$ unless $Q$ is constant, and in fact
\[ Q(x) = \left[ U(x) - U(\cos 1\dg) \right] \wt Q(x) + Q(\cos(1\dg)) \]
so if one applies the induction hypothesis to $\wt Q(x)$
(which has degree $\deg Q - 8$), the proof is completed.
|
sols-TSTST-2014_4
|
Let $P(x)$ and $Q(x)$ be arbitrary polynomials with real coefficients,
with $P \neq 0$, and let $d = \deg P$.
Prove that there exist polynomials $A(x)$ and $B(x)$,
not both zero, such that $\max \{ \deg A, \deg B \} \le d/2$
and $P(x) \mid A(x) + Q(x) \cdot B(x)$.
|
Let $V$ be the vector space of real polynomials with degree at most $d/2$.
Consider maps of linear spaces
\begin{align*}
V^{\oplus 2} &\to \RR[x] / (P(x)) \\
\text{by} \qquad (A,B) &\mapsto A+QB \pmod P.
\end{align*}
The domain has dimension
\[ 2 \left( \left\lfloor d/2 \right\rfloor + 1 \right) \]
while the codomain has dimension $d$.
For dimension reasons it has nontrivial kernel.
|
sols-TSTST-2014_5
|
Find the maximum number $E$ such that the following holds:
there is an edge-colored graph with $60$ vertices and $E$ edges,
with each edge colored either red or blue, such that in that coloring,
there is no monochromatic cycles of length $3$
and no monochromatic cycles of length $5$.
|
The answer is $E = 30^2 + 2 \cdot 15^2 = 6 \cdot 15^2 = 1350$.
First, we prove $E \le 1350$. Observe that:
\begin{claim*}
$G$ contains no $K_5$.
\end{claim*}
\begin{proof}
It's a standard fact that the only triangle-free two-coloring
of the edges of $K_5$ is the union of two monochromatic $C_5$'s.
\end{proof}
Hence by Tur\'{a}n theorem we have $E \le \binom42 \cdot 15^2 = 1350$.
To show this is achievable, take a red $K_{30,30}$,
and on each side draw a blue $K_{15,15}$.
This graph has no monochromatic odd cycles at all as desired.
|
sols-TSTST-2014_6
|
Suppose we have distinct positive integers $a$, $b$, $c$, $d$
and an odd prime $p$ not dividing any of them,
and an integer $M$ such that if one considers the infinite sequence
\begin{align*}
ca &- db \\
ca^2 &- db^2 \\
ca^3 &- db^3 \\
ca^4 &- db^4 \\
&\vdots
\end{align*}
and looks at the highest power of $p$ that divides each of them,
these powers are not all zero, and are all at most $M$.
Prove that there exists some $T$ (which may depend on $a,b,c,d,p,M$)
such that whenever $p$ divides an element of this sequence,
the maximum power of $p$ that divides that element is exactly $p^T$.
\end{enumerate}
|
By orders, the indices of terms divisible by $p$
is an arithmetic subsequence of $\NN$:
say they are $\kappa$, $\kappa + \lambda$, $\kappa + 2\lambda$, \dots,
where $\lambda$ is the order of $a/b$.
That means we want
\[ \nu_p \left( c a^{\kappa + n\lambda} - d b^{\kappa + n \lambda} \right)
= \nu_p \left(
\left( \frac{a^\lambda}{b^\lambda} \right)^n
- \frac{d a^\kappa}{c b^{\kappa}} \right)
\]
to be constant.
Thus, we have reduced the problem to the following proposition:
\begin{proposition*}
Let $p$ be an odd prime.
Let $x, y \in \QQ_{>0}$ such that $x \equiv y \equiv 1 \pmod p$.
If the sequence $\nu_p\left( x^n - y \right)$ of positive integers
is nonconstant, then it is unbounded.
\end{proposition*}
For this it would be sufficient to prove the following claim.
\begin{claim*}
Let $p$ be an odd prime.
Let $x, y \in \QQ_{>0}$ such that $x \equiv y \equiv 1 \pmod p$.
Suppose $m$ and $n$ are positive integers such that
\[ d = \nu_p(x^n-y) < \nu_p(x^m-y) = e. \]
Then there exists $\ell$ such that $\nu_p(x^\ell - y) \ge e + 1$.
\end{claim*}
\begin{proof}
First, note that
$\nu_p(x^m - x^n) = \nu_p\left( (x^m-y) - (x^n-y) \right) = d$
and so by exponent lifting we can find \emph{some} $k$ such that
\[ \nu_p(x^k-1) = e \]
namely $k = p^{e-d}|m-n|$.
(In fact, one could also choose more carefully
$k = p^{e-d} \cdot \gcd(m-n, p^\infty)$,
so that $k$ is a power of $p$.)
Suppose we set $x^k = p^e u + 1$ and $x^m = p^e v + y$
where $u,v \in \QQ$ aren't divisible by $p$.
Now for any integer $1 \le r \le p-1$ we consider
\begin{align*}
x^{kr+m}-y &= (p^e u + 1)^r \cdot (p^e v + y) - y \\
&= p^e \left( v + yu \cdot r \right) + p^{2e} \left( \dots \right).
\end{align*}
By selecting $r$ with $r \equiv -v/u \pmod p$,
we ensure $p^{e+1} \mid x^{kr+m} - y$,
hence $\ell = kr+m$ is as desired.
\end{proof}
\begin{remark*}
One way to motivate the proof of the claim is as follows.
Suppose we are given $\nu_p(x^m-y) = e$,
and we wish to find $\ell$ such that $\nu_p(x^\ell - y) > e$.
Then, it is necessary (albeit insufficient) that
\[ x^{\ell-m} \equiv 1 \pmod{p^e} \text{ but }
x^{\ell-m} \not\equiv 1 \pmod{p^{e+1}}. \]
In particular, we need $\nu_p(x^{\ell-m}-1) = e$ exactly.
So the $k$ in the claim must exist if we are going to succeed.
On the other hand, if $k$ is \emph{some} integer for which
$\nu_p(x^k-1) = e$, then by choosing $\ell-m$ to be some
multiple of $k$ with no extra factors of $p$,
we hope that we can get $\nu_p(x^\ell-y) = e+1$.
That's why we write $\ell = kr+m$ and see what happens when we expand.
\end{remark*}
|
sols-TSTST-2015_1
|
Let $a_1$, $a_2$, \dots, $a_n$ be a sequence of real numbers,
and let $m$ be a fixed positive integer less than $n$.
We say an index $k$ with $1 \le k \le n$ is \textit{good} if there exists
some $\ell$ with $1 \le \ell \le m$ such that
\[ a_k + a_{k + 1} + \dotsb + a_{k + \ell - 1} \ge 0, \]
where the indices are taken modulo $n$.
Let $T$ be the set of all good indices. Prove that
\[ \sum_{k \in T} a_k \ge 0. \]
|
First we prove the result if the indices are not taken modulo $n$.
Call a number $\ell$-good if $\ell$ is the \emph{smallest} number
such that $a_k + a_{k+1} + \dots + a_{k+\ell-1} \ge 0$, and $\ell \le m$.
