problem_id
stringlengths 16
19
| problem
stringlengths 69
2.04k
| solution
stringlengths 60
23.9k
|
|---|---|---|
JMO-2021-notes_6
|
Let $n \ge 4$ be an integer.
Find all positive real solutions to the following
system of $2n$ equations:
\begin{align*}
a_1 &= \frac{1}{a_{2n}} + \frac{1}{a_{2}}, & a_2 &= a_1 + a_3, \\[1ex]
a_3 &= \frac{1}{a_{2}} + \frac{1}{a_{4}}, & a_4 &= a_3 + a_5, \\[1ex]
a_5 &= \frac{1}{a_{4}} + \frac{1}{a_{6}}, & a_6 &= a_5 + a_7, \\[1ex]
&\vdotswithin{=} &&\vdotswithin{=} \\
a_{2n-1} &= \frac{1}{a_{2n-2}} + \frac{1}{a_{2n}}, & a_{2n} &= a_{2n-1} + a_1.
\end{align*}
\end{enumerate}
|
The answer is that the only solution is
$(1,2,1,2,\dots,1,2)$ which works.
We will prove $a_{2k}$ is a constant sequence,
at which point the result is obvious.
\paragraph{First approach (Andrew Gu).}
Apparently, with indices modulo $2n$, we should have
\[ a_{2k} = \frac{1}{a_{2k-2}}
+ \frac{2}{a_{2k}} + \frac{1}{a_{2k+2}} \]
for every index $k$ (this eliminates all $a_{\text{odd}}$'s).
Define
\[ m = \min_k a_{2k} \qquad\text{and}\qquad M = \max_k a_{2k}. \]
Look at the indices $i$ and $j$
achieving $m$ and $M$ to respectively get
\begin{align*}
m &= \frac2m + \frac{1}{a_{2i-2}} + \frac{1}{a_{2i+2}}
\ge \frac2m + \frac1M + \frac1M = \frac2m + \frac2M \\[1ex]
M &= \frac2M + \frac{1}{a_{2j-2}} + \frac{1}{a_{2j+2}}
\le \frac2M + \frac1m + \frac1m = \frac2m + \frac2M.
\end{align*}
Together this gives $m \ge M$, so $m = M$.
That means $a_{2i}$ is constant as $i$ varies,
solving the problem.
\paragraph{Second approach (author's solution).}
As before, we have
\[ a_{2k} = \frac{1}{a_{2k-2}}
+ \frac{2}{a_{2k}} + \frac{1}{a_{2k+2}} \]
The proof proceeds in three steps.
\begin{itemize}
\ii Define
\[ S = \sum_k a_{2k},
\quad\text{and}\quad
T = \sum_k \frac{1}{a_{2k}}. \]
Summing gives $S = 4T$.
On the other hand, Cauchy-Schwarz says $S \cdot T \ge n^2$,
so $T \ge \half n$.
\ii On the other hand,
\[ 1 = \frac{1}{a_{2k-2} a_{2k}}
+ \frac{2}{a_{2k}^2} + \frac{1}{a_{2k} a_{2k+2}} \]
Sum this modified statement to obtain
\[
n = \sum_k \left( \frac{1}{a_{2k}} + \frac{1}{a_{2k+2}} \right)^2
\overset{\text{QM-AM}}{\ge}
\frac 1n \left( \sum_k \frac{1}{a_{2k}} + \frac{1}{a_{2k+2}} \right)^2
= \frac 1n \left( 2T \right)^2
\]
So $T \le \half n$.
\ii Since $T \le \half n$ and $T \ge \half n$,
we must have equality everywhere above.
This means $a_{2k}$ is a constant sequence.
\end{itemize}
\begin{remark*}
The problem is likely intractable over $\CC$,
in the sense that one gets a high-degree polynomial
which almost certainly has many complex roots.
So it seems likely that most solutions must involve
some sort of inequality,
using the fact we are over $\RR_{>0}$ instead.
\end{remark*}
|
JMO-2022-notes_1
|
For which positive integers $m$ does there exist an infinite sequence in $\ZZ/m\ZZ$
which is both an arithmetic progression and a geometric progression,
but is nonconstant?
|
Answer: $m$ must \emph{not} be squarefree.
The problem is essentially asking when there exists
a nonconstant arithmetic progression in $\ZZ/m\ZZ$
which is also a geometric progression.
Now,
\begin{itemize}
\ii If $m$ is squarefree, then consider three $(s-d, d, s+d)$ in arithmetic progression.
It's geometric if and only if $d^2 = (s-d)(s+d) \pmod m$, meaning $d^2 \equiv 0 \pmod m$.
Then $d \equiv 0 \pmod m$.
So any arithmetic progression which is also geometric is constant in this case.
\ii Conversely if $p^2 \mid m$ for some prime $p$,
then any arithmetic progression with common difference $m/p$ is geometric
by the same calculation.
\end{itemize}
|
JMO-2022-notes_2
|
Let $a$ and $b$ be positive integers.
Every cell of an $(a+b+1)\times (a+b+1)$ grid is colored either amber or bronze
such that there are at least $a^2+ab-b$ amber cells
and at least $b^2+ab-a$ bronze cells.
Prove that it is possible to choose $a$ amber cells and $b$ bronze cells
such that no two of the $a+b$ chosen cells lie in the same row or column.
|
\begin{claim*}
There exists a transversal $T_a$ with at least $a$ amber cells.
Analogously, there exists a transversal $T_b$ with at least $b$ bronze cells.
\end{claim*}
\begin{proof}
If one picks a random transversal, the expected value
of the number of amber cells is at least
\[ \frac{a^2+ab-b}{a+b+1} = (a-1) + \frac{1}{a+b+1} > a-1. \qedhere \]
\end{proof}
Now imagine we transform $T_a$ to $T_b$ in some number of steps,
by repeatedly choosing cells $c$ and $c'$ and swapping them
with the two other corners of the rectangle formed by their row/column,
as shown in the figure.
\begin{center}
\begin{asy}
filldraw(shift(0,0)*unitsquare, lightcyan, blue);
filldraw(shift(3,2)*unitsquare, lightcyan, blue);
draw(shift(3,0)*unitsquare, dotted+blue);
draw(shift(0,2)*unitsquare, dotted+blue);
label("$c$", (0.5,0.5));
label("$c'$", (3.5,2.5));
label(scale(2.5)*"$\implies$", (5.5, 1.5));
filldraw(shift(7,2)*unitsquare, lightcyan, blue);
filldraw(shift(10,0)*unitsquare, lightcyan, blue);
draw(shift(7,0)*unitsquare, dotted+blue);
draw(shift(10,2)*unitsquare, dotted+blue);
\end{asy}
\end{center}
By ``discrete intermediate value theorem'',
the number of amber cells will be either $a$ or $a+1$
at some point during this transformation.
This completes the proof.
|
JMO-2022-notes_3
|
Let $b\geq 2$ and $w\geq 2$ be fixed integers, and $n=b+w$.
Given are $2b$ identical black rods and $2w$ identical white rods,
each of side length $1$.
We assemble a regular $2n$-gon using these rods
so that parallel sides are the same color.
Then, a convex $2b$-gon $B$ is formed by translating the black rods,
and a convex $2w$-gon $W$ is formed by translating the white rods.
An example of one way of doing the assembly when $b=3$ and $w=2$ is shown below,
as well as the resulting polygons $B$ and $W$.
\begin{center}
\begin{asy}
size(10cm);
real w = 2*Sin(18);
real h = 0.10 * w;
real d = 0.33 * h;
picture wht;
picture blk;
draw(wht, (0,0)--(w,0)--(w+d,h)--(-d,h)--cycle);
fill(blk, (0,0)--(w,0)--(w+d,h)--(-d,h)--cycle, black);
// draw(unitcircle, blue+dotted);
// Original polygon
add(shift(dir(108))*blk);
add(shift(dir(72))*rotate(324)*blk);
add(shift(dir(36))*rotate(288)*wht);
add(shift(dir(0))*rotate(252)*blk);
add(shift(dir(324))*rotate(216)*wht);
add(shift(dir(288))*rotate(180)*blk);
add(shift(dir(252))*rotate(144)*blk);
add(shift(dir(216))*rotate(108)*wht);
add(shift(dir(180))*rotate(72)*blk);
add(shift(dir(144))*rotate(36)*wht);
// White shifted
real Wk = 1.2;
pair W1 = (1.8,0.1);
pair W2 = W1 + w*dir(36);
pair W3 = W2 + w*dir(108);
pair W4 = W3 + w*dir(216);
path Wgon = W1--W2--W3--W4--cycle;
draw(Wgon);
pair WO = (W1+W3)/2;
transform Wt = shift(WO)*scale(Wk)*shift(-WO);
draw(Wt * Wgon);
label("$W$", WO);
/*
draw(W1--Wt*W1);
draw(W2--Wt*W2);
draw(W3--Wt*W3);
draw(W4--Wt*W4);
*/
// Black shifted
real Bk = 1.10;
pair B1 = (1.5,-0.1);
pair B2 = B1 + w*dir(0);
pair B3 = B2 + w*dir(324);
pair B4 = B3 + w*dir(252);
pair B5 = B4 + w*dir(180);
pair B6 = B5 + w*dir(144);
path Bgon = B1--B2--B3--B4--B5--B6--cycle;
pair BO = (B1+B4)/2;
transform Bt = shift(BO)*scale(Bk)*shift(-BO);
fill(Bt * Bgon, black);
fill(Bgon, white);
label("$B$", BO);
\end{asy}
\end{center}
Prove that the difference of the areas of $B$ and $W$
depends only on the numbers $b$ and $w$,
and not on how the $2n$-gon was assembled.
|
We are going to prove that one may swap a black rod with an adjacent white rod
(as well as the rods parallel to them)
without affecting the difference in the areas of $B-W$.
Let $\vec u$ and $\vec v$ denote the originally black and white vectors
that were adjacent on the $2n$-gon and are now going to be swapped.
Let $\vec x$ denote the sum of all the other black vectors between
$\vec u$ and $-\vec u$, and define $\vec y$ similarly.
See the diagram below,
where $B_0$ and $W_0$ are the polygons before the swap,
and $B_1$ and $W_1$ are the resulting changed polygons.
\begin{center}
\begin{asy}
size(12cm);
picture B0;
picture W0;
picture B1;
picture W1;
pen br = black + 2;
pen bp = black + dashed;
path barc = (0,0)..(1,1)..(6,1)..(7,0);
draw(B0, (3,-1)--(4,1), br);
draw(B0, (-4,-1)--(-3,1), br);
draw(B0, shift(-3,1)*barc, bp);
draw(B0, shift(3,-1)*rotate(180)*barc, bp);
label(B0, "$B_0$", (0,0));
draw(B0, (-3,1)--(4,1), red, EndArrow(TeXHead), Margins);
draw(B0, (3,-1)--(-4,-1), red, EndArrow(TeXHead), Margins);
label(B0, "$\vec x$", (-3,1)--(4,1), dir(90), red);
label(B0, "$-\vec x$", (3,-1)--(-4,-1), dir(-90), red);
label(B0, "$\vec u$", (-3.5,0), dir(160));
draw(B0, (-4,-1)--(-3,1), black+1, EndArrow(TeXHead));
draw(B1, (4,-1)--(3,1), br);
draw(B1, (-3,-1)--(-4,1), br);
label(B1, "$B_1$", (0,0));
draw(B1, shift(-4,1)*barc, bp);
draw(B1, shift(4,-1)*rotate(180)*barc, bp);
draw(B1, (-4,1)--(3,1), red, EndArrow(TeXHead), Margins);
draw(B1, (4,-1)--(-3,-1), red, EndArrow(TeXHead), Margins);
label(B1, "$\vec x$", (-4,1)--(3,1), dir(90), red);
label(B1, "$-\vec x$", (4,-1)--(-3,-1), dir(-90), red);
label(B1, "$\vec v$", (-3.5,0), dir(210));
draw(B1, (-3,-1)--(-4,1), black+1, EndArrow(TeXHead));
path warc = (0,0)..(4,1)..(9,-2);
draw(W0, shift(-5,1)*warc, bp);
draw(W0, shift(5,-3)*rotate(180)*warc, bp);
draw(W0, (5,-3)--(4,-1), br);
draw(W0, (-4,-1)--(-5,1), br);
draw(W0, (-5,1)--(4,-1), red, EndArrow(TeXHead), Margins);
draw(W0, (5,-3)--(-4,-1), red, EndArrow(TeXHead), Margins);
draw(W0, (5,-3)--(4,-1), white+1.2);
draw(W0, (-4,-1)--(-5,1), white+1.2);
label(W0, "$W_0$", (0,-1));
label(W0, "$\vec v$", (-4.5,0), dir(200));
label(W0, "$\vec y$", (-5,1)--(4,-1), dir(70), red);
label(W0, "$-\vec y$", (5,-3)--(-4,-1), dir(250), red);
draw(W1, shift(-4,1)*warc, bp);
draw(W1, shift(4,-3)*rotate(180)*warc, bp);
draw(W1, (4,-3)--(5,-1), br);
draw(W1, (-5,-1)--(-4,1), br);
draw(W1, (4,-3)--(5,-1), white+1.2);
draw(W1, (-5,-1)--(-4,1), white+1.2);
draw(W1, (-4,1)--(5,-1), red, EndArrow(TeXHead), Margins);
draw(W1, (4,-3)--(-5,-1), red, EndArrow(TeXHead), Margins);
label(W1, "$W_1$", (0,-1));
label(W1, "$\vec u$", (-4.5,0), dir(140));
label(W1, "$\vec y$", (0.5,0), dir(70), red);
label(W1, "$-\vec y$", (-0.5,-2), dir(250), red);
add(B0);
add(shift(12,0)*B1);
add(shift(0,-6)*W0);
add(shift(12,-6)*W1);
\end{asy}
\end{center}
Observe that the only change in $B$ and $W$
is in the parallelograms shown above in each diagram.
Letting $\wedge$ denote the wedge product,
we need to show that
\[ \vec u \wedge \vec x - \vec v \wedge \vec y
= \vec v \wedge \vec x - \vec u \wedge \vec y \]
which can be rewritten as
\[ (\vec u - \vec v) \wedge (\vec x + \vec y) = 0. \]
In other words, it would suffice to show
$\vec u - \vec v$ and $\vec x + \vec y$ are parallel.
(Students not familiar with wedge products can replace
every $\wedge$ with the cross product $\times$ instead.)
\begin{claim*}
Both $\vec u - \vec v$ and $\vec x + \vec y$
are perpendicular to vector $\vec u + \vec v$.
\end{claim*}
\begin{proof}
We have $(\vec u - \vec v) \perp (\vec u + \vec v)$
because $\vec u$ and $\vec v$ are the same length.
For the other perpendicularity, note that
$\vec u + \vec v + \vec x + \vec y$
traces out a diameter of the circumcircle of the original $2n$-gon;
call this diameter $AB$, so
\[ A + \vec u + \vec v + \vec x + \vec y = B. \]
Now point $A + \vec u + \vec v$ is a point on this semicircle,
which means (by the inscribed angle theorem)
the angle between $\vec u + \vec v$ and $\vec x + \vec y$ is $90^\circ$.
\end{proof}
|
JMO-2022-notes_4
|
Let $ABCD$ be a rhombus, and let $K$ and $L$ be points
such that $K$ lies inside the rhombus, $L$ lies outside the rhombus,
and $KA = KB = LC = LD$.
Prove that there exist points $X$ and $Y$ on lines $AC$ and $BD$
such that $KXLY$ is also a rhombus.
|
To start, notice that $\triangle AKB \cong \triangle DLC$ by SSS.
Then by the condition $K$ lies inside the rhombus while $L$ lies outside it,
we find that the two congruent triangles are just translations of each other
(i.e.\ they have the same orientation).
\paragraph{First solution.}
Let $M$ be the midpoint of $\ol{KL}$ and is $O$ the center of the rhombus.
\begin{claim*}
$\ol{MO} \perp \ol{AB}$.
\end{claim*}
\begin{proof}
Let $U$ and $V$ denote the midpoint of $\ol{AB}$ and $\ol{CD}$ respectively.
Then $\ol{KU}$ and $\ol{LV}$ are obviously translates,
and perpendicular to $\ol{AB} \parallel \ol{CD}$.
Since $M$ is the midpoint of $\ol{KL}$ and $O$ is the midpoint of $\ol{UV}$,
the result follows.
\end{proof}
We choose $X$ and $Y$ to be the intersections of
the perpendicular bisector of $\ol{KL}$ with $\ol{AC}$ and $\ol{BD}$.
\begin{center}
\begin{asy}
pair A = (-4,0);
pair B = (0,-3);
pair C = -A;
pair D = -B;
pair K = midpoint(A--B) + dir(90)*(B-A)*0.2;
pair L = D+K-A;
pair X = -A+2*foot(K,A,C);
pair Y = -D+2*foot(L,B,D);
pair O = origin;
pair M = midpoint(K--L);
pair U = midpoint(A--B);
pair V = midpoint(C--D);
filldraw(A--B--C--D--cycle, opacity(0.1)+lightred, red);
draw(A--C, red);
draw(B--D, red);
filldraw(A--K--B--cycle, opacity(0.1)+cyan, blue);
filldraw(C--L--D--cycle, opacity(0.1)+cyan, blue);
draw(X--Y, deepgreen);
draw(L--K, deepgreen);
draw(K--U, black+1.5);
draw(L--V, black+1.5);
draw(M--O, black+1.5);
dot("$A$", A, dir(210));
dot("$B$", B, dir(340));
dot("$C$", C, dir(-30));
dot("$D$", D, dir(160));
dot("$K$", K, dir(100));
dot("$L$", L, dir(90));
dot("$X$", X, dir(305));
dot("$Y$", Y, dir(200));
dot("$M$", M, dir(-5));
dot("$O$", O, dir(225));
dot("$U$", U, dir(250));
dot("$V$", V, dir(250));
\end{asy}
\end{center}
\begin{claim*}
The midpoint of $\ol{XY}$ coincides with the midpoint of $\ol{KL}$.
\end{claim*}
\begin{proof}
Because
\begin{align*}
\ol{XY} &\perp \ol{KL} \parallel \ol{BC} \\
\ol{MO} &\perp \ol{AB} \\
\ol{BD} &\perp \ol{AC}
\end{align*}
it follows that $\triangle MOY$,
which was determined by the three lines $\ol{XY}$, $\ol{MO}$, $\ol{BD}$,
is similar to $\triangle ABC$.
In particular, it is isosceles with $MY = MO$.
Analogously, $MX = MO$.
\end{proof}
\begin{remark*}
It is also possible to simply use coordinates to prove both claims.
\end{remark*}
\paragraph{Second solution (author's solution).}
In this solution, we instead define
$X$ and $Y$ as the intersections of the circles
centered at $K$ and $L$ of equal radii $KA$,
which will be denoted $\omega_K$ and $\omega_L$.
It is clear that $KXLY$ is a rhombus under this construction,
so it suffices to show that $X$ and $Y$ lie on $AC$ and $BD$ (in some order).
\begin{center}
\begin{asy}
pair A = (-4,0);
pair B = (0,-3);
pair C = -A;
pair D = -B;
pair K = midpoint(A--B) + dir(90)*(B-A)*0.2;
pair L = D+K-A;
pair X = -A+2*foot(K,A,C);
pair Y = -D+2*foot(L,B,D);
filldraw(A--B--C--D--cycle, opacity(0.1)+lightred, red);
draw(A--C, red);
draw(B--D, red);
filldraw(A--K--B--cycle, opacity(0.1)+cyan, blue);
filldraw(C--L--D--cycle, opacity(0.1)+cyan, blue);
draw(X--L--Y--K--cycle, deepgreen);
draw(circle(K,abs(K-A)), gray);
draw(circle(L,abs(L-C)), gray);
dot("$A$", A, dir(210));
dot("$B$", B, dir(340));
dot("$C$", C, dir(-30));
dot("$D$", D, dir(160));
dot("$K$", K, dir(250));
dot("$L$", L, dir(90));
dot("$X$", X, dir(305));
dot("$Y$", Y, dir(200));
label("$\omega_L$", L + abs(L-C)*dir(40), dir(40));
label("$\omega_K$", K + abs(K-B)*dir(220), dir(220));
\end{asy}
\end{center}
To see this, let $\ol{AC}$ meet $\omega_K$ again at $X'$.
We have
\[ \dang CX'D = \dang BX'C = \dang BX'A = \half \opname{m} \arc{AB} = \opname{m} \arc{CD} \]
where the arcs are directed modulo $360\dg$;
here $\arc{AB}$ is the arc of $\omega_K$ cut out by $\dang AXB$,
and $\arc{DC}$ is the analogous arc of $\omega_L$.
This implies $X'$ lies on $\omega_L$ by the inscribed angle theorem.
Hence $X = X'$, and it follows $X$ lies on $\ol{AC}$.
Analogously $Y$ lies on $BD$.
\begin{remark*}
The angle calculation above can also be replaced with a length calculation, as follows.
Let $M$ and $N$ be the projections of $K$ and $L$ onto $\ol{AC}$, respectively.
Then $X'$ is the reflection of $A$ across $M$;
analogously, the second intersection $X''$ with $\ol{AC}$
should be the reflection of $C$ across $N$.
So to get $X = X' = X''$, we would need to show $AC = 2MN$.
However, note that $AKLD$ is a parallelogram.
As $MN$ was the projection of $\ol{KL}$ onto $\ol{AC}$,
its length should be the same as the projection of $\ol{AD}$ onto $\ol{AC}$,
which is obviously $\half AC$ because the projection of $D$ onto $\ol{AC}$ is
exactly the midpoint of $\ol{AC}$ (i.e.\ the center of the rhombus).
\end{remark*}
|
JMO-2022-notes_5
|
Find all pairs of primes $(p, q)$ for which $p-q$ and $pq-q$ are both perfect squares.
|
The answer is $(3,2)$ only.
This obviously works so we focus on showing it is the only one.
\paragraph{Approach using difference of squares (from author).}
Set
\begin{align*}
a^2 &= p-q \\
b^2 &= pq-q.
\end{align*}
Note that $0 < a < p$, and $0 < b < p$ (because $q \le p$).
Now subtracting gives
\[ \underbrace{(b-a)}_{<p} \underbrace{(b+a)}_{<2p} = b^2-a^2 = p(q-1) \]
The inequalities above now force $b+a = p$.
Hence $q-1 = b-a$.
This means $p$ and $q-1$ have the same parity,
which can only occur if $q = 2$.
Finally, taking mod $3$ shows $p \equiv 0 \pmod 3$.
So $(3,2)$ is the only possibility (and it does work).
\paragraph{Divisibility approach (Aharshi Roy).}
Since $pq-q = q(p-1)$ is a square, it follows that $q$ divides $p-1$
and that $\frac{p-1}{q}$ is a perfect square too.
Hence the number
\[ s^2 \coloneq (p-q) \cdot \frac{p-1}{q} = \frac{p^2-qp-p+q}{q} \]
is also a perfect square.
Rewriting this equation gives
\[ q = \frac{p^2-p}{s^2 + (p-1)}. \]
In particular, $s^2 + (p-1)$ divides $p^2-p$, and in particular $s \leq p$.
We consider two cases:
\begin{itemize}
\ii If $s^2+(p-1)$ is not divisible by $p$, then it must divide $p-1$,
which can only happen if $s^2 = 0$, or $p = q$.
However, it's easy to check there are no solutions in this case.
\ii Otherwise, we should have $s^2 \equiv 1 \pmod p$,
so either $s = 1$ or $s = p-1$.
If $s = p-1$ we get $q=1$ which is absurd.
On the other hand, if $s=1$ we conclude $q = p-1$ and hence $q=2$, $p=3$.
\end{itemize}
|
JMO-2022-notes_6
|
Let $a_0$, $b_0$, $c_0$ be complex numbers, and define
\begin{align*}
a_{n+1} &= a_n^2 + 2b_nc_n \\
b_{n+1} &= b_n^2 + 2c_na_n \\
c_{n+1} &= c_n^2 + 2a_nb_n
\end{align*}
for all nonnegative integers $n$.
Suppose that $\max{\{|a_n|, |b_n|, |c_n|\}} \leq 2022$ for all $n \ge 0$.
Prove that \[ |a_0|^2 + |b_0|^2 + |c_0|^2 \leq 1. \]
\end{enumerate}
|
For brevity, set $s_n \coloneq |a_n|^2 + |b_n|^2 + |c_n|^2$.
Note that the $s_n$ are real numbers.
\begin{claim*}
[Key miraculous identity]
We have
\[ s_{n+1} - s_n^2 = 2 |a_n \ol{b_n} + b_n \ol{c_n} + c_n \ol{a_n}|^2. \]
\end{claim*}
\begin{proof}
We prove this by mechanical calculation. First,
\begin{align*}
s_{n+1} &= \left\lvert a_n^2+2b_nc_n \right\rvert^2 +
\left\lvert b_n^2+2c_na_n \right\rvert^2 +
\left\lvert c_n^2+2a_nb_n \right\rvert^2 \\
&= \sum_{\text{cyc}} \left\lvert a_n^2+2b_nc_n \right\rvert^2 \\
&= \sum_{\text{cyc}} (a_n^2+2b_nc_n)(\ol{a_n}^2 + 2\ol{b_n}\ol{c_n}) \\
&= \sum_{\text{cyc}} \left( |a_n|^4 + 2 \ol{a_n}^2 b_nc_n
+ 2a_n^2 \ol{b_n} \ol{c_n} + 4 |b_n|^2 |c_n|^2 \right) \\
&= s_n^2 + 2 \sum_{\text{cyc}}
\left( \ol{a_n}^2 b_nc_n
+ a_n^2 \ol{b_n} \ol{c_n} + |b_n|^2 |c_n|^2 \right).
\end{align*}
Meanwhile,
\begin{align*}
|a_n \ol{b_n} + b_n \ol{c_n} + c_n \ol{a_n}|^2
&= \left( a_n \ol{b_n} + b_n \ol{c_n} + c_n \ol{a_n} \right)
\left( \ol{a_n} b_n + \ol{b_n} c_n + \ol{c_n} a_n \right) \\
&= |a_n|^2 |b_n^2| + a_n \ol{b_n}^2 c_n + a_n^2 \ol{b_n} \ol{c_n} \\
&\qquad + \ol{a_n} b_n^2 \ol{c_n} + |b_n|^2 |c_n|^2 + a_n b_n \ol{c_n}^2 \\
&\qquad + \ol{a_n}^2 b_n c_n + \ol{a_n} \ol{b_n} c_n^2 + |a_n|^2 |c_n|^2
\end{align*}
which exactly matches the earlier sum, term for term.
\end{proof}
In particular, $s_{n+1} \ge s_n^2$, so applying repeatedly,
\[ s_n \ge s_0^{2^n}. \]
Hence if $s_0 > 1$, it follows $s_n$ is unbounded, contradicting
$\max{\{|a_n|, |b_n|, |c_n|\}} \leq 2022$.
\begin{remark*}
The originally intended solution was to capture all three
recursions in the following way.
First, change the recursion to
\begin{align*}
a_{n+1} &= a_n^2 + 2b_nc_n \\
c_{n+1} &= b_n^2 + 2c_na_n \\
b_{n+1} &= c_n^2 + 2a_nb_n
\end{align*}
which is OK because we are just rearranging the terms in each triple.
Then if $\omega$ is any complex number with $\omega^3 = 1$, and we define
\[ z_n \coloneq a_n + b_n \omega + c_n \omega^2, \]
the recursion amounts to saying that $z_{n+1} = z_n^2$.
This allows us to analyze $|z_n|$ in a similar way as above,
as now $|z_n| = |z_0|^{2^n}$.
\end{remark*}
|
JMO-2023-notes_1
|
Find all triples of positive integers $(x,y,z)$ satisfying
\[ 2(x+y+z+2xyz)^2 = (2xy+2yz+2zx+1)^2 + 2023. \]
|
Answer: $(3,3,2)$ and permutations.
The solution hinges upon the following claim:
\begin{claim*}
The identity
\[ 2(x+y+z+2xyz)^2 - (2xy+2yz+2zx+1)^2 = (2x^2-1)(2y^2-1)(2z^2-1) \]
is true.
\end{claim*}
\begin{proof}
This can be proved by manually expanding; we show where it ``came from''.
In algebraic number theory, there is a \emph{norm function}
$\opname{Norm} \colon \QQ(\sqrt2) \to \QQ$
defined by
\[ \opname{Norm}(a+b\sqrt2) = a^2-2b^2 \]
which is \emph{multiplicative}, meaning
\[ \opname{Norm}(u \cdot v) = \opname{Norm}(u) \cdot \opname{Norm}(v). \]
This means that for any rational numbers $x$, $y$, $z$, we should have
\begin{align*}
&\phantom{=} \opname{Norm}
\left( (1+\sqrt2x)(1+\sqrt2y)(1+\sqrt2z) \right) \\
&= \opname{Norm}(1+\sqrt2x)
\cdot \opname{Norm}(1+\sqrt2y)
\cdot \opname{Norm}(1+\sqrt2z).
\end{align*}
But $(1+\sqrt2x)(1+\sqrt2y)(1+\sqrt2z) = (2xy+2yz+2zx+1) + (x+y+z+2xyz)\sqrt2$
so the above equation is the negative of the desired identity.
\end{proof}
We are thus reduced to find positive integers $x$, $y$, $z$ satisfying
\[ (2x^2-1)(2y^2-1)(2z^2-1) = 2023 = 7 \cdot 17^2. \]
Each of the factors is a positive integer greater than $1$.
The only divisors of $2023$ of the form $2t^2-1$ are $1$, $7$, $17$.
This gives the answers claimed.
|
JMO-2023-notes_2
|
In an acute triangle $ABC$, let $M$ be the midpoint of $\ol{BC}$.
Let $P$ be the foot of the perpendicular from $C$ to $AM$.
Suppose that the circumcircle of triangle $ABP$
intersects line $BC$ at two distinct points $B$ and $Q$.
Let $N$ be the midpoint of $\ol{AQ}$.
Prove that $NB = NC$.
|
We show several different approaches.
In all solutions, let $D$ denote the foot of the altitude from $A$.
\begin{center}
\begin{asy}
size(10cm);
pair A = dir(115);
pair B = dir(210);
pair C = dir(330);
pair D = foot(A, B, C);
pair M = midpoint(B--C);
pair P = foot(C, A, M);
pair Q = 2*M-D;
filldraw(A--B--C--cycle, opacity(0.1)+cyan, blue);
filldraw(circumcircle(A, B, P), opacity(0.1)+yellow, orange+dashed);
draw(A--D, blue);
draw(A--P--C, deepgreen);
draw(A--Q, red+dashed);
pair N = midpoint(A--Q);
draw(B--N--C, red+dotted);
pair R = foot(B, A, M);
draw(B--R, dashed);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$M$", M, dir(45));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(315));
dot("$N$", N, dir(45));
dot("$R$", R, dir(30));
/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
!size(10cm);
A = dir 115
B = dir 210
C = dir 330
D = foot A B C
M 45 = midpoint B--C
P = foot C A M
Q 315 = 2*M-D
A--B--C--cycle / 0.1 cyan / blue
circumcircle A B P / 0.1 yellow / orange dashed
A--D blue
A--P--C deepgreen
A--Q / red dashed
N 45 = midpoint A--Q
B--N--C / red dotted
R 30 = foot B A M
B--R / dashed
*/
\end{asy}
\end{center}
\paragraph{Most common synthetic approach.}
The solution hinges on the following claim:
\begin{claim*}
$Q$ coincides with the reflection of $D$ across $M$.
\end{claim*}
\begin{proof}
Note that $\dang ADC = \dang APC = 90\dg$, so $ADPC$ is cyclic.
Then by power of a point (with the lengths directed),
\[ MB \cdot MQ = MA \cdot MP = MC \cdot MD. \]
Since $MB = MC$, the claim follows.
\end{proof}
It follows that $\ol{MN} \parallel \ol{AD}$,
as $M$ and $N$ are respectively the midpoints of $\ol{AQ}$ and $\ol{DQ}$.
Thus $\ol{MN} \perp \ol{BC}$,
and so $N$ lies on the perpendicular bisector of $\ol{BC}$, as needed.
\begin{remark*}
[David Lin]
One can prove the main claim without power of a point as well, as follows:
Let $R$ be the foot from $B$ to $\ol{AM}$, so $BRCP$ is a parallelogram.
Note that $ABDR$ is cyclic, and hence
\[ \dang DRM = \dang DBA = QBA = \dang QPA = \dang QPM. \]
Thus, $\ol{DR} \parallel \ol{PQ}$, so $DRQP$ is also a parallelogram.
\end{remark*}
\paragraph{Synthetic approach with no additional points at all.}
\begin{claim*}
$\triangle BPC \sim \triangle ANM$ (oppositely oriented).