Then if $a_k$ is $\ell$-good,
the numbers $a_{k+1}, \dots, a_{k+\ell-1}$ are good as well.
Then by greedy from left to right,
we can group all the good numbers into blocks with nonnegative sums.
Repeatedly take the first good number,
if $\ell$-good, group it with the next $\ell$ numbers.
An example for $m = 3$:
\[ \langle 4 \rangle
\quad \langle -1 \quad -2 \quad 6 \rangle
\quad -9
\quad -7
\quad \langle 3 \rangle
\quad \langle -2 \quad 4 \rangle
\quad \langle -1. \]
We can now return to the original problem.
Let $N$ be a large integer; applying the algorithm to
$N$ copies of the sequence, we deduce that
\[ N \sum_{k \in T} a_k + c_N \ge 0 \]
where $c_N$ represents some ``error'' from left-over terms.
As $|c_N| \le \sum |a_i|$, by taking $N$ large enough we deduce the problem.
\begin{remark*}
This solution was motivated by looking at the case $m = 1$,
realizing how dumb it was, then looking at $m = 2$,
and realizing it was equally dumb.
\end{remark*}
|
sols-TSTST-2015_2
|
Let $ABC$ be a scalene triangle. Let $K_a$, $L_a$, and
$M_a$ be the respective intersections with $BC$ of the internal
angle bisector, external angle bisector, and the median from
$A$. The circumcircle of $AK_aL_a$ intersects $AM_a$ a second time
at a point $X_a$ different from $A$. Define $X_b$ and $X_c$
analogously. Prove that the circumcenter of $X_aX_bX_c$ lies on
the Euler line of $ABC$.
|
The main content of the problem:
\begin{claim*}
$\angle H X_a G = 90\dg$.
\end{claim*}
This implies the result,
since then the desired circumcenter is the midpoint of $\ol{GH}$.
(This is the main difficulty; the Euler line is a red herring.)
In what follows, we abbreviate $K_a$ $L_a$, $M_a$, $X_a$
to $K$, $L$, $M$, $X$.
\begin{proof}
[First proof by Brocard]
To do this, it suffices to show that $M$ has the
same power with respect to the circle
with diameter $\ol{AH}$ and the circle with diameter $\ol{K L}$.
In fact I claim both circles are orthogonal
to the circle with diameter $\ol{BC}$!
The former follows from Brocard's theorem, noting that $A$
is on the polar of $H$,
and the latter follows from the harmonic bundle.
\begin{center}
\begin{asy}
size(11cm);
pair A = dir(150);
pair B = dir(200);
pair C = dir(340);
filldraw(A--B--C--cycle, opacity(0.1)+lightred, red);
pair M = midpoint(B--C);
pair H = orthocenter(A, B, C);
pair G = centroid(A, B, C);
pair X = foot(H, M, A);
pair K = extension(A, incenter(A, B, C), B, C);
pair L = extension(A, A+dir(90)*(K-A), B, C);
draw(K--A--L--B, red);
draw(circumcircle(A, K, L), orange+dashed);
draw(A--M, red);
draw(X--H, orange);
dot("$A$", A, dir(A));
dot("$B$", B, dir(270));
dot("$C$", C, dir(270));
dot("$M$", M, dir(M));
dot("$H$", H, dir(270));
dot("$G$", G, dir(10));
dot("$X$", X, dir(10));
dot("$K$", K, dir(K));
dot("$L$", L, dir(L));
/* TSQ Source:
!size(11cm);
A = dir 150
B = dir 200 R270
C = dir 340 R270
A--B--C--cycle 0.1 lightred / red
M = midpoint B--C
H = orthocenter A B C R270
G = centroid A B C R10
X = foot H M A R10
K = extension A incenter A B C B C
L = extension A A+dir(90)*(K-A) B C
K--A--L--B red
circumcircle A K L orange dashed
A--M red
X--H orange
*/
\end{asy}
\end{center}
Then $\ol{AM}$ is the radical axis, so $X$ lies on both circles.
\end{proof}
\begin{proof}
[Second proof by orthocenter reflection, Bendit Chan]
As before, we know $MX \cdot MA = MK \cdot ML = MB \cdot MC$,
but $X$ lies inside segment $AM$.
Construct parallelogram $ABA'C$.
Then $MX \cdot MA' = MB \cdot MC$,
so $XBA'C$ is concyclic.
However, it is well-known the circumcircle of $\triangle BA'C$
(which is the reflection of $(ABC)$ across $\ol{BC}$)
passes through $H$ and in fact has diameter $\ol{A'H}$.
So this gives $\angle HXA' = 90\dg$ as needed.
\end{proof}
\begin{proof}
[Third proof by barycentric coordinates]
Alternatively we may just compute $X = (a^2:2S_A:2S_A)$.
Let $F = (0 : S_C : S_B)$ be the foot from $H$.
Then we check that $X H F M$ is cyclic,
which is power of a point from $A$.
\end{proof}
|
sols-TSTST-2015_3
|
Let $P$ be the set of all primes, and let $M$ be a
non-empty subset of $P$. Suppose that for any non-empty subset
$\{p_1, p_2, \dots, p_k\}$ of $M$, all prime factors of
$p_1 p_2 \dots p_k + 1$ are also in $M$. Prove that $M = P$.
|
The following solution was found by user \texttt{Aiscrim} on AOPS.
Obviously $|M| = \infty$.
Assume for contradiction $p \notin M$.
We say a prime $q \in M$ is \emph{sparse}
if there are only finitely many elements of $M$
which are $q \pmod p$
(in particular there are finitely many sparse primes).
Now let $C$ be the product of all sparse primes (note $p \nmid C$).
First, set $a_0 = 1$.
For $k \ge 0$, consider then the prime factorization of the number
\[ Ca_k + 1. \]
No prime in its factorization is sparse,
so consider the number $a_{k+1}$ obtained by
\textbf{replacing each prime in its factorization
with some arbitrary representative of that prime's residue class}.
In this way we select a number $a_{k+1}$ such that
\begin{itemize}
\ii $a_{k+1} \equiv Ca_k + 1 \pmod p$, and
\ii $a_{k+1}$ is a product of distinct primes in $M$.
\end{itemize}
In particular, $a_k \equiv C^k + C^{k-1} + \dots + 1 \pmod p$
But since $C \not\equiv 0 \pmod p$,
we can find a $k$ such that $a_k \equiv 0 \pmod p$
(namely, $k=p-1$ if $C \equiv 1$ and $k=p-2$ else)
which is clearly impossible since $a_k$ is a product of primes in $M$!
|
sols-TSTST-2015_4
|
Let $x,y,z$ be real numbers (not necessarily positive)
such that $x^4 + y^4 + z^4 + xyz = 4$.
Prove that $x \le 2$ and \[ \sqrt{2-x} \ge \frac{y+z}{2}. \]
|
We prove that the condition $x^4+y^4+z^4+xyz = 4$ implies
\[ \sqrt{2-x} \geq \frac{y+z}{2}. \]
We first prove the easy part.
\begin{claim*}
We have $x \le 2$.