\end{claim*}
\begin{proof}
We have $\triangle BMP \sim \triangle AMQ$ from the given concyclicity of $ABPQ$.
Then
\[ \frac{BM}{BP} = \frac{AM}{AQ} \implies
\frac{2BM}{BP} = \frac{AM}{AQ/2} \implies
\frac{BC}{BP} = \frac{AM}{AN} \]
implying the similarity (since $\dang MAQ = \dang BPM$).
\end{proof}
This similarity gives us the equality of directed angles
\[ \dang \left( BC, MN \right) = -\dang \left( PC, AM \right) = 90\dg \]
as desired.
\paragraph{Synthetic approach using only the point $R$.}
Again let $R$ be the foot from $B$ to $\ol{AM}$, so $BRCP$ is a parallelogram.
\begin{claim*}
$ARQC$ is cyclic; equivalently, $\triangle MAQ \sim \triangle MCR$.
\end{claim*}
\begin{proof}
$MR \cdot MA = MP \cdot MA = MB \cdot MQ = MC \cdot MQ$.
\end{proof}
Note that in $\triangle MCR$, the $M$-median is parallel to $\ol{CP}$
and hence perpendicular to $\ol{RM}$.
The same should be true in $\triangle MAQ$ by the similarity,
so $\ol{MN} \perp \ol{MQ}$ as needed.
\paragraph{Cartesian coordinates approach with power of a point.}
Suppose we set $B = (-1,0)$, $M = (0,0)$, $C = (1,0)$, and $A = (a,b)$.
One may compute:
\begin{align*}
\overleftrightarrow{AM} : 0 &= bx - ay \iff y = \frac ba x \\
\overleftrightarrow{CP} : 0 &= a(x-1) + by
\iff y = -\frac ab (x-1) = -\frac ab x + \frac ab. \\
P &= \left( \frac{a^2}{a^2+b^2}, \frac{ab}{a^2+b^2} \right) \\
\end{align*}
Now note that
\[ AM = \sqrt{a^2+b^2}, \qquad PM = \frac{a}{\sqrt{a^2+b^2}} \]
together with power of a point
\[ AM \cdot PM = BM \cdot QM \]
to immediately deduce that $Q = (a,0)$.
Hence $N = (0, b/2)$ and we're done.
\paragraph{Cartesian coordinates approach without power of a point (outline).}
After computing $A$ and $P$ as above, one could also directly calculate
\begin{align*}
\text{Perpendicular bisector of $\ol{AB}$}:
y &= -\frac{a+1}{b} x + \frac{a^2+b^2-1}{2b} \\
\text{Perpendicular bisector of $\ol{PB}$}:
y &= - \left( \frac{2a}{b} + \frac{b}{a} \right) x - \frac{b}{2a} \\
\text{Perpendicular bisector of $\ol{PA}$}:
y &= - \frac{a}{b} x + \frac{a+a^2+b^2}{2b}. \\
\text{Circumcenter of $\triangle PAB$}
&= \left( -\frac{a+1}{2}, \frac{2a^2+2a+b^2}{2b} \right).
\end{align*}
This is enough to extract the coordinates of $Q = (\bullet, 0)$,
because $B = (-1,0)$ is given, and the $x$-coordinate
of the circumcenter should be the average of the $x$-coordinates of $B$ and $Q$.
In other words, $Q = (-a,0)$.
Hence, $N = \left( 0, \frac b2 \right)$, as needed.
\paragraph{Ill-advised barycentric approach (outline).}
Use reference triangle $ABC$.
The $A$-median is parametrized by $(t:1:1)$ for $t \in \RR$.
So because of $\ol{CP} \perp \ol{AM}$, we are looking for $t$ such that
\[
\left( \frac{t \vec A + \vec B + \vec C}{t+2} - \vec C \right)
\perp \left( A - \frac{\vec B + \vec C}{2} \right).
\]
This is equivalent to
\[
\left( t \vec A + \vec B - (t+1) \vec C \right)
\perp \left( 2 \vec A - \vec B - \vec C \right).
\]
By the perpendicularity formula for barycentric coordinates (EGMO 7.16),
this is equivalent to
\begin{align*}
0 &= a^2t - b^2 \cdot (3t+2) + c^2 \cdot (2-t) \\
&= \left( a^2-3b^2-c^2 \right) t - 2(b^2-c^2) \\
\implies t &= \frac{2(b^2-c^2)}{a^2-3b^2-c^2}.
\end{align*}
In other words,
\[ P = \left( 2(b^2-c^2) : a^2-3b^2-c^2 : a^2-3b^2-c^2 \right). \]
A long calculation gives
$a^2 y_P z_P + b^2 z_P x_P + c^2 x_P y_P
= (a^2-3b^2-c^2)(a^2-b^2+c^2)(a^2-2b^2-2c^2)$.
Together with $x_P+y_P+z_P=2a^2-4b^2-4c^2$,
this makes the equation of $(ABP)$ as
\[
0=-a^2yz-b^2zx-c^2xy
+ \frac{a^2-b^2+c^2}{2} z(x+y+z).
\]
To solve for $Q$, set $x=0$ to get to get
\[ a^2yz = \frac{a^2-b^2+c^2}{2} z(y+z)
\implies \frac yz = \frac{a^2-b^2+c^2}{a^2+b^2-c^2}. \]
In other words,
\[ Q = \left( 0 : a^2-b^2+c^2 : a^2+b^2-c^2 \right). \]
Taking the average with $A = (1,0,0)$ then gives
\[ N = \left( 2a^2 : a^2-b^2+c^2 : a^2+b^2-c^2 \right). \]
The equation for the perpendicular bisector of $\ol{BC}$
is given by (see EGMO 7.19)
\[ 0 = a^2(z-y)+x(c^2-b^2) \]
which contains $N$, as needed.
\paragraph{Extremely ill-advised complex numbers approaches (outline).}
Suppose we pick $a$, $b$, $c$ as the unit circle, and let $m = (b+c)/2$.
Using the fully general ``foot'' formula, one can get
\[
p = \frac{(a-m) \ol c + (\ol a - \ol m)c
+ \ol a m - a \ol m}{2(\ol a - \ol m)}
= \frac{a^2 b - a^2 c - a b^2 - 2 a b c - a c^2 + b^2 c + 3 b c^2}
{4bc-2a(b+c)}
\]
Meanwhile, an extremely ugly calculation will eventually yield
\[ q = \frac{\frac{bc}{a}+b+c-a}{2} \]
so
\[ n = \frac{a+q}{2} = \frac{a+b+c+\frac{bc}{a}}{4}
= \frac{(a+b)(a+c)}{2a}. \]
There are a few ways to then verify $NB = NC$.
The simplest seems to be to verify that
\[ \frac{n - \frac{b+c}{2}}{b-c} = \frac{a-b-c+\frac{bc}{a}}{4(b-c)}
= \frac{(a-b)(a-c)}{2a(b-c)} \]
is pure imaginary, which is clear.
|
JMO-2023-notes_3
|
Consider an $n$-by-$n$ board of unit squares for some odd positive integer $n$.
We say that a collection $C$ of identical dominoes is a
maximal grid-aligned configuration on the board if $C$ consists of $(n^2-1)/2$
dominoes where each domino covers exactly two neighboring squares
and the dominoes don't overlap: $C$ then covers all but one square on the board.
We are allowed to slide (but not rotate) a domino on the board to
cover the uncovered square, resulting in a new maximal grid-aligned configuration
with another square uncovered. Let $k(C)$ be the number of distinct maximal
grid-aligned configurations obtainable from $C$ by repeatedly sliding dominoes.
Find the maximum possible value of $k(C)$ as a function of $n$.
|
The answer is that
\[ k(C) \le \left( \frac{n+1}{2} \right)^2. \]
\begin{remark*}
[Comparison with USAMO version]
In the USAMO version of the problem,
students instead are asked to find all possible values of $k(C)$.
The answer is
$k(C) \in \left\{ 1, 2, \dots, \left( \frac{n-1}{2} \right)^2 \right\}
\cup \left\{ \left( \frac{n+1}{2} \right)^2 \right\}$.
\end{remark*}
Index the squares by coordinates $(x,y) \in \{1,2,\dots,n\}^2$.
We say a square is \emph{special} if it is empty or
it has the same parity in both coordinates as the empty square.
Construct a directed graph $G = G(C)$ whose vertices are special squares as follows:
for each domino on a special square $s$, we draw a directed edge
from $s$ to the special square that domino points to, if any.
(If the special square has both odd coordinates,
all special squares have an outgoing edge except the empty cell.
In the even-even case, some arrows may point ``off the board''
and not be drawn.)
\begin{center}
\begin{asy}
size(6cm);
pen border = gray+1.5;
filldraw(box((0, 1), (1, 3)), lightgreen, border);
filldraw(box((0, 3), (2, 4)), lightcyan, border);
filldraw(box((0, 4), (2, 5)), lightgreen, border);
filldraw(box((1, 0), (2, 3)), lightred, border);
filldraw(box((1, 2), (3, 3)), lightblue, border);
filldraw(box((2, 3), (3, 5)), palered, border);
filldraw(box((2, 0), (4, 1)), lightcyan, border);
filldraw(box((2, 1), (4, 2)), lightgreen, border);
filldraw(box((4, 0), (5, 2)), lightred, border);
filldraw(box((3, 3), (5, 4)), lightblue, border);
filldraw(box((3, 2), (5, 3)), palered, border);
filldraw(box((3, 4), (5, 5)), lightcyan, border);
filldraw(box((0, 0), (1, 1)), gray, border);
pair A = (0.5, 4.5);
pair B = (2.5, 4.5);
pair C = (4.5, 4.5);
pair D = (0.5, 2.5);
pair E = (2.5, 2.5);
pair F = (4.5, 2.5);
pair X = (0.5, 0.5);
pair Y = (2.5, 0.5);
pair Z = (4.5, 0.5);
dotfactor *= 2;
dot(A); dot(B); dot(C);
dot(D); dot(E); dot(F);
dot(X); dot(Y); dot(Z);
draw(A--B, EndArrow, Margin(3,3));
draw(C--B, EndArrow, Margin(3,3));
draw(B--E, EndArrow, Margin(3,3));
draw(E--D, EndArrow, Margin(3,3));
draw(F--E, EndArrow, Margin(3,3));
draw(Z--F, EndArrow, Margin(3,3));
draw(Y--Z, EndArrow, Margin(3,3));
draw(D--X, EndArrow, Margin(3,3));
\end{asy}
\end{center}
Now focus specifically on the weakly connected component $T$ of $G$
(i.e.\ the connected component of the undirected version of $G$)
containing the empty square.
\begin{claim*}
The graph $T$ has no cycles, even undirected.
Hence, the undirected version of $T$ is tree.
\end{claim*}
\begin{proof}
Assume for contradiction $T$ had an undirected cycle.
Then if we look at the direction of arrows along the cycle,
because every vertex of $T$ had outdegree at most $1$,
the arrows must all point in the same direction
(i.e.\ we actually have a directed cycle).
But then $T$ must consist solely of this cycle.
Yet the empty square has outdegree $0$, contradiction.
\end{proof}
Notice that all the arrows along $T$ point towards the empty cell,
and moving a domino corresponds to flipping an arrow.
Therefore:
\begin{claim*}
$k(C)$ is exactly the number of vertices of $T$.
\end{claim*}
\begin{proof}
Starting with the underlying tree,
the set of possible graphs is described by picking one vertex to be the sink
(the empty cell) and then directing all arrows towards it.
\end{proof}
This implies that $k(C) \le \left( \frac{n+1}{2} \right)^2$,
the total number of vertices of $G$
(this could only occur if the special squares are odd-odd, not even-even).
Equality is achieved as long as $T$ is a spanning tree;
one example of a way to achieve this is using the snake configuration below.
\begin{center}
\begin{asy}
unitsize(0.9cm);
dotfactor *= 2;
pen border = gray+1.5;
pair A = (0.5, 4.5);
pair B = (2.5, 4.5);
pair C = (4.5, 4.5);
pair D = (0.5, 2.5);
pair E = (2.5, 2.5);
pair F = (4.5, 2.5);
pair X = (0.5, 0.5);
pair Y = (2.5, 0.5);
pair Z = (4.5, 0.5);
filldraw(box((3, 4), (5, 5)), palecyan, border);
filldraw(box((1, 4), (3, 5)), palecyan, border);
filldraw(box((0, 3), (1, 5)), palecyan, border);
filldraw(box((0, 2), (2, 3)), palecyan, border);
filldraw(box((2, 2), (4, 3)), palecyan, border);
filldraw(box((4, 1), (5, 3)), palecyan, border);
filldraw(box((3, 0), (5, 1)), palecyan, border);
filldraw(box((1, 0), (3, 1)), palecyan, border);
fill(unitsquare, mediumgray);
filldraw(box((3, 3), (5, 4)), lightred, border);
filldraw(box((1, 3), (3, 4)), lightred, border);
filldraw(box((0, 1), (2, 2)), lightred, border);
filldraw(box((2, 1), (4, 2)), lightred, border);
dot(A); dot(B); dot(C);
dot(D); dot(E); dot(F);
dot(X); dot(Y); dot(Z);
draw(C--B, EndArrow, Margin(3,3));
draw(B--A, EndArrow, Margin(3,3));
draw(A--D, EndArrow, Margin(3,3));
draw(D--E, EndArrow, Margin(3,3));
draw(E--F, EndArrow, Margin(3,3));
draw(F--Z, EndArrow, Margin(3,3));
draw(Z--Y, EndArrow, Margin(3,3));
draw(Y--X, EndArrow, Margin(3,3));
\end{asy}
\end{center}
|
JMO-2023-notes_5
|
Positive integers $a$ and $N$ are fixed,
and $N$ positive integers are written on a blackboard.
Alice and Bob play the following game.
On Alice's turn, she must replace some integer $n$ on the board with $n+a$,
and on Bob's turn he must replace some even integer $n$ on the board with $n/2$.
Alice goes first and they alternate turns.
If on his turn Bob has no valid moves, the game ends.
After analyzing the $N$ integers on the board, Bob realizes that,
regardless of what moves Alice makes,
he will be able to force the game to end eventually.
Show that, in fact, for this value of $a$ and these $N$ integers on the board,
the game is guaranteed to end regardless of Alice's or Bob's moves.
|
For $N=1$, there is nothing to prove since each player has only one option each turn.
We address $N \ge 2$ only henceforth.
Let $S$ denote the numbers on the board.
\begin{claim*}
When $N \ge 2$, if $\nu_2(x) < \nu_2(a)$ for all $x \in S$,
the game must terminate no matter what either player does.
\end{claim*}
\begin{proof}
The $\nu_2$ of a number is unchanged by Alice's move
and decreases by one on Bob's move.
The game ends when every $\nu_2$ is zero.
Hence, in fact the game will always terminate in exactly
$\sum_{x \in S} \nu_2(x)$ moves in this case,
regardless of what either player does.
\end{proof}
\begin{claim*}
When $N \ge 2$, if there exists a number $x$ on the board such that
$\nu_2(x) \ge \nu_2(a)$, then Alice can cause the game to go on forever.
\end{claim*}
\begin{proof}
Denote by $x$ the first entry of the board (its value changes over time).
Then Alice's strategy is to:
\begin{itemize}
\ii Operate on the first entry if $\nu_2(x) = \nu_2(a)$
(the new entry thus has $\nu_2(x+a) > \nu_2(a)$);
\ii Operate on any other entry besides the first one, otherwise.
\end{itemize}
A double induction then shows that
\begin{itemize}
\ii Just before each of Bob's turns, $\nu_2(x) > \nu_2(a)$ always holds; and
\ii After each of Bob's turns, $\nu_2(x) \ge \nu_2(a)$ always holds.
\end{itemize}
In particular Bob will never run out of legal moves,
since halving $x$ is always legal.
\end{proof}
|
JMO-2023-notes_6
|
Isosceles triangle $ABC$, with $AB=AC$, is inscribed in circle $\omega$.
Let $D$ be an arbitrary point inside $BC$ such that $BD \neq DC$.
Ray $AD$ intersects $\omega$ again at $E$ (other than $A$).
Point $F$ (other than $E$) is chosen on $\omega$ such that $\angle DFE = 90\dg$.
Line $FE$ intersects rays $AB$ and $AC$ at points $X$ and $Y$, respectively.
Prove that $\angle XDE = \angle EDY$.
\end{enumerate}
|
We present three solutions.
\paragraph{Angle chasing solution.}
Note that $(BDA)$ and $(CDA)$ are congruent,
since $BA=CA$ and $\angle BDA + \angle CDA = 180\dg$.
So these two circles are reflections around line $ED$.
Moreover, $(DEF)$ is obviously also symmetric around line $ED$.
\begin{center}
\begin{asy}
size(10cm);
pair E = dir(100);
pair B = dir(160);
pair C = dir(0);
pair A = dir(260);
draw(unitcircle, gray);
draw(E--B--C--cycle, dashed+gray);
pair D = extension(E, A, B, C);
pair W = -E;
pair F = foot(E, W, D);
pair X = extension(A, B, E, F);
pair Y = extension(A, C, E, F);
draw(E--A, blue);
draw(X--Y--A--cycle, dashed+gray);
draw(X--D--Y, red);
pair X_prime = foot(E, D, X);
pair Y_prime = foot(E, D, Y);
filldraw(circumcircle(E, X_prime, Y_prime), opacity(0.1)+cyan, blue);
filldraw(circumcircle(B, D, A), opacity(0.1)+cyan, blue);
filldraw(circumcircle(C, D, A), opacity(0.1)+cyan, blue);
dot("$E$", E, dir(E));
dot("$B$", B, dir(160));
dot("$C$", C, dir(340));
dot("$A$", A, dir(A));
dot("$D$", D, dir(315));
dot("$F$", F, dir(F));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
dot(X_prime);
dot(Y_prime);
clip(scale(1.7)*unitcircle);
dot("$Y$", D+2*dir(Y-D), dir(Y-D));
/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
!size(14cm);
E = dir 100
B 160 = dir 160
C 340 = dir 0
A = dir 260
unitcircle / gray
E--B--C--cycle / dashed gray
D 315 = extension E A B C
W := -E
F = foot E W D
X = extension A B E F
Y = extension A C E F
E--A blue
X--Y--A--cycle / dashed gray
X--D--Y / red
X' 70 .= foot E D X
Y' 130 .= foot E D Y
circumcircle E X' Y' / 0.1 cyan / blue
circumcircle B D A / 0.1 cyan / blue
circumcircle C D A / 0.1 cyan / blue
*/
\end{asy}
\end{center}
Hence, the radical axis of $(BDA)$ and $(DEF)$,
and the radical axis of $(CDA)$ and $(DEF)$,
should be symmetric about line $DE$.
But these radical axii are exactly lines $XD$ and $YD$, so we're done.
\begin{remark*}
[Motivation]
The main idea is that you can replace $DX$ and $DY$ with the radical axii,
letting $X'$ and $Y'$ be the second intersections of the blue circles.
Then for the problem to be true, you'd need $X'$ and $Y'$ to be reflections.
That's equivalent to $(BDA)$ and $(CDA)$ being congruent;
you check it and it's indeed true.
\end{remark*}
\paragraph{Harmonic solution (mine).}
Let $T$ be the point on line $\ol{XFEY}$ such that $\angle EDT = 90\dg$,
and let $\ol{AT}$ meet $\omega$ again at $K$.
Then
\[ TD^2 = TF \cdot TE = TK \cdot TA \implies \angle DKT = 90\dg \]
so line $DK$ passes through the antipode $M$ of $A$.
\begin{center}
\begin{asy}
size(14cm);
pair E = dir(110);
pair B = dir(160);
pair C = dir(20);
pair A = dir(270);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
filldraw(E--B--C--cycle, opacity(0.1)+yellow, blue);
pair D = extension(E, A, B, C);
pair W = -E;
pair F = foot(E, W, D);
pair X = extension(A, B, E, F);
pair Y = extension(A, C, E, F);
draw(E--A, blue);
draw(X--Y--A--cycle, blue);
draw(X--D--Y, deepgreen);
pair M = dir(90);
pair T = extension(E, F, D, D+M-E);
draw(X--T, blue);
draw(D--T, red);
draw(E--M, red);
pair K = extension(D, M, T, A);
draw(T--A, red);
draw(K--M, red);
dot("$E$", E, dir(E));
dot("$B$", B, dir(200));
dot("$C$", C, dir(340));
dot("$A$", A, dir(A));
dot("$D$", D, dir(315));
dot("$F$", F, dir(F));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
dot("$M$", M, dir(35));
dot("$T$", T, dir(T));
dot("$K$", K, dir(K));
/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
!size(14cm);
E = dir 110
B 200 = dir 160
C 340 = dir 20
A = dir 270
unitcircle / 0.1 lightcyan / blue
E--B--C--cycle / 0.1 yellow / blue
D 315 = extension E A B C
W := -E
F = foot E W D
X = extension A B E F
Y = extension A C E F
E--A blue
X--Y--A--cycle / blue
X--D--Y / deepgreen
M 35 = dir 90
T = extension E F D D+M-E
X--T / blue
D--T / red
E--M / red
K = extension D M T A
T--A / red
K--M / red
*/
\end{asy}
\end{center}
Thus,
\[ -1 = (AM;CB)_\omega \overset{D}{=} (EK;BC)_\omega \overset{A}{=} (TE;XY) \]
and since $\angle EDT = 90\dg$ we're done.
\begin{remark*}
[Motivation]
The idea is to kill the points $X$ and $Y$
by reinterpreting the desired condition as $(TD;XY)=-1$
and then projecting through $A$ onto $\omega$.
This eliminates points $X$ and $Y$ altogether
and reduces the problem to showing that $\ol{TA}$ passes
through the harmonic conjugate of $E$ with respect to $BC$ on $\omega$.
The labels on the diagram are slightly misleading in that
$\triangle EBC$ should probably be thought of as the ``reference'' triangle.
\end{remark*}
\paragraph{Pascal solution (Zuming Feng).}
Extend ray $FD$ to the antipode $T$ of $E$ on $\omega$. Then,
\begin{itemize}
\ii By Pascal's theorem on $EFTABC$,
the points $X$, $D$, and $P \coloneq \ol{EC} \cap \ol{AT}$ are collinear.
\ii Similarly by Pascal's theorem on $EFTACB$, the points
the points $Y$, $D$, and $Q \coloneq \ol{EB} \cap \ol{AT}$ are collinear.
\end{itemize}
\begin{center}
\begin{asy}
size(14cm);
pair E = dir(100);
pair B = dir(160);
pair C = dir(0);
pair A = dir(260);
filldraw(unitcircle, opacity(0.1)+yellow, blue);
draw(E--B--C--cycle, dashed+gray);
pair D = extension(E, A, B, C);
pair W = -E;
pair F = foot(E, W, D);
pair X = extension(A, B, E, F);
pair Y = extension(A, C, E, F);
draw(E--A, blue);
draw(X--Y--A--cycle, dashed+gray);
draw(X--D--Y, red);
pair T = -E;
pair P = extension(C, E, T, A);
pair Q = extension(B, E, T, A);
draw(P--D--Q, red);
draw(P--Q--E--cycle, deepgreen);
draw(F--T, blue);
filldraw(circumcircle(E, D, F), opacity(0.1)+cyan, blue);
dot("$E$", E, dir(E));
dot("$B$", B, dir(160));
dot("$C$", C, dir(0));
dot("$A$", A, dir(A));
dot("$D$", D, dir(315));
dot("$F$", F, dir(F));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
dot("$T$", T, dir(T));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
!size(14cm);
E = dir 100
B 160 = dir 160
C 0 = dir 0
A = dir 260
unitcircle / 0.1 yellow / blue
E--B--C--cycle / dashed gray
D 315 = extension E A B C
W := -E
F = foot E W D
X = extension A B E F
Y = extension A C E F
E--A blue
X--Y--A--cycle / dashed gray
X--D--Y / red
T = -E
P = extension C E T A
Q = extension B E T A
P--D--Q / red
P--Q--E--cycle / deepgreen
F--T / blue
circumcircle E D F / 0.1 cyan / blue
*/
\end{asy}
\end{center}
Now it suffices to prove $\ol{ED}$ bisects $\angle QDP$.
However, $\ol{ED}$ is the angle bisector of $\angle QEP = \angle BEC$,
but also $\ol{EA} \perp \ol{QP}$.
Thus triangle $QEP$ is isosceles with $QE=PE$, and $\ol{EA}$ cuts it in half.
Since $D$ is on $\ol{EA}$, the result follows now.
|
JMO-2024-notes_1
|
Let $ABCD$ be a cyclic quadrilateral with $AB=7$ and $CD=8$.
Points $P$ and $Q$ are selected on line segment $AB$ so that $AP=BQ=3$.
Points $R$ and $S$ are selected on line segment $CD$ so that $CR=DS=2$.
Prove that $PQRS$ is a cyclic quadrilateral.
|
Here are three possible approaches.
\paragraph{The one-liner.}
The four points $P$, $Q$, $R$, $S$ have equal power $-12$ with respect to $(ABCD)$.
So in fact they're on a circle concentric with $(ABCD)$.
\paragraph{The external power solution.}
We distinguish between two cases.
\subparagraph{Case where $AB$ and $CD$ are not parallel.}
We let lines $AB$ and $CD$ meet at $T$.
Without loss of generality,
$A$ lies between $B$ and $T$ and $D$ lies between $C$ and $T$.
Let $x=TA$ and $y=TD$, as shown below.
\begin{center}
\begin{asy}
size(8cm);
pair A = dir(170);
pair B = dir(80);
pair C = dir(323.912853);
pair D = dir(216.087147);
pair T = extension(A, B, C, D);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
draw(B--T--C--B, blue);
draw(A--D, blue);
pair P = 4*A/7+3*B/7;
pair Q = 3*A/7+4*B/7;
pair R = 6*C/8+2*D/8;
pair S = 2*C/8+6*D/8;
filldraw(circumcircle(P, Q, R), opacity(0.1)+yellow, dashed+red);
label("$x$", midpoint(T--A), dir(A+B), deepgreen);
label("$3$", midpoint(A--P), dir(A+B), deepgreen);
label("$1$", midpoint(P--Q), -dir(A+B), deepgreen);
label("$3$", midpoint(Q--B), dir(A+B), deepgreen);
label("$y$", midpoint(T--D), dir(C+D), deepgreen);
label("$2$", midpoint(D--S), dir(C+D), deepgreen);
label("$4$", midpoint(S--R), -dir(C+D), deepgreen);
label("$2$", midpoint(R--C), dir(C+D), deepgreen);
dot("$A$", A, dir(120));
dot("$B$", B, dir(B));
dot("$C$", C, dir(300));
dot("$D$", D, dir(240));
dot("$T$", T, dir(T));
dot("$P$", P, dir(160));
dot("$Q$", Q, dir(80));
dot("$R$", R, dir(270));
dot("$S$", S, dir(270));
/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
A 120 = dir 170
B = dir 80
C 300 = dir 323.912853
D 240 = dir 216.087147
T = extension A B C D
unitcircle / 0.1 lightcyan / blue
B--T--C--B / blue
A--D / blue
P 160 = 4*A/7+3*B/7
Q 80 = 3*A/7+4*B/7
R 270 = 6*C/8+2*D/8
S 270 = 2*C/8+6*D/8
circumcircle P Q R / 0.1 yellow / dashed red
!label("$x$", midpoint(T--A), dir(A+B), deepgreen);
!label("$3$", midpoint(A--P), dir(A+B), deepgreen);
!label("$1$", midpoint(P--Q), -dir(A+B), deepgreen);
!label("$3$", midpoint(Q--B), dir(A+B), deepgreen);
!label("$y$", midpoint(T--D), dir(C+D), deepgreen);
!label("$2$", midpoint(D--S), dir(C+D), deepgreen);
!label("$4$", midpoint(S--R), -dir(C+D), deepgreen);
!label("$2$", midpoint(R--C), dir(C+D), deepgreen);
*/
\end{asy}
\end{center}
By power of a point,
\begin{align*}
\text{$ABCD$ cyclic} \iff x(x+7) &= y(y+8) \\
\text{$PQRS$ cyclic} \iff (x+3)(x+4) &= (y+2)(y+6).
\end{align*}
However, the latter equation is just the former with $12$ added to both sides.
(That is, $(x+3)(x+4) = x(x+7)+12$ while $(y+2)(y+6)=y(y+8)+12$.)
So the conclusion is immediate.
\subparagraph{Case where $AB$ and $CD$ are parallel.}
In that case $ABCD$ is an isosceles trapezoid.
Then the entire picture is symmetric around the common perpendicular bisector
of the lines $AB$ and $CD$.
Now $PQRS$ is also an isosceles trapezoid, so it's cyclic too.
\begin{center}
\begin{asy}
size(4cm);
pair A = dir(135);
pair B = dir(45);
pair C = dir(323.912853);
pair D = dir(216.087147);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
draw(A--B--C--D--cycle, blue);
pair P = 4*A/7+3*B/7;
pair Q = 3*A/7+4*B/7;
pair R = 6*C/8+2*D/8;
pair S = 2*C/8+6*D/8;
filldraw(circumcircle(P, Q, R), opacity(0.1)+yellow, dashed+red);
dot("$A$", A, dir(120));
dot("$B$", B, dir(B));
dot("$C$", C, dir(300));
dot("$D$", D, dir(240));
dot("$P$", P, dir(90));
dot("$Q$", Q, dir(80));
dot("$R$", R, dir(270));
dot("$S$", S, dir(270));
\end{asy}
\end{center}
\paragraph{The Pythagorean bash.}
Let $\rho$ and $O$ denote the radius and center of the circle circumscribing $ABCD$.
We will show that in fact, $P$, $Q$, $R$, $S$ lie on a circle centered at $O$,
i.e.\ that the lengths $PO$, $QO$, $RO$, $SO$ are all equal.
\begin{center}
\begin{asy}
size(9cm);
pair A = dir(170);
pair B = dir(80);
pair C = dir(323.912853);
pair D = dir(216.087147);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
draw(A--B--C--D--cycle, blue);
pair P = 4*A/7+3*B/7;
pair Q = 3*A/7+4*B/7;
pair R = 6*C/8+2*D/8;
pair S = 2*C/8+6*D/8;
pair O = origin;
pair M = midpoint(P--Q);
pair N = midpoint(R--S);
draw(N--O--M, dashed);
filldraw(circumcircle(P, Q, R), opacity(0.1)+yellow, dashed+red);
label("$3$", midpoint(A--P), dir(A+B), deepgreen);
label("$3$", midpoint(Q--B), dir(A+B), deepgreen);
label("$2$", midpoint(D--S), dir(C+D), deepgreen);
label("$2$", midpoint(R--C), dir(C+D), deepgreen);
label("$0.5$", midpoint(P--M), dir(285), deepgreen);
label("$0.5$", midpoint(Q--M), dir(330), deepgreen);
label("$2$", midpoint(S--N), -dir(C+D), deepgreen);
label("$2$", midpoint(R--N), -dir(C+D), deepgreen);
dot("$A$", A, dir(120));
dot("$B$", B, dir(B));
dot("$C$", C, dir(300));
dot("$D$", D, dir(240));
dot("$P$", P, dir(160));
dot("$Q$", Q, dir(80));
dot("$R$", R, dir(270));
dot("$S$", S, dir(270));
dot("$O$", O, dir(0));
dot("$M$", M, dir(M));
dot("$N$", N, dir(N));
\end{asy}
\end{center}
Our proof will use repeated applications of the Pythagorean theorem.
Let $M$ denote the midpoint of $\overline{PQ}$,
which is also the midpoint of $\overline{AB}$,
as $AM = MB = 3.5$ and $PM = QM = 0.5$.
Assuming $M$ is distinct from $O$,
it follows $\overline{OM}$ is the perpendicular bisector of $\overline{AB}$,
and so the Pythagorean theorem we get
\begin{align*}
PO^2 &= PM^2 + MO^2 = PM^2 + (AO^2 - AM^2) = 0.5^2 + \rho^2 - 3.5^2 \\
QO^2 &= QM^2 + MO^2 = QM^2 + (BO^2 - BM^2) = 0.5^2 + \rho^2 - 3.5^2.
\end{align*}
And in the case $O = M$, the same equations are valid too.
Similarly let $N$ denote the midpoint of $\overline{RS}$,
which is also the midpoint of $\overline{CD}$,
as $CN = DN = 4$ and $RN = SN = 2$.
Repeating the same calculation gives
\begin{align*}
RO^2 &= RN^2 + NO^2 = RN^2 + (CO^2 - CN^2) = 2^2 + \rho^2 - 4^2 \\
SO^2 &= SN^2 + NO^2 = SN^2 + (DO^2 - DN^2) = 2^2 + \rho^2 - 4^2.
\end{align*}
From this it follows that
\[ PO^2 = QO^2 = RO^2 = SO^2 = \rho^2 - 12. \]
Hence the problem is solved.
|
JMO-2024-notes_2
|
Let $m$ and $n$ be positive integers.
Let $S$ be the set of lattice points $(x,y)$ with $1 \le x \le 2m$ and $1 \le y \le 2n$.
A configuration of $mn$ axis-parallel rectangles is called \emph{happy}
if each point of $S$ is the vertex of exactly one rectangle.
Prove that the number of happy configurations is odd.
|
There are several possible approaches to the problem;
most of them involve pairing some of the happy configurations in various ways,
leaving only a few configurations which remain fixed.
We present the original proposer's solution and Evan's more complicated one.
\paragraph{Original proposer's solution.}
To this end, let's denote by $f(2m, 2n)$ the number of happy configurations
for a $2m \times 2n$ grid of lattice points
(not necessarily equally spaced --- this doesn't change the count).
We already have the following easy case.
\begin{claim*}
We have $f(2, 2n) = (2n-1)!{!} = (2n-1) \cdot (2n-3) \cdot \dots \cdot 3 \cdot 1$.
\end{claim*}
\begin{proof}
The top row is the top edge of some rectangle and there are $2n-1$ choices
for the bottom edge of that rectangle.
It then follows $f(2, 2n) = (2n-1) \cdot f(2, 2n-2)$
and the conclusion follows by induction on $n$, with $f(2,2) = 1$.
\end{proof}
We will prove that:
\begin{claim*}
Assume $m,n \ge 1$. When $f(2m, 2n) \equiv f(2m-2, 2n) \pmod 2$.
\end{claim*}
\begin{proof}
Given a happy configuration $\mathcal C$, let $\tau(\mathcal C)$
be the happy configuration obtained by swapping the last two columns.
Obviously $\tau(\tau(\mathcal C)) = \mathcal C$ for every happy $\mathcal C$.
So in general, we can consider two different kinds of configurations $\mathcal C$,
those for which $\tau(\mathcal C) \neq \mathcal C$,
so we get pairs $\{\mathcal C, \tau(\mathcal C)\}$,
and those with $\tau(\mathcal C) = \mathcal C$.
Now configurations fixed by $\tau$ can be described readily:
this occurs if and only if the last two columns are self-contained,
meaning every rectangle with a vertex in these columns is
completely contained in these two columns.
\begin{center}
\begin{asy}
real eps = 0.6;
filldraw(box((8.4, 0.4), (10.6, 8.6)), opacity(0.2)+yellow, black+dashed);
for (int x=1; x<=10; ++x) {
for (int y=1; y<=8; ++y) {
pair P = (x,y);
if (x==9 || x==10) { fill(circle(P, 0.1), blue); }
else { dot(P); }
}
}
draw((9,0.3)..(9.5,-0.2)..(10,0.3), deepgreen+1.2, Arrows(TeXHead));
label("$\tau$", (9.5,-0.2), dir(-90), deepgreen);
label("$f(2,2n)$", (9.5,8.6), dir(90), black);
label("$f(2m-2,2n)$", (4.5,8.6), dir(90), black);
draw((0.8,8.4)--(0.8,8.6)--(8.2,8.6)--(8.2,8.4), gray);
\end{asy}
\end{center}
Hence it follows that
\[ f(2m, 2n) = 2 (\text{number of pairs}) + f(2m-2, 2n) \cdot f(2, 2n). \]
Taking modulo $2$ gives the result.
\end{proof}
By the same token $f(2m, 2n) \equiv f(2m, 2n-2) \pmod 2$.
So all $f$-values have the same parity, and from $f(2,2)=1$ we're done.
\begin{remark*}
There are many variations of the solution using different kinds of $\tau$.
The solution with $\tau$ swapping two rows seems to be the simplest.
\end{remark*}
\paragraph{Evan's permutation-based solution.}
Retain the notation $f(2m, 2n)$ from before.
Given a happy configuration, consider all the rectangles
whose left edge is in the first column.
Highlight every column containing the right edge of such a rectangle.
For example, in the figure below, there are two highlighted columns.
(The rectangles are drawn crooked so one can tell them apart.)
\begin{center}
\begin{asy}
real eps = 0.6;
for (int x=1; x<=10; ++x) {
for (int y=1; y<=6; ++y) {
pair P = (x,y);
dot(P);
}
}
dotfactor *= 1;
real eps = 0.3;
void rect(real a, real b, real u, real v, pen p) {
draw( (a,b)--((a+u)/2, b-eps)--(u,b)--(u+eps,(b+v)/2)--(u,v)
--((a+u)/2, v+eps)--(a,v)--(a-eps, (b+v)/2)--cycle, p);
fill(circle((a,b), 0.2), p);
fill(circle((u,b), 0.2), p);
fill(circle((a,v), 0.2), p);
fill(circle((u,v), 0.2), p);
}
rect(1,1,8,3, blue + 1.5);
rect(1,2,5,6, deepgreen + 1.5);
rect(1,4,8,5, red + 1.5);
filldraw(box((4.5, 0.6), (5.5, 6.4)), opacity(0.2)+yellow, black+dashed);
filldraw(box((7.5, 0.6), (8.5, 6.4)), opacity(0.2)+yellow, black+dashed);
\end{asy}
\end{center}
We organize happy configurations based on the set of highlighted columns.
Specifically, define the relation $\sim$ on configurations by saying that
$\mathcal C \sim \mathcal C'$ if they differ by any permutation of the highlighted columns.
This is an equivalence relation.
And in general, if there are $k$ highlighted columns,
its equivalence class under $\sim$ has $k!$ elements.
Then
\begin{claim*}
$f(2m, 2n)$ has the same parity as the number of happy configurations
with \emph{exactly} one highlighted column.
\end{claim*}
\begin{proof}
Since $k!$ is even for all $k \ge 2$, but odd when $k=1$.
\end{proof}
There are $2m-1$ ways to pick a single highlighted column,
and then $f(2, 2n) = (2n-1){!}{!}$ ways to use the left column and highlighted column.
So the count in the claim is exactly given by
\[ (2m-1) \cdot (2n-1){!}{!} f(2m-2, 2n). \]
This implies $f(2m, 2n) \equiv f(2m-2,2n) \pmod 2$
and proceeding by induction as before solves the problem.
|
JMO-2024-notes_3
|
A sequence $a_1$, $a_2$, \dots\ of positive integers is defined recursively by
$a_1 = 2$ and \[ a_{n+1} = a_n^{n+1} - 1 \qquad \text{ for } n \ge 1. \]
Prove that for every odd prime $p$ and integer $k$,
some term of the sequence is divisible by $p^k$.
|
We start with the following.
\begin{claim*}
Assume $n$ is a positive integer divisible by $p-1$.
Then either $a_{n-1} \equiv 0 \pmod p$ or $a_n \equiv 0 \pmod p$.
\end{claim*}
\begin{proof}
Suppose that $a_{n-1} \not\equiv 0 \pmod p$.
Then by Fermat's little theorem,
\[ a_n = a_{n-1}^n - 1 \equiv x^{p-1} - 1 \equiv 0 \pmod p \]
where $x \coloneq a_{n-1}^{\frac{n}{p-1}}$ is an integer not divisible by $p$.
\end{proof}
\begin{claim*}
If $n \ge 2$ is even, then
\[ a_{n}^{n+1} \mid a_{n+2}. \]
\end{claim*}
\begin{proof}
Note $a_{n+2} = a_{n+1}^{n+2} - 1$.
The right-hand side is a difference of $(n+2)$\ts{nd} powers,
which for even $n$ is divisible by $a_{n+1} + 1 = a_{n}^{n+1}$.
\end{proof}
By considering multiples $n$ of $p-1$ which are larger than $k$,
we see that if $a_n \equiv 0 \pmod p$ ever happens,
we are done by combining the two previous claims.
So the difficult case of the problem is the bad situation
where $a_{n-1} \equiv 0 \pmod p$ occurs for almost all $n \equiv 0 \pmod{p-1}$.
To resolve the difficult case and finish the problem,
we zoom in on specific $n$ that will let us use lifting the exponent on $a_{n-1}$.
\begin{claim*}
Suppose $n$ is an (even) integer satisfying
\begin{align*}
n &\equiv 0 \pmod{p-1} \\
n &\equiv 1 \pmod{p^{k-1}}.
\end{align*}
If $a_{n-1} \equiv 0 \pmod p$, then in fact $p^k \mid a_{n-1}$.
\end{claim*}
\begin{proof}
Write $a_{n-1} = a_{n-2}^{n-1} - 1$.
We know $a_{n-2} \not\equiv 0 \pmod p$, and so $a_{n-2}^n \equiv 1 \pmod p$.
Taking modulo $p$ gives
\[ 0 \equiv \frac{1}{a_{n-2}} - 1 \pmod p \implies a_{n-2} \equiv 1 \pmod p. \]
Hence lifting the exponent applies and we get
\[ \nu_p(a_{n-1}) \equiv \nu_p \left( a_{n-2}^{n-1} - 1 \right)
= \nu_p(a_{n-2}-1) + \nu_p(n-1) \geq 1 + (k-1) = k \]
as desired.
\end{proof}
\begin{remark*}
The first few terms are $a_1=2$, $a_2=3$, $a_3=26$, $a_4=456975$,
and $a_5 = 19927930852449199486318359374$, \dots.
No element of the sequence is divisible by $4$:
the residues modulo $4$ alternate between $2$ and $3$.
\end{remark*}
\begin{remark*}
The second claim is important for the solution to work once $k \ge 2$.
One could imagine a variation of the first claim that states
if $n$ is divisible by $\varphi(p^k) = p^{k-1}(p-1)$,
then either $a_{n-1} \equiv 0 \pmod p$ or $a_n \equiv 0 \pmod{p^k}$.
However this gives an obstruction (for $k \ge 2$)
where we are guaranteed to have $n-1 \not\equiv 0 \pmod p$ now,
so lifting the exponent will never give additional factors of $p$ we want.
\end{remark*}
|
JMO-2024-notes_4
|
Let $n \geq 3$ be an integer.
Rowan and Colin play a game on an $n \times n$ grid of squares,
where each square is colored either red or blue.
Rowan is allowed to permute the rows of the grid and
Colin is allowed to permute the columns.
A grid coloring is orderly if:
\begin{itemize}
\ii no matter how Rowan permutes the rows of the coloring,
Colin can then permute the columns to restore the original grid coloring; and
\ii no matter how Colin permutes the columns of the coloring,
Rowan can then permute the rows to restore the original grid coloring.
\end{itemize}
In terms of $n$, how many orderly colorings are there?
|
The answer is $2n!+2$.
In fact, we can describe all the orderly colorings as follows:
\begin{itemize}
\ii The all-blue coloring.
\ii The all-red coloring.
\ii Each of the $n!$ colorings where every row/column has exactly one red cell.
\ii Each of the $n!$ colorings where every row/column has exactly one blue cell.
\end{itemize}
These obviously work; we turn our attention to proving these are the only ones.
For the other direction, fix a orderly coloring $\mathcal A$.
Consider any particular column $C$ in $\mathcal A$
and let $m$ denote the number of red cells that $C$ has.
Any row permutation (say $\sigma$) that Rowan chooses will transform $C$ into
some column $\sigma(C)$, and our assumption requires $\sigma(C)$
has to appear somewhere in the original assignment $\mathcal A$.
An example for $n = 7$, $m = 2$, and a random $\sigma$ is shown below.
\begin{center}
\begin{asy}
defaultpen(fontsize(14pt));
pen bo = black + 1.4;
filldraw(shift(0,0)*unitsquare, lightcyan, bo);
filldraw(shift(0,1)*unitsquare, lightred, bo);
filldraw(shift(0,2)*unitsquare, lightcyan, bo);
filldraw(shift(0,3)*unitsquare, lightcyan, bo);
filldraw(shift(0,4)*unitsquare, lightred, bo);
filldraw(shift(0,5)*unitsquare, lightcyan, bo);
filldraw(shift(0,6)*unitsquare, lightcyan, bo);
label("$C$", (0.5,0), dir(-90));
pen gr = deepgreen;
real h = 8;
draw((1,1.5)--(h,4.5), gr, EndArrow, Margins);
draw((1,4.5)--(h,3.5), gr, EndArrow, Margins);
draw((1,2.5)--(h,1.5), gr, EndArrow, Margins);
draw((1,0.5)--(h,0.5), gr, EndArrow, Margins);
draw((1,3.5)--(h,6.5), gr, EndArrow, Margins);
draw((1,6.5)--(h,5.5), gr, EndArrow, Margins);
draw((1,5.5)--(h,2.5), gr, EndArrow, Margins);
filldraw(shift(h,0)*unitsquare, lightcyan, bo);
filldraw(shift(h,1)*unitsquare, lightcyan, bo);
filldraw(shift(h,2)*unitsquare, lightcyan, bo);
filldraw(shift(h,3)*unitsquare, lightred, bo);
filldraw(shift(h,4)*unitsquare, lightred, bo);
filldraw(shift(h,5)*unitsquare, lightcyan, bo);
filldraw(shift(h,6)*unitsquare, lightcyan, bo);
label("$\sigma(C)$", (h+0.5,0), dir(-90));
label("$\sigma$", ((1+h)/2, 0), dir(-90), gr);
\end{asy}
\end{center}
On the other hand, the number of possible patterns of $\sigma(C)$ is easily seen
to be exactly $\binom{n}{m}$ --- and they must all appear.
In particular, if $m \notin \{0, 1, n-1, n\}$,
then we immediately get a contradiction because $\mathcal A$ would need
too many columns (there are only $n$ columns in $\mathcal A$,
which is fewer than $\binom{n}{m}$).
Moreover, if either $m=1$ or $m=n-1$,
then these columns are all the columns of $\mathcal A$;
this leads to the $2n!$ main cases we found before.
The only remaining case is when $m \in \{0,n\}$ for every column,
i.e.\ every column is monochromatic.
It's easy to see in that case the columns must all be the same color.
|
JMO-2024-notes_5
|
Solve over $\RR$ the functional equation
\[ f(x^2-y)+2yf(x) = f(f(x))+f(y). \]
|
The answer is $f(x) \equiv x^2$, $f(x) \equiv 0$, $f(x) \equiv -x^2$,
which obviously work.
Let $P(x,y)$ be the usual assertion.
\begin{claim*}
We have $f(0) = 0$ and $f$ even.
\end{claim*}
\begin{proof}
Combine $P(1,1/2)$ with $P(1,0)$ to get $f(0) = 0$.
Use $P(0,y)$ to deduce $f$ is even.
\end{proof}
\begin{claim*}
$f(x) \in \{-x^2, 0, x^2\}$ for every $x \in \RR$.
\end{claim*}
\begin{proof}
Note that $P(x,x^2/2)$ and $P(x,0)$ respectively give
\[ x^2f(x) = f(x^2) = f(f(x)). \]
Repeating this key identity several times gives
\begin{align*}
f(f(f(x))) &= f(f(x^2)) = f(x^4) = x^4 f(x^2) \\
&= f(x)^2 \cdot f(f(x)) = f(x)^2 f(x^2) = f(x)^3 x^2
\end{align*}
Suppose $t \neq 0$ is such that $f(t^2) \neq 0$.
Then the above equalities imply
\[ t^4 f(t^2) = f(t)^2 f(t^2) \implies f(t) = \pm t^2 \]
and then
\[ f(t)^2 f(t^2) = f(t)^3 t^2 \implies f(t^2) = \pm t^4. \]
Together with $f$ even, we get the desired result.
\end{proof}
\begin{remark*}
Another proof is possible here that doesn't use as iterations of $f$:
the idea is to ``show $f$ is injective up to sign outside its kernel''.
Specifically, if $f(a) = f(b) \neq 0$,
then $a^2f(a) = f(f(a)) = f(f(b)) = b^2f(b) \implies a^2=b^2$.
But we also have $f(f(x))=f(x^2)$,
so we are done except in the case $f(f(x))=f(x^2)=0$.
That would imply $x^2f(x) = 0$, so the claim follows.
\end{remark*}
Now, note that $P(1,y)$ gives
\[ f(1-y) + 2y \cdot f(1) = f(1) + f(y). \]
We consider cases on $f(1)$ and show that $f$ matches the desired form.
\begin{itemize}
\ii If $f(1) = 1$, then $f(1-y) + (2y-1) = f(y)$.
Consider the nine possibilities that arise:
\[
\begin{array}{lll}
(1-y)^2 + (2y-1) = y^2 &
0 + (2y-1) = y^2 &
-(1-y)^2 + (2y-1) = y^2 \\
(1-y)^2 + (2y-1) = 0 &
0 + (2y-1) = 0 &
-(1-y)^2 + (2y-1) = 0 \\
(1-y)^2 + (2y-1) = -y^2 &
0 + (2y-1) = -y^2 &
-(1-y)^2 + (2y-1) = -y^2.
\end{array}
\]
Each of the last eight equations is a nontrivial polynomial equation.
Hence, there is some constant $C > 100$ such that the latter eight equations
are all false for any real number $y > C$.
Consequently, $f(y) = y^2$ for $y > C$.
Finally, for any real number $z > 0$, take $x,y > C$ such that $x^2-y=z$;
then $P(x,y)$ proves $f(z) = z^2$ too.
\ii Note that (as $f$ is even), $f$ works iff $-f$ works,
so the case $f(1) = -1$ is analogous.
\ii If $f(1) = 0$, then $f(1-y) = f(y)$.
Hence for any $y$ such that $|1-y| \neq |y|$, we conclude $f(y) = 0$.
Then take $P(2, 7/2) \implies f(1/2) = 0$.
\end{itemize}
\begin{remark*}
There is another clever symmetry approach possible after the main claim.
The idea is to write
\[ P(x,y^2) \implies f(x^2-y^2) + 2y^2f(x) = f(f(x)) + f(f(y)). \]
Since $f$ is even gives $f(x^2-y^2) = f(y^2-x^2)$,
one can swap the roles of $x$ and $y$ to get $2y^2f(x) = 2x^2f(y)$.
Set $y=1$ to finish.
\end{remark*}
|
JMO-2024-notes_6
|
Point $D$ is selected inside acute triangle $ABC$
so that $\angle DAC = \angle ACB$ and $\angle BDC = 90^\circ + \angle BAC$.
Point $E$ is chosen on ray $BD$ so that $AE = EC$.
Let $M$ be the midpoint of $BC$.
Show that line $AB$ is tangent to the circumcircle of triangle $BEM$.
\end{enumerate}
|
This problem has several approaches and we showcase a collection of them.
\paragraph{The author's original solution.}
Complete isosceles trapezoid $ABQC$ (so $D \in \ol{AQ}$).
Reflect $B$ across $E$ to point $F$.
\begin{center}
\begin{asy}
size(7cm);
pair A = dir(185);
pair C = dir(355);
pair B = dir(115);
pair Q = A*C/B;
pair X = extension(A, B, Q, C);
pair Y = 2*X-A;
filldraw(A--B--C--cycle, opacity(0.1)+lightred, red);
draw(unitcircle, red);
pair D = IP(A--Q, circumcircle(B, C, Y));
pair N = midpoint(A--C);
pair M = midpoint(B--C);
draw(A--Q, mediumblue);
pair E = extension(N, origin, B, D);
pair F = 2*E-B;
draw(F--B, deepgreen);
pair S = dir(90);
draw(S--N, brown);
draw(F--Q, brown);
draw(M--E, gray);
draw(F--C--Q, gray);
filldraw(circumcircle(Q, D, C), opacity(0.1)+yellow, orange);
markangle(n=2, radius=6, M, E, B, deepgreen);
markangle(n=2, radius=6, C, F, B, deepgreen);
markangle(n=1, radius=6, D, Q, C, deepgreen);
markangle(n=1, radius=6, A, B, C, deepgreen);
dot("$A$", A, dir(225));
dot("$C$", C, dir(315));
dot("$B$", B, dir(160));
dot("$Q$", Q, dir(100));
dot("$D$", D, dir(180));
dot(N);
dot("$M$", M, dir(70));
dot("$E$", E, dir(225));
dot("$F$", F, dir(F));
/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
!size(7cm);
A 225 = dir 185
C 315 = dir 355
B 160 = dir 115
Q 100 = A*C/B
X := extension A B Q C
Y := 2*X-A
A--B--C--cycle / 0.1 lightred / red
unitcircle / red
D 180 = IP A--Q (circumcircle B C Y)
N = midpoint A--C
M 70 = midpoint B--C
A--Q / mediumblue
E 225 = extension N origin B D
F = 2*E-B
F--B / deepgreen
S := dir 90
S--N / brown
F--Q / brown
M--E / gray
F--C--Q / gray
circumcircle Q D C / 0.1 yellow / orange
!markangle(n=2, radius=6, M, E, B, deepgreen);
!markangle(n=2, radius=6, C, F, B, deepgreen);
!markangle(n=1, radius=6, D, Q, C, deepgreen);
!markangle(n=1, radius=6, A, B, C, deepgreen);
*/
\end{asy}
\end{center}
\begin{claim*}
We have $DQCF$ is cyclic.
\end{claim*}
\begin{proof}
Since $EA=EC$, we have $\ol{QF} \perp \ol{AC}$
as line $QF$ is the image of the perpendicular bisector of $\ol{AC}$
under a homothety from $B$ with scale factor $2$.
Then
\begin{align*}
\dang FDC &= -\dang CDB = 180\dg - (90\dg + \dang CAB) = 90\dg - \dang CAB \\
&= 90\dg - \dang QCA = \dang FQC. \qedhere
\end{align*}
\end{proof}
To conclude, note that
\[ \dang BEM = \dang BFC = \dang DFC = \dang DQC = \dang AQC = \dang ABC = \dang ABM. \]
\begin{remark*}
[Motivation]
Here is one possible way to come up with the construction of point $F$
(at least this is what led Evan to find it).
If one directs all the angles in the obvious way,
there are really two points $D$ and $D'$ that are possible,
although one is outside the triangle;
they give corresponding points $E$ and $E'$.
The circles $BEM$ and $BE'M$ must then actually coincide since they
are both alleged to be tangent to line $AB$.
See the figure below.
\begin{center}
\begin{asy}
size(11cm);
pair A = dir(185);
pair C = dir(355);
pair B = dir(115);
pair Q = A*C/B;
pair X = extension(A, B, Q, C);
pair Y = 2*X-A;
filldraw(A--B--C--cycle, opacity(0.1)+lightred, red);
draw(unitcircle, red);
draw(circumcircle(B, C, Y), gray);
pair D = IP(A--Q, circumcircle(B, C, Y));
pair D_prime = -D+2*foot(circumcenter(B, C, Y), A, D);
pair N = midpoint(A--C);
pair M = midpoint(B--C);
draw(A--D_prime, mediumblue);
pair E = extension(N, origin, B, D);
pair E_prime = extension(N, origin, B, D_prime);
pair F = 2*E-B;
pair F_prime = 2*E_prime-B;
draw(F--B--D_prime, deepgreen);
draw(E_prime--N, brown);
draw(F--F_prime, brown);
filldraw(circumcircle(B, E, E_prime), opacity(0.2)+gray, red+1.1);
filldraw(M--E--E_prime--cycle, opacity(0.1)+yellow, brown+1.1);
filldraw(D--C--D_prime--cycle, opacity(0.1)+cyan, heavycyan+1.1);
filldraw(F--C--F_prime--cycle, opacity(0.1)+yellow, brown+1.1);
dot("$A$", A, dir(225));
dot("$C$", C, dir(315));
dot("$B$", B, dir(160));
dot("$Q$", Q, dir(130));
dot("$D$", D, dir(180));
dot("$D'$", D_prime, dir(D_prime));
dot(N);
dot("$M$", M, dir(65));
dot("$E$", E, dir(225));
dot("$E'$", E_prime, dir(E_prime));
dot("$F$", F, dir(F));
dot("$F'$", F_prime, dir(F_prime));
/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
!size(11cm);
A 225 = dir 185
C 315 = dir 355
B 160 = dir 115
Q 130 = A*C/B
X := extension A B Q C
Y := 2*X-A
A--B--C--cycle / 0.1 lightred / red
unitcircle / red
circumcircle B C Y / gray
D 180 = IP A--Q (circumcircle B C Y)
D' = -D+2*foot (circumcenter B C Y) A D
N .= midpoint A--C
M 65 = midpoint B--C
A--D' / mediumblue
E 225 = extension N origin B D
E' = extension N origin B D'
F = 2*E-B
F' = 2*E'-B
F--B--D' / deepgreen
E'--N / brown
F--F' / brown
circumcircle E B E' / 0.2 gray / red 1.1
M--E--E'--cycle / 0.1 yellow / brown 1.1
D--C--D'--cycle / 0.1 cyan / heavycyan 1.1
F--C--F'--cycle / 0.1 yellow / brown 1.1
*/
\end{asy}
\end{center}
One can already prove using angle chasing that $\ol{AB}$ is tangent to $(BEE')$.
So the point of the problem is to show that $M$ lies on this circle too.
However, from looking at the diagram, one may realize that in fact it seems
\[ \triangle MEE' \overset{+}{\sim} \triangle CDD' \]
is going to be true from just those marked in the figure
(and this would certainly imply the desired concyclic conclusion).
Since $M$ is a midpoint, it makes sense to dilate $\triangle EME'$
from $B$ by a factor of $2$ to get $\triangle FCF'$
so that the desired similarity is actually a spiral similarity at $C$.
Then the spiral similarity lemma says that the desired similarity
is equivalent to requiring $\ol{DD'} \cap \ol{FF'} = Q$
to lie on both $(CDF)$ and $(CD'F')$.
Hence the key construction and claim from the solution are both discovered naturally,
and we find the solution above.
(The points $D'$, $E'$, $F'$ can then be deleted to hide the motivation.)
\end{remark*}
\paragraph{Another short solution.}
Let $Z$ be on line $BDE$ such that $\angle BAZ = 90\dg$.
This lets us interpret the angle condition as follows:
\begin{claim*}
Points $A$, $D$, $Z$, $C$ are cyclic.
\end{claim*}
\begin{proof}
Because $\dang ZAC = 90\dg - A = 180\dg - \dang CDB = \dang ZDC$.
\end{proof}
\begin{center}
\begin{asy}
size(8cm);
pair A = dir(185);
pair C = dir(355);
pair B = dir(110);
pair Q = A*C/B;
pair X = extension(A, B, Q, C);
pair Y = 2*X-A;
filldraw(A--B--C--cycle, opacity(0.1)+lightred, red);
draw(unitcircle, red);
pair D = IP(A--Q, circumcircle(B, C, Y));
pair N = midpoint(A--C);
pair M = midpoint(B--C);
draw(A--Q, mediumblue);
pair E = extension(N, origin, B, D);
pair S = dir(90);
draw(S--N, brown);
draw(M--E, blue);
pair Z = extension(B, D, A, -B);
draw(A--Z--B, deepgreen);
filldraw(circumcircle(A, D, Z), opacity(0.1)+palecyan, deepgreen);
pair W = midpoint(B--Z);
pair O = origin;
draw(M--W--O--cycle, deepgreen);
filldraw(circumcircle(E, O, M), opacity(0.1)+palecyan, deepgreen);
dot("$A$", A, dir(180));
dot("$C$", C, dir(0));
dot("$B$", B, dir(140));
dot("$D$", D, dir(160));
dot("$N$", N, dir(N));
dot("$M$", M, dir(90));
dot("$E$", E, dir(225));
dot("$Z$", Z, dir(Z));
dot("$W$", W, dir(250));
dot("$O$", O, dir(180));
/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
!size(8cm);
A 180 = dir 185
C 0 = dir 355
B 140 = dir 110
Q := A*C/B
X := extension A B Q C
Y := 2*X-A
A--B--C--cycle / 0.1 lightred / red
unitcircle / red
D 160 = IP A--Q (circumcircle B C Y)
N = midpoint A--C
M 90 = midpoint B--C
A--Q / mediumblue
E 225 = extension N origin B D
S := dir 90
S--N / brown
M--E / blue
Z = extension B D A -B
A--Z--B / deepgreen
circumcircle A D Z / 0.1 palecyan / deepgreen
W 250 = midpoint B--Z
O 180 = origin
M--W--O--cycle / deepgreen
circumcircle E O M / 0.1 palecyan / deepgreen
*/
\end{asy}
\end{center}
Define $W$ as the midpoint of $\ol{BZ}$, so $\ol{MW} \parallel \ol{CZ}$.
And let $O$ denote the center of $(ABC)$.
\begin{claim*}
Points $M$, $E$, $O$, $W$ are cyclic.
\end{claim*}
\begin{proof}
Note that
\begin{align*}
\dang MOE
&= \dang(\ol{OM},\ol{BC}) + \dang(\ol{BC},\ol{AC}) + \dang(\ol{AC},\ol{OE}) \\
&= 90\dg + \dang BCA + 90\dg \\
&= \dang BCA = \dang CAD = \dang CZD = \dang MWD = \dang MWE. \qedhere
\end{align*}
\end{proof}
To finish, note
\begin{align*}
\dang MEB &= \dang MEW = \dang MOW \\
&= \dang(\ol{MO}, \ol{BC}) + \dang(\ol{BC}, \ol{AB}) + \dang(\ol{AB}, \ol{OW}) \\
&= 90\dg + \dang CBA + 90\dg = \dang CBA = \dang MBA.
\end{align*}
This implies the desired tangency.
\paragraph{A Menelaus-based approach (Kevin Ren).}
Let $P$ be on $\ol{BC}$ with $AP=PC$.
Let $Y$ be the point on line $AB$ such that $\angle ACY = 90\dg$;
as $\angle AYC = 90\dg - A$ it follows $BDYC$ is cyclic.
Let $K = \ol{AP} \cap \ol{CY}$,
so $\triangle ACK$ is a right triangle with $P$ the midpoint of its hypotenuse.
\begin{center}
\begin{asy}
size(12cm);
pair A = dir(185);
pair C = dir(355);
pair B = dir(115);
pair Q = A*C/B;
pair X = extension(A, B, Q, C);
pair Y = 2*X-A;
filldraw(A--B--C--cycle, opacity(0.1)+lightred, red);
draw(unitcircle, red);
filldraw(circumcircle(B, C, Y), opacity(0.1)+palecyan, deepcyan);
pair D = IP(A--Q, circumcircle(B, C, Y));
pair N = midpoint(A--C);
pair M = midpoint(B--C);
pair E = extension(N, origin, B, D);
pair P = extension(A, Q, B, C);
pair Y = extension(A, B, C, -A);
draw(B--Y--C, red);
pair K = extension(A, P, C, Y);
draw(A--K, blue);
filldraw(B--E--M--cycle, opacity(0.1)+yellow, deepgreen);
filldraw(Y--D--C--cycle, opacity(0.1)+yellow, deepgreen);
draw(Y--P, gray);
draw(P--N, brown);
dot("$A$", A, dir(225));
dot("$C$", C, dir(315));
dot("$B$", B, dir(160));
dot(Q);
dot("$D$", D, dir(180));
dot(N);
dot("$M$", M, dir(35));
dot("$E$", E, dir(225));
dot("$P$", P, dir(100));
dot("$Y$", Y, dir(Y));
dot("$K$", K, dir(20));
/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
!size(11cm);
A 225 = dir 185
C 315 = dir 355
B 160 = dir 115
Q 130 .= A*C/B
X := extension A B Q C
Y := 2*X-A
A--B--C--cycle / 0.1 lightred / red
unitcircle / red
circumcircle B C Y / 0.1 palecyan / deepcyan
D 180 = IP A--Q (circumcircle B C Y)
N .= midpoint A--C
M 35 = midpoint B--C
E 225 = extension N origin B D
P 100 = extension A Q B C
Y = extension A B C -A
B--Y--C / red
K 20 = extension A P C Y
A--K blue
B--E--M--cycle / 0.1 yellow / deepgreen
Y--D--C--cycle / 0.1 yellow / deepgreen
Y--P / gray
P--N / brown
*/
\end{asy}
\end{center}
\begin{claim*}
Triangles $BPE$ and $DYK$ are similar.
\end{claim*}
\begin{proof}
We have $\dang MPE = \dang CPE = \dang KCP = \dang PKC$
and $\dang EBP = \dang DBC = \dang DYC = \dang DYK$.
\end{proof}
\begin{claim*}
Triangles $BEM$ and $YDC$ are similar.
\end{claim*}
\begin{proof}
By Menelaus $\triangle PCK$ with respect to collinear points $A$, $B$, $Y$ that
\[ \frac{BP}{BC} \frac{YC}{YK} \frac{AK}{AP} = 1. \]
Since $AK/AP = 2$ (note that $P$ is the midpoint of the hypotenuse of right triangle $ACK$)
and $BC = 2BM$, this simplifies to
\[ \frac{BP}{BM} = \frac{YK}{YC}. \qedhere \]
\end{proof}
To finish, note that
\[ \dang DBA = \dang DBY = \dang DCY = \dang BME \]
implying the desired tangency.