\end{claim*}
\begin{proof}
Indeed, AM-GM gives that
\begin{align*}
5 =x^4+y^4+(z^4+1)+xyz
&= \frac{3x^4}{4}+\left(\frac{x^4}{4}+y^4\right)+(z^4+1)+xyz \\
&\geq \frac{3x^4}{4}+x^2y^2+2z^2+xyz.
\end{align*}
We evidently have that $x^2y^2+2z^2+xyz \geq 0$ because the
quadratic form $a^2+ab+2b^2$ is positive definite, so $x^4 \leq
\frac{20}{3} \implies x \leq 2$.
\end{proof}
Now, the desired statement is
implied by its square, so it suffices to show that
\[ 2-x \geq \left(\frac{y+z}{2}\right)^2 \]
We are going to proceed by contradiction
(it seems that many solutions do this)
and assume that
\[ 2-x < \left( \frac{y+z}{2} \right)^2
\iff 4x + y^2 + 2yz + z^2 > 8. \]
By AM-GM,
\begin{align*}
x^4 + 3 & \ge 4x \\
\tfrac{y^4+1}{2} &\ge y^2 \\
\tfrac{z^4+1}{2} &\ge z^2
\end{align*}
which yields that
\[ x^4+\frac{y^4+z^4}{2}+2yz+4 > 8. \]
If we replace $x^4 = 4 - (y^4+z^4+xyz)$ now,
this gives
\[ -\frac{y^4+z^4}{2}+(2-x)yz > 0 \implies (2-x)yz > \frac{y^4+z^4}{2}. \]
Since $2-x$ and the right-hand side are positive,
we have $yz \geq 0$.
Now
\[ \frac{y^4+z^4}{2yz} < 2-x < \left(\frac{y+z}{2}\right)^2
\implies 2y^4+2z^4 < yz(y+z)^2 = y^3z+2y^2z^2+yz^3. \]
This is clearly false by AM-GM, so we have a contradiction.
|
sols-TSTST-2015_5
|
Let $\varphi(n)$ denote the number of positive integers
less than $n$ that are relatively prime to $n$.
Prove that there exists a positive integer $m$ for which the equation
$\varphi(n) = m$ has at least $2015$ solutions in $n$.
|
Here are two explicit solutions.
\paragraph{First solution with ad-hoc subsets, by Evan Chen.}
I consider the following eleven prime numbers:
\[ S = \left\{ 11, 13, 17, 19, 29, 31, 37, 41, 43, 61, 71 \right\}. \]
This has the property that for any $p \in S$,
all prime factors of $p-1$ are one digit.
Let $N = (210)^{\text{billion}}$,
and consider $M = \varphi\left( N \right)$.
For any subset $T \subset S$, we have
\[ M = \varphi\left(
\frac{N}{\prod_{p\in T} (p-1)} \prod_{p \in T} p \right). \]
Since $2^{\left\lvert S \right\rvert} > 2015$ we're done.
\begin{remark*}
This solution is motivated by the deep fact that
$\varphi(11 \cdot 1000) = \varphi(10 \cdot 1000)$,
for example.
\end{remark*}
\paragraph{Second solution with smallest primes, by Yang Liu.}
Let $2 = p_1 < p_2 < \dots < p_{2015}$
be the \emph{smallest} $2015$ primes.
Then the $2015$ numbers
\begin{align*}
n_1 &= (p_1-1) p_2 \dots p_{2015} \\
n_2 &= p_1 (p_2-1) \dots p_{2015} \\
&\vdotswithin= \\
n_{2015} &= p_1 p_2 \dots \left( p_{2015}-1 \right)
\end{align*}
all have the same phi value, namely
\[ \varphi(p_1 p_2 \dots p_{2015}) = \prod_{i=1}^{2015} (p_i-1). \]
|
sols-TSTST-2015_6
|
A \emph{Nim-style game} is defined as follows. Two
positive integers $k$ and $n$ are specified, along with a finite
set $S$ of $k$-tuples of integers (not necessarily positive). At
the start of the game, the $k$-tuple $(n,0,0,\dots,0)$ is written
on the blackboard.
A legal move consists of erasing the tuple $(a_1,a_2,\dots,a_k)$
which is written on the blackboard and replacing it with
$(a_1+b_1,a_2+b_2,\dots,a_k+b_k)$, where $(b_1,b_2,\dots,b_k)$
is an element of the set $S$. Two players take turns making legal
moves, and the first to write a negative integer loses. In the
event that neither player is ever forced to write a negative
integer, the game is a draw.
Prove that there is a choice of $k$ and $S$ with the following
property: the first player has a winning strategy if $n$ is a
power of $2$, and otherwise the second player has a winning
strategy.
\end{enumerate}
|
Here we present a solution with $14$ registers and $22$ moves.
Initially $X=n$ and all other variables are zero.
\begin{center}
\scriptsize
\setlength{\tabcolsep}{3pt}
\begin{tabular}{l|rr|r|rrr|rrr|r|rr|rr}
&$X$&$Y$&Go&$S_X^0$&$S_X$&$S_X'$&$S_Y^0$&$S_Y$&$S_Y'$&Cl&$A$&$B$&Die&Die'\\\hline
Init&-1&&1&&&&&&&&&1&1&1\\
Begin&1&&-1&1&&&&&&&-1&1&&\\\hline
Sleep&&&&&&&&&&&1&-1&&\\\hline
StartX&&&&-1&1&&&&&&-1&1&&\\
WorkX&-1&&&&-1&1&&&&&-1&1&&\\
WorkX'&-1&1&&&1&-1&&&&&-1&1&&\\
DoneX&&&&&-1&&1&&&&-1&1&&\\
WrongX&-1&&&&&&-1&&&&&-1&&\\\hline
StartY&&&&&&&-1&1&&&-1&1&&\\
WorkY&&-1&&&&&&-1&1&&-1&1&&\\
WorkY'&1&-1&&&&&&1&-1&&-1&1&&\\
DoneY&&&&1&&&&-1&&&-1&1&&\\
WrongY&&-1&&-1&&&&&&&&-1&&\\\hline
ClaimX&-1&&&-1&&&&&&1&-1&1&&\\
ClaimY&&-1&&&&&-1&&&1&-1&1&&\\
FakeX&-1&&&&&&&&&-1&&-1&&\\
FakeY&&-1&&&&&&&&-1&&-1&&\\
Win&&&&&&&&&&-1&-1&&&\\\hline
PunA&&&&&&&&&&&&-2&&\\
PunB&&&&&&&&&&&-1&-1&&\\
Kill&&&&&&&&&&&&-1&-2&1\\
Kill'&&&&&&&&&&&&-1&1&-2
\end{tabular}
\end{center}
Now, the ``game'' is played as follows.
The mechanics are controlled by the \emph{turn counters} $A$ and $B$.
Observe the game starts with Alice playing Init.
Thereafter, we say that the game is
\begin{itemize}
\ii In the \emph{main part} if $A+B = 1$,
and no one has played Init a second time.
\ii In the \emph{death part} otherwise.