\paragraph{A spiral similarity approach (Hans Yu).}
As in the previous solution,
let $Y$ be the point on line $AB$ such that $\angle ACY = 90\dg$;
so $BDYC$ is cyclic.
Let $\Gamma$ be the circle through $B$ and $M$ tangent to $\ol{AB}$,
and let $\Omega \coloneq (BCYD)$.
We need to show $E \in \Gamma$.
\begin{center}
\begin{asy}
size(11cm);
pair A = dir(185);
pair C = dir(355);
pair B = dir(110);
pair Q = A*C/B;
pair H = extension(A, B, Q, C);
pair Y = 2*H-A;
filldraw(A--B--C--cycle, opacity(0.1)+lightred, red);
draw(unitcircle, red);
filldraw(circumcircle(B, C, Y), opacity(0.1)+palecyan, deepcyan+1.2);
pair D = IP(A--Q, circumcircle(B, C, Y));
pair N = midpoint(A--C);
pair M = midpoint(B--C);
pair E = extension(N, origin, B, D);
pair P = extension(A, Q, B, C);
pair Y = extension(A, B, C, -A);
draw(C--Y, gray);
draw(B--Y, red);
draw(A--Q, red);
filldraw(circumcircle(B, M, E), opacity(0.1)+palecyan, deepcyan+1.2);
draw(B--E, red);
pair S = -B+2*foot(B, circumcenter(B, M, E), circumcenter(B, C, Y));
pair O = circumcenter(A, B, C);
filldraw(O--B--M--cycle, opacity(0.2)+yellow, deepgreen+1.2);
draw(P--O, red);
pair O_1 = circumcenter(B, M, E);
pair O_2 = circumcenter(B, C, Y);
label("$\Gamma$", O_1+abs(O_1-B)*dir(80), dir(80), deepcyan);
label("$\Omega$", O_2+abs(O_2-B)*dir(320), dir(320), deepcyan);
dot("$A$", A, dir(225));
dot("$C$", C, dir(315));
dot("$B$", B, dir(160));
dot(Q);
dot("$D$", D, dir(80));
dot("$M$", M, dir(20));
dot("$E$", E, dir(290));
dot("$P$", P, dir(90));
dot("$Y$", Y, dir(Y));
dot("$S$", S, dir(80));
dot("$O$", O, dir(180));
/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
!size(11cm);
A 225 = dir 185
C 315 = dir 355
B 160 = dir 110
Q 130 .= A*C/B
H := extension A B Q C
Y := 2*H-A
A--B--C--cycle / 0.1 lightred / red
unitcircle / red
circumcircle B C Y / 0.1 palecyan / deepcyan 1.2
D 80 = IP A--Q (circumcircle B C Y)
N := midpoint A--C
M 20 = midpoint B--C
E 290 = extension N origin B D
P 90 = extension A Q B C
Y = extension A B C -A
C--Y / gray
B--Y / red
A--Q / red
circumcircle B M E / 0.1 palecyan / deepcyan 1.2
B--E / red
S 80 = -B+2*foot B (circumcenter B M E) (circumcenter B C Y)
O 180 = circumcenter A B C
O--B--M--cycle / 0.2 yellow / deepgreen 1.2
P--O / red
O_1 := circumcenter B M E
O_2 := circumcenter B C Y
!label("$\Gamma$", O_1+abs(O_1-B)*dir(80), dir(80), deepcyan);
!label("$\Omega$", O_2+abs(O_2-B)*dir(320), dir(320), deepcyan);
*/
\end{asy}
\end{center}
Denote by $S$ the second intersection of $\Gamma$ and $\Omega$.
The main idea behind is to consider the spiral similarity
\[ \Psi : \Omega \to \Gamma \qquad C \mapsto M \text{ and } Y \mapsto B \]
centered at $S$ (due to the spiral similarity lemma), and show that $\Psi(D) = E$.
The spiral similarity lemma already promises $\Psi(D)$ lies on line $BD$.
\begin{claim*}
We have $\Psi(A) = O$, the circumcenter of $ABC$.
\end{claim*}
\begin{proof}
Note $\triangle OBM \overset{+}{\sim} \triangle AYC$;
both are right triangles with $\dang BAC = \dang BOM$.
\end{proof}
\begin{claim*}
$\Psi$ maps line $AD$ to line $OP$.
\end{claim*}
\begin{proof}
If we let $P$ be on $\ol{BC}$ with $AP=PC$ as before,
\[ \dang(\ol{AD}, \ol{OP}) = \dang APO = \dang OPC = \dang YCP = \dang(\ol{YC}, \ol{BM}). \]
As $\Psi$ maps line $YC$ to line $BM$ and $\Psi(A) = O$, we're done.
\end{proof}
Hence $\Psi(D)$ should not only lie on $BD$ but also line $OP$.
This proves $\Psi(D) = E$, so $E \in \Gamma$ as needed.
|
JMO-2025-notes_1
|
Prove that if $f \colon \ZZ \to \ZZ$ is any function,
then there are infinitely many integers $c$ such that the function
$g(x) = f(x)+cx$ is not a bijection.
|
Assume for contradiction that there exists a finite ``bad'' set $S$
such that $f(x)+cx$ is bijective for all $c \notin S$.
The first observation is basically that given just $f(0)$ and $f(1)$,
or any two consecutive $f$-values, we can already find bad values of $c$.
\begin{claim*}
The numbers $f(0)-f(1)$, $f(1)-f(2)$, $f(2)-f(3)$, \dots\ are all in $S$.
That is, consecutive differences of $f$ only take finitely many values.
\end{claim*}
\begin{proof}
To see that $f(0) - f(1) \in S$, just note that
\[ x \mapsto f(x) + \left( f(0) - f(1) \right) \cdot x \]
is obviously not a bijective function,
since it takes the same values at $x=0$ and $x=1$, namely $f(0)$.
The same is true for other elements.
\end{proof}
In particular, suppose $M > \max_{s \in S} |s|$.
Then the function
\[ g(x) = f(x) + (M+100) x \]
is supposed to be a bijection (because $M+100 \notin S$),
and yet $g(x+1) - g(x) > 100$ for all $x$, which is a contradiction.
|
JMO-2025-notes_2
|
Fix positive integers $k$ and $d$.
Prove that for all sufficiently large odd positive integers $n$,
the digits of the base-$2n$ representation of $n^k$ are all greater than $d$.
|
The problem actually doesn't have much to do with digits:
the idea is to pick any length $\ell \le k$,
and look at the rightmost $\ell$ digits of $n^k$;
that is, the remainder upon division by $(2n)^\ell$.
We compute it exactly:
\begin{claim*}
Let $n \ge 1$ be an odd integer, and $k \ge \ell \ge 1$ integers.
Then
\[ n^k \bmod{(2n)^i} = c(k,\ell) \cdot n^i \]
for some odd integer $1 \le c(k,\ell) \le 2^\ell-1$.
\end{claim*}
\begin{proof}
This follows directly by the Chinese remainder theorem,
with $c(k,\ell)$ being the residue class of $n^{k-i} \pmod{2^\ell}$
(which makes sense because $n$ was odd).
\end{proof}
We can now stake the required threshold:
\begin{claim*}
The problem statement holds once $n \ge (d+1) \cdot 2^{k-1}$.
\end{claim*}
\begin{proof}
Suppose $n$ is that large.
Then $n^k$ has $k$ digits in base-$2n$.
Moreover, for each $1 \le \ell \le k$ we have
\[ c(k,\ell) \cdot n^\ell \ge (d+1) \cdot (2n)^{\ell-1} \]
because $n$ is large enough;
that implies the $\ell$\ts{th} digit from the right is at least $d+1$.
Hence the problem is solved.
\end{proof}
\begin{remark*}
Note it doesn't really matter that $c(k,i)$ is odd \emph{per se};
we only need that $c(k,i) \ge 1$.
\end{remark*}
|
JMO-2025-notes_3
|
Let $m$ and $n$ be positive integers,
and let $\mathcal R$ be a $2m\times{2n}$ grid of unit squares.
A \emph{domino} is a $1\times2$ or $2\times{1}$ rectangle.
An \emph{up-right} path is a path from the lower-left corner of $\mathcal R$
to the upper-right corner of $\mathcal R$ formed by exactly $2m+2n$ edges of the grid squares.
In terms of $m$ and $n$, find the number of up-right paths that divide $\mathcal R$
into two subsets (possibly empty), each of which can be tiled with dominoes.
|
Apply the usual black/white checkerboard coloring.
We define a \emph{staircase} to be the below above an up-right path
(equivalently, a Young diagram rotated $180\dg$).
The proof is composed of two steps.
One is rewriting the ``domino-tileable'' condition to one that is actually usable;
the other is the counting part.
\paragraph{Determining when a staircase can be domino-tileable.}
It turns out the obvious guess is right.
\begin{lemma*}
A staircase can be tiled with dominoes if and only if
the number of black and white cells in it is equal.
\end{lemma*}
\begin{proof}
It's obvious equal black and white cells is necessary; we prove it's sufficient.
Assume the number of black and white cells is equal.
The proof is by induction with empty staircases being vacuous.
We consider the following four cases (which could overlap, pick one arbitrarily if so).
Note that the latter two cases are just mirrors of the first two.
\begin{description}
\ii[Case 1]
If the rightmost two columns of the staircase have the same height,
tile those two columns.
\ii[Case 2]
If some two consecutive columns of the staircase differ by more than
two cells in height, say the $(i-1)$\ts{st} and the $i$\ts{th},
then tile the top two cells in the $i$\ts{th} column and rightwards.
\ii[Case 3]
If the bottom two rows of the staircase have the same length,
tile those two rows.
\ii[Case 4]
If some two consecutive rows of the staircase differ by more than
two cells in length, say the $(j-1)$\ts{st} and the $j$\ts{th},
then tile the leftmost two cells in the $j$\ts{th} column and below.
\end{description}
Needless to say, each induction step preserves that the number of black squares
equals the number of white squares.
Hence the remaining cells can then be tiled by the induction hypothesis.
See the image below for an illustration of all four cases.
\begin{center}
\begin{asy}
pen border = blue + 1.4;
void draw_staircase(picture pic, string s, int[] heights) {
int h;
for (int i=0; i<heights.length; ++i) {
for (int j=0; j<heights[i]; ++j) {
draw(pic, shift(i,j)*unitsquare, gray);
}
h = heights[i];
int prev_height = (i>0) ? heights[i-1] : 0;
draw(pic, (i,prev_height)--(i,h)--(i+1,h), border);
}
// h is now the longest height
int w = heights.length; // max width
draw(pic, (0,0)--(w,0)--(w,h), border);
label(pic, s, (w/2,0), dir(-90));
}
real eps = 0.2;
void draw_domino(picture pic, int x, int y, bool vertical) {
filldraw(pic,
shift(x,y) * box((eps,eps), vertical ? (1-eps,2-eps) : (2-eps,1-eps)),
opacity(0.2)+yellow, red);
}
picture case1, case2, case3, case4;
draw_staircase(case1, "Case 1", new int[] {2,3,3,4,4,4});
draw_domino(case1, 4, 0, false);
draw_domino(case1, 4, 1, false);
draw_domino(case1, 4, 2, false);
draw_domino(case1, 4, 3, false);
draw_staircase(case2, "Case 2", new int[] {1,1,3,4,4,5});
draw_domino(case2, 5, 3, true);
draw_domino(case2, 4, 2, true);
draw_domino(case2, 3, 2, true);
draw_domino(case2, 2, 1, true);
draw_staircase(case3, "Case 2", new int[] {2,2,3,4,4,5});
draw_domino(case3, 0, 0, true);
draw_domino(case3, 1, 0, true);
draw_domino(case3, 2, 0, true);
draw_domino(case3, 3, 0, true);
draw_domino(case3, 4, 0, true);
draw_domino(case3, 5, 0, true);
draw_staircase(case4, "Case 4", new int[] {1,2,3,3,4,5});
draw_domino(case4, 0, 0, false);
draw_domino(case4, 1, 1, false);
draw_domino(case4, 2, 2, false);
add(shift(0,7)*case1);
add(shift(8,7)*case2);
add(shift(0,0)*case3);
add(shift(8,0)*case4);
\end{asy}
\end{center}
I claim that at least one of these four cases must apply unless the staircase is empty.
This could only occur if the heights of the staircase are $1,2,\dots,k$ in that order.
However, in that case the number of black cells and white cells
are obviously not equal (unless $k=0$), so this can never occur.
This completes the induction.
\end{proof}
\paragraph{Counting.}
Back to the main problem.
The point is that we need to find the number of up-right paths
for which the two resulting regions have equal numbers of black and white squares;
call such paths \emph{balanced}.
Since the overall $2m \times 2n$ grid has $2mn$ black and $2mn$ white squares,
it's sufficient for just the bottom-right staircase to have equal black and white cells.
Then in general, an up-right path is characterized by integer sequences
\[ 0 \le h_1 \le h_2 \le \dots \le h_{2n} \le 2m \]
corresponding to the heights of the staircase.
\begin{claim*}
The $(h_i)$ correspond to a balanced up-right path if and only if
$\{ i : h_i \equiv 1 \pmod 2\}$ has an equal number of even and odd indices.
\end{claim*}
\begin{proof}
WLOG let's fix our coloring so that the bottom-left square is black.
Then in the $i$\ts{th} column, for $i = 1, \dots, 2n$, has
(i) one more black square than white if $h_i$ is odd and $i \equiv 1 \pmod 2$;
(ii) one more white square than black if $h_i$ is odd and $i \equiv 0 \pmod 2$;
(iii) equal black and white squares if $h_i$ is even.
The conclusion follows immediately.
\end{proof}
The trick is to instead define
\[ t_i \coloneq h_i + i. \]
That is, we will instead describe the up-right paths by integer sequences $t_i$ with
\[ 1 \le t_1 < t_2 < \dots < t_{2n} \le 2n+2m. \]
\begin{claim*}
The $(t_i)$ correspond to a balanced up-right path if and only if
the set $\{t_1, \dots, t_{2n}\}$ has an equal number of even and odd elements.
\end{claim*}
\begin{proof}
Translating the four cases between the two notations gives the following table:
\[
\begin{array}{c|cc}
& i \text{ even} & i \text{ odd} \\ \hline
h_i \equiv 1 \pmod 2 & a & b \\
h_i \equiv 0 \pmod 2 & 2n-a & 2n-b
\end{array}
\iff
\begin{array}{c|cc}
& i \text{ even} & i \text{ odd} \\ \hline
t_i \equiv 1 \pmod 2 & a & 2n-b \\
t_i \equiv 0 \pmod 2 & 2n-a & b.
\end{array}
\]
Then we have
\[ \text{balanced} \iff a=b \iff a+(2n-b) = (2n-a)+b
\iff \#\{\text{odd } t_i\} = \#\{\text{even } t_i\} \]
as desired.
\end{proof}
Hence we must count the number of $2m$-element subsets of $\{1, 2, \dots, 2n+2m\}$
with $m$ even and $m$ odd terms.
Since the even and odd terms can be chosen separately,
this gives an answer of $\binom{n+m}{m}^2$.
|
JMO-2025-notes_4
|
Let $n$ be a positive integer, and let $a_0 \ge a_1 \ge \dots \ge a_n \ge 0$ be integers.
Prove that
\[ \sum_{i=0}^n i\binom{a_i}{2} \le \half \binom{a_0+a_1+\dots+a_n}{2}. \]
|
For $n=0$ (which we permit) there is nothing to prove.
Hence to prove by induction on $n$, it would be sufficient to verify
\[ 2n \binom{a_n}{2} \le \binom{a_0 + a_1 + \dots + a_n}{2}
- \binom{a_0 + a_1 + \dots + a_{n-1}}{2}. \]
Rearranging the terms around, that's equivalent to proving
\begin{align*}
\iff 2n(a_n^2 - a_n) &\le a_n^2 + a_n \cdot (2(a_0 + \dots + a_{n-1}) - 1) \\
\iff 0 &\le 2a_n(a_0 + \dots + a_{n-1} - na_n) + a_n(a_n + 2n - 1).
\end{align*}
However, the last line is obvious because $\min(a_0, \dots, a_{n-1}) \ge a_n$,
and $a_n \ge 0$.
\begin{remark*}
The only equality case is when $a_0 \in \{0,1\}$ and $a_i = 0$ for $i \ge 1$.
The bound in the problem is extremely loose and pretty much anything will work.
\end{remark*}
|
JMO-2025-notes_5
|
Let $H$ be the orthocenter of an acute triangle $ABC$,
let $F$ be the foot of the altitude from $C$ to $AB$,
and let $P$ be the reflection of $H$ across $BC$.
Suppose that the circumcircle of triangle $AFP$ intersects line $BC$
at two distinct points $X$ and $Y$.
Prove that $CX = CY$.
|
Let $Q$ be the antipode of $B$.
\begin{claim*}
$AHQC$ is a parallelogram, and $APCQ$ is an isosceles trapezoid.
\end{claim*}
\begin{proof}
As $\ol{AH} \perp \ol{BC} \perp \ol{CQ}$ and $\ol{CF} \perp \ol{AB} \perp \ol{AQ}$.
\end{proof}
\begin{center}
\begin{asy}
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
pair P = -B*C/A;
pair H = orthocenter(A, B, C);
pair F = foot(C, A, B);
pair Q = -B;
filldraw(A--B--C--cycle, opacity(0.1)+lightblue, blue);
draw(unitcircle, lightblue);
filldraw(A--H--C--Q--cycle, opacity(0.1)+lightred, red);
draw(H--P, blue);
pair O = origin;
draw(B--Q, dotted+blue);
draw(F--H, blue);
pair M = midpoint(Q--C);
pair N = midpoint(A--F);
draw(M--N, red);
draw(A--M--P, deepgreen+dashed);
draw(M--F, deepgreen+dashed);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$P$", P, dir(P));
dot("$H$", H, dir(220));
dot("$F$", F, dir(F));
dot("$Q$", Q, dir(Q));
dot("$O$", O, dir(270));
dot("$M$", M, dir(M));
dot("$N$", N, dir(N));
/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
A = dir 110
B = dir 210
C = dir 330
P = -B*C/A
H 220 = orthocenter A B C
F = foot C A B
Q = -B
A--B--C--cycle / 0.1 lightblue / blue
unitcircle / lightblue
A--H--C--Q--cycle / 0.1 lightred / red
H--P / blue
O 270 = origin
B--Q / dotted blue
F--H / blue
M = midpoint Q--C
N = midpoint A--F
M--N / red
A--M--P / deepgreen dashed
M--F / deepgreen dashed
*/
\end{asy}
\end{center}
Let $M$ be the midpoint of $\ol{QC}$.
\begin{claim*}
Point $M$ is the circumcenter of $\triangle AFP$.
\end{claim*}
\begin{proof}
It's clear that $MA = MP$ from the isosceles trapezoid.
As for $MA = MF$, let $N$ denote the midpoint of $\ol{AF}$;
then $\ol{MN}$ is a midline of the parallelogram, so $\ol{MN} \perp \ol{AF}$.
\end{proof}
Since $\ol{CM} \perp \ol{BC}$ and $M$ is the center of $(AFP)$, it follows $CX = CY$.
|
JMO-2025-notes_6
|
Let $S$ be a set of integers with the following properties:
\begin{itemize}
\ii $\{ 1, 2, \dots, 2025 \} \subseteq S$.
\ii If $a, b \in S$ and $\gcd(a, b) = 1$, then $ab \in S$.
\ii If $s + 1$ is composite for some $s \in S$, then all positive divisors of $s + 1$ are in $S$.
\end{itemize}
Prove that $S$ contains all positive integers.
\end{enumerate}
|
We prove by induction on $N$ that $S$ contains $\{1, \dots, N\}$
with the base cases being $N = 1, \dots, 2025$ already given.
For the inductive step, to show $N+1 \in S$:
\begin{itemize}
\ii If $N+1$ is composite we're already done from the third bullet.
\ii Otherwise, assume $N+1 = p \ge 2025$ is an (odd) prime number.
We say a number is \emph{good} if the prime powers in its prime factorization
are all less than $p$.
Hence by the second bullet (repeatedly), good numbers are in $S$.
Now our proof is split into three cases:
\begin{description}
\ii[Case 1.]
Suppose neither $p-1$ nor $p+1$ is a power of $2$ (but both are still even).
We claim that the number \[ s \coloneq p^2-1 = (p-1)(p+1) \] is good.
Indeed, one of the numbers has only a single factor of $2$,
and the other by hypothesis is not a power of $2$ (but still even).
So the largest power of $2$ dividing $p^2-1$ is certainly less than $p$.
And every other prime power divides at most one of $p-1$ and $p+1$.
Hence $s \coloneq p^2-1$ is good. As $s+1 = p^2$, Case 1 is done.
\ii[Case 2.]
Suppose $p+1$ is a power of $2$; that is $p = 2^q-1$.
Since $p > 2025$, we assume $q \ge 11$ is odd.
First we contend that the number
\[ s' \coloneq 2^{q+1} - 1 =
\left( 2^{(q+1)/2}-1 \right) \left( 2^{(q+1)/2}+1 \right) \]
is good.
Indeed, this follows from the two factors being coprime and both less than $p$.
Hence $s'+1 = 2^{q+1}$ is in $S$.
Thus, we again have
\[ s \coloneq p^2-1 = (p-1)(p+1) \in S \]
as we did in the previous case, because the largest power of $2$ dividing $p^2-1$
will be exactly $2^{q+1}$ which is known to be in $S$.
And since $s+1=p^2$, Case 2 is done.
\ii[Case 3.]
Finally suppose $p-1$ is a power of $2$; that is $p = 2^{2^e}+1$ is a Fermat prime.
Then in particular, $p \equiv 2 \pmod 3$.
Now observe that
\[ s \coloneq 2p-1 \equiv 0 \pmod 3 \]
and moreover $2p-1$ is not a power of $3$
(it would imply $2^{2^e+1} + 1 = 3^k$, which is impossible for $k \ge 3$
by Zsigmondy/Mihailescu/etc.).
So $s$ is good, and since $s+1 = 2p$, Case 3 is done.
\end{description}
Having finished all the cases, we conclude $p \in S$ and the induction is done.
\end{itemize}
\begin{remark*}
In fact just $2025 \in S$ is sufficient as a base case;
however this requires a bit more work to check.
Here is how:
\begin{itemize}
\ii From $2025 \in S$ we get $2026 = 2 \cdot 1013$, so $2,1013 \in S$.
\ii From $1013+1 = 1014 = 2 \cdot 3 \cdot 13^2$ we get $3,13 \in S$.
\ii From $3+1 = 4$ we get $4 \in S$.
\ii $3 \cdot 13 + 1 = 40 = 2^3 \cdot 5$ we get $5 \in S$.
\ii Once $\{1,2,\dots,5\} \subseteq S$,
the induction above actually works fine;
that is, $N \le 5$ are sufficient as base cases
for the earlier cases to finish the rest of the problem.
(Case 2 works once $q \ge 3$, and Case 3 works once $e \ge 2$.)
\end{itemize}
However, $\{1,2,3,4\} \subseteq S$ is not sufficient;
for example $S = \{1,2,3,4,6,12\}$ satisfies all the problem conditions.
\end{remark*}
|
USAMO-1996-notes_1
|
Prove that the average of the numbers $n \sin n^{\circ}$ for
$n = 2,4,6,\dots,180$ is $\cot 1^{\circ}$.
|
Because
\[ n \sin n\dg + (180-n) \sin(180\dg-n\dg) = 180 \sin n\dg \]
So enough to show that
\[ \sum_{n=0}^{89} \sin (2n)\dg = \cot1\dg \]
Let $\zeta = \cos2\dg + i \sin 2 \dg$ be a primitive root.
Then
\begin{align*}
\sum_{n=0}^{89} \frac{\zeta^n - \zeta^{-n}}{2i}
&= \frac{1}{2i}
\left[ \frac{\zeta^{90}-1}{\zeta-1} - \frac{\zeta^{-90} - 1}{\zeta^{-1}-1} \right] \\
&= \frac{1}{2i}
\left[ \frac{-2}{\zeta-1} - \frac{-2}{\zeta^{-1}-1} \right] \\
&= \frac{1}{-i} \frac{\zeta\inv-\zeta}{(\zeta-1)(\zeta\inv-1)}
= i \cdot \frac{\zeta+1}{\zeta-1}.
\end{align*}
Also,
\begin{align*}
\cot 1\dg &= \frac{\cos 1\dg}{\sin 1\dg}
= \frac{(\cos1\dg)^2}{\cos1\dg\sin1\dg} \\
&= \frac{\frac{\cos2\dg+1}{2}}{\frac{\sin2\dg}{2}}
= \frac{\half(\zeta+\zeta\inv)+1}{\frac{1}{2i}(\zeta-\zeta\inv)} \\
&= i \cdot \frac{(\zeta+1)^2}{\zeta^2-1} = i \cdot \frac{\zeta+1}{\zeta-1}.
\end{align*}
So we're done.
|
USAMO-1996-notes_2
|
For any nonempty set $S$ of real numbers,
let $\sigma(S)$ denote the sum of the elements of $S$.
Given a set $A$ of $n$ positive integers,
consider the collection of all distinct sums $\sigma(S)$ as $S$ ranges
over the nonempty subsets of $A$.
Prove that this collection of sums can be
partitioned into $n$ classes so that in each class,
the ratio of the largest sum to the smallest sum does not exceed $2$.
|
By induction on $n$ with $n=1$ being easy.
For the inductive step, assume
\[ A = \left\{ a_1 > a_2 > \dots > a_n \right\}. \]
Fix any index $k$ with the property that
\[ a_k > \frac{\sigma(A)}{2^k} \]
(which must exist since $\half+\frac14+\dots+\frac{1}{2^k} < 1$).
Then
\begin{itemize}
\ii We make $k$ classes for the sums between $\frac{\sigma(A)}{2^k}$
and $\sigma(A)$; this handles every set
which has any element in $\{a_1, \dots, a_k\}$.
\ii We make $n-k$ classes via induction hypothesis
on $\{a_{k+1}, \dots, a_n\}$.
\end{itemize}
This solves the problem.
\begin{remark*}
An easy small case to motivate this is the situation $k=1$,
which means that the largest element $a_1 \in A$
is enormous compared to all the other elements of $A$,
specifically $a_1 > a_2 + a_3 + \dots + a_n$.
In that case, all the sums involving this giant $a_1$ need to be in their own class.
The idea of the solution is to make this argument work even if $a_1$
is not quite that enormous but the sum of the largest few numbers
is big compared to the rest of $A$.
\end{remark*}
|
USAMO-1996-notes_3
|
Let $ABC$ be a triangle. Prove that there is a line $\ell$ (in the plane
of triangle $ABC$) such that the intersection of the interior of
triangle $ABC$ and the interior of its reflection $A'B'C'$ in $\ell$ has
area more than $\frac23$ the area of triangle $ABC$.
|
All that's needed is:
\begin{claim*}
If $ABC$ is a triangle where $\half < \frac{AB}{AC} < 1$, then
the $\angle A$ bisector works.
\end{claim*}
\begin{proof}
Let the $\angle A$-bisector meet $BC$ at $D$.
The overlapped area is $2[ABD]$ and
\[ \frac{[ABD]}{[ABC]} = \frac{BD}{BC} = \frac{AB}{AB+AC} \]
by angle bisector theorem.
\end{proof}
In general, suppose $x < y < z$ are sides of a triangle.
Then $\half < \frac yz < 1$ by triangle inequality as needed.
|
USAMO-1996-notes_4
|
An $n$-term sequence $(x_1, x_2, \dots, x_n)$
in which each term is either $0$ or $1$ is called a binary sequence of length $n$.
Let $a_n$ be the number of binary sequences of length $n$ containing
no three consecutive terms equal to $0$, $1$, $0$ in that order.
Let $b_n$ be the number of binary sequences of length $n$ that
contain no four consecutive terms equal to 0, 0, 1, 1 or 1, 1, 0, 0 in that order.
Prove that $b_{n+1} = 2a_n$ for all positive integers $n$.
|
Consider the map from
sequences of the latter form to sequences of the first form by
\[ (y_1, \dots, y_{n+1}) \mapsto (y_1 + y_2, y_2 + y_3, \dots, y_n + y_{n+1}). \]
It is $2$-to-$1$. The end.
|
USAMO-1996-notes_5
|
Let $ABC$ be a triangle, and $M$ an interior point such that
$\angle MAB=10^\circ$, $\angle MBA=20^\circ$, $\angle MAC=40^\circ$ and $\angle MCA=30^\circ$.
Prove that the triangle is isosceles.
|
Let $\theta = \angle MBC < 80\dg$.
By trig Ceva, we get
\[
\frac{\sin 10\dg}{\sin 40\dg} \cdot \frac{\sin \theta}{\sin 20\dg}
\cdot \frac{\sin 30\dg}{\sin (80\dg-\theta)} = 1.
\]
This simplifies to
\[ \sin \theta = 4 \sin(80\dg- \theta) \sin 40\dg \cos 10\dg.\]
\begin{claim*}
We have $\theta=60\dg$.
\end{claim*}
\begin{proof}
The left-hand side is increasing in $\theta$ and the right-hand
side is decreasing in $\theta$,
so at most one value of $\theta$ works.
But we also have
\begin{align*}
4\sin20\dg \sin40\dg \cos10\dg
&= 2\left( \cos20\dg-\cos60\dg \right) \cos 10\dg \\
&= 2\cos20\dg \cos 10\dg + \sin 80\dg \\
&= (\cos 30\dg - \cos 10\dg) + \sin 80\dg = \cos 30\dg
\end{align*}
as desired.
\end{proof}
|
USAMO-1996-notes_6
|
Determine with proof whether there is a subset $X \subseteq \ZZ$
with the following property: for any $n \in \ZZ$,
there is exactly one solution to $a+2b = n$, with $a,b \in X$.
\end{enumerate}
|
The idea is generating functions, but extra care is required
since exponents will be in $\ZZ$ rather than in $\ZZ_{\ge 0}$.
However, consider formally the limit
\[ f(x) = \prod_{k \ge 0} \left( 1 + x^{(-4)^k} \right). \]
For size reasons, this indeed converges formally to a power series,
in the sense that the coefficient of any $x^k$
is eventually zero or one for all partial sums.
We claim $X = \{ n : [x^n] f(x) = 1 \}$ works.
For a given $n$, we can truncate the sum at some large $N$ again for size reasons.
For convenience assume $N$ is even.
Now set
\[ f_N(x) = \prod_{k = 0}^N \left( 1 + x^{(-4)^k} \right). \]
Next, we compute
\[
f_N(x) f_N(x^2)
= \frac{(1+x)(1+x^2) \dots ( 1+x^{2^{2N+1}} )}%
{x^{2+8+\dots+2^{2N-1}}}
= \dots + x^{-2} + x^{-1} + 1 + x + x^2 + \dots
\]
as desired.
\begin{remark*}
The analogous problem with $\ZZ$ replaced by $\ZZ_{\ge 0}$ is easier to think about;
this has the same generating function but it's easier to think about.
In that case, the desired $X'$ is also equivalently be described as the set of numbers
whose base-$4$ representation has only $0$ and $1$.
Then every base $4$-number can be written uniquely as $a+2b$ for $a,b \in X'$,
by looking digit-by-digit.
\end{remark*}
|
USAMO-1997-notes_1
|
Let $p_1, p_2, p_3, \dots$ be the prime numbers
listed in increasing order,
and let $0 < x_0 < 1$ be a real number between 0 and 1.
For each positive integer $k$, define
\[ x_k = \begin{cases} 0 & \mbox{if} \; x_{k-1} = 0, \\[.1in]
{\displaystyle \left\{ \frac{p_k}{x_{k-1}} \right\}}
& \mbox{if} \; x_{k-1} \neq 0,
\end{cases} \]
where $\{x\}$ denotes the fractional part of $x$.
Find, with proof, all $x_0$ satisfying $0 < x_0 < 1$
for which the sequence $x_0, x_1, x_2, \dots$ eventually becomes $0$.
|
The answer is $x_0$ rational.
If $x_0$ is irrational, then all $x_i$ are irrational by induction.
So the sequence cannot become zero.
If $x_0$ is rational, then all are.
Now one simply observes that the denominators of $x_n$
are strictly decreasing, until we reach $0 = \frac 01$.
This concludes the proof.
\begin{remark*}
The sequence $p_k$ could have been any sequence of integers.
\end{remark*}
|
USAMO-1997-notes_2
|
Let $ABC$ be a triangle.
Take noncollinear points $D$, $E$, $F$ on the perpendicular bisectors
of $BC$, $CA$, $AB$ respectively.
Show that the lines through $A$, $B$, $C$
perpendicular to $EF$, $FD$, $DE$ respectively are concurrent.
|
The three lines are the radical axii of the three circles
centered at $D$, $E$, $F$, so they concur.
|
USAMO-1997-notes_3
|
Prove that for any integer $n$,
there exists a unique polynomial $Q$ with coefficients
in $\{0,1,\dots,9\}$ such that $Q(-2) = Q(-5) = n$.
|
If we let
\[ Q(x) = \sum_{k \ge 0} a_k x^k \]
then $a_k$ is uniquely determined by $n \pmod{2^k}$ and $n \pmod{5^k}$.
Indeed, we can extract the coefficients of $Q$
exactly by the following algorithm:
\begin{itemize}
\ii Define $b_0 = c_0 = n$.
\ii For $i \ge 0$, let $a_i$ be the unique digit
satisfying $a_i \equiv b_i \pmod 2$, $a_i \equiv c_i \pmod 5$.
Then, define
\[ b_{i+1} = \frac{b_i - a_i}{-2}, \qquad
c_{i+1} = \frac{c_i - a_i}{-5}. \]
\end{itemize}
The proof is automatic by Chinese remainder theorem,
so this shows uniqueness already.
The tricky part is to show that all $a_i$ are eventually zero
(i.e.\ the ``existence'' step is nontrivial
because a polynomial may only have finitely many nonzero terms).
In fact, we will prove the following claim:
\begin{claim*}
Suppose $b_0$ and $c_0$ are any integers such that
\[ b_0 \equiv c_0 \pmod 3.\]
Then defining $b_i$ and $c_i$ as above,
we have $b_i \equiv c_i \pmod 3$ for all $i$,
and $b_N = c_N = 0$ for large enough $N$.
\end{claim*}
\begin{proof}
Dropping the subscripts for ease of notation,
we are looking at the map
\[ (b, c) \mapsto \left( \frac{b-a}{-2},
\frac{c-a}{-5} \right) \]
for some $0 \le a \le 9$ (a function in $b$ and $c$).
The $b \equiv c \pmod 3$ is clearly preserved.
Also, examining the size,
\begin{itemize}
\ii If $|c| > 2$,
we have $\left\lvert \frac{c-a}{-5} \right\rvert
\le \frac{|c|+9}{5} < |c|$.
Thus, we eventually reach a pair with $|c| \le 2$.
\ii Similarly, if $|b| > 9$,
we have $\left\lvert \frac{b-a}{-2} \right\rvert
\le \frac{|b|+9}{2} < |b|$,
so we eventually reach a pair with $|b| \le 9$.
\end{itemize}
this leaves us with $5 \cdot 19 = 95$ ordered pairs to check
(though only about one third have $b \equiv c \pmod 3$).
This can be done by the following code:
\begin{lstlisting}[language=Python]
import functools
@functools.lru_cache()
def f(x0, y0):
if x0 == 0 and y0 == 0:
return 0
if x0 % 2 == (y0 % 5) % 2:
d = y0 % 5
else:
d = (y0 % 5) + 5
x1 = (x0 - d) // (-2)
y1 = (y0 - d) // (-5)
return 1 + f(x1, y1)
for x in range(-9, 10):
for y in range(-2, 3):
if (x % 3 == y % 3):
print(f"({x:2d}, {y:2d}) finished in {f(x,y)} moves")
\end{lstlisting}
As this gives the output
\begin{lstlisting}
(-9, 0) finished in 5 moves
(-8, -2) finished in 5 moves
(-8, 1) finished in 5 moves
(-7, -1) finished in 5 moves
(-7, 2) finished in 5 moves
(-6, 0) finished in 3 moves
(-5, -2) finished in 3 moves
(-5, 1) finished in 3 moves
(-4, -1) finished in 3 moves
(-4, 2) finished in 3 moves
(-3, 0) finished in 3 moves
(-2, -2) finished in 3 moves
(-2, 1) finished in 3 moves
(-1, -1) finished in 3 moves
(-1, 2) finished in 3 moves
( 0, 0) finished in 0 moves
( 1, -2) finished in 2 moves
( 1, 1) finished in 1 moves
( 2, -1) finished in 2 moves
( 2, 2) finished in 1 moves
( 3, 0) finished in 2 moves
( 4, -2) finished in 2 moves
( 4, 1) finished in 2 moves
( 5, -1) finished in 2 moves
( 5, 2) finished in 2 moves
( 6, 0) finished in 4 moves
( 7, -2) finished in 4 moves
( 7, 1) finished in 4 moves
( 8, -1) finished in 4 moves
( 8, 2) finished in 4 moves
( 9, 0) finished in 4 moves
\end{lstlisting}
we are done.
\end{proof}
|
USAMO-1997-notes_4
|
To clip a convex $n$-gon means to choose a pair of consecutive sides $AB$, $BC$
and to replace them by the three segments $AM$, $MN$, and $NC$,
where $M$ is the midpoint of $AB$ and $N$ is the midpoint of $BC$.
In other words, one cuts off the triangle $MBN$ to obtain a convex $(n+1)$-gon.
A regular hexagon $\mathcal{P}_6$ of area 1 is clipped to obtain a heptagon $\mathcal{P}_7$.
Then $\mathcal{P}_7$ is clipped (in one of the seven possible ways)
to obtain an octagon $\mathcal{P}_8$, and so on.
Prove that no matter how the clippings are done,
the area of $\mathcal{P}_n$ is greater than $\frac 13$, for all $n \geq 6$.
|
Call the original hexagon $ABCDEF$.
We show the area common to triangles $ACE$ and $BDF$ is in every
$\mathcal{P}_n$; this solves the problem since the area is $1/3$.
For every side of a clipped polygon,
we define its \emph{foundation} recursively as follows:
\begin{itemize}
\ii $AB$, $BC$, $CD$, $DE$, $EF$, $FA$
are each their own foundation (we also call these \emph{original sides}).
\ii When a new clipped edge is added, its foundation is the union of the foundations
of the two edges it touches.
\end{itemize}
Hence, any foundations are nonempty subsets of original sides.
\begin{claim*}
All foundations are in fact at most two-element sets of adjacent original sides.
\end{claim*}
\begin{proof}
It's immediate by induction
that any two adjacent sides have at most two elements
in the union of their foundations, and if there are two,
they are two adjacent original sides.
\end{proof}
Now, if a side has foundation contained in $\{AB,BC\}$, say,
then the side should be contained within triangle $ABC$.
Hence the side does not touch $AC$. This proves the problem.
|
USAMO-1997-notes_5
|
If $a,b,c > 0$ prove that
\[
\frac{1}{a^3+b^3+abc}
+ \frac{1}{b^3+c^3+abc}
+ \frac{1}{c^3+a^3+abc}
\le \frac{1}{abc}.
\]
|
From $a^3 + b^3 \ge ab(a+b)$, the left-hand side becomes
\[
\sum_{\text{cyc}} \frac{1}{a^3+b^3+abc}
\le \sum_{\text{cyc}} \frac{1}{ab(a+b+c)}
= \frac{1}{abc} \sum_{\text{cyc}} \frac{c}{a+b+c}
= \frac{1}{abc}.
\]
|
USAMO-1997-notes_6
|
Suppose the sequence of nonnegative integers
$a_1, a_2, \dots, a_{1997}$ satisfies
\[ a_i + a_j \leq a_{i+j} \leq a_i + a_j + 1 \]
for all $i,j \geq 1$ with $i + j \leq 1997$.
Show that there exists a real number $x$ such that
$a_n = \lfloor nx \rfloor$ for all $1 \leq n \leq 1997$.
\end{enumerate}
|
We are trying to show there exists an $x \in \RR$
such that
\[ \frac{a_n}{n} \le x < \frac{a_n+1}{n} \qquad \forall n. \]
This means we need to show
\[ \max_i \frac{a_i}{i} < \min_j \frac{a_j+1}{j}. \]
Replace 1997 by $N$.
We will prove this by induction,
but we will need some extra hypotheses on the indices $i,j$
which are used above.
\begin{claim*}
Suppose that
\begin{itemize}
\ii Integers $a_1$, $a_2$, \dots, $a_N$ satisfy the given conditions.
\ii Let $i = \opname{argmax}_n \frac{a_n}{n}$;
if there are ties, pick the smallest $i$.
\ii Let $j = \opname{argmin}_n \frac{a_n+1}{n}$;
if there are ties, pick the smallest $j$.
\end{itemize}
Then \[ \frac{a_i}{i} < \frac{a_j+1}{j}. \]
Moreover, these two fractions are in lowest terms,
and are adjacent in the Farey sequence of order $\max(i,j)$.
\end{claim*}
\begin{proof}
By induction on $N \ge 1$ with the base case clear.
So suppose we have the induction hypothesis
with numbers $a_1$, \dots, $a_{N-1}$,
with $i$ and $j$ as promised.
Now, consider the new number $a_N$.
We have two cases:
\begin{itemize}
\ii Suppose $i+j > N$.
Then, no fraction with denominator $N$
can lie strictly inside the interval;
so we may write for some integer $b$
\[ \frac bN \le \frac{a_i}{i}
< \frac{a_j+1}{j} \le \frac{b+1}{N}. \]
For purely algebraic reasons we have
\[ \frac{b-a_i}{N-i} \le \frac bN \le \frac{a_i}{i}
< \frac{a_j+1}{j} \le \frac{b+1}{N}
\le \frac{b-a_j}{N-j}. \]
Now,
\begin{align*}
a_N &\ge a_i + a_{N-i}
\ge a_i + (N-i) \cdot \frac{a_i}{i} \\
&\ge a_i + (b-a_i) = b \\
a_N &\le a_j + a_{N-j} + 1
\le (a_j+1) + (N-j) \cdot \frac{a_j+1}{j} \\
&= (a_j+1) + (b-a_j) = b+1.
\end{align*}
Thus $a_N \in \{b,b+1\}$.
This proves that $\frac{a_N}{N} \le \frac{a_i}{i}$
while $\frac{a_N+1}{N} \ge \frac{a_j+1}{j}$.
Moreover, the pair $(i,j)$ does not change,
so all inductive hypotheses carry over.
\ii On the other hand, suppose $i+j = N$.
Then we have
\[ \frac{a_i}{i} < \frac{a_i + a_j + 1}{N} < \frac{a_j+1}{j}. \]
Now, we know $a_N$ could be either $a_i + a_j$ or $a_i + a_j + 1$.
If it's the former, then $(i,j)$ becomes $(i,N)$.
If it's the latter, then $(i,j)$ becomes $(N,j)$.
The properties of Farey sequences ensure that
the $\frac{a_i + a_j + 1}{N}$ is reduced, either way.
\end{itemize}
\end{proof}
|
USAMO-1998-notes_1
|
Suppose that the set $\{1,2,\dots, 1998\}$
has been partitioned into disjoint pairs $\{a_i,b_i\}$ ($1\leq i\leq 999$)
so that for all $i$, $|a_i-b_i|$ equals $1$ or $6$.
Prove that the sum
\[ |a_1-b_1|+|a_2-b_2|+\dotsb +|a_{999}-b_{999}| \]
ends in the digit $9$.
|
Let $S$ be the sum.
Modulo $2$,
\[ S = \sum |a_i-b_i| \equiv \sum (a_i+b_i)
= 1 + 2 + \dots + 1998 \equiv 1 \pmod 2. \]
Modulo $5$,
\[ S = \sum |a_i-b_i| = 1 \cdot 999 \equiv 4 \pmod 5. \]
So $S \equiv 9 \pmod{10}$.
|
USAMO-1998-notes_2
|
Let $\mathcal C_1$ and $\mathcal C_2$ be concentric circles, with
$\mathcal C_2$ in the interior of $\mathcal C_1$.
From a point $A$ on $\mathcal C_1$ one draws the tangent $AB$
to $\mathcal C_2$ ($B\in \mathcal C_2$).
Let $C$ be the second point of intersection of ray $AB$ and
$\mathcal C_1$, and let $D$ be the midpoint of $\ol{AB}$.
A line passing through $A$ intersects $\mathcal C_2$ at $E$ and $F$
in such a way that the perpendicular bisectors of $\ol{DE}$ and $\ol{CF}$
intersect at a point $M$ on line $AB$.
Find, with proof, the ratio $AM/MC$.
|
By power of a point we have
\[ AE \cdot AF = AB^2
= \left( \half AB \right) \cdot \left( 2AB \right)
= AD \cdot AC \]
and hence $CDEF$ is cyclic.
Then $M$ is the circumcenter of quadrilateral $CDEF$.
\begin{center}
\begin{asy}
pair A = dir(170);
pair C = dir(40);
pair B = midpoint(A--C);
pair D = midpoint(A--B);
pair M = midpoint(C--D);
pair O = origin;
filldraw(unitcircle, opacity(0.1)+lightcyan, heavycyan);
draw(CP(O, B), heavycyan);
pair E = IP(CP(M, D), CP(O, B));
pair F = OP(CP(M, D), CP(O, B));
filldraw(CP(M, D), opacity(0.1)+lightred, red);
draw(A--C, heavygreen);
draw(A--F, heavygreen);
draw(D--E, red);
draw(C--F, red);
pair U = midpoint(D--E);
pair V = midpoint(C--F);
draw(U--M--V, red+dotted);
dot("$A$", A, dir(A));
dot("$C$", C, dir(C));
dot("$B$", B, dir(B));
dot("$D$", D, dir(D));
dot("$M$", M, dir(M));
dot("$O$", O, dir(45));
dot("$E$", E, dir(220));
dot("$F$", F, dir(F));
/* TSQ Source:
A = dir 170
C = dir 40
B = midpoint A--C
D = midpoint A--B
M = midpoint C--D
O = origin R45
unitcircle 0.1 lightcyan / heavycyan
CP O B heavycyan
E = IP CP M D CP O B R220
F = OP CP M D CP O B
CP M D 0.1 lightred / red
A--C heavygreen
A--F heavygreen
D--E red
C--F red
U := midpoint D--E
V := midpoint C--F
U--M--V red dotted
*/
\end{asy}
\end{center}
Thus $M$ is the midpoint of $\ol{CD}$
(and we are given already that $B$ is the midpoint of $\ol{AC}$,
$D$ is the midpoint of $\ol{AB}$).
Thus a quick computation along $\ol{AC}$ gives $AM/MC = 5/3$.
|
USAMO-1998-notes_3
|
Let $a_0,a_1,\dots ,a_n$ be numbers from the interval $(0,\pi/2)$
such that $\tan (a_0-\frac{\pi}{4})+ \tan (a_1-\frac{\pi}{4})
+ \dotsb +\tan (a_n-\frac{\pi}{4}) \ge n-1$.
Prove that
\[ \tan a_0\tan a_1 \dotsm \tan a_n \ge n^{n+1}. \]
|
Let $x_i = \tan(a_i - \frac{\pi}{4})$.
Then we have that
\[ \tan a_i = \tan(a_i - 45\dg + 45\dg) = \frac{x_i + 1}{1 - x_i}. \]
If we further substitute $y_i = \frac{1 - x_i}{2} \in (0,1)$,
then we have to prove that the following statement:
\begin{claim*}
If $\sum_0^n y_i \le 1$ and $y_i \ge 0$, we have
\[ \prod_{i=1}^n \left( \frac{1}{y_i}-1 \right) \ge n^{n+1}. \]
\end{claim*}
\begin{proof}
Homogenizing, we have to prove that
\[ \prod_{i=1}^n \left( \frac{y_0+y_1+y_2+\dots+y_n}{y_i}-1 \right) \ge n^{n+1}. \]
By AM-GM, we have
\[ \frac{y_1+y_2+y_3+\dots+y_n}{y_0}
\ge n \sqrt[n]{\frac{y_1y_2y_3\dots y_n}{y_1}}. \]
Cyclic product works.
\end{proof}
\begin{remark*}
Alternatively, the function $x \mapsto \log(1/x-1)$
is a convex function on $(0,1)$ so Jensen inequality should also work.
\end{remark*}
|
USAMO-1998-notes_4
|
A computer screen shows a $98 \times 98$ chessboard, colored in the usual way.
One can select with a mouse any rectangle with sides on the lines
of the chessboard and click the mouse button:
as a result, the colors in the selected rectangle switch
(black becomes white, white becomes black).
Find, with proof, the minimum number of mouse clicks
needed to make the chessboard all one color.
|
The answer is $98$.
One of several possible constructions is to toggle
all columns and rows with even indices.
In the other direction, let $n = 98$ and
suppose that $k$ rectangles are used,
none of which are $n \times n$ (else we may delete it).
Then, for any two orthogonally adjacent cells,
the edge between them must be contained in the edge of
one of the $k$ rectangles.
We define a \emph{gridline} to be a line segment
that runs in the interior of the board from
one side of the board to the other.
Hence there are $2n-2$ gridlines exactly.
Moreover, we can classify these rectangles into two types:
\begin{itemize}
\ii \emph{Full length rectangles}:
these span from one edge of the board to the other.
The two long sides completely cover two gridlines,
but the other two sides of the rectangle do not.
\ii \emph{Partial length rectangles}:
each of four sides can partially cover ``half a'' gridline.
\end{itemize}
See illustration below for $n = 6$.
\begin{center}
\begin{asy}
unitsize(0.5cm);
picture base;
for (int i=1; i<=5; ++i) {
draw(base, (0,i)--(6,i), black );
draw(base, (i,0)--(i,6), black );
}
picture full;
picture partial;
add(full, base);
add(partial, base);
label(full, "Full length", (3,0), dir(-90), red);
label(partial, "Partial length", (3,0), dir(-90), blue);
filldraw(full, box( (0,3), (6,1) ), lightred+opacity(0.3), red+1.4 );
filldraw(partial, box( (2,2), (4,5) ), lightcyan+opacity(0.3), blue+1.4 );
add(shift(0,0)*base);
add(shift(7,0)*full);
add(shift(14,0)*partial);
\end{asy}
\end{center}
Since there are $2n-2$ gridlines;
and each rectangle can cover at most two gridlines in total
(where partial-length rectangles are ``worth $\half$''
on each of the four sides),
we immediately get the bound $2k \ge 2n-2$, or $k \ge n-1$.
To finish, we prove that:
\begin{claim*}
If equality holds and $k = n-1$, then $n$ is odd.
\end{claim*}
\begin{proof}
If equality holds, then look at the horizontal gridlines
and say two gridlines are \emph{related} if some rectangle
has horizontal edges along both gridlines.
(Hence, the graph has degree either $1$ or $2$ at each vertex,
for equality to hold.)
The reader may verify the resulting graph consists only of
even length cycles and single edges,
which would mean $n-1$ is even.
\end{proof}
Hence for $n = 98$ the answer is indeed $98$ as claimed.
|
USAMO-1998-notes_5
|
Prove that for each $n\geq 2$,
there is a set $S$ of $n$ integers
such that $(a-b)^2$ divides $ab$ for every distinct $a,b\in S$.
|
This is a direct corollary of the more difficult USA TST 2015/2, reproduced below.
\begin{quote}
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 $S$.
\end{quote}
|
USAMO-1998-notes_6
|
Let $n \geq 5$ be an integer.
Find the largest integer $k$ (as a function of $n$)
such that there exists a convex $n$-gon $A_{1}A_{2}\dots A_{n}$
for which exactly $k$ of the quadrilaterals
$A_{i}A_{i+1}A_{i+2}A_{i+3}$ have an inscribed circle,
where indices are taken modulo $n$.
\end{enumerate}
|
The main claim is the following:
\begin{claim*}
We can't have both $A_1 A_2 A_3 A_4$
and $A_2 A_3 A_4 A_5$ be circumscribed.
\end{claim*}
\begin{proof}
If not, then we have the following diagram,
where $a = A_1A_2$, $b = A-2 A_3$, $c = A_3 A_4$, $d = A_4 A_5$.
\begin{center}
\begin{asy}
size(9cm);
pair A1 = dir(10);
pair A2 = dir(40);
pair A3 = dir(90);
pair A4 = dir(130);
pair A5 = dir(200);
draw(A1--A2--A3--A4--A5, blue);
label("$a$", midpoint(A1--A2), dir(A1+A2), blue);
label("$b$", midpoint(A2--A3), dir(A2+A3), blue);
label("$c$", midpoint(A3--A4), dir(A3+A4), blue);
label("$d$", midpoint(A4--A5), dir(A4+A5), blue);
draw(A1--A4, red);
label(rotate(-20)*"$c+a-b$", 0.6*A4+0.4*A1, dir(A1+A4), red);
draw(A2--A5, orange);
label(rotate(30)*"$b+d-c$", 0.5*A5+0.5*A2, dir(A2+A5), orange);
dot("$A_1$", A1, dir(A1), blue);
dot("$A_2$", A2, dir(A2), blue);
dot("$A_3$", A3, dir(A3), blue);
dot("$A_4$", A4, dir(A4), blue);
dot("$A_5$", A5, dir(A5), blue);
\end{asy}
\end{center}
Then $A_1 A_4 = c+a-b$ and $A_5A_2 = b+d-c$.
But now
\[ A_1 A_4 + A_2 A_5 = (c+a-b) + (b+d-c) = a+d = A_1 A_2 + A_4 A_5 \]
but in the picture we have an obvious violation
of the triangle inequality.
\end{proof}
This immediately gives an upper bound of
$\left\lfloor n/2 \right\rfloor$.
For the construction, one can construct a suitable cyclic $n$-gon by
using a continuity argument (details to be added).
|
USAMO-1999-notes_1
|
Some checkers placed on an $n \times n$ checkerboard satisfy the following conditions:
\begin{enumerate}
\ii[(a)] every square that does not contain a checker shares a side with one that does;
\ii[(b)] given any pair of squares that contain checkers,
there is a sequence of squares containing checkers,
starting and ending with the given squares,
such that every two consecutive squares of the sequence share a side.
\end{enumerate}
Prove that at least $(n^{2}-2)/3$ checkers have been placed on the board.
|
Take a spanning tree on the set $V$ of checkers
where the $|V|-1$ edges of the tree are given by orthogonal adjacency.
By condition (a) we have
\[ \sum_{v \in V} (4-\deg v) \ge n^2 - |V| \]
and since $\sum_{v \in V} \deg v = 2(|V|-1)$ we get
\[ 4|V| - \left( 2|V|-2 \right) \ge n^2 - |V| \]
which implies $|V| \ge \frac{n^2-2}{3}$.
|
USAMO-1999-notes_2
|
Let $ABCD$ be a convex cyclic quadrilateral.
Prove that \[ |AB - CD| + |AD - BC| \geq 2|AC - BD|. \]
|
Let the diagonals meet at $P$,
and let $AP = pq$, $DP = pr$, $BP = qs$, $CP = rs$.
Then set $AB = qx$, $CD = rx$, $AD = py$, $BC = sy$.
In this way we compute
\[ \left\lvert AC-BD \right\rvert
= \left\lvert (p-s)(q-r) \right\rvert \]
and
\[ \left\lvert AB-CD \right\rvert
= \left\lvert q-r \right\rvert x. \]
By triangle inequality on $\triangle AXB$,
we have $x \ge \left\lvert p-s \right\rvert$.
So $\left\lvert AB-CD \right\rvert \ge \left\lvert AC-BD \right\rvert$.
|
USAMO-1999-notes_3
|
Let $p > 2$ be a prime and let $a$, $b$, $c$, $d$ be integers not divisible by $p$,
such that
\[ \left\{ \dfrac{ra}{p} \right\} + \left\{ \dfrac{rb}{p} \right\}
+ \left\{ \dfrac{rc}{p} \right\} + \left\{ \dfrac{rd}{p} \right\} = 2 \]
for any integer $r$ not divisible by $p$.
(Here, $\{t\} = t - \left\lfloor t \right\rfloor$ is the fractional part.)
Prove that at least two of the numbers
$a+b$, $a+c$, $a+d$, $b+c$, $b+d$, $c+d$ are divisible by $p$.
|
First of all, we apparently have $r(a+b+c+d) \equiv 0 \pmod p$
for every prime $p$, so it automatically follows that $a+b+c+d \equiv 0 \pmod p$.
By scaling appropriately, and also replacing each number with its remainder modulo $p$,
we are going to assume that
\[ 1 = a \le b \le c \le d < p. \]
We are going to prove that $d = p-1$, which will solve the problem.
\begin{claim*}
For each integer $r = 1, 2, \dots, p-1$ we have
\[ 2(r-1) = \left\lfloor \frac{rb}{p} \right\rfloor
+ \left\lfloor \frac{rc}{p} \right\rfloor
+ \left\lfloor \frac{rd}{p} \right\rfloor. \]
\end{claim*}
\begin{proof}
By plugging in $r=1$ to the given we have $a+b+c+d=2p$.
Now, we have
\[ 2 = \sum_{\text{cyc}} \left( \frac{ra}{p}
- \left\lfloor \frac{ra}{p} \right\rfloor \right) \]
and since $a+b+c+d=2p$ the conclusion follows.
\end{proof}
%We now define $x = b/p$, $y = c/p$, $z = d/p$.
%Hence $0 < x \le y \le z < 1$.
%Our goal it show that $z = (p-1)/p$.
We vaguely outline the approach now, before giving a formalization.
Imagine the interval $[0,1]$.
One by one, for each $r = 1, 2, 3, \dots, p-1$,
we mark the fractions with denominator $r$ on this number line;
the resulting pictures may be better known as \emph{Farey fractions}.
At each step, we can place the three numbers $b/p$, $c/p$, $d/p$
into one of the resulting sub-intervals.
Our goal is to show that $d/p$ is always in the rightmost interval,
while $b/p$ and $c/p$ are always to the right of symmetrically mirrored points.
An example of a possible diagram is shown below (not to scale).
\begin{center}
\begin{asy}
unitsize(11cm);
real h = 0.05;
draw( (0,0)--(1,0) );
void tick(string s, real x, real t, pen p) {
label(s, (x,-t), dir(-90), p);
draw((x,-t)--(x,t), p);
}
tick("$0$", 0, h, black);
tick("$1$", 1, h, black);
tick("$\frac{1}{2}$", 1/2, h, black);
tick("$\frac{1}{3}$", 1/3, 0.7*h, blue);
tick("$\frac{2}{3}$", 2/3, 0.7*h, blue);
tick("$\frac{1}{4}$", 1/4, 0.5*h, deepgreen);
tick("$\frac{3}{4}$", 3/4, 0.5*h, deepgreen);
tick("$\frac{1}{5}$", 1/6, 0.3*h, gray); // white lie
tick("$\frac{2}{5}$", 2/5, 0.3*h, gray);
tick("$\frac{3}{5}$", 3/5, 0.3*h, gray);
tick("$\frac{4}{5}$", 5/6, 0.3*h, gray); // white lie
dot("$\boxed{\frac{b}{p}}$", (0.3,0), dir(90), red);
dot("$\boxed{\frac{c}{p}}$", (0.72,0), dir(90), red);
dot("$\boxed{\frac{d}{p}}$", (0.95,0), dir(90), red);
\end{asy}
\end{center}
In symbols, it will be enough to prove the following.
\begin{claim*}
For each $r = 1, 2, \dots, p-2$ we have
$\frac{r-1}{r} < \frac dp < 1$.
Equivalently, for each $r = 1, 2, \dots, p-2$
we have $\left\lfloor \frac{rb}{p} \right\rfloor
+ \left\lfloor \frac{rc}{p} \right\rfloor = r-1$.
\end{claim*}
\begin{proof}
Assume this is not true and take the minimal counterexample $r > 1$.
Then evidently
\[ r-1 > \left\lfloor \frac{rd}{p} \right\rfloor
\ge \left\lfloor \frac{(r-1)d}{p} \right\rfloor
= r-2. \]
Now, we have that
\[ 2(r-1) = \left\lfloor \frac{rb}{p} \right\rfloor
+ \left\lfloor \frac{rc}{p} \right\rfloor
+ \underbrace{\left\lfloor \frac{rd}{p} \right\rfloor}_{= r-2}. \]
Thus
$\left\lfloor \frac{rb}{p} \right\rfloor >
\left\lfloor \frac{(r-1)b}{p} \right\rfloor$,
and $\left\lfloor \frac{rc}{p} \right\rfloor >
\left\lfloor \frac{(r-1)b}{p} \right\rfloor$.
An example of this situation is illustrated below with $r = 7$
(not to scale).
\begin{center}
\begin{asy}
unitsize(11cm);
real h = 0.05;
draw( (0,0)--(1,0) );
void tick(string s, real x, real t, pen p) {
label(s, (x,-t), dir(-90), p);
draw((x,-t)--(x,t), p);
}
tick("$0$", 0, h, black);
tick("$1$", 1, h, black);
tick("$\frac{1}{2}$", 1/2, h, black);
tick("$\frac{1}{3}$", 1/3, 0.7*h, blue);
tick("$\frac{2}{3}$", 2/3, 0.7*h, blue);
tick("$\frac{1}{4}$", 1/4, 0.5*h, deepgreen);
tick("$\frac{3}{4}$", 3/4, 0.5*h, deepgreen);
tick("$\frac{1}{5}$", 1/6, 0.3*h, gray);
tick("$\frac{2}{5}$", 2/5, 0.3*h, gray);
tick("$\frac{3}{5}$", 3/5, 0.3*h, gray);
tick("$\frac{4}{5}$", 5/6, 0.3*h, gray);
tick("$\frac{1}{6}$", 1/8, 0.3*h, gray);
tick("$\frac{5}{6}$", 7/8, 0.3*h, gray);
dot("$\boxed{\frac{b}{p}}$", (0.31,0), dir(90), red);
dot("$\boxed{\frac{c}{p}}$", (0.725,0), dir(90), red);
dot("$\boxed{\frac{d}{p}}$", (0.92,0), dir(90), red);
tick("$\dfrac{2}{7}$", 2/7, 1.5*h, red+1);
tick("$\dfrac{5}{7}$", 0.7, 1.5*h, red+1);
tick("$\dfrac{6}{7}$", 0.9, 1.5*h, red+1);
tick("$\dfrac{7}{8}$", 0.95, 2*h, orange+1);
tick("$\dfrac{3}{8}$", 0.375, 2*h, orange+1);
tick("$\dfrac{6}{8}$", 0.75, 2*h, orange+1);
\end{asy}
\end{center}
Right now, $\frac bp$ and $\frac cp$
are just to the right of
$\frac ur$ and $\frac vr$ for some $u$ and $v$ with $u+v=r$.
The issue is that the there is some fraction just to the right of
$\frac bp$ and $\frac cp$ from an earlier value of $r$,
and by hypothesis its denominator is going to be strictly greater than $1$.
It is at this point we are going to use the properties of Farey sequences.
When we consider the fractions with denominator $r+1$,
they are going to lie outside of the interval
they we have constrained $\frac bp$ and $\frac cp$ to lie in.
Indeed, our minimality assumption on $r$ guarantees that there is no
fraction with denominator less than $r$ between $\frac ur$ and $\frac bp$.
So if $\frac ur < \frac bp < \frac st$
(where $\frac ur$ and $\frac st$ are the closest fractions
with denominator at most $r$ to $\frac bp$)
then Farey theory says the next fraction
inside the interval $[\frac ur, \frac st]$
is $\frac{u+s}{r+t}$, and since $t > 1$, we have $r+t > r+1$.
In other words, we get an inequality of the form
\[ \frac ur < \frac bp < \underbrace{\text{something}}_{=s/t} \le \frac{u+1}{r+1}. \]
The same holds for $\frac cp$ as
\[ \frac vr < \frac cp < \text{something} \le \frac{v+1}{r+1}. \]
Finally,
\[ \frac dp < \frac{r-1}{r} < \frac{r}{r+1}. \]
So now we have that
\[ \left\lfloor \frac{(r+1)b}{p} \right\rfloor
+ \left\lfloor \frac{(r+1)c}{p} \right\rfloor
+ \left\lfloor \frac{(r+1)d}{p} \right\rfloor
\le u + v + (r-1) = 2r-1 \]
which is a contradiction.
\end{proof}
Now, since
\[ \frac{p-3}{p-2} < \frac dp
\implies d > \frac{p(p-3)}{p-2} = p - 1 - \frac{2}{p-2} \]
which for $p > 2$ gives $d = p-1$.
|
USAMO-1999-notes_4
|
Let $a_1$, $a_2$, \dots, $a_n$ be a sequence of $n > 3$ real numbers
such that
\[ a_1 + \dots + a_n \ge n \quad\text{and}\quad
a_1^2 + \dots + a_n^2 \ge n^2. \]
Prove that $\max(a_1, \dots, a_n) \ge 2$.
|
Proceed by contradiction, assuming $a_i < 2$ for all $i$.
If all $a_i \ge 0$, then $ n^2 \le \sum_i a_i^2 < n \cdot 2^2$,
contradiction.
Otherwise, assume at least one $a_i$ is negative.
Note that if $-x$ and $-y$ are both present in the sequence ($x,y>0$),
then we can replace them with $-(x+y)$ and $0$.
So we may assume that there is exactly one negative term, say $a_n = -M$.
Now, smooth all the nonnegative $a_i$ to be $2$,
making all inequalities strict.
Now, we have that
\begin{align*}
2(n-1) - M &> n \\
4(n-1) + M^2 &> n^2.
\end{align*}
This gives $n-2 < M < n-2$, contradiction.