\end{itemize}
Observe that in the main state,
on Alice's turn we always have $(A,B) = (1,0)$
and on Bob's turn we always have $(A,B) = (0,1)$.
\begin{claim*}
A player who plays Init a second time must lose.
In particular, a player who makes a move when $A=B=0$ must lose.
\end{claim*}
\begin{proof}
Situations with $A+B \ge 2$ cannot occur during main part,
so there are only a few possibilities.
\begin{itemize}
\ii Suppose the offending player is in a situation where $A=B=0$.
Then he/she must play Init.
At this point, the opposing player can respond by playing Kill.
Then the offending player must play Init again.
The opposing player now responds with Kill'.
This iteration continues until $X$ reaches a negative number
and the offending player loses.
\ii Suppose Alice has $(A,B) = (1,0)$ but plays Init again anyways.
Then Bob responds with PunB to punish her; he then wins as in the first case.
\ii Suppose Bob has $(A,B) = (0,1)$ but plays Init again anyways.
Alice responds with PunA in the same way. \qedhere
\end{itemize}
\end{proof}
Thus we may assume that players avoid the death part at all costs.
Hence the second moves consist of Bob playing Sleep,
and then Alice playing Begin
(thus restoring the value of $n$ in $X$), then Bob playing Sleep.
Now we return to analysis of the main part.
We say the game is in \emph{state} $S$ for
$S \in \{S_X^0, S_X, S_X', S_Y^0, S_Y, S_Y', \text{Cl} \}$
if $S = 1$ and all other variables are zero.
By construction, this is always the case.
From then on the main part is divided into several phases:
\begin{itemize}
\ii An \emph{$X$-phase}: this begins with Alice at $S_X^0$, and
ends when the game is in a state other than $S_X$ and $S_X'$.
(She can never return to $S_X^0$ during an $X$-phase.)
\ii A \emph{$Y$-phase}: this begins with Alice at $S_Y^0$, and
ends when the game is in a state other than $S_Y$ and $S_Y'$.
(She can never return to $S_Y^0$ during a $Y$-phase.)
\end{itemize}
\begin{claim*}
Consider an $X$-phase in which $(X,Y) = (x,0)$, $x > 1$.
Then Alice can complete the phase without losing if and only if $x$ is even;
if so she begins a $Y$-phase with $(X,Y) = (0,x/2)$.
\end{claim*}
\begin{proof}
As $x >1$, Alice cannot play ClaimX since Bob will respond with FakeX and win.
Now by alternating between WorkX and WorkX',
Alice can repeatedly deduct $2$ from $X$ and add $1$ to $Y$,
leading to $(x-2,y+1)$, $(x-4,y+2)$, and so on.
(During this time, Bob can only play Sleep.)
Eventually, she must stop this process by playing DoneX, which begins a $Y$-phase.
Now note that unless $X=0$, Bob now has a winning move WrongX.
Conversely he may only play Sleep if $X=0$.
\end{proof}
We have an analogous claim for $Y$-phases.
Thus if $n$ is not a power of $2$, we see that Alice eventually loses.
Now suppose $n = 2^k$;
then Alice reaches $(X,Y) = (0,2^{k-1}), \; (2^{k-2}, 0), \; \dots$
until either reaching $(1,0)$ or $(0,1)$.
At this point she can play ClaimX or ClaimY, respectively;
the game is now in state Cl.
Bob cannot play either FakeX or FakeY, so he must play Sleep,
and then Alice wins by playing Win.
Thus Alice has a winning strategy when $n = 2^k$.
|
sols-TSTST-2016_1
|
Let $A = A(x,y)$ and $B = B(x,y)$ be
two-variable polynomials with real coefficients.
Suppose that $A(x,y)/B(x,y)$ is a polynomial in $x$
for infinitely many values of $y$,
and a polynomial in $y$ for infinitely many values of $x$.
Prove that $B$ divides $A$, meaning there exists a third polynomial $C$
with real coefficients such that $A = B \cdot C$.
|
This is essentially an application of the division algorithm,
but the details require significant care.
First, we claim that $A/B$ can be written as a polynomial in $x$
whose coefficients are rational functions in $y$.
To see this, use the division algorithm to get
\[ A = Q \cdot B + R \qquad Q,R \in (\RR(y))[x] \]
where $Q$ and $R$ are polynomials in $x$
whose coefficients are rational functions in $y$,
and moreover $\deg_x B > \deg_x R$.
Now, we claim that $R \equiv 0$.
Indeed, we have by hypothesis that for infinitely many values of $y_0$
that $B(x,y_0)$ divides $A(x, y_0)$,
which means $B(x,y_0) \mid R(x,y_0)$ as polynomials in $\RR[x]$.
Now, we have $\deg_x B(x,y_0) > \deg_x R(x,y_0)$
outside of finitely many values of $y_0$ (but not all of them!);
this means for infinitely many $y_0$ we have $R(x,y_0) \equiv 0$.
So each coefficient of $x^i$ (in $\RR(y)$)
has infinitely many roots, hence is a zero polynomial.
Consequently, we are able to write $A / B = F(x,y) / M(y)$
where $F \in \RR[x,y]$ and $M \in \RR[y]$ are each polynomials.
Repeating the same argument now gives
\[ \frac AB = \frac{F(x,y)}{M(y)} = \frac{G(x,y)}{N(x)}. \]
Now, by unique factorization of polynomials in $\RR[x,y]$,
we can discuss GCD's.
So, we tacitly assume $\gcd(F,M) = \gcd(G,N) = (1)$.
Also, we obviously have $\gcd(M,N) = (1)$.
But $F \cdot N = G \cdot M$, so $M \mid F \cdot N$,
thus we conclude $M$ is the constant polynomial.
This implies the result.
\begin{remark*} This fact does not generalize to arbitrary functions that are separately polynomial:
see e.g. \url{http://aops.com/community/c6h523650p2978180}. \end{remark*}
|
sols-TSTST-2016_2
|
Let $ABC$ be a scalene triangle with orthocenter $H$ and circumcenter $O$
and denote by $M$, $N$ the midpoints of $\ol{AH}$, $\ol{BC}$.
Suppose the circle $\gamma$ with diameter $\ol{AH}$ meets
the circumcircle of $ABC$ at $G \neq A$,
and meets line $\ol{AN}$ at $Q \neq A$.
The tangent to $\gamma$ at $G$ meets line $OM$ at $P$.
Show that the circumcircles of $\triangle GNQ$
and $\triangle MBC$ intersect on $\ol{PN}$.
|
We present two solutions,
one using essentially only power of a point,
and the other more involved.
\paragraph{First solution (found by contestants).}
Denote by $\triangle DEF$ the orthic triangle.
Observe $\ol{PA}$ and $\ol{PG}$ are tangents to $\gamma$,
since $\ol{OM}$ is the perpendicular bisector of $\ol{AG}$.
Also note that $\ol{AG}$, $\ol{EF}$, $\ol{BC}$ are concurrent at some point $R$
by radical axis on $(ABC)$, $\gamma$, $(BFEC)$.