Equality in the original occurs when $n-1$ of the $a_i$ are equal to $2$
and the last one is equal to $-(n-2)$.
|
USAMO-1999-notes_5
|
The Y2K Game is played on a $1 \times 2000$ grid as follows.
Two players in turn write either an S or an O in an empty square.
The first player who produces three consecutive boxes that spell SOS wins.
If all boxes are filled without producing SOS then the game is a draw.
Prove that the second player has a winning strategy.
|
The main insight is that a construct of the form
\[ S \; \square \; \square \; S \]
(here the $\square$ is blank)
will kill any player which plays inside it.
We call this a \emph{trap} accordingly.
\begin{claim*}
The second player can force a trap to exist;
in this case the game will never end in a draw.
\end{claim*}
\begin{proof}
Actually the second player can construct a trap
on her second turn by playing an $S$ far enough away
from the edges of the board and the first player's initial move.
\end{proof}
\begin{claim*}
The second player always has a move which prevents
her from losing.
\end{claim*}
\begin{proof}
Since there are an odd number of empty squares at the start of
the second player's turn,
there must be a square which is bordered by either
two filled or two empty squares.
The second player can then play $O$ in this square,
which is always safe.
\end{proof}
Together these two claims finish the problem.
\begin{remark*}
Actually, one can show that the ``only'' way to lose
is to be forced to play inside a trap.
Indeed, suppose playing in a certain cell $c$ loses.
If we wrote $O$, that means $c$ is bordered by exactly one $S$,
with a blank cell on the neighbor.
But we could also write $S$; checking cases we find $c$
is part of a trap.
Thus a player can lose only if all blank cells
are in traps; ergo, the number of blank cells is even.
This never happens for the second player.
Thus this gives an alternative solution,
and moreover a reason to believe that all
correct solutions must involve traps.
A similar proof shows that for large $n$, with a $1 \times n$ board,
the first player has a winning strategy for odd $n$,
and the second player has a winning strategy for even $n$.
\end{remark*}
|
USAMO-1999-notes_6
|
Let $ABCD$ be an isosceles trapezoid with $AB \parallel CD$.
The inscribed circle $\omega$ of triangle $BCD$ meets $CD$ at $E$.
Let $F$ be a point on the (internal) angle bisector of $\angle DAC$
such that $EF \perp CD$.
Let the circumscribed circle of triangle $ACF$
meet line $CD$ at $C$ and $G$.
Prove that the triangle $AFG$ is isosceles.
\end{enumerate}
|
Note $E$ is contact point of $A$-excircle of $\triangle ACD$,
so $F$ is $A$-excenter.
Hence $CF$ is external angle bisector of $\angle ACG$
which implies $FA = FG$
(since $F$ is the arc midpoint on the circumcircle of $AFG$).
|
USAMO-2000-notes_1
|
Call a real-valued function $f$ \emph{very convex} if
\[ \frac{f(x)+f(y)}{2}
\ge f\left( \frac{x+y}{2} \right) + \left\lvert x-y \right\rvert \]
holds for all real numbers $x$ and $y$.
Prove that no very convex function exists.
|
For $C \ge 0$, we say a function $f$ is \emph{$C$-convex}
\[ \frac{f(x)+f(y)}{2}
\ge f\left( \frac{x+y}{2} \right) + C\left\lvert x-y \right\rvert. \]
Suppose $f$ is $C$-convex.
Let $a < b < c < d < e$ be any arithmetic progression,
such that $t = |e-a|$.
Observe that
\begin{align*}
f(a) + f(c) &\ge 2f(b) + C \cdot \half t \\
f(c) + f(e) &\ge 2f(d) + C \cdot \half t \\
f(b) + f(d) &\ge 2f(c) + C \cdot \half t
\end{align*}
Adding the first two to twice the third gives
\[ f(a) + f(e) \ge 2f(c) + 2C \cdot t. \]
So we conclude $C$-convex function is also $2C$-convex.
This is clearly not okay for $C > 0$.
|
USAMO-2000-notes_2
|
Let $S$ be the set of all triangles $ABC$ for which
\[ 5 \left( \frac{1}{AP} + \frac{1}{BQ} + \frac{1}{CR} \right)
- \frac{3}{\min\{ AP, BQ, CR \}} = \frac{6}{r}, \]
where $r$ is the inradius and $P$, $Q$, $R$ are the points of tangency
of the incircle with sides $AB$, $BC$, $CA$ respectively.
Prove that all triangles in $S$ are isosceles and similar to one another.
|
We will prove the inequality
\[ \frac{2}{AP} + \frac{5}{BQ} + \frac{5}{CR} \ge \frac 6r \]
with equality when $AP : BQ : CR = 1 : 4 : 4$.
This implies the problem statement.
Letting $x= AP$, $y = BQ$, $z = CR$, the inequality becomes
\[ \frac2x + \frac5y + \frac5z \ge 6\sqrt{\frac{x+y+z}{xyz}}. \]
Squaring both sides and collecting terms gives
\[
\frac{4}{x^2} + \frac{25}{y^2} + \frac{25}{z^2} + \frac{14}{yz}
\ge \frac{16}{xy} + \frac{16}{xz}.
\]
If we replace $x=1/a$, $y=4/b$, $z=4/c$,
then it remains to prove the inequality
\[ 64a^2 + 25(b+c)^2 \ge 64a(b+c) + 36bc \]
where equality holds when $a=b=c$.
This follows by two applications of AM-GM:
\begin{align*}
16 \left( 4a^2 + (b+c)^2 \right) &\ge 64a(b+c) \\
9(b+c)^2 &\ge 36bc.
\end{align*}
Again one can tell this is an inequality by counting
degrees of freedom.
|
USAMO-2000-notes_3
|
A game of solitaire is played with $R$ red cards,
$W$ white cards, and $B$ blue cards.
A player plays all the cards one at a time.
With each play he accumulates a penalty.
If he plays a blue card, then he is charged a penalty
which is the number of white cards still in his hand.
If he plays a white card, then he is charged a penalty
which is twice the number of red cards still in his hand.
If he plays a red card, then he is charged a penalty
which is three times the number of blue cards still in his hand.
Find, as a function of $R$, $W$, and $B$, the minimal total penalty
a player can amass and the number of ways in which this minimum can be achieved.
|
The minimum penalty is
\[ f(B,W,R) = \min (BW, 2WR, 3RB) \]
or equivalently, the natural guess of
``discard all cards of one color first''
is actually optimal (though not necessarily unique).
This can be proven directly by induction.
Indeed the base case $BWR = 0$
(in which case zero penalty is clearly achievable).
The inductive step follows from
\[ f(B,W,R) = \min
\begin{cases}
f(B-1,W,R) + W \\
f(B,W-1,R) + 2R \\
f(B,W,R-1) + 3B.
\end{cases}
\]
It remains to characterize the strategies.
This is an annoying calculation, so we just state the result.
\begin{itemize}
\ii If any of the three quantities $BW$, $2WR$, $3RB$
is strictly smaller than the other three,
there is one optimal strategy.
\ii If $BW = 2WR < 3RB$,
there are $W+1$ optimal strategies,
namely discarding from $0$ to $W$ white cards,
then discarding all blue cads.
(Each white card discarded still preserves $BW = 2WR$.)
\ii If $2WR = 3RB < BW$,
there are $R+1$ optimal strategies,
namely discarding from $0$ to $R$ red cards,
and then discarding all white cards.
\ii If $3RB = RB < 2WR$,
there are $B+1$ optimal strategies,
namely discarding from $0$ to $B$ blue cards,
and then discarding all red cards.
\ii Now suppose $BW = 2WR = 3RB$.
Discarding a card of one color
ends up in exactly one of the previous three cases.
This gives an answer of $R+W+B$ strategies.
\end{itemize}
|
USAMO-2000-notes_4
|
Find the smallest positive integer $n$ such
that if $n$ squares of a $1000 \times 1000$ chessboard are colored,
then there will exist three colored squares
whose centers form a right triangle with sides
parallel to the edges of the board.
|
The answer is $n = 1999$.
For a construction with $n = 1998$,
take a punctured L as illustrated below (with $1000$ replaced by $4$):
\[ \begin{bmatrix} 1 \\ 1 \\ 1 \\ & 1 & 1 & 1 \end{bmatrix}. \]
We now show that if there is no right triangle,
there are at most $1998$ tokens (colored squares).
In every column with more than two tokens,
we have token emit a bidirectional horizontal
death ray (laser) covering its entire row:
the hypothesis is that the death ray won't hit any other tokens.
\begin{center}
\begin{asy}
size(6cm);
for (int i=1; i<=8; ++i) {
if ((i-1)*(i-3)*(i-5) != 0) draw( (0.5,i)--(8.5,i), red+1, Arrows(TeXHead) );
for (int j=1; j<=8; ++j) {
if ( (i-j)/2 == (i-j)#2 )
filldraw(shift(i-0.5,j-0.5)*unitsquare, opacity(0.2)+gray, gray);
else
draw(shift(i-0.5,j-0.5)*unitsquare, gray);
}
}
draw( (3,2)--(3,7), blue);
draw( (5,4)--(5,8), blue);
void token(pair P) { fill(circle(P, 0.2), black); }
token( (1,1) );
token( (2,1) );
token( (3,2) );
token( (3,6) );
token( (3,7) );
token( (4,1) );
token( (5,4) );
token( (5,8) );
token( (6,5) );
token( (7,3) );
token( (8,1) );
\end{asy}
\end{center}
Assume there are $n$ tokens and that $n > 1000$.
Then obviously some column has more than two tokens,
so at most $999$ tokens don't emit a death ray
(namely, any token in its own column).
Thus there are at least $n-999$ death rays.
On the other hand, we can have at most $999$ death rays total
(since it would not be okay for the whole board to have death rays,
as some row should have more than two tokens).
Therefore, $n \le 999 + 999 = 1998$ as desired.
|
USAMO-2000-notes_5
|
Let $A_1 A_2 A_3$ be a triangle,
and let $\omega_1$ be a circle in its plane
passing through $A_1$ and $A_2$.
Suppose there exists circles $\omega_2$, $\omega_3$, \dots, $\omega_7$
such that for $k = 2, 3, \dots, 7,$
circle $\omega_k$ is externally tangent to $\omega_{k-1}$ and passes
through $A_k$ and $A_{k+1}$ (indices mod $3$).
Prove that $\omega_7 = \omega_1$.
|
The idea is to keep track of the subtended arc
$\widehat{A_i A_{i+1}}$ of $\omega_i$ for each $i$.
To this end, let $\beta = \dang A_1 A_2 A_3$,
$\gamma = \dang A_2 A_3 A_1$
and $\alpha = \dang A_1 A_2 A_3$.
\begin{center}
\begin{asy}
size(8cm);
defaultpen(fontsize(9pt));
pair A2 = (0,0);
pair O1 = (-8,0);
pair O2 = (5,0);
pair A1 = abs(A2-O1)*dir(285)+O1;
pair A3 = abs(A2-O2)*dir(245)+O2;
pair O3 = extension(O2, A3, midpoint(A1--A3), midpoint(A1--A3)+dir(A1-A3)*dir(90));
filldraw(circle(O1,abs(A1-O1)), opacity(0.1)+lightgreen, deepgreen);
filldraw(circle(O2,abs(A2-O2)), opacity(0.1)+lightgreen, deepgreen);
filldraw(circle(O3,abs(A3-O3)), opacity(0.1)+lightgreen, deepgreen);
filldraw(A1--A2--A3--cycle, opacity(0.1)+orange, red);
pair J = (0,-3.2);
draw(O1--O2--O3, deepcyan+dotted);
dot("$O_1$", O1, dir(90), deepcyan);
dot("$O_2$", O2, dir(90), deepcyan);
dot("$O_3$", O3, dir(-90), deepcyan);
dot("$A_1$", A1, dir(A1-J), red);
dot("$A_2$", A2, dir(A2-J), red);
dot("$A_3$", A3, dir(A3-J), red);
label("$\alpha$", A1, 2.4*dir(J-A1), red);
label("$\beta$", A2, 2*dir(J-A2), red);
label("$\gamma$", A3, 2*dir(J-A3), red);
\end{asy}
\end{center}
Initially, we set $\theta = \dang O_1 A_2 A_1$.
Then we compute
\begin{align*}
\dang O_1 A_2 A_1 &= \theta \\
\dang O_2 A_3 A_2 &= -\beta-\theta \\
\dang O_3 A_1 A_3 &= \beta-\gamma+\theta \\
\dang O_4 A_2 A_1 &= (\gamma-\beta-\alpha)-\theta \\
\end{align*}
and repeating the same calculation another round gives
\[ \dang O_7 A_2 A_1 = k-(k-\theta) = \theta \]
with $k = \gamma-\beta-\alpha$.
This implies $O_7 = O_1$, so $\omega_7 = \omega_1$.
|
USAMO-2000-notes_6
|
Let $a_1, b_1, a_2, b_2, \dots , a_n, b_n$ be nonnegative real numbers.
Prove that
\[ \sum_{i, j = 1}^{n} \min\{a_i a_j, b_i b_j\}
\le \sum_{i, j = 1}^{n} \min\{a_i b_j, a_j b_i\}. \]
\end{enumerate}
|
We present two solutions.
\paragraph{First solution by creating a single min (Vincent Huang and Ravi Boppana).}
Let $b_i = r_i a_i$ for each $i$,
and rewrite the inequality as
\[ \sum_{i,j} a_i a_j \left[
\min (r_i, r_j) - \min(1, r_i r_j) \right] \ge 0. \]
We now do the key manipulation to convert the double min
into a separate single min.
Let $\eps_i = +1$ if $r_i \ge 1$, and $\eps_i = -1$ otherwise,
and let $s_i = |r_i-1|$.
Then we pass to absolute values:
\begin{align*}
2\min(r_i,r_j) - 2\min(1,r_i r_j)
&= |r_i r_j-1| - |r_i-r_j| - (r_i-1)(r_j-1) \\
&= |r_i r_j-1| - |r_i-r_j| - \eps_i \eps_j s_i s_j \\
&= \eps_i \eps_j \min \left( |1-r_i r_j \pm (r_i-r_j)| \right)
- \eps_i \eps_j s_i s_j \\
&= \eps_i \eps_j \min \left( s_i(r_j+1), s_j(r_i+1) \right)
- \eps_i \eps_j s_i s_j \\
&= (\eps_i s_i)(\eps_j s_j)
\min \left( \frac{r_j+1}{s_j}-1, \frac{r_i+1}{s_i}-1 \right).
\end{align*}
So let us denote $x_i = a_i \eps_i s_i \in \RR$,
and $t_i = \frac{r_i+1}{s_i}-1 \in \RR_{\ge 0}$.
Thus it suffices to prove that:
\begin{claim*}
We have
\[ \sum_{i,j} x_i x_j \min(t_i, t_j) \ge 0 \]
for arbitrary $x_i \in \RR$, $t_i \in \RR_{\ge 0}$.
\end{claim*}
\begin{proof}
One can just check this ``by hand''
by assuming $t_1 \le t_2 \le \dots \le t_n$;
then the left-hand side becomes
\[ \sum_i t_i x_i^2 + 2 \sum_{i < j} 2 t_i x_i x_j
= \sum_i (t_i - t_{i-1}) (x_i + x_{i+1} + \dots + x_n)^2 \ge 0. \]
There is also a nice proof using the integral identity
\[ \min(t_i, t_j) = \int_0^\infty \mathbf1(u \le t_i) \mathbf1(u \le t_j) \; du \]
where the $\mathbf1$ are indicator functions.
Indeed,
\begin{align*}
\sum_{i,j} x_i x_j \min(t_i, t_j)
&= \sum_{i,j} x_i x_j \int_0^\infty \mathbf1(u \le t_i) \mathbf1(u \le t_j) \; du \\
&= \int_0^\infty \sum_i x_i \mathbf1(u \le t_i)
\sum_j x_j \mathbf1(u \le t_j) \; du \\
&= \int_0^\infty \left( \sum_i x_i \mathbf1(u \le t_i) \right)^2 \; du \\
&\ge 0. \qedhere
\end{align*}
\end{proof}
\paragraph{Second solution by smoothing (Alex Zhai).}
The case $n=1$ is immediate, so we'll proceed by induction on $n \ge 2$.
Again, let $b_i = r_i a_i$ for each $i$, and
write the inequality as
\[ L_n(a_1, \dots, a_n, r_1, \dots, r_n)
\coloneq \sum_{i,j} a_i a_j \left[
\min (r_i, r_j) - \min(1, r_i r_j) \right] \ge 0. \]
First note that if $r_1 = r_2$ then
\[ L_n(a_1, a_2, a_3, \dots, r_1, r_1, r_3 \dots)
= L_{n-1}(a_1+a_2, a_3, \dots, r_1, r_3, \dots) \]
and so our goal is to smooth to a situation
where two of the $r_i$'s are equal,
so that we may apply induction.
On the other hand, $L_n$ is a \emph{piecewise linear} function in $r_1 \ge 0$.
Let us smooth $r_1$ then.
Note that if the minimum is attained at $r_1 = 0$,
we can ignore $a_1$ and reduce to the $(n-1)$-variable case.
On the other hand, the minimum must be achieved at a cusp
which opens upward, which can only happen if $r_i r_j = 1$ for some $j$.
(The $r_i = r_j$ cusps open downward, sadly.)
In this way, whenever some $r_i$ is not equal to the reciprocal
of any other $r_\bullet$, we can smooth it.
This terminates; so we may smooth until we reach a situation for which
\[ \{ r_1, \dots, r_n \} = \{ 1/r_1, \dots, 1/r_n \}. \]
Now, assume WLOG that $r_1 = \max_i r_i$ and $r_2 = \min_i r_i$,
hence $r_1 r_2 = 1$ and $r_1 \ge 1 \ge r_2$.
We isolate the contributions from $a_1$, $a_2$, $r_1$ and $r_2$.
\begin{align*}
L_n(\dots) &= a_1^2 \left[ r_1 - 1 \right] + a_2^2 \left[ r_2 - r_2^2 \right]
+ 2a_1a_2 \left[ r_2 - 1 \right] \\
&+ 2a_1 \left[ (a_3r_3+\dots+a_n r_n) - (a_3+\dots+a_n) \right] \\
&+ 2a_2r_2 \left[ (a_3+\dots+a_n) - (a_3r_3+\dots+a_n r_n) \right] \\
&+ \sum_{i=3}^n \sum_{j=3}^n a_i a_j \left[
\min (r_i, r_j) - \min(1, r_i r_j) \right].
\end{align*}
The idea now is to smooth via
\[ (a_1, a_2, r_1, r_2)
\longrightarrow
\left( a_1, \frac 1t a_2, \frac 1t r_1, tr_2 \right) \]
where $t \ge 1$ is such that
$\frac 1t r_1 \ge \max(1, r_3, \dots, r_n)$ holds.
(This choice is such that $a_1$ and $a_2 r_2$ are unchanged,
because we don't know the sign of $\sum_{i \ge 3} (1-r_i) a_i$
and so the post-smoothing value is still at least the max.)
Then,
\begin{align*}
&\phantom{ {} = {} } L_n(a_1, a_2, \dots, r_1, r_2, \dots)
- L_n\left(a_1, \frac 1t a_2, \dots, \frac1t r_1, tr_2\right) \\
&= a_1^2 \left( r_1 - \frac 1t r_1 \right)
+ a_2^2 \left( r_2 - \frac 1t r_2 \right)
+ 2a_1a_2 \left( \frac 1t - 1 \right) \\
&= \left( 1 - \frac 1t \right) \left( r_1 a_1^2 + r_2 a_2^2 - 2a_1a_2 \right)
\ge 0
\end{align*}
the last line by AM-GM.
Now pick $t = \frac{r_1}{\max(1, r_3, \dots, r_n)}$,
and at last we can induct down.
|
USAMO-2001-notes_1
|
Each of eight boxes contains six balls.
Each ball has been colored with one of $n$ colors,
such that no two balls in the same box are the same color,
and no two colors occur together in more than one box.
Find with proof the smallest possible $n$.
|
The answer is $n = 23$.
Shown below is a construction using that many colors,
which we call $\{1,2,\dots,15,a,\dots,f,X,Y\}$.
\[
\begin{bmatrix}
X & X & X & 1 & 2 & 3 & 4 & 5 \\
1 & 6 & 11 & 6 & 7 & 8 & 9 & 10 \\
2 & 7 & 12 & 11 & 12 & 13 & 14 & 15 \\
3 & 8 & 13 & Y & Y & Y & a & b \\
4 & 9 & 14 & a & c & e & c & d \\
5 & 10 & 15 & b & d & f & e & f
\end{bmatrix}
\]
We present now two proofs that $n = 23$ is best possible.
I think the first is more motivated --- it will
actually show us how we could come up with the example above.
\paragraph{First solution (hands-on).}
We say a color $x$ is \emph{overrated}
if it is used at least three times.
First we make the following smoothing argument.
\begin{claim*}
Suppose some box contains a ball of overrated color $x$
plus a ball of color $y$ used only once.
Then we can change one ball of color $x$ to color $y$
while preserving all the conditions.
\end{claim*}
\begin{proof}
Obvious.
(Though the color $x$ could cease to be overrated after this operation.)
\end{proof}
By applying this operation as many times as possible,
we arrive at a situation in which whenever we have
a box with an overrated color,
the other colors in the box are used twice or more.
Assume now $n \le 23$ and the assumption;
we will show the equality case must of the form we gave.
Since there are a total of $48$ balls,
at least two colors are overrated.
Let $X$ be an overrated color and take three boxes where it appears.
Then there are $15$ more distinct colors, say $\{1, \dots, 15\}$
lying in those boxes.
Each of them must appear at least once more,
so we arrive at the situation
\[
\begin{bmatrix}
X & X & X & 1 & 2 & 3 & 4 & 5 \\
1 & 6 & 11 & 6 & 7 & 8 & 9 & 10 \\
2 & 7 & 12 & 11 & 12 & 13 & 14 & 15 \\
3 & 8 & 13 \\
4 & 9 & 14 \\
5 & 10 & 15 \\
\end{bmatrix}
\]
up to harmless permutation of the color names.
Now, note that none of these $15$ colors can reappear.
So it remains to fill up the last five boxes.
Now, there is at least one more overrated color,
distinct from any we have seen; call it $Y$.
In the three boxes $Y$ appears in,
there must be six new colors,
and this gives the lower bound $n \ge 1 + 15 + 1 + 6 = 23$
which we sought,
with equality occurring as we saw above.
\begin{remark*}
[Partial progresses]
The fact that $\binom{16}{2} = 120 = 8 \binom{6}{2}$
(suggesting the bound $n \ge 16$) is misleading
and not that helpful.
There is a simple argument showing that $n$
should be much larger than $16$.
Imagine opening the boxes in any order.
The first box must contain six new colors.
The second box must contain five new colors,
and so on; thus $n \ge 6 + 5 + 4 + 3 + 2 + 1 = 21$.
This is sharp for seven boxes, as the example below shows.
\[
\begin{bmatrix}
1 & 1 & 2 & 3 & 4 & 5 & 6 \\
2 & 7 & 7 & 8 & 9 & 10 & 11 \\
3 & 8 & 12 & 12 & 13 & 14 & 15 \\
4 & 9 & 13 & 16 & 16 & 17 & 18 \\
5 & 10 & 14 & 17 & 19 & 19 & 20 \\
6 & 11 & 15 & 18 & 20 & 21 & 21
\end{bmatrix}
\]
However, one cannot add an eight box,
suggesting the answer should be a little larger than $21$.
One possible eight box is $\{1,12,19,a,b,c\}$
which gives $n \le 24$; but the true answer is a little trickier.
\end{remark*}
\paragraph{Second solution (slick).}
Here is a short proof from the official solutions of the bound.
Consider the $8 \times 6$ grid of colors as before.
For each ball $b$, count the number of times $n_b$ its color is used,
and write the fraction $\frac{1}{n_b}$.
On the one hand, we should have
\[ n = \sum_{\text{all $48$ balls $b$}} \frac{1}{n_b}. \]
On the other hand, for any given box $B$, we have
$\sum_{b \in B} (n_b-1) \le 7$, as among the other seven boxes
at most one color from $B$ appears.
Therefore, $\sum_{b \in B} n_b \le 13$,
and a smoothing argument this implies
\[ \sum_{b \in B} \frac{1}{n_b} \ge \frac13 \cdot 1 + \frac12 \cdot 5
= \frac{17}{6}. \]
Thus, $n \ge 8 \cdot \frac{17}{6} = 22.66\dots$, so $n \ge 23$.
|
USAMO-2001-notes_2
|
Let $ABC$ be a triangle and let $\omega$ be its incircle.
Denote by $D_1$ and $E_1$ the points where $\omega$
is tangent to sides $BC$ and $AC$, respectively.
Denote by $D_2$ and $E_2$ the points on sides $BC$ and $AC$,
respectively, such that $CD_2=BD_1$ and $CE_2=AE_1$,
and denote by $P$ the point of intersection of segments $AD_2$ and $BE_2$.
Circle $\omega$ intersects segment $AD_2$ at two points,
the closer of which to the vertex $A$ is denoted by $Q$.
Prove that $AQ=D_2P$.
|
We have that $P$ is the Nagel point
\[ P = \left( s-a : s-b : s-c \right). \]
Therefore,
\[ \frac{PD_2}{AD_2} = \frac{s-a}{(s-a)+(s-b)+(s-c)} = \frac{s-a}{s}. \]
Meanwhile, $Q$ is the antipode of $D_1$.
The classical homothety at $A$ mapping $Q$ to $D_1$
(by mapping the incircle to the $A$-excircle)
has ratio $\frac{s-a}{s}$ as well
(by considering the length of the tangents from $A$),
so we are done.
|
USAMO-2001-notes_3
|
Let $a, b, c$ be nonnegative real numbers
such that $a^2+b^2+c^2 + abc = 4$.
Show that \[ 0 \le ab + bc + ca - abc \le 2. \]
|
The left-hand side of the inequality is trivial;
just note that $\min \left\{ a,b,c \right\} \le 1$.
Hence, we focus on the right side.
We use Lagrange Multipliers.
Define \[ U = \left\{ (a,b,c) \mid a,b,c > 0
\text{ and } a^2+b^2+c^2 < 1000 \right\}. \]
This is an intersection of open sets, so it is open.
Its closure is
\[ \ol U = \left\{ (a,b,c) \mid a,b,c \ge 0
\text{ and } a^2+b^2+c^2 \le 1000 \right\}. \]
Hence the constraint set
\[ \ol S = \left\{ \mathbf{x} \in \ol U \mid g(\mathbf{x}) = 4 \right\} \]
is compact, where $g(a,b,c) = a^2+b^2+c^2+abc$.
Define \[ f(a,b,c) = a^2+b^2+c^2+ab+bc+ca. \]
It's equivalent to show that $f \le 6$ subject to $g$.
Over $\ol S$, it must achieve a global maximum.
Now we consider two cases.
If $\mathbf{x}$ lies on the boundary,
that means one of the components is zero
(since $a^2+b^2+c^2=1000$ is clearly impossible).
WLOG $c=0$, then we wish to show $a^2+b^2+ab \le 6$
for $a^2+b^2=4$, which is trivial.
Now for the interior $U$, we may use the method of Lagrange Multipliers.
Consider a local maximum $\mathbf{x} \in U$.
Compute \[ \nabla f = \left<2a+b+c, 2b+c+a, 2c+a+b \right> \]
and \[ \nabla g = \left<2a+bc, 2b+ca, 2c+ab\right>. \]
Of course, $\nabla g \neq \mathbf 0$ everywhere,
so introducing our multiplier yields
\[ \left<2a+b+c,a+2b+c,a+b+2c\right> = \lambda \left<2a+bc,2b+ca,2c+ab\right>. \]
Note that $\lambda \neq 0$ since $a,b,c > 0$.
Subtracting $2a+b+c = \lambda(2a+bc)$ from $a+2b+c = \lambda(2b+ca)$
implies that \[ (a-b)(\left[ 2\lambda - 1 \right] - \lambda c) = 0. \]
We can derive similar equations for the others.
Hence, we have three cases.
\begin{enumerate}
\ii If $a=b=c$, then $a=b=c=1$, and this satisfies $f(1,1,1) \le 6$.
\ii If $a$, $b$, $c$ are pairwise distinct,
then we derive $a = b = c = 2 - \lambda^{-1}$, contradiction.
\ii Now suppose that $a=b \neq c$.
Meanwhile, the constraint (with $a=b$) reads
\begin{align*}
a^2+b^2+c^2+abc=4
&\iff c^2 + a^2c + (2a^2-4) = 0 \\
&\iff (c+2)(c-(2-a^2)) = 0
\end{align*}
which since $c > 0$ gives $c = 2-a^2$.
Noah Walsh points out that at this point,
we don't need to calculate the critical point;
we just directly substitute $a=b$ and $c=2-a^2$ into
the desired inequality:
\[ a^2+2a(2-a^2)-a^2(2-a^2) = (a-1)^2(a^2-2) + 2 \le 2 \]
since $a \le \sqrt 2$.
So any point here satisfies the inequality anyways.
\end{enumerate}
\begin{remark*}
It can actually be shown that the critical point
in the third case we skipped is pretty close: it is given by
\[ a = b = \frac{1+\sqrt{17}}{4}
\quad c = \frac{1}{8} \left( 7 - \sqrt{17} \right). \]
This satisfies
\[ f(a,b,c) = 3a^2+2ac+c^2
= \frac{1}{32} \left( 121 + 17\sqrt{17} \right)
\approx 5.97165 \]
which is just a bit less than $6$.
\end{remark*}
\begin{remark*}
Equality holds for the upper bound if $(a,b,c)=(1,1,1)$
or $(a,b,c)=(\sqrt2,\sqrt2,0)$ and permutations.
The lower bound is achieved if $(a,b,c)=(2,0,0)$ and permutations.
\end{remark*}
|
USAMO-2001-notes_4
|
Let $ABC$ be a triangle and $P$ any point
such that $PA$, $PB$, $PC$ are the sides of an obtuse triangle,
with $PA$ the longest side.
Prove that $\angle BAC$ is acute.
|
Using Ptolemy's inequality and Cauchy-Schwarz,
\begin{align*}
PA \cdot BC
&\le PB \cdot AC + PC \cdot AB \\
&\le \sqrt{(PB^2+PC^2)(AB^2+AC^2)} \\
&< \sqrt{PA^2 \cdot (AB^2+AC)^2} = PA \cdot \sqrt{AB^2+AC^2}
\end{align*}
meaning $BC^2 < AB^2+AC^2$, so $\angle BAC$ is acute.
\begin{remark*}
[Lokman G\"{o}k\c{c}e]
Here is another approach using Euler's quadrilateral formula.
Let $M$ and $N$ be midpoints of $AP$ and $BC$, respectively.
For the points $A$, $B$, $P$, $C$; apply Euler's quadrilateral formula to get
\[ AB^2 + BP^2 + PC^2 + CA^2 = AP^2 + BC^2 + 4MN^2 \geq AP^2 + BC^2. \]
We are given that $AP^2 > BP^2 + PC^2$, so $AB^2 + AC^2 > BC^2$,
and we get $\angle BAC$ is acute.
\end{remark*}
|
USAMO-2001-notes_5
|
Let $S \subseteq \ZZ$ be such that:
\begin{enumerate}
\ii[(a)] there exist $a,b \in S$ with
$\gcd(a,b) = \gcd(a-2, b-2) = 1$;
\ii[(b)] if $x$ and $y$ are elements of $S$ (possibly equal),
then $x^2-y$ also belongs to $S$.
\end{enumerate}
Prove that $S = \ZZ$.
|
Call an integer $d > 0$ \emph{shifty} if
$S = S+d$ (meaning $S$ is invariant under shifting by $d$).
First, note that if $u, v \in S$, then for any $x \in S$,
\[ v^2 - (u^2-x) = (v^2-u^2) + x \in S. \]
Since we can easily check that $|S| > 1$
and $S \neq \{n, -n\}$ we conclude there exists a shifty integer.
We claim $1$ is shifty, which implies the problem.
Assume for contradiction that $1$ is not shifty.
Then for GCD reasons the set of shifty integers must be $d \ZZ$ for some $d \ge 2$.
\begin{claim*}
We have $S \subseteq \left\{ x : x^2 \equiv m \pmod d \right\}$ for some fixed $m$.
\end{claim*}
\begin{proof}
Otherwise if we take any $p,q \in S$ with distinct squares modulo $d$,
then $q^2-p^2 \not\equiv 0 \pmod d$ is shifty, which is impossible.
\end{proof}
Now take $a,b \in S$ as in (a).
In that case we need to have
\[ a^2 \equiv b^2 \equiv (a^2-a)^2 \equiv (b^2-b)^2 \pmod d. \]
Passing to a prime $p \mid d$, we have the following:
\begin{itemize}
\ii Since $a^2 \equiv (a^2-a)^2 \pmod p$ or equivalently $a^3(a-2) \equiv 0 \pmod p$,
either $a \equiv 0 \pmod p$ or $a \equiv 2 \pmod p$.