Now, consider circles $(PAGM)$, $(MFDNE)$, and $(MBC)$.
We already saw the point $R$ satisfies
\[ RA \cdot RG = RE \cdot RF = RB \cdot RC \]
and hence has equal powers to all three circles;
but since the circles at $M$ already, they must actually be coaxial.
Assume they meet again at $T \in \ol{RM}$, say.
Then $\angle PTM$ and $\angle MTN$ are both right angles, hence $T$ lies on $\ol{PN}$.
Finally $H$ is the orthocenter of $\triangle ARN$,
and thus the circle with diameter $\ol{RN}$ passes through $G$, $Q$, $N$.
\begin{center}
\begin{asy}
size(11cm);
pair A = dir(120);
pair B = dir(210);
pair C = dir(330);
pair N = midpoint(B--C);
pair H = orthocenter(A, B, C);
pair M = midpoint(A--H);
pair O = circumcenter(A, B, C);
pair P = extension(O, M, A, B-C+A);
pair T = foot(M, P, N);
pair D = foot(A, B, C);
pair E = foot(B, C, A);
pair F = foot(C, A, B);
filldraw(A--B--C--cycle, opacity(0.1)+lightblue, blue);
draw(unitcircle, blue);
draw(A--D, heavygreen);
draw(B--E, heavygreen);
draw(C--F, heavygreen);
draw(E--F, heavygreen);
draw(B--M--C, lightred);
// filldraw(circumcircle(M, B, C), opacity(0.05)+lightred, red);
filldraw(circumcircle(D, E, F), opacity(0.05)+lightred, red);
draw(O--P--N, heavycyan);
draw(A--P, heavycyan);
filldraw(circumcircle(A, E, F), opacity(0.1)+lightgreen, heavygreen);
pair Q = foot(H, A, N);
pair G = -A+2*foot(A, M, P);
pair R = extension(E, F, B, C);
draw(R--F, heavygreen);
draw(R--B, blue);
draw(R--Q, heavygreen);
draw(A--R, heavygreen);
draw(A--N, blue);
draw(P--G, heavycyan);
draw(G--N, blue);
draw(M--R, heavygreen);
filldraw(circumcircle(R, G, T), opacity(0.05)+lightred, red);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$N$", N, dir(N));
dot("$H$", H, dir(H));
dot("$M$", M, dir(M));
dot("$O$", O, dir(225));
dot("$P$", P, dir(P));
dot("$T$", T, dir(T));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$Q$", Q, dir(Q));
dot("$G$", G, dir(G));
dot("$R$", R, dir(R));
/* TSQ Source:
!size(11cm);
A = dir 120
B = dir 210
C = dir 330
N = midpoint B--C
H = orthocenter A B C
M = midpoint A--H
O = circumcenter A B C R225
P = extension O M A B-C+A
T = foot M P N
D = foot A B C
E = foot B C A
F = foot C A B
A--B--C--cycle 0.1 lightblue / blue
unitcircle blue
A--D heavygreen
B--E heavygreen
C--F heavygreen
E--F heavygreen
B--M--C lightred
// circumcircle M B C 0.05 lightred / red
circumcircle D E F 0.1 lightred / red
O--P--N heavycyan
A--P heavycyan
circumcircle A E F 0.1 lightgreen / heavygreen
Q = foot H A N
G = -A+2*foot A M P
R = extension E F B C
R--F heavygreen
R--B blue
R--Q heavygreen
A--R heavygreen
A--N blue
P--G heavycyan
G--N blue
M--R heavygreen
circumcircle R G T 0.1 lightred / red
*/
\end{asy}
\end{center}
\paragraph{Alternate solution (by proposer).}
Let $L$ be diametrically opposite $A$ on the circumcircle.
Denote by $\triangle DEF$ the orthic triangle.
Let $X = \ol{AH} \cap \ol{EF}$.
Finally, let $T$ be the second intersection of $(MFDNE)$ and $(MBC)$.
\begin{center}
\begin{asy}
size(11cm);
pair A = dir(120);
pair B = dir(210);
pair C = dir(330);
pair N = midpoint(B--C);
pair H = orthocenter(A, B, C);
pair M = midpoint(A--H);
pair O = circumcenter(A, B, C);
pair P = extension(O, M, A, B-C+A);
pair X = extension(P, N, A, H);
pair T = foot(M, P, N);
pair D = foot(A, B, C);
pair E = foot(B, C, A);
pair F = foot(C, A, B);
pair L = -A;
pair K = extension(H, N, A, P);
filldraw(A--B--C--cycle, opacity(0.1)+lightblue, blue);
filldraw(B--L--C--cycle, opacity(0.1)+lightblue, blue);
draw(unitcircle, blue);
draw(A--L, blue);
draw(A--D, heavygreen);
draw(B--E, heavygreen);
draw(C--F, heavygreen);
draw(E--F, heavygreen);
draw(B--M--C, lightred);
filldraw(circumcircle(M, B, C), opacity(0.05)+lightred, red);
filldraw(circumcircle(D, E, F), opacity(0.05)+lightred, red);
draw(A--K--L, heavycyan);
draw(O--P--N, heavycyan);
filldraw(circumcircle(A, E, F), opacity(0.1)+lightgreen, heavygreen);
pair Q = foot(H, A, N);
pair G = -A+2*foot(A, M, P);
pair R = extension(E, F, B, C);
draw(R--F, heavygreen);
draw(R--B, blue);
draw(R--Q, heavygreen);
draw(A--R, heavygreen);
draw(A--N, blue);
draw(P--G, heavycyan);
draw(M--R, heavygreen);
filldraw(circumcircle(R, G, T), opacity(0.05)+lightred, red);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$N$", N, dir(N));
dot("$H$", H, dir(H));
dot("$M$", M, dir(M));
dot("$O$", O, dir(225));
dot("$P$", P, dir(P));
dot("$X$", X, dir(X));
dot("$T$", T, dir(T));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$L$", L, dir(L));
dot("$K$", K, dir(K));
dot("$Q$", Q, dir(Q));
dot("$G$", G, dir(G));
dot("$R$", R, dir(R));
/* TSQ Source:
!size(11cm);
A = dir 120
B = dir 210
C = dir 330
N = midpoint B--C
H = orthocenter A B C
M = midpoint A--H
O = circumcenter A B C R225
P = extension O M A B-C+A
X = extension P N A H
T = foot M P N
D = foot A B C
E = foot B C A
F = foot C A B
L = -A
K = extension H N A P
A--B--C--cycle 0.1 lightblue / blue
B--L--C--cycle 0.1 lightblue / blue
unitcircle blue
A--L blue
A--D heavygreen
B--E heavygreen
C--F heavygreen
E--F heavygreen
B--M--C lightred
circumcircle M B C 0.05 lightred / red
circumcircle D E F 0.05 lightred / red
A--K--L heavycyan
O--P--N heavycyan
circumcircle A E F 0.1 lightgreen / heavygreen
Q = foot H A N
G = -A+2*foot A M P
R = extension E F B C
R--F heavygreen
R--B blue
R--Q heavygreen
A--R heavygreen
A--N blue
P--G heavycyan
M--R heavygreen
circumcircle R G T 0.05 lightred / red
*/
\end{asy}
\end{center}
We begin with a few easy observations.
First, points $H$, $G$, $N$, $L$ are collinear and $\angle AGL = 90\dg$.
Also, $Q$ is the foot from $H$ to $\ol{AN}$.
Consequently, lines $AG$, $EF$, $HQ$, $BC$, $TM$
concur at a point $R$ (radical axis).
Moreover, we already know $\angle MTN = 90\dg$.
This implies $T$ lies on the circle with diameter $\ol{RN}$,
which is exactly the circumcircle of $\triangle GQN$.
Note by Brocard's Theorem on $AFHE$, the point $X$ is the orthocenter of $\triangle MBC$.
But $\angle MTN = 90\dg$ already, and $N$ is the midpoint of $\ol{BC}$.
Consequently, points $T$, $X$, $N$ are collinear.
Finally, we claim $P$, $X$, $N$ are collinear, which solves the problem.
Note $P = \ol{GG} \cap \ol{AA}$.
Set $K = \ol{HNL} \cap \ol{AP}$.
Then by noting \[ -1 = (D,X;A,H) \overset{N}{=} (\infty, \ol{NX} \cap \ol{AK}; A, K) \]
we see that $\ol{NX}$ bisects segment $\ol{AK}$, as desired.
(A more projective finish is to show that $\ol{PXN}$ is the polar of $R$ to $\gamma$).
\begin{remark*}
The original problem proposal reads as follows:
\begin{quote}
Let $ABC$ be a triangle with orthocenter $H$ and circumcenter $O$
and denote by $M$, $N$ the midpoints of $\ol{AH}$, $\ol{BC}$.
Suppose ray $OM$ meets the line parallel to $\ol{BC}$ through $A$ at $P$.
Prove that the line through the circumcenter of $\triangle MBC$
and the midpoint of $\ol{OH}$ is parallel to $\ol{NP}$.
\end{quote}
The points $G$ and $Q$ were added to the picture later
to prevent the problem from being immediate by coordinates.
\end{remark*}
|
sols-TSTST-2016_3
|
Decide whether or not there exists a nonconstant polynomial $Q(x)$
with integer coefficients with the following property:
for every positive integer $n > 2$, the numbers
\[ Q(0), \; Q(1), Q(2), \; \dots, \; Q(n-1) \]
produce at most $0.499n$ distinct residues when taken modulo $n$.
|
We claim that
\[ Q(x) = 420(x^2-1)^2 \]
works.
Clearly, it suffices to prove the result when $n=4$ and when $n$ is an odd prime $p$.
The case $n=4$ is trivial, so assume now $n=p$ is an odd prime.
First, we prove the following easy claim.
\begin{claim*}
For any odd prime $p$, there are at least $\frac12(p-3)$
values of $a$ for which $\left( \frac{1-a^2}{p} \right) = +1$.
\end{claim*}
\begin{proof}
Note that if $k \neq 0$ and $k^2 \neq -1$, then $a = \frac{1-k^2}{k^2+1}$ works.
\end{proof}
\begin{remark*}
The above identity comes from starting with the equation $1-a^2=b^2$,
and writing it as $\left( \frac 1b \right)^2 - \left( \frac ab \right)^2 = 1$.
Then solve $\frac 1b - \frac ab = k$ and $\frac 1b + \frac ab = 1/k$ for $a$.
\end{remark*}
Let $F(x) = (x^2-1)^2$.
The range of $F$ modulo $p$ is contained within the $\half(p+1)$ quadratic residues modulo $p$.
On the other hand, if for some $t$ neither of $1 \pm t$ is a quadratic residue,
then $t^2$ is omitted from the range of $F$ as well.
Call such a value of $t$ \emph{useful}, and let $N$ be the number of useful residues.
We aim to show $N \ge \frac14 p - 2$.
We compute a lower bound on the number $N$ of useful $t$ by writing
\begin{align*}
N &= \frac{1}{4} \left( \sum_t \left[ \left(1 - \left(\frac{1-t}{p} \right) \right)
\left(1 - \left(\frac{1+t}{p} \right) \right) \right]
- \left( 1- \left( \frac2p \right) \right)
- \left( 1- \left( \frac{-2}p \right) \right)
\right) \\
&\ge \frac{1}{4} \sum_t \left[ \left(1 - \left(\frac{1-t}{p} \right) \right)
\left(1 - \left(\frac{1+t}{p} \right) \right) \right] -1 \\
&= \frac{1}{4} \left(p + \sum_t \left(\frac{1-t^2}{p} \right) \right) -1 \\
&\ge \frac14 \left( p + (+1) \cdot \tfrac12(p-3) + 0 \cdot 2
+ (-1) \cdot ( (p-2) - \tfrac12(p-3)) \right) - 1 \\
&\ge \frac14 \left( p - 5 \right).
\end{align*}
Thus, the range of $F$ has size at most
\[ \half(p+1) - \half N \le \frac38(p+3). \]
This is less than $0.499p$ for any $p \ge 11$.
\begin{remark*}
In fact, the computation above is essentially an equality.
There are only two points where terms are dropped:
one, when $p \equiv 3 \pmod 4$ there are no $k^2 = -1$ in the lemma,
and secondly, the terms $1-\left( 2/p \right)$ and $1-\left( -2/p \right)$
are dropped in the initial estimate for $N$.
With suitable modifications, one can show that in fact,
the range of $F$ is exactly equal to
\[
\half(p+1) - \half N =
\begin{cases}
\frac18(3p+5) & p \equiv 1 \pmod 8 \\
\frac18(3p+7) & p \equiv 3 \pmod 8 \\
\frac18(3p+9) & p \equiv 5 \pmod 8 \\
\frac18(3p+3) & p \equiv 7 \pmod 8.
\end{cases}
\]
\end{remark*}
|
sols-TSTST-2016_4
|
Prove that if $n$ and $k$ are positive integers
satisfying $\varphi^k(n) = 1$, then $n \le 3^k$.
(Here $\varphi^k$ denotes $k$ applications of the Euler phi function.)
|
The main observation is that the exponent of $2$ decreases
by at most $1$ with each application of $\varphi$.
This will give us the desired estimate.
Define the \emph{weight} function $w$ on positive integers as follows: it satisfies
\begin{align*}
w(ab) &= w(a)+w(b); \\
w(2) &= 1; \quad\text{and} \\
w(p) &= w(p-1) \quad \text{for any prime $p > 2$}.
\end{align*}
By induction, we see that $w(n)$ counts the powers of $2$
that are produced as $\varphi$ is repeatedly applied to $n$.
In particular, $k \ge w(n)$.
From $w(2) = 1$, it suffices to prove that $w(p) \ge \log_3 p$ for every $p > 2$.
We use strong induction and note that
\[
w(p) = w(2) + w\left( \frac{p-1}{2} \right)
\ge 1 + \log_3(p-1) - \log_3 2 \ge \log_3 p
\]
for any $p > 2$. This solves the problem.
\begin{remark*}
One can motivate this solution through small cases $2^x 3^y$
like $2^x 17^w$, $2^x 3^y 7^z$, $2^x 11^t$.