\ii Similarly, either $b \equiv 0 \pmod p$ or $b \equiv 2 \pmod p$.
\ii Since $a^2 \equiv b^2 \pmod p$, or $a \equiv \pm b \pmod p$, we find either
$a \equiv b \equiv 0 \pmod p$ or $a \equiv b \equiv 2 \pmod p$ (even if $p=2$).
\end{itemize}
This is a contradiction.
\begin{remark*}
The condition (a) cannot be dropped, since otherwise we may take
$S = \left\{ 2 \pmod p \right\}$ or $S = \left\{ 0 \pmod p \right\}$, say.
\end{remark*}
|
USAMO-2001-notes_6
|
Each point in the plane is assigned a real number.
Suppose that for any nondegenerate triangle, the number at its incenter
is the arithmetic mean of the three numbers at its vertices.
Prove that all points in the plane were assigned the same number.
\end{enumerate}
|
In this solution, we denote points by capital letters
and the associated values by lowercase letters
(e.g.\ $p$ denotes the value of our function at $P$).
\begin{claim*}
If $ABCD$ is an isosceles trapezoid, then $a+c=b+d$.
\end{claim*}
\begin{proof}
Indeed, suppose WLOG that rays $BA$ and $CD$ meet at $X$.
Then triangles $XAC$ and $XBD$ share an incircle, proving the claim.
\end{proof}
Now, given any two points $A$ and $B$, construct regular pentagon $ABCDE$.
We have \[ a+c=b+d=c+e=d+a=e+b \implies a=b=c=d=e \]
as desired.
|
USAMO-2002-notes_1
|
Let $S$ be a set with $2002$ elements, and let $N$ be an integer with $0 \leq N \leq 2^{2002}$.
Prove that it is possible to color every subset of $S$
either black or white so that the following conditions hold:
\begin{enumerate}
\ii[(a)] the union of any two white subsets is white;
\ii[(b)] the union of any two black subsets is black;
\ii[(c)] there are exactly $N$ white subsets.
\end{enumerate}
|
We will solve the problem with $2002$
replaced by an arbitrary integer $n \ge 0$.
In other words, we prove:
\begin{claim*}
For any nonnegative integers $n$ and $N$ with $0 \le N \le 2^n$,
it is possible to color the $2^n$ subsets
of $\{1, \dots, n\}$ black and white
satisfying the conditions of the problem.
\end{claim*}
The proof is by induction on $n$.
When $n = 1$ the problem is easy to do by hand,
so this gives us a base case.
For the inductive step, we divide into two cases:
\begin{itemize}
\ii If $N \le 2^{n-1}$,
then we take a coloring of subsets of $\{1, \dots, n-1\}$
with $N$ white sets;
then we color the remaining $2^{n-1}$ sets
(which contain $n$) black.
\ii If $N > 2^{n-1}$,
then we take a coloring of subsets of $\{1, \dots, n-1\}$
with $N - 2^{n-1}$ white sets;
then we color the remaining $2^{n-1}$ sets
(which contain $n$) white.
\end{itemize}
|
USAMO-2002-notes_2
|
Let $ABC$ be a triangle such that
\[
\left( \cot \frac{A}{2} \right)^2
+ \left( 2\cot \frac{B}{2} \right)^2
+ \left( 3\cot \frac{C}{2} \right)^2
= \left( \frac{6s}{7r} \right)^2,
\]
where $s$ and $r$ denote its semiperimeter and its inradius, respectively.
Prove that triangle $ABC$ is similar to a triangle $T$ whose side lengths are all positive integers
with no common divisors and determine these integers.
|
Let $x = s-a$, $y = s-b$, $z = s-c$ in the usual fashion, then the equation reads
\[ x^2 + 4y^2 + 9z^2 = \left( \frac67(x+y+z) \right)^2. \]
However, by Cauchy-Schwarz, we have
\[ \left( 1 + \tfrac14 + \tfrac19 \right)\left( x^2 + 4y^2 + 9z^2 \right) \ge \left( x+y+z \right)^2 \]
with equality if and only if $1 : \tfrac12 : \tfrac13 = x : 2y : 3z$,
id est $x : y : z = 1 : \tfrac14 : \tfrac19 = 36 : 9 : 4$.
This is equivalent to $y+z : z+x : x+y = 13 : 40 : 45$.
%Wow.
%I was laughing at how un-geometric this problem looked,
%as it was the only geometry problem on the USAMO 2002.
%Then I actually tried it and realized it was not geometry.
\begin{remark*}
You can tell this is not a geometry problem
because you eliminate the cotangents right away to get an algebra problem\dots
and then you realize the problem claims that one equation
can determine three variables up to scaling,
at which point you realize it has to be an inequality
(otherwise degrees of freedom don't work).
So of course, Cauchy-Schwarz\dots
\end{remark*}
|
USAMO-2002-notes_3
|
Prove that any monic polynomial (a polynomial with leading coefficient $1$)
of degree $n$ with real coefficients is the average
of two monic polynomials of degree $n$ with $n$ real roots.
|
First,
\begin{lemma*}
If $p$ is a monic polynomial of degree $n$,
and $p(1)p(2) < 0$, $p(2)p(3) < 0$, \dots, $p(n-1)p(n) < 0$
then $p$ has $n$ real roots.
\end{lemma*}
\begin{proof}
The intermediate value theorem already guarantees
the existence of $n-1$ real roots.
The last root is obtained by considering cases on $n \pmod 2$.
\begin{itemize}
\ii If $n$ is even, then $p(1)$ and $p(n)$ have opposite sign,
while we must have either
\[ \lim_{x \to -\infty} p(x) = \lim_{x \to \infty} p(x) = \pm \infty \]
so we get one more root.
\ii The $n$ odd case is similar,
with $p(1)$ and $p(n)$ now having the same sign,
but $\lim_{x \to -\infty} p(x) = -\lim_{x \to \infty} p(x)$ instead. \qedhere
\end{itemize}
\end{proof}
Let $f(n)$ be the monic polynomial and let
$M > 1000\max_{t=1, \dots, n} |f(t)|+1000$.
Then we may select reals $a_1, \dots, a_n$ and $b_1, \dots, b_n$
such that for each $k = 1, \dots, n$, we have
\begin{align*}
a_k + b_k &= 2f(k) \\
(-1)^k a_k & > M \\
(-1)^{k+1} b_k & > M.
\end{align*}
We may interpolate monic polynomials $g$ and $h$ through the $a_k$ and $b_k$
(if the $a_k$, $b_k$ are selected ``generically'' from each other).
Then one can easily check $f = \tfrac12(g+h)$ works.
|
USAMO-2002-notes_4
|
Determine all functions $f \colon \RR \to \RR$ such that
\[ f(x^2 - y^2) = x f(x) - y f(y) \]
for all pairs of real numbers $x$ and $y$.
|
The answer is $f(x) = cx$, $c \in \RR$ (these obviously work).
First, by putting $x=0$ and $y=0$ respectively
we have \[ f(x^2) = xf(x) \quad\text{and}\quad f(-y^2) = -yf(y). \]
From this we deduce that $f$ is odd, in particular $f(0) = 0$.
Then, we can rewrite the given as $f(x^2-y^2) + f(y^2) = f(x^2)$.
Combined with the fact that $f$ is odd, we deduce that $f$ is additive
(i.e. $f(a+b)=f(a)+f(b)$).
\begin{remark*}
[Philosophy]
At this point we have $f(x^2) \equiv xf(x)$ and $f$ additive,
and everything we have including the given equation is a direct corollary of these two.
So it makes sense to only focus on these two conditions.
\end{remark*}
Then
\begin{align*}
f( (x+1)^2 ) &= (x+1)f(x+1) \\
\implies f(x^2) + 2f(x) + f(1)
&= (x+1)f(x) + (x+1)f(1)
\end{align*}
which readily gives $f(x) = f(1)x$.
|
USAMO-2002-notes_5
|
Let $a$, $b$ be integers greater than $2$.
Prove that there exists a positive integer $k$
and a finite sequence $n_1, n_2, \dots, n_k$ of
positive integers such that $n_1 = a$, $n_k = b$,
and $n_i n_{i+1}$ is divisible by $n_i + n_{i+1}$
for each $i$ ($1 \leq i < k$).
|
Consider a graph $G$ on the vertex set $\{3, 4, \dots\}$
and with edges between $v$, $w$ if $v + w \mid vw$;
the problem is equivalent to showing that $G$ is connected.
First, note that $n$ is connected to $n(n-1)$, $n(n-1)(n-2)$, etc.\
up to $n!$.
But for $n > 2$, $n!$ is connected to $(n+1)!$ too:
\begin{itemize}
\ii $n! \to (n+1)!$ if $n$ is even
\ii $n! \to 2n! \to (n+1)!$ if $n$ is odd.
\end{itemize}
This concludes the problem.
|
USAMO-2002-notes_6
|
I have an $n \times n$ sheet of stamps, from which I've been asked
to tear out blocks of three adjacent stamps in a single row or column.
(I can only tear along the perforations separating adjacent stamps,
and each block must come out of the sheet in one piece.)
Let $b(n)$ be the smallest number of blocks I can tear out
and make it impossible to tear out any more blocks.
Prove that there are real constants $c$ and $d$ such that
\[ \frac{1}{7} n^2 - cn \leq b(n) \leq \frac{1}{5} n^2 + dn \]
for all $n > 0$.
\end{enumerate}
|
For the lower bound: there are $2n(n-2)$ places one could put a block.
Note that each block eliminates at most $14$ such places.
For the upper bound, the construction of $\frac15$ is easy to build.
Here is an illustration of one possible construction
for $n=9$ which generalizes readily,
using only vertical blocks.
\[
\begin{bmatrix}
A & & E & & I & L & & P & \\
A & & E & G & & L & & P & R \\
A & C & & G & & L & N & & R \\
& C & & G & J & & N & & R \\
& C & F & & J & & N & Q & \\
B & & F & & J & M & & Q & \\
B & & F & H & & M & & Q & S \\
B & D & & H & & M & O & & S \\
& D & & H & K & & O & & S \\
\end{bmatrix}
\]
Actually, for the lower bound,
one may improve $1/7$ to $1/6$.
Count the number $A$ of pairs of adjacent squares
one of which is torn out and the other which is not:
\begin{itemize}
\ii For every deleted block, there are eight neighboring squares,
at least two on each long edge which have been deleted too.
Hence $N \le 6b(n)$.
\ii For every block still alive and not on the border, there are
four neighboring squares, and clearly at least two are deleted.
Hence $N \ge 2\left( (n-2)^2 - 3b(n) \right)$.
\end{itemize}
Collating these solves the problem.
|
USAMO-2003-notes_1
|
Prove that for every positive integer $n$
there exists an $n$-digit number divisible by $5^n$
all of whose digits are odd.
|
This is immediate by induction on $n$.
For $n = 1$ we take $5$;
moving forward if $M$ is a working $n$-digit number then exactly one of
\begin{align*}
N_1 &= 10^n + M \\
N_3 &= 3 \cdot 10^n + M \\
N_5 &= 5 \cdot 10^n + M \\
N_7 &= 7 \cdot 10^n + M \\
N_9 &= 9 \cdot 10^n + M
\end{align*}
is divisible by $5^{n+1}$;
as they are all divisible by $5^n$
and $N_k/5^n$ are all distinct.
|
USAMO-2003-notes_2
|
A convex polygon $\mathcal{P}$ in the plane
is dissected into smaller convex polygons by drawing all of its diagonals.
The lengths of all sides and all diagonals of the polygon $\mathcal{P}$ are rational numbers.
Prove that the lengths of all sides of all polygons
in the dissection are also rational numbers.
|
Suppose $AB$ is a side of a polygon in the dissection,
lying on diagonal $XY$, with $X$, $A$, $B$, $Y$ in that order.
Then \[ AB = XY - XA - YB. \]
In this way, we see that it actually just suffices to
prove the result for a quadrilateral.
We present two approaches to this end.
\paragraph{First approach (trig).}
Consider quadrilateral $ABCD$.
There are twelve angles one can obtain using three of its four vertices,
three at each vertex; denote this set of $12$ angles by $S$
Note that:
\begin{itemize}
\ii The law of cosines implies $\cos \theta \in \QQ$ for each $\theta \in S$.
\ii Hence, $(\sin \theta)^2 \in \QQ$ for $\theta \in S$.
(This is because $\sin^2 \theta + \cos^2 \theta = 1$.)
\end{itemize}
We say two angles $\theta_1$ and $\theta_2$ are
\emph{equivalent} if $\frac{\sin \theta_1}{\sin \theta_2} \in \QQ$.
This is the same as saying,
when $\sin\theta_1$ and $\sin\theta_2$ are written in simplest radical form,
the part under the square root is the same.
Now we contend:
\begin{claim*}
The angles $\angle BAC$, $\angle CAD$, $\angle BAD$ are equivalent.
\end{claim*}
\begin{proof}
Note that
\[ \QQ \ni \cos(\angle BAD) = \cos \angle BAC \cos \angle CAD - \sin \angle BAC \sin \angle CAD \]
so $\angle BAC$ and $\angle CAD$ are equivalent.
Then
\[ \sin (\angle BAD) = \sin \angle BAC \cos \angle CAD + \cos \angle BAC \sin \angle CAD \]
implies $\angle BAD$ is equivalent to those two.
\end{proof}
\begin{claim*}
The angles $\angle BAD$, $\angle DBA$, $\angle ADB$ are equivalent.
\end{claim*}
\begin{proof}
Law of sines on $\triangle BAD$.
\end{proof}
Iterating the argument implies that all angles are equivalent.
Now, if $AB$ and $CD$ meet at $E$,
the law of sines on $\triangle AEB$, etc.\ implies the result.
\paragraph{Second approach (barycentric coordinates).}
To do this, we apply barycentric coordinates.
Consider quadrilateral $ABDC$
(note the changed order of vertices),
with $A=(1,0,0)$, $B=(0,1,0)$, $C=(0,0,1)$.
Let $D = (x,y,z)$, with $x+y+z=1$.
By hypothesis, each of the numbers
\begin{align*}
-a^2yz + b^2(1-x)z + c^2(1-x)y &= AD^2 \\
a^2(1-y)z + b^2zx + c^2(1-y)x &= BD^2 \\
-a^2(1-z)y - b^2(1-z)x + c^2xy &= CD^2
\end{align*}
is rational. Let $W = a^2yz + b^2zx + c^2xy$. Then,
\begin{align*}
b^2z + c^2y &= AD^2 + W \\
a^2z + c^2x &= BD^2 + W \\
a^2y + b^2x &= CD^2 + W.
\end{align*}
This implies that $AD^2 + BD^2 + 2W - c^2 = 2S_C z$ and cyclically
(as usual $2S_C = a^2+b^2-c^2$).
If any of $S_A$, $S_B$, $S_C$ are zero, then we deduce $W$ is rational.
Otherwise, we have that
\[ 1 = x+y+z = \sum_{\text{cyc}} \frac{AD^2 + BD^2 + 2W - c^2}{2S_C} \]
which implies that $W$ is rational,
because it appears with coefficient
$\frac{1}{S_A} + \frac{1}{S_B} + \frac{1}{S_C} \neq 0$
(since $S_{BC} + S_{CA} + S_{AB}$ is actually the area of $ABC$).
Hence from the rationality of $W$,
we deduce that $x$ is rational as long as $S_A \neq 0$, and similarly for the others.
So at most one of $x$, $y$, $z$ is irrational,
but since $x+y+z=1$ this implies they are all rational.
Finally, if $P = \ol{AD} \cap \ol{BC}$
then $AP = \frac{1}{y+z} AD$, so $AP$ is rational too,
completing the proof.
\begin{remark*}
After the reduction to quadrilateral,
a third alternate approach goes by quoting Putnam 2018 A6, reproduced below:
\begin{quote}
Four points are given in the plane, with no three collinear,
such that the squares of the $\binom42=6$ pairwise distances are all rational.
Show that the ratio of the areas between any two of the
$\binom43=4$ triangles determined by these points is also rational.
\end{quote}
If $ABCD$ is the quadrilateral,
the heights from $C$ and $D$ to $AB$ have rational ratio.
Letting $P = AC \cap BD$, we see $AP/AB$ can be shown as rational via coordinates, as needed.
\end{remark*}
|
USAMO-2003-notes_3
|
Let $n$ be a positive integer.
For every sequence of integers
\[ A = (a_0, \; a_1, \; a_2, \; \dots, a_n) \]
satisfying $0 \le a_i \le i$, for $i=0,\dots,n$,
we define another sequence
\[ t(A)= (t(a_0), \; t(a_1), \; t(a_2), \; \dots, \; t(a_n)) \]
by setting $t(a_i)$ to be the number of terms in the
sequence $A$ that precede the term $a_i$ and are different from $a_i$.
Show that, starting from any sequence $A$ as above,
fewer than $n$ applications of the transformation $t$
lead to a sequence $B$ such that $t(B) = B$.
|
We go by strong induction on $n$ with the base cases $n=1$ and $n=2$ done by hand.
Consider two cases:
\begin{itemize}
\ii If $a_0 = 0$ and $a_1 = 1$, then $1 \le t(a_i) \le i$ for $i \ge 1$;
now apply induction to
\[ \left(t(a_1)-1, \; t(a_2)-1, \; \dots, \; t(a_n)-1\right). \]
\ii Otherwise, assume that $a_0 = a_1 = \dots = a_{k-1} = 0$
but $a_k \neq 0$, where $k \ge 2$.
Assume $k < n$ or it's obvious.
Then $t(a_i) \neq 0$ for $i \ge k$, thus $t(t(a_i)) \ge k$ for $i \ge k$,
and we can apply induction hypothesis to
\[ \left( t(t(a_k))-k, \; \dots, \; t(t(a_n))-k \right). \]
\end{itemize}
\begin{remark*}
I think the set of sequences with $t(B) = B$, i.e.\ the possible fixed points,
can be described as blocks of equal numbers such as
\[ B = (0,0,0,0,0,5,5,7,7,7,7,7,7,7,7,7,7,7,18,18) \]
where the number in each block is the index of the first term in it.
\end{remark*}
|
USAMO-2003-notes_4
|
Let $ABC$ be a triangle.
A circle passing through $A$ and $B$
intersects segments $AC$ and $BC$ at $D$ and $E$, respectively.
Lines $AB$ and $DE$ intersect at $F$,
while lines $BD$ and $CF$ intersect at $M$.
Prove that $MF = MC$ if and only if $MB \cdot MD = MC^2$.
|
Ceva theorem plus the similar triangles.
\begin{center}
\begin{asy}
size(7.5cm);
pair C = Drawing("C", (3,0), dir(-45));
pair D = Drawing("D", (0,0));
pair M = Drawing("M", (1,-0.7));
pair F = Drawing("F", 2*M-C);
pair B = Drawing("B", conj(C/M) * (M-C)+C, dir(130));
pair E = Drawing("E", extension(D,F,B,C), dir(40));
pair A = Drawing("A", extension(D,C,B,F), dir(225));
draw(A--B--C--cycle);
draw(circumcircle(B,D,E), dashed);
draw(A--F, dotted);
draw(E--F--C, dotted);
draw(B--M, dotted);
\end{asy}
\end{center}
We know unconditionally that
\[ \dang CBD = \dang EBD = \dang EAD = \dang EAC. \]
Moreover, by Ceva's theorem on $\triangle BCF$,
we have $MF = MC \iff \ol{FC} \parallel \ol{AE}$.
So we have the equivalences
\begin{align*}
MF = MC &\iff \ol{FC} \parallel \ol{AE} \\
&\iff \dang FCA = \dang EAC \\
&\iff \dang MCD = \dang CBD \\
&\iff MC^2 = MB \cdot MD.
\end{align*}
|
USAMO-2003-notes_5
|
Let $a$, $b$, $c$ be positive real numbers.
Prove that
\[ \frac{(2a+b+c)^2}{2a^2+(b+c)^2}
+ \frac{(2b+c+a)^2}{2b^2+(c+a)^2}
+ \frac{(2c+a+b)^2}{2c^2+(a+b)^2} \le 8. \]
|
This is a canonical example of tangent line trick.
Homogenize so that $a + b + c = 3$.
The desired inequality reads
\[ \sum_{\text{cyc}} \frac{(a+3)^2}{2a^2+(3-a)^2} \le 8. \]
This follows from
\[ f(x) = \frac{(x+3)^2}{2x^2+(3-x)^2}
\le \frac{1}{3} (4x + 4) \]
which can be checked as
$\frac 13 (4x+4)(2x^2+(3-x)^2) - (x+3)^2
= (x-1)^2 (4x+3) \ge 0$.
|
USAMO-2003-notes_6
|
At the vertices of a regular hexagon are written six nonnegative integers
whose sum is $2003^{2003}$.
Bert is allowed to make moves of the following form:
he may pick a vertex and replace the number written
there by the absolute value of the difference between the numbers
written at the two neighboring vertices.
Prove that Bert can make a sequence of moves,
after which the number $0$ appears at all six vertices.
\end{enumerate}
|
If $a \le b \le c$ are \emph{odd} integers,
the configuration which has $(a,b-a,b,c-b,c,c-a)$ around the hexagon
in some order (up to cyclic permutation and reflection)
is said to be \emph{great} (picture below).
\begin{claim*}
We can reach a great configuration
from any configuration with odd sum.
\end{claim*}
\begin{proof}
We should be able to find an equilateral triangle
whose vertices have odd sum.
If all three vertices are odd, then we are already done.
Otherwise, operate as in the following picture (modulo $2$).
\begin{center}
\begin{asy}
size(10cm);
picture[] h = new picture[4];
h[0] = new picture;
h[1] = new picture;
h[2] = new picture;
h[3] = new picture;
pen cu = blue + fontsize(9pt);
pen cd = deepgreen + fontsize(7pt);
pen mu = lightred + fontsize(9pt);
pen md = orange + fontsize(7pt);
for (int i=0; i<=3; ++i) {
draw(h[i], dir(30)--dir(150)--dir(270)--cycle, cd);
draw(h[i], dir(90)--dir(210)--dir(330)--cycle, cu);
draw(h[i], dir(30)--dir(90)--dir(150)--dir(210)--dir(270)--dir(330)--cycle, lightgray);
}
label(h[0], "$1$", dir(90), dir(90), cu);
label(h[0], "$\ast$", dir(150), dir(150), cd);
label(h[0], "$0$", dir(210), dir(210), cu);
label(h[0], "$\ast$", dir(270), dir(270), cd);
label(h[0], "$0$", dir(330), dir(330), cu);
label(h[0], "$\ast$", dir(30), dir(30), cd);
label(h[1], "$1$", dir(90), dir(90), cu);
label(h[1], "$1$", dir(150), dir(150), md);
label(h[1], "$0$", dir(210), dir(210), cu);
label(h[1], "$0$", dir(270), dir(270), md);
label(h[1], "$0$", dir(330), dir(330), cu);
label(h[1], "$1$", dir(30), dir(30), md);
label(h[2], "$1$", dir(90), dir(90), cu);
label(h[2], "$1$", dir(150), dir(150), cd);
label(h[2], "$1$", dir(210), dir(210), mu);
label(h[2], "$0$", dir(270), dir(270), cd);
label(h[2], "$1$", dir(330), dir(330), mu);
label(h[2], "$1$", dir(30), dir(30), cd);
label(h[3], "$1$", dir(90), dir(90), cu);
label(h[3], "$0$", dir(150), dir(150), md);
label(h[3], "$1$", dir(210), dir(210), cu);
label(h[3], "$0$", dir(270), dir(270), md);
label(h[3], "$1$", dir(330), dir(330), cu);
label(h[3], "$0$", dir(30), dir(30), md);
add(shift(0,0)*h[0]);
add(shift(3,0)*h[1]);
add(shift(6,0)*h[2]);
add(shift(9,0)*h[3]);
\end{asy}
\end{center}
Thus we arrived at a great configuration.
\end{proof}
\begin{claim*}
Bert's goal is possible for all great configurations.
\end{claim*}
\begin{proof}
[Proof, suggested by Haoran Chen]
If $a=b=c$ then we have $(t,0,t,0,t,0)$ which is obviously winnable.
Otherwise, we can perform the following three operations
shown in the figure below,
which yield a great configuration whose odd entries
are $a$, $b$, $|c-2a|$.
\begin{center}
\begin{asy}
size(12.5cm);
picture[] h = new picture[5];
h[0] = new picture;
h[1] = new picture;
h[2] = new picture;
h[3] = new picture;
h[4] = new picture;
pen cu = blue + fontsize(9pt);
pen cd = deepgreen + fontsize(7pt);
pen mu = red + fontsize(9pt);
pen md = red + fontsize(9pt);
for (int i=0; i<=4; ++i) {
draw(h[i], dir(30)--dir(150)--dir(270)--cycle, cd);
draw(h[i], dir(90)--dir(210)--dir(330)--cycle, cu);
draw(h[i], dir(30)--dir(90)--dir(150)--dir(210)--dir(270)--dir(330)--cycle, lightgray);
}
label(h[0], "$a$", dir(90), dir(90), cu);
label(h[0], "$b-a$", dir(150), dir(150), cd);
label(h[0], "$b$", dir(210), dir(210), cu);
label(h[0], "$c-b$", dir(270), dir(270), cd);
label(h[0], "$c$", dir(330), dir(330), cu);
label(h[0], "$c-a$", dir(30), dir(30), cd);
label(h[1], "$a$", dir(90), dir(90), cu);
label(h[1], "$b-a$", dir(150), dir(150), cd);
label(h[1], "$b$", dir(210), dir(210), cu);
label(h[1], "$c-b$", dir(270), dir(270), cd);
label(h[1], "$b-a$", dir(330), dir(330), mu);
label(h[1], "$c-a$", dir(30), dir(30), cd);
label(h[2], "$a$", dir(90), dir(90), cu);
label(h[2], "$b-a$", dir(150), dir(150), cd);
label(h[2], "$b$", dir(210), dir(210), cu);
label(h[2], "$a$", dir(270), dir(270), md);
label(h[2], "$b-a$", dir(330), dir(330), cu);
label(h[2], "$c-a$", dir(30), dir(30), cd);
label(h[3], "$a$", dir(90), dir(90), cu);
label(h[3], "$b-a$", dir(150), dir(150), cd);
label(h[3], "$b$", dir(210), dir(210), cu);
label(h[3], "$a$", dir(270), dir(270), cd);
label(h[3], "$|c-2a|$", dir(330), dir(330), mu);
label(h[3], "$c-a$", dir(30), dir(30), cd);
label(h[4], "$a$", dir(90), dir(90), cu);
label(h[4], "$b$", dir(210), dir(210), cu);
label(h[4], "$|c-2a|$", dir(330), dir(330), cu);
add(shift(0,3)*h[0]);
add(shift(4,3)*h[1]);
add(shift(8,3)*h[2]);
add(shift(2,0)*h[3]);
add(shift(6,0)*h[4]);
\end{asy}
\end{center}
Since $|c-2a| < c$ unless $a=b=c$,
this decreases the sum.
So an induction now completes the problem.
% This is annoying, but straightforward.
% Our standing assumption is $a \neq c$ (but possibly $b=c$).
% It's already obvious that $|c-2a| < c$,
% so focus on the last term.
% If $c > 2b$, then $\left\lvert (c-2b)-(c-a) \right\rvert =
% \left\lvert 2b-a \right\rvert < c$ as well for $a \neq c$.
% When $c \le 2b$ we instead have
% $\left\lvert (2b-c)-(c-a) \right\rvert
% \le \max \left( 2b-c, c-a \right)$
% with equality if and only if $c-a = 0$; but $2b-c \le c$ as needed.
% Thus, in all situations we have
% \[ c \neq a \implies
% \max\left( \left\lvert \left\lvert c-2b \right\rvert - (c-a)
% \right\rvert, \left\lvert c-2a \right\rvert \right) < c. \]
% Now denote the new odd entries by $a' \le b' \le c'$ (in some order).
% If $b < c$ then $c' < c$,
% while if $b = c$ then $c' = b$ but $b' < c = b$.
% Thus $(c', b', a')$ precedes $(c, b, a)$ lexicographically, and we
% can induct down.
\end{proof}
\begin{remark*}
One simple idea might be to try to overwrite
the maximum number at each point, decreasing the sum.
However, this fails on the arrangement $(t,t,0,t,t,0)$.
Unfortunately, this issue is actually fatal,
as the problem has a hidden parity obstruction.
The configuration $(1,1,0,1,1,0) \bmod 2$ is invariant modulo $2$,
and so Bert can walk into a ``fatal death-trap'' of this shape
long before the numbers start becoming equal/zero/etc.
In other words, you can mess up on the first move!
This is why the initial sum is given to be odd;
however, it's not possible for Bert to win
so one essentially has to ``tip-toe'' around the $110110$ trap
any time one leaves the space of odd sum.
That's why the great configurations defined above serve as an anchor,
making sure we never veer too far from the
safe $101010$ configuration.
\end{remark*}
\begin{remark*}
On the other hand, many other approaches are possible
which anchor around a different parity configuration,
like $100000$ for example.
The choice of $101010$ by me is due to symmetry --- ostensibly,
if it worked, there should be fewer cases.
\end{remark*}
|
USAMO-2004-notes_1
|
Let $ABCD$ be a quadrilateral circumscribed about a circle,
whose interior and exterior angles are at least $60$ degrees.
Prove that
\[ \frac{1}{3}|AB^3 - AD^3|
\le |BC^3 - CD^3| \le 3|AB^3 - AD^3|. \]
When does equality hold?
|
Clearly it suffices to show the left inequality.
Since $AB+CD = BC+AD \implies |AB-AD| = |BC-CD|$, it suffices to prove
\[ \frac13(AB^2 + AB \cdot AD + AD^2)
\le BC^2 + BC \cdot CD + CD^2. \]
This follows by noting that
\begin{align*}
BC^2 + BC \cdot CD + CD^2
&\ge BC^2 + CD^2 - 2(BC)(CD)\cos(\angle BCD) \\
&= BD^2 \\
&= AB^2 + AD^2 - 2(AB)(AD)\cos(\angle BAD) \\
&\ge AB^2 + AD^2 - AB \cdot AD \\
&\ge \tfrac13(AB^2 + AD^2 + AB \cdot AD)
\end{align*}
the last line following by AM-GM.
The equality holds iff $ABCD$ is a kite with $AB=AD$, $CB=CD$.
|
USAMO-2004-notes_2
|
Let $a_1$, $a_2$, \dots, $a_n$ be integers whose greatest common divisor is $1$.
Let $S$ be a set of integers with the following properties:
\begin{enumerate}
\ii[(a)] $a_i \in S$ for $i = 1, \dots, n$.
\ii[(b)] $a_i - a_j \in S$ for $i, j = 1, \dots, n$, not necessarily distinct.
\ii[(c)] If $x, y \in S$ and $x+y \in S$, then $x-y \in S$ too.
\end{enumerate}
Prove that $S = \ZZ$.
|
The idea is to show any linear combination of the $a_i$ are in $S$,
which implies (by Bezout) that $S = \ZZ$.
This is pretty intuitive, but the details require some care
(in particular there is a parity obstruction at the second lemma).
First, we make the following simple observations:
\begin{itemize}
\ii $0 \in S$, by putting $i=j=1$ in (b).
\ii $s \in S \iff -s \in S$, by putting $x=0$ in (c).
\end{itemize}
Now, we prove that:
\begin{lemma*}
For any integers $c$, $d$, and indices $i$, $j$, we have $ca_i + da_j \in S$.
\end{lemma*}
\begin{proof}
We will assume $c,d > 0$ since the other cases are analogous.
In that case it follows by induction on $c+d$;
for example $ca_i + (d-1)a_j$, $a_j$, $ca_i + da_j$ in $S$
implies $ca_i + (d+1)a_j \in S$.
\end{proof}
\begin{lemma*}
For any nonzero integers $c_1$, $c_2$, \dots, $c_m$,
and any distinct indices $\{i_1 , i_2 , \dots , i_m\}$, we have
\[ \sum_k c_k a_{i_k} \in S. \]
\end{lemma*}
\begin{proof}
By induction on $m$, with base case $m \le 2$ already done.
For the inductive step,
we will assume that $i_1 = 1$, $i_2 = 2$, et cetera, for notational convenience.
The proof is then split into two cases.
\begin{itemize}
\ii \textbf{First Case}: some $c_i$ is even.
WLOG $c_1 \neq 0$ is even and note that
\begin{align*}
x &\coloneq \half c_1 a_1 + \sum_{k \ge 3} c_k a_k \in S \\
y &\coloneq -\half c_1 a_1 - c_2 a_2 \in S \\
x+y &= -c_2 a_2 + \sum_{k \ge 3} c_k a_k \in S \\
\implies x-y &= \sum_{k \ge 1} c_k a_k \in S.