Moreover, the stronger bound \[ n \le 2 \cdot 3^{k-1} \] is true and best possible.
\end{remark*}
|
sols-TSTST-2016_5
|
In the coordinate plane are finitely many \emph{walls},
which are disjoint line segments, none of which are parallel to either axis.
A bulldozer starts at an arbitrary point and moves in the $+x$ direction.
Every time it hits a wall, it turns at a right angle to its path,
away from the wall, and continues moving.
(Thus the bulldozer always moves parallel to the axes.)
Prove that it is impossible for the bulldozer
to hit both sides of every wall.
|
We say a wall $v$ is \emph{above} another wall $w$ if some point on
$v$ is directly above a point on $w$.
(This relation is anti-symmetric, as walls do not intersect).
The critical claim is as follows:
\begin{claim*}
There exists a lowest wall,
i.e.\ a wall not above any other walls.
\end{claim*}
\begin{proof}
Assume not.
Then we get a directed cycle of some length $n \ge 3$:
it's possible to construct a series of points $P_i$, $Q_i$,
for $i = 1, \dots, n$ (indices modulo $n$), such that
the point $Q_i$ is directly above $P_{i+1}$ for each $i$,
the segment $\ol{Q_i P_{i+1}}$ does not intersect any wall in its interior,
and finally each segment $\ol{P_i Q_i}$ is contained inside a wall.
This gives us a broken line on $2n$ vertices which is not self-intersecting.
Now consider the leftmost vertical segment $\ol{Q_i P_{i+1}}$
and the rightmost vertical segment $\ol{Q_j P_{j+1}}$.
The broken line gives a path from $P_{i+1}$ to $Q_j$,
as well as a path from $P_{j+1}$ to $Q_i$.
These clearly must intersect, contradiction.
\end{proof}
\begin{remark*}
This claim is Iran TST 2010.
\end{remark*}
Thus if the bulldozer eventually moves upwards indefinitely,
it may never hit the bottom side of the lowest wall.
Similarly, if the bulldozer eventually moves downwards indefinitely,
it may never hit the upper side of the highest wall.
|
sols-TSTST-2016_6
|
Let $ABC$ be a triangle with incenter $I$, and whose incircle is tangent
to $\ol{BC}$, $\ol{CA}$, $\ol{AB}$ at $D$, $E$, $F$, respectively.
Let $K$ be the foot of the altitude from $D$ to $\ol{EF}$.
Suppose that the circumcircle of $\triangle AIB$
meets the incircle at two distinct points $C_1$ and $C_2$,
while the circumcircle of $\triangle AIC$ meets the incircle
at two distinct points $B_1$ and $B_2$.
Prove that the radical axis of the circumcircles of
$\triangle BB_1B_2$ and $\triangle CC_1C_2$ passes through the midpoint $M$ of $\ol{DK}$.
\end{enumerate}
|
\paragraph{First solution (Allen Liu).}
Let $X$, $Y$, $Z$ be midpoints of $EF$, $FD$, $DE$, and let $G$ be the Gergonne point.
By radical axis on $(AEIF)$, $(DEF)$, $(AIC)$ we see that $B_1$, $X$, $B_2$ are collinear.
Likewise, $B_1$, $Z$, $B_2$ are collinear, so lines $B_1B_2$ and $XZ$ coincide.
Similarly, lines $C_1C_2$ and $XY$ coincide.
In particular lines $B_1B_2$ and $C_1C_2$ meet at $X$.
\begin{center}
\begin{asy}
size(12cm);
pair A = dir(70);
pair B = dir(240);
pair C = dir(300);
pair I = incenter(A, B, C);
pair D = foot(I, B, C);
pair E = foot(I, C, A);
pair F = foot(I, A, B);
pair G = extension(A, D, B, E);
filldraw(A--B--C--cycle, opacity(0.1)+lightcyan, blue);
draw(D--E--F--cycle, blue);
draw(incircle(A, B, C), blue);
draw(A--D, blue);
draw(B--E, blue);
draw(C--F, blue);
pair Q = extension(G, G+F-D, E, F);
pair S = extension(G, G+F-D, E, D);
pair R = extension(G, G+D-E, F, D);
pair U = extension(G, G+D-E, F, E);
pair X = midpoint(E--F);
pair Y = midpoint(F--D);
pair Z = midpoint(D--E);
pair Qp = 2*Q-G;
pair Rp = 2*R-G;
pair Sp = 2*S-G;
pair Up = 2*U-G;
pair B_1 = OP(circumcircle(B, Qp, Sp), incircle(A, B, C));
pair B_2 = IP(circumcircle(B, Qp, Sp), incircle(A, B, C));
pair C_1 = OP(circumcircle(C, Rp, Up), incircle(A, B, C));
pair C_2 = IP(circumcircle(C, Rp, Up), incircle(A, B, C));
draw(circumcircle(B, B_1, B_2), red);
draw(circumcircle(C, C_1, C_2), red);
pair V = extension(X, Y, A, B);
pair W = extension(X, Z, A, C);
pair T = extension(B, W, C, V);
draw(X--T, yellow);
draw(unitcircle, blue);
draw(B--W, heavygreen);
draw(C--V, heavygreen);
draw(B--V, blue);
draw(C--W, blue);
draw(C_2--V, heavygreen);
draw(B_2--W, heavygreen);
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(170));
dot("$G$", G, dir(G));
dot("$X$", X, dir(110));
dot("$Y$", Y, dir(350));
dot("$Z$", Z, dir(Z));
dot("$B_1$", B_1, dir(B_1));
dot("$B_2$", B_2, dir(100));
dot("$C_1$", C_1, dir(C_1));
dot("$C_2$", C_2, dir(40));
dot("$V$", V, dir(V));
dot("$W$", W, dir(W));
dot("$T$", T, dir(T));
/* Source generated by TSQ */
\end{asy}
\end{center}
Note $G$ is the symmedian point of $DEF$, so
it is well-known that $XG$ passes through the midpoint of $DK$.
So we just have to prove $G$ lies on the radical axis.
First, note that $\triangle DEF$ is the cevian triangle of the Gergonne point $G$.
Set $V = \ol{XY} \cap \ol{AB}$, $W = \ol{XZ} \cap \ol{AC}$, and $T = \ol{BW} \cap \ol{CV}$.
We begin with the following completely projective claim.
\begin{claim*}
The points $X$, $G$, $T$ are collinear.
\end{claim*}
\begin{proof}
It suffices to view $\triangle XYZ$ as any cevian triangle of $\triangle DEF$
(which is likewise any cevian triangle of $\triangle ABC$).
Then
\begin{itemize}
\ii By Cevian Nest on $\triangle ABC$,
it follows that $\ol{AX}$, $\ol{BY}$, $\ol{CZ}$ are concurrent.
\ii Hence $\triangle BYV$ and $\triangle CZW$ are perspective.
\ii Hence $\triangle BZW$ and $\triangle CYV$ are perspective too.
\ii Hence we deduce by Desargues theorem that $T$, $X$,
and $\ol{BZ} \cap \ol{CY}$ are collinear.
\ii Finally, the Cevian Nest theorem applied on $\triangle GBC$
(which has cevian triangles $\triangle DFE$, $\triangle XZY$)
we deduce $G$, $X$, and $\ol{BZ} \cap \ol{CY}$, proving the claim.
\end{itemize}
One could also proceed by using barycentric coordinates on $\triangle DEF$.
\end{proof}
\begin{remark*}
[Eric Shen]
The first four bullets can be replaced by non-projective means:
one can check that $\ol{BZ} \cap \ol{CY}$ is the radical center
of $(BIC)$, $(BB_1B_2)$, $(CC_1C_2)$ and therefore it lies on line $\ol{XT}$.
\end{remark*}
Now, we contend point $V$ is the radical center $(CC_1C_2)$, $(ABC)$ and $(DEF)$.
To see this, let $V' = \ol{ED} \cap \ol{AB}$;
then $(FV';AB)$ is harmonic, and $V$ is the midpoint of $\ol{FV'}$,
and thus $VA \cdot VB = VF^2 = VC_1 \cdot VC_2$.
So in fact $\ol{CV}$ is the radical axis of $(ABC)$ and $(CC_1C_2)$.
Similarly, $\ol{BW}$ is the radical axis of $(ABC)$ and $(BB_1B_2)$.
Thus $T$ is the radical center of $(ABC)$, $(BB_1B_2)$, $(CC_1C_2)$.
This completes the proof, as now $\ol{XT}$ is the desired radical axis.
\paragraph{Second solution (Evan Chen).}
As before, we just have to prove $G$ lies on the radical axis.
\begin{center}
\begin{asy}
size(14cm);
pair A = dir(50);
pair B = dir(220);
pair C = dir(320);
pair I = incenter(A, B, C);
pair D = foot(I, B, C);
pair E = foot(I, C, A);
pair F = foot(I, A, B);
pair G = extension(A, D, B, E);
filldraw(A--B--C--cycle, opacity(0.1)+lightcyan, blue);
draw(D--E--F--cycle, blue);
draw(incircle(A, B, C), blue);
draw(A--D, blue);
draw(B--E, blue);
draw(C--F, blue);
pair P = extension(G, G+E-F, D, F);
pair T = extension(G, G+E-F, D, E);
pair Q = extension(G, G+F-D, E, F);
pair S = extension(G, G+F-D, E, D);
pair R = extension(G, G+D-E, F, D);
pair U = extension(G, G+D-E, F, E);
pair X = midpoint(E--F);
pair Y = midpoint(F--D);
pair Z = midpoint(D--E);
pair Pp = 2*P-G;
pair Qp = 2*Q-G;
pair Rp = 2*R-G;
pair Sp = 2*S-G;
pair Tp = 2*T-G;
pair Up = 2*U-G;
draw(Pp--Tp, heavygreen);
draw(Rp--Up, heavygreen);
draw(Qp--Sp, heavygreen);
pair V = extension(B, X, Sp, Z);
pair W = extension(B, R, Qp, X);
pair B_1 = OP(circumcircle(B, Qp, Sp), incircle(A, B, C));
pair B_2 = IP(circumcircle(B, Qp, Sp), incircle(A, B, C));
pair C_1 = OP(circumcircle(C, Rp, Up), incircle(A, B, C));
pair C_2 = IP(circumcircle(C, Rp, Up), incircle(A, B, C));
pair K = foot(D, E, F);
pair M = midpoint(D--K);
draw(D--K, blue);
draw(C_1--C_2, red);
draw(B_1--B_2, red);
filldraw(circumcircle(P, Q, R), opacity(0.1)+lightgreen, green+1.4);
filldraw(circumcircle(Pp, Qp, Rp), opacity(0.1)+lightgreen, green);
draw(circumcircle(Sp, Q, P), orange);
draw(circumcircle(B, B_1, B_2), red);
draw(circumcircle(C, C_1, C_2), red);
draw(X--M, yellow);
draw(B--V, orange);
draw(B--W, orange);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$I$", I, dir(I));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$G$", G, dir(G));
dot("$P$", P, dir(200));
dot("$T$", T, dir(100));
dot("$Q$", Q, dir(70));
dot("$S$", S, dir(260));
dot("$R$", R, dir(250));
dot("$U$", U, dir(U));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(350));
dot("$Z$", Z, dir(Z));
dot("$P'$", Pp, dir(Pp));
dot("$Q'$", Qp, dir(Qp));
dot("$R'$", Rp, dir(240));
dot("$S'$", Sp, dir(Sp));
dot("$T'$", Tp, dir(Tp));
dot("$U'$", Up, dir(60));
dot("$V$", V, dir(V));
dot("$W$", W, dir(W));
dot("$B_1$", B_1, dir(B_1));
dot("$B_2$", B_2, dir(B_2));
dot("$C_1$", C_1, dir(C_1));
dot("$C_2$", C_2, dir(C_2));
dot("$K$", K, dir(200));
dot("$M$", M, dir(M));
\end{asy}
\end{center}
Construct parallelograms $GPFQ$, $GRDS$, $GTUE$
such that $P,R \in DF$, $S,T \in DE$, $Q,U \in EF$.
As $FG$ bisects $PQ$ and is isogonal to $FZ$,
we find $PQED$, hence $PQRU$, is cyclic.
Repeating the same logic and noticing $PR$, $ST$, $QU$ not concurrent,
all six points $PQRSTU$ are cyclic.
Moreover, since $PQ$ bisects $GF$, we see that
a dilation with factor $2$ at $G$ sends $PQ$ to $P', Q' \in AB$, say,
with $F$ the midpoint of $P'Q'$.
Define $R', S' \in BC$ similarly now and $T', U' \in CA$.
Note that $EQPDS'$ is in cyclic too,
as $\dang DS'Q = \dang DRS = \dang DEF$.
By homothety through $B$, points $B$, $P$, $X$ are collinear;
assume they meet $(EQPDS')$ again at $V$.
Thus $EVQPDS'$ is cyclic, and now
\[ \dang BVS' = \dang PVS' = \dang PQS = \dang PTS = \dang FED = \dang XEZ = \dang XVZ \]
hence $V$ lies on $(BQ'S')$.
Since $FB \parallel QP$, we get $EVFB$ is cyclic too,
so $XV \cdot XB = XE \cdot XF$ now;
thus $X$ lies on the radical axis of $(BS'Q')$ and $(DEF)$.
By the same argument with $W \in BZ$, we get $Z$ lies on the radical axis too.
Thus the radical axis of $(BS'Q')$ and $(DEF)$ must be line $XZ$,
which coincides with $B_1B_2$; so $(BB_1B_2) = (BS'Q')$.
Analogously, $(CC_1C_2) = (CR'U')$.
Since $G = Q'S' \cap R'U'$, we need only prove that $Q'R'S'U'$ is cyclic.
But $QRSU$ is cyclic, so we are done.
The circle $(PQRSTU)$ is called the \emph{Lemoine circle} of $ABC$.
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