\end{align*}
\ii \textbf{Second Case}: all $c_i$ are odd.
We reduce this to the first case as follows.
Let $u = \frac{a_1}{\gcd(a_1, a_2)}$ and $v = \frac{a_2}{\gcd(a_1, a_2)}$.
Then $\gcd(u,v) = 1$ and so WLOG $u$ is odd.
Then
\[
c_1 a_1 + c_2 a_2
= (c_1 + v)a_1 + (c_2 - u) a_2
\]
and so we can replace our given combination
by $(c_1+v)a_1 + (c_2-u)a_2 + c_3a_3 + \dots$
which now has an even coefficient for $a_2$. \qedhere
\end{itemize}
\end{proof}
We then apply the lemma at $m = n$; this finishes the problem since Bezout's lemma
says that $\sum c_i a_i = 1$ for some choice of $c_i \in \ZZ$.
|
USAMO-2004-notes_3
|
For what real values of $k > 0$ is it possible to
dissect a $1 \times k$ rectangle into two similar but noncongruent polygons?
|
Answer: the dissection is possible for every $k > 0$ except for $k = 1$.
\paragraph{Construction.}
By symmetry it suffices to give a construction for $k > 1$
(since otherwise we replace $k$ by $k\inv$).
For every integer $n \ge 2$ and real number $r > 1$,
we define a shape $\mathcal R(n,r)$ as follows.
\begin{itemize}
\ii We start with a rectangle of width $1$ and height $r$.
To its left, we glue a rectangle of height $r$ and width $r^2$ to its left.
\ii Then, we glue a rectangle of width $1+r^2$ and height $r^3$ below our figure,
followed by a rectangle of height $r+r^3$ and width $r^4$ to the left of our figure.
\ii Next, we glue a rectangle of width $1+r^2+r^4$ and height $r^5$ below our figure,
followed by a rectangle of height $r+r^3+r^5$ and width $r^6$ to the left of our figure.
\end{itemize}
\dots and so on, up until we have put $2n$ rectangles together.
The picture $\mathcal R(3,r)$ is shown below as an example.
\begin{center}
\begin{asy}
size(8cm);
defaultpen(fontsize(10pt));
real r = 1.6;
pair A = (0,r^3+r);
pair B = (r^4,r^3+r);
pair C = (r^4+r^2,r^3+r);
pair D = (r^4+r^2+1,r^3+r);
pair E = (r^4,r^3);
pair F = (r^4+r^2, r^3);
pair G = (r^4+r^2+1, r^3);
pair O = (0,0);
pair H = (r^4,0);
pair I = (r^4+r^2+1,0);
pair W = (-r^6, A.y);
pair X = (W.x,-r^5);
pair Y = (0,-r^5);
pair Z = (I.x,-r^5);
pen p = black+1.3;
pen q = red;
fill(O--A--B--H--cycle, palecyan);
fill(B--C--F--E--cycle, palecyan);
fill(W--A--Y--X--cycle, palecyan);
fill(C--D--G--F--cycle, palegreen);
fill(E--G--I--H--cycle, palegreen);
fill(O--Y--Z--I--cycle, palegreen);
draw(O--A, q);
draw(B--E, q);
draw(F--G, q);
draw(H--I, q);
draw(W--D--Z--X--cycle, p);
draw(C--F--E--H--O--Y, p);
label("$1$", (C+D)/2, dir(90));
label("$r^2$", (B+C)/2, dir(90));
label("$r^4$", (A+B)/2, dir(90));
label("$r^6$", (W+A)/2, dir(90));
label("$r^1$", (D+G)/2, dir(0));
label("$r^3$", (G+I)/2, dir(0));
label("$r^5$", (I+Z)/2, dir(0));
\end{asy}
\end{center}
Observe that by construction,
the entire shape $\mathcal R(n,r)$ is a rectangle
which consists of two similar ``staircase'' polygons
(which are not congruent, since $r>1$).
Note that $\mathcal R(n,r)$ is similar to a $1 \times f_n(r)$ rectangle
where $f_n(r)$ is the aspect ratio of $\mathcal R(n,r)$, defined by
\[ f_n(r) = \frac{1+r^2+\dots+r^{2n}}{r+r^3+\dots+r^{2n-1}}
= r + \frac{1}{r+r^3+\dots+r^{2n-1}}. \]
We claim that this is enough.
Indeed for each fixed $n$, note that
\[ \lim_{r \to 1^+} f_n(r) = 1 + \frac1n
\; \text{ and } \;
\lim_{r \to \infty} f_n(r) = \infty. \]
Since $f_n$ is continuous, it achieves all values greater than $1+\frac1n$.
Thus by taking sufficiently large $n$ (such that $k > 1+\frac1n$),
we obtain a valid construction for any $k > 1$.
\paragraph{Proof of impossibility for a square.}
Now we show that $k = 1$ is impossible (the tricky part!).
Suppose we have a squared dissected into
two similar polygons $\mathcal P \sim \mathcal Q$.
Let $\Gamma$ be their common boundary.
By counting the number of sides of $\mathcal P$ and $\mathcal Q$
we see $\Gamma$ must run from one side of the square to an opposite side
(possibly ending at a corner of the square).
We orient the figure so $\Gamma$ runs from north to south,
with $\mathcal P$ to the west and $\mathcal Q$ to the east.
\begin{center}
\begin{asy}
size(4cm);
pair A = (0,0);
pair B = (1,0);
pair C = (1,1);
pair D = (0,1);
pair W = (0.4,0);
pair X = (0.3,0.7);
pair Y = (0.7,0.3);
pair Z = (0.6,1);
filldraw(A--W--X--Y--Z--D--cycle, palecyan, black);
filldraw(B--W--X--Y--Z--C--cycle, palegreen, black);
draw( W--X--Y--Z, black+1 );
label("$\Gamma$", Y, dir(15));
label("$\mathcal P$", (0.1,0.5), blue);
label("$\mathcal Q$", (0.9,0.5), deepgreen);
\end{asy}
\end{center}
Let $s$ be the longest length of a segment in $\Gamma$.
\begin{claim*}
The longest side length of $\mathcal P$ is $\max(s,1)$.
Similarly, the longest side length of $\mathcal Q$ is $\max(s,1)$ as well.
\end{claim*}
\begin{proof}
The only edges of $\mathcal P$ not in $\Gamma$
are the west edge of our original square, which has length $1$,
and the north/south edges of $\mathcal P$ (if any),
which have length at most $1$.
An identical argument works for $\mathcal Q$.
\end{proof}
It follows the longest sides of $\mathcal P$ and $\mathcal Q$ have the same length!
Hence the two polygons are in fact congruent, ending the proof.
|
USAMO-2004-notes_4
|
Alice and Bob play a game on a $6$ by $6$ grid.
On his turn, a player chooses a rational number not yet appearing in the grid
and writes it in an empty square of the grid.
Alice goes first and then the players alternate.
When all squares have numbers written in them, in each row,
the square with the greatest number in that row is colored black.
Alice wins if he can then draw a line from the top of the grid to the bottom of the grid
that stays in black squares, and Bob wins if he can't.
(If two squares share a vertex,
Alice can draw a line from one to the other that stays in those two squares.)
Find, with proof, a winning strategy for one of the players.
|
Bob can win.
Label the first two rows as follows:
\[
\begin{bmatrix}
a & b & c & d & e & f \\
d' & e' & f' & a' & b' & c'
\end{bmatrix}
\]
These twelve boxes thus come in six \emph{pairs}, $(a, a')$, $(b, b')$ and so on.
\begin{claim*}
Bob can ensure that the order relation
of the labels is the same between the two rows,
meaning that $a < b$ if and only if $a' < b'$, and so on.
\end{claim*}
\begin{proof}
If Alice plays $q$ in some box in the first two rows, then Bob can plays $q + \eps$
in the corresponding box in the same pair, for some sufficiently small $\eps$
(in terms of the existing numbers).
When Alice writes a number in any other row, Bob writes anywhere in rows $3$ to $6$.
\end{proof}
Under this strategy the black squares in the first two rows will be a pair
and therefore will not touch, so Bob wins.
|
USAMO-2004-notes_5
|
Let $a$, $b$, $c$ be positive reals.
Prove that
\[ ( a^5-a^2+3 )( b^5-b^2+3 )( c^5-c^2+3 )
\ge \left( a+b+c \right)^3. \]
|
Observe that for all real numbers $a$, the inequality
\[ a^5-a^2+3 \ge a^3+2 \]
holds.
Then the problem follows by H\"{o}lder in the form
\[ (a^3+1+1)(1+b^3+1)(1+1+c^3) \ge (a+b+c)^3. \]
|
USAMO-2004-notes_6
|
A circle $\omega$ is inscribed in a quadrilateral $ABCD$.
Let $I$ be the center of $\omega$.
Suppose that \[ (AI + DI)^2 + (BI + CI)^2 = (AB + CD)^2. \]
Prove that $ABCD$ is an isosceles trapezoid.
\end{enumerate}
|
Here's a completely algebraic solution.
WLOG $\omega$ has radius $1$, and let $a$, $b$, $c$, $d$ be the
lengths of the tangents from $A$, $B$, $C$, $D$ to $\omega$.
It is known that
\[ a+b+c+d = abc+bcd+cda+dab \qquad (\star) \]
which can be proved by, say $\tan$-addition formula.
Then, the content of the problem is to show that
\[
(\sqrt{a^2+1}+\sqrt{d^2+1})^2
+ (\sqrt{b^2+1}+\sqrt{c^2+1})^2
\le (a+b+c+d)^2
\]
subject to $(\star)$,
with equality only when $a=d=\tfrac1b=\tfrac1c$.
Let $S = ab+bc+cd+da+ac+bd$. Then the inequality is
\[ \sqrt{(a^2+1)(d^2+1)} + \sqrt{(b^2+1)(c^2+1)} \le S - 2. \]
Now, by \textbf{USAMO 2014 Problem 1} and the condition $(\star)$,
we have that $(a^2+1)(b^2+1)(c^2+1)(d^2+1) = (S - abcd - 1)^2$.
So squaring both sides, the inequality becomes
\[ (ad)^2+(bc)^2 + a^2+b^2+c^2+d^2 \le S^2 - 6S + 2abcd + 4. \]
To simplify this, we use the identities
\begin{align*}
S^2 &= 6abcd + \sum_{\text{sym}} a^2bc
+ \frac14\sum_{\text{sym}} a^2b^2 \\
(a+b+c+d)^2 &= (abc+bcd+cda+dab)(a+b+c+d) \\
&= 4abcd + \frac12\sum_{\text{sym}} a^2bc
\end{align*}
So $S^2+2abcd = \frac14\textstyle\sum_{\text{sym}}
a^2b^2 + 2(a^2+b^2+c^2+d^2) + 4S$
and the inequality we want to prove reduces to
\[ 2S \le (ab)^2+(ac)^2+(bd)^2+(cd)^2 + 4 + a^2 + b^2 + c^2 + d^2. \]
This follows by AM-GM since
\begin{align*}
(ab)^2 + 1 &\ge 2ab \\
(ac)^2 + 1 &\ge 2ac \\
(bd)^2 + 1 &\ge 2bd \\
(cd)^2 + 1 &\ge 2cd \\
a^2 + d^2 &\ge 2ad \\
b^2 + c^2 &\ge 2bc.
\end{align*}
The equality case is when $ab=ac=bd=cd=1$, $a=d$, $b=c$,
as needed to imply an isosceles trapezoid.
\begin{remark*}
Note that a priori one expects an inequality.
Indeed,
\begin{itemize}
\ii Quadrilaterals with incircles have four degrees of freedom.
\ii There is one condition imposed.
\ii Isosceles trapezoid with incircles have two degrees of freedom.
\end{itemize}
\end{remark*}
|
USAMO-2005-notes_1
|
Determine all composite positive integers $n$ for which
it is possible to arrange all divisors of $n$ that are greater than $1$
in a circle so that no two adjacent divisors are relatively prime.
|
The only bad ones are $n = pq$, products of two distinct primes.
Clearly they can't be so arranged, so we show all others work.
\begin{itemize}
\ii If $n$ is a power of a prime, the result is obvious; any arrangement works.
\ii If $n = p_1^{e_1} \dots p_k^{e_k}$ for some $k \ge 3$,
then first situate $p_1p_2$, $p_2p_3$, \dots, $p_kp_1$ on the circle.
Then we can arbitrarily place any multiples of $p_i$
between $p_{i-1} p_i$ and $p_i p_{i+1}$.
This finishes this case.
\ii Finally suppose $n = p^a q^b$.
If $a > 1$, say, we can repeat the argument by first placing
$pq$ and $p^2q$ and then placing multiples of $p$ in one arc
and multiples of $q$ in the other arc.
On the other hand the case $a=b=1$ is seen to be impossible.
\end{itemize}
|
USAMO-2005-notes_2
|
Prove that the system of equations
\begin{align*}
x^6 + x^3 + x^3y + y &= 147^{157} \\
x^3 + x^3y + y^2 + y + z^9 &= 157^{147}
\end{align*}
has no integer solutions.
|
Sum the equations and add $1$ to both sides to get
\[ (x^3+y+1)^2 + z^9 = 147^{157} + 157^{147} + 1
\equiv 14 \pmod{19} \]
But $a^2 + b^9 \not\equiv 14 \pmod{19}$
for any integers $a$ and $b$,
since the ninth powers modulo $19$ are $0$, $\pm 1$
and none of $\{13, 14, 15\}$ are squares modulo $19$.
Therefore, there are no integer solutions.
\begin{remark*}
In fact, $a^2 + b^3 \not\equiv 14 \pmod{19}$
has no solutions modulo $19$ either.
\end{remark*}
\begin{remark*}
It can also be checked that the original system has no equations modulo $13$,
although this requires using both equations rather than just their sum.
(As in the modulo $19$ situation, $z^9$ may be replaced by $z^3$
and this remark still holds.)
\end{remark*}
|
USAMO-2005-notes_3
|
Let $ABC$ be an acute-angled triangle,
and let $P$ and $Q$ be two points on side $BC$.
Construct a point $C_1$ in such a way that the convex quadrilateral $APBC_1$ is cyclic,
$\ol{QC_1} \parallel \ol{CA}$, and $C_1$ and $Q$ lie on opposite sides of line $AB$.
Construct a point $B_1$ in such a way that the convex quadrilateral $APCB_1$ is cyclic,
$\ol{QB_1} \parallel \ol{BA}$, and $B_1$ and $Q$ lie on opposite sides of line $AC$.
Prove that the points $B_1$, $C_1$, $P$, and $Q$ lie on a circle.
|
It is enough to prove that $A$, $B_1$, and $C_1$ are collinear,
since then $\dang C_1QP = \dang ACP = \dang AB_1P = \dang C_1B_1P$.
\begin{center}
\begin{asy}
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
pair P = 0.7*C+0.3*B;
draw(A--B--C--cycle);
pair O_1 = circumcenter(A, P, C);
pair B_1 = 2*O_1-C;
pair Q = extension(B, C, B_1, B_1+B-A);
pair C_1 = extension(A, B_1, Q, Q+C-A);
draw(circumcircle(A, B, P));
draw(circumcircle(A, C, P));
draw(C_1--Q--B_1);
draw(B_1--C_1, dotted);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$P$", P, dir(P));
dot("$B_1$", B_1, dir(B_1));
dot("$Q$", Q, dir(Q));
dot("$C_1$", C_1, dir(C_1));
/* Source generated by TSQ */
\end{asy}
\end{center}
\paragraph{First solution.}
Let $T$ be the second intersection of $\ol{AC_1}$ with $(APC)$.
Then readily $\triangle PC_1T \sim \triangle ABC$.
Consequently, $\ol{QC_1} \parallel \ol{AC}$ implies $TC_1QP$ cyclic.
Finally, $\ol{TQ} \parallel \ol{AB}$ now follows from the cyclic condition,
so $T = B_1$ as desired.
\paragraph{Second solution.}
One may also use barycentric coordinates.
Let $P = (0,m,n)$ and $Q = (0,r,s)$ with $m+n=r+s=1$.
Once again, \[ (APB) : -a^2yz-b^2zx-c^2xy + (x+y+z)(a^2m \cdot z) = 0. \]
Set $C_1 = (s-z, r, z)$, where $C_1Q \parallel AC$ follows by $(s-z) + r + z = 1$.
We solve for this $z$.
\begin{align*}
0 &= -a^2rz + (s-z)(-b^2z-c^2r) + a^2mz \\
&= b^2z^2 + (-sb^2+rc^2)z - a^2rz + a^2mz - c^2rs \\
&= b^2z^2 + (-sb^2+rc^2+a^2(m-r))z - c^2rs \\
\implies 0 &= rb^2 \left( \frac{z}{r} \right)^2 + (-sb^2+rc^2 + a^2(m-r)) \left(\frac{z}{r}\right) - c^2s.
\end{align*}
So the quotient of the $z$ and $y$ coordinates of $C_1$ satisfies this quadratic.
Similarly, if $B_1 = (r-y, y, s)$ we obtain that
\[
0 = sc^2 \left( \frac{y}{s} \right)^2 + (-rc^2 + sb^2 + a^2(n-s)) \left(\frac{y}{s}\right) - b^2r
\]
Since these two quadratics are the same when one is written backwards (and negated),
it follows that their roots are reciprocals.
But the roots of the quadratics represent $\tfrac{z}{y}$ and $\tfrac{y}{z}$
for the points $C_1$ and $B_1$, respectively.
This implies that the points $B_1$ and $C_1$ are collinear with $A=(1,0,0)$
(in some line of the form $\tfrac{y}{z} = k$), as desired.
|
USAMO-2005-notes_4
|
Legs $L_1$, $L_2$, $L_3$, $L_4$ of a square table each have length $n$,
where $n$ is a positive integer.
For how many ordered $4$-tuples $(k_1, k_2, k_3, k_4)$ of nonnegative integers
can we cut a piece of length $k_i$ from the end of leg $L_i$
and still have a stable table?
(The table is \emph{stable} if it can be placed
so that all four of the leg ends touch the floor.
Note that a cut leg of length $0$ is permitted.)
|
Flip the table upside-down so that that the table's surface
rests on the floor.
Then, we see that we want the truncated legs
to have endpoints $A$, $B$, $C$, $D$ which are coplanar (say).
\begin{claim*}
This occurs if and only if $ABCD$ is a parallelogram.
\end{claim*}
\begin{proof}
Obviously $ABCD$ being a parallelogram is sufficient.
Conversely, if they are coplanar,
we let $D'$ be such that $ABCD'$ is a parallelogram.
Then $D'$ also lies in the same plane as $ABCD$,
but is situated directly above $D$
(since the table was a square).
This implies $D' = D$, as needed.
\end{proof}
In still other words, we are counting the number of solutions to
\[ (n-k_1) + (n-k_3) = (n-k_2) + (n-k_4)
\iff k_1 + k_3 = k_2 + k_4. \]
Define
\[ a_r = \# \{(a,b) \mid a+b=r, \; 0 \le a,b \le n \} \]
so that the number of solutions to $k_1 + k_3 = k_2 + k_4 = r$
is just given by $a_r^2$.
We now just compute
\begin{align*}
\sum_{r=0}^{2n} a_r^2 &= 1^2 + 2^2 + \dots + n^2 + (n+1)^2 + n^2 + \dots + 1^2 \\
&= \frac{1}{3}(n+1)(2n^2 + 4n + 3).
\end{align*}
|
USAMO-2005-notes_5
|
Let $n > 1$ be an integer.
Suppose $2n$ points are given in the plane, no three of which are collinear.
Suppose $n$ of the given $2n$ points are colored blue and the other $n$ colored red.
A line in the plane is called a \emph{balancing line}
if it passes through one blue and one red point
and, for each side of the line, the number of blue points on that
side is equal to the number of red points on the same side.
Prove that there exist at least two balancing lines.
|
Consider the convex hull $\mathcal H$ of the polygon. There are two cases.
\paragraph{Easy case: the convex hull has both colors.}
If the convex hull $\mathcal H$ is not all the same color,
there exist two edges of $\mathcal H$ (at least) which have differently colored endpoints.
The extensions of those sides form balancing lines;
indeed given any such line $\ell$ one side of $\ell$ has no points,
the other has $n-1$ red and $n-1$ blue points.
\paragraph{Harder case: the convex hull is all one color.}
Now assume $\mathcal H$ is all blue (WLOG).
We will prove there are at least $|\mathcal H|$ balancing lines in the following way.
\begin{claim*}
For any vertex $B$ of $\mathcal H$ there is a balancing line through it.
\end{claim*}
\begin{proof}
Assume $A$, $B$, $C$ are three consecutive
blue vertices of $\mathcal H$.
Imagine starting with line $\ell$ passing through $B$ and $A$,
then rotating it through $B$ until it coincides with line $BC$,
through the polygon.
\begin{center}
\begin{asy}
size(5cm);
filldraw(dir(90)--dir(210)--dir(330)--cycle,
rgb(1,1,0.9), blue);
label(scale(2)*"$\mathcal H$", midpoint(dir(90)--dir(330)), dir(30), blue);
dot("$B$", dir(90), dir(130), blue);
dot("$A$", dir(210), dir(210), blue);
dot("$C$", dir(330), dir(330), blue);
dot( (0.2,-0.4), red);
dot( (-0.1, 0.3), red);
dot( (0.4, 0.1), red);
dot( (-0.5, -0.1), red);
dot( (0.3, -0.2), blue);
draw(dir(90)--dir(234), dashed, EndArrow);
label("$\ell$", dir(234), dir(234));
\end{asy}
\end{center}
During this process, we consider the set of points on the same side of $\ell$ as $C$,
and let $x$ be the number of such red points minus the number of such blue points.
Note that:
\begin{itemize}
\ii Every time $\ell$ touches a blue point, $x$ increases by $1$.
\ii Every time $\ell$ touches a red point, $x$ decreases by $1$.
\ii Initially, $x = +1$.
\ii Just before reaching the end we have $x = -1$.
\end{itemize}
So at the moment where $x$ first equals zero, we have found our balancing line.
\end{proof}
\begin{remark*}
Some students may hope that in fact for every red point there is balancing line through it.
This claim is false even for $n=2$.
For example, if $ABC$ is an equilateral triangle with center $O$,
if we color $A$ and $O$ red but $B$ and $C$ blue,
then no balancing line passes through $O$.
On the other hand, in \emph{On the number of balanced lines},
by J.\ Pach and R.\ Pinchasi, (Discrete and Computational Geometry, 2001, p.\ 611-628)
it is shown that in fact there are at least $n$ balancing lines.
\end{remark*}
|
USAMO-2005-notes_6
|
For a positive integer $m$,
let $s(m)$ denote the sum of the decimal digits of $m$.
A set $S$ positive integers is \emph{$k$-stable}
if $s(\sum_{x \in X} x) = k$ for any nonempty subset $X \subseteq S$.
For each integer $n \ge 2$ let $f(n)$ be the minimal $k$
for which there exists a $k$-stable set with $n$ integers.
Prove that there are constants $0 < C_1 < C_2$ with
\[ C_1 \log_{10} n \le f(n) \le C_2 \log_{10} n. \]
\end{enumerate}
|
\paragraph{Construction showing $f(n) \le 9 \left\lceil \log_{10} \binom{n+1}{2} \right\rceil$.}
Let $n \ge 1$ and $e \ge 1$ be integers satisfying $1 + 2 + \dots + n < 10^e$.
Consider the set
\[ S = \left\{ 10^e-1, \; 2(10^e-1), \; \dots, \; n(10^e-1) \right\}. \]
For example, if $n = 6$ and $e = 3$,
we have $S = \{999, 1998, 2997, 3996, 4995, 5994\}$.
The set $S$ here is easily seen to be $9e$-stable.
Thus $f(n) \le 9 \left\lceil \log_{10} \binom{n+1}{2} \right\rceil$,
proving one direction.
\begin{remark*}
I think the problem is actually more natural
with a multiset $S$ rather than a vanilla set,
in which case $S = \{10^e-1, 10^e-1, \dots, 10^e-1\}$ works fine,
and is easier to think of.
In some sense the actual construction is obtained
by starting with this one,
and then pushing together the terms together
in order to get the terms to be distinct,
hence the $1+2+\dots+n$ appearance.
\end{remark*}
\paragraph{Proof that $f(n) \ge \log_{12} n$.}
We are going to prove the following, which obviously sufficient.
\begin{claim*}
Let $k$ be a positive integer.
In any (multi)set $S$ of more than $12^k$ integers,
there exists a subset whose sum of decimal digits exceeds $k$.
\end{claim*}
\begin{proof}
Imagine writing entries of $S$ on a blackboard,
while keeping a running sum $\Sigma$ initially set to zero.
For $i = 1, 2, \dots$ we will have a process such that
at the end of the $i$th step all entries on the board
are divisible by $10^i$.
It goes as follows:
\begin{itemize}
\ii If the $i$th digit from the right of $\Sigma$ is nonzero,
then arbitrarily partition the numbers on the board
into groups of $10$, erasing any leftover numbers.
Within each group of $10$, we can find
a nonempty subset with sum $0 \bmod{10^i}$;
we then erase each group and replace it with that sum.
\ii If the $i$th digit from the right of $\Sigma$ is zero,
but some entry on the board is not divisible by $10^i$,
then we erase that entry and add it to $\Sigma$.
Then we do the grouping as in the previous step.
\ii If the $i$th digit from the right of $\Sigma$ is zero,
and all entries on the board are divisible by $10^i$,
we do nothing and move on to the next step.
\end{itemize}
This process ends when no numbers remain on the blackboard.
The first and second cases occur at least $k+1$ times
(the number of entries decreases by a factor of at most $12$ each step),
and each time $\Sigma$ gets some nonzero digit,
which is never changed at later steps.
Therefore $\Sigma$ has sum of digits at least $k+1$ as needed.
\end{proof}
\begin{remark*}
The official solutions contain a slicker proof:
it turns out that any multiple of $10^e-1$ has
sum of decimal digits at least $9e$.
However, if one does not know this lemma it seems
nontrivial to imagine coming up with it.
\end{remark*}
|
USAMO-2006-notes_1
|
Let $p$ be a prime number and let $s$ be an integer with $0 < s < p$.
Prove that there exist integers $m$ and $n$ with $0 < m < n < p$ and
\[ \left \{\frac{sm}{p} \right\} < \left \{\frac{sn}{p} \right \} < \frac{s}{p} \]
if and only if $s$ is not a divisor of $p-1$.
|
It's equivalent to $ms \bmod p < ns \bmod p < s$,
where $x \bmod p$ means the remainder when $x$ is divided by $p$,
by slight abuse of notation.
We will assume $s \ge 2$ for simplicity,
since the case $s = 1$ is clear.
For any $x \in \{1, 2, \dots, s-1\}$
we define $f(x)$ to be the unique number in $\{1, \dots, p-1\}$
such that $s \cdot f(x) \bmod p = x$.
Then, $m$ and $n$ fail to exist exactly when
\[ f(s-1) < f(s-2) < \dots < f(1). \]
We give the following explicit description of $f$:
choose $t \equiv -s^{-1} \pmod p$, $0 < t < p$.
Then $f(x) = 1 + (s-x) \cdot t \bmod p$.
So our displayed inequality is equivalent to
\[ (1+t) \bmod p < (1+2t) \bmod p < (1+3t) \bmod p
< \dots < (1+(s-1)t) \bmod p. \]
This just means that the sequence $1+kt$
never ``wraps around'' modulo $p$
as we take $k=1,2,\dots,s-1$.
Since we assumed $s \neq 1$, we have $0 < 1+t < p$.
Now since $1+kt$ never wraps around as $k=1,2,\dots,s-1$,
and increases in increments of $t$,
it follows that $1+kt < p$ for all $k = 1, 2, \dots, s-1$.
Finally, as $1+st \equiv 0 \pmod p$ we get $1+st=p$.
In summary, $m$, $n$ fail to exist
precisely when $1 + st = p$.
That is of course equivalent to $s \mid p-1$.
|
USAMO-2006-notes_2
|
Let $k > 0$ be a fixed integer.
Compute the minimum integer $N$ (in terms of $k$) for which
there exists a set of $2k+1$ distinct positive integers
that has sum greater than $N$,
but for which every subset of size $k$ has sum at most $N/2$.
|
The answer is $N = k(2k^2+3k+3)$ given by
\[ S = \left\{ k^2+1, k^2+2, \dots, k^2+2k+1 \right\}. \]
To show this is best possible,
let the set be $S = \{ a_0 < a_1 < \dots < a_{2k} \}$
so that the hypothesis becomes
\begin{align*}
N + 1 &\le a_0 + a_1 + \dots + a_{2k} \\
N/2 &\ge a_{k+1} + \dots + a_{2k}.
\end{align*}
Subtracting twice the latter from the former gives
\begin{align*}
a_0 &\ge 1 + (a_{k+1}-a_1) + (a_{k+2}-a_2) + \dots
+ (a_{2k} - a_k) \\
&\ge 1 + \underbrace{k + k + \dots + k}_{k \text{ terms}} = 1 + k^2.
\end{align*}
Now, we have
\begin{align*}
N/2 &\ge a_{k+1} + \dots + a_{2k} \\
&\ge (a_0 + (k+1)) + (a_0 + (k+2)) + \dots + (a_0 + 2k) \\
&= k \cdot a_0 + \left( (k+1) + \dots + 2k \right) \\
&\ge k(k^2+1) + k \cdot \frac{3k+1}{2}
\end{align*}
so $N \ge k(2k^2+3k+3)$.
\begin{remark*}
The exact value of $N$ is therefore very superficial.
From playing with these concrete examples
we find out we are essentially
just trying to find an increasing set $S$ obeying
\[ a_0 + a_1 + \dots + a_k > a_{k+1} + \dots + a_{2k} \qquad (\star) \]
and indeed given a sequence satisfying these properties
one simply sets $N = 2(a_{k+1} + \dots + a_{2k})$.
Therefore we can focus almost entirely on $a_i$ and not $N$.
\end{remark*}
\begin{remark*}
It is relatively straightforward to figure out what is going on
based on the small cases.
For example, one can work out by hand that
\begin{itemize}
\ii $\{2,3,4\}$ is optimal for $k=1$
\ii $\{5,6,7,8,9\}$ is optimal for $k=2$,
\ii $\{10,11,12,13,14,15,16\}$ is optimal for $k=3$.
\end{itemize}
In all the examples,
the $a_i$ are an arithmetic progression of difference $1$,
so that $a_j - a_i \ge j-i$ is a sharp for all $i<j$,
and thus this estimate may be used freely without loss of sharpness;
applying it in $(\star)$ gives a lower bound on $a_0$
which is then good enough to get a lower bound on $N$
matching the equality cases we found empirically.
\end{remark*}
|
USAMO-2006-notes_3
|
For integral $m$, let $p(m)$ be the greatest prime divisor of $m$.
By convention, we set $p(\pm 1) = 1$ and $p(0) = \infty$.
Find all polynomials $f$ with integer coefficients
such that the sequence
\[ \{ p(f(n^2)) - 2n \}_{n \geq 0} \]
is bounded above.
(In particular, this requires $f(n^2) \neq 0$ for $n \ge 0$.)
|
If $f$ is the (possibly empty)
product of linear factors of the form $4n-a^2$,
then it satisfies the condition.
We will prove no other polynomials work.
In what follows, assume $f$ is irreducible and nonconstant.
It suffices to show for every positive integer $c$,
there exists a prime $p$ and a nonnegative integer $n$
such that $n \le \tfrac{p-1}{2} - c$ and $p$ divides $f(n^2)$.
Firstly, recall there are infinitely many odd primes $p$,
with $p>c$, such that $p$ divides some $f(n^2)$, by Schur's Theorem.
Looking mod such a $p$ we can find $n$ between $0$
and $\tfrac{p-1}{2}$ (since $n^2 \equiv (-n)^2 \pmod p$).
We claim that only finitely many $p$ from this set can fail now.
For if a $p$ fails, then its $n$ must be between
$\tfrac{p-1}{2} - c$ and $\tfrac{p-1}{2}$.
That means for some $0 \le k \le c$ we have
\[ 0 \equiv f\left( \left(\frac{p-1}{2}-k\right)^2 \right)
\equiv f\left( \left( k + \frac12 \right)^2 \right) \pmod{p}. \]
There are only finitely many $p$ dividing
\[ \prod_{k=1}^c f\left( \left( k+\frac12 \right)^2 \right) \]
unless one of the terms in the product is zero;
this means that $4n-(2k+1)^2$ divides $f(n)$.
This establishes the claim and finishes the problem.
|
USAMO-2006-notes_4
|
Find all positive integers $n$ for which there exist an integer $k \ge 2$
and positive rational numbers $a_1$, \dots, $a_k$ satisfying
$a_1 + a_2 + \dots + a_k = a_1 a_2 \dots a_k = n$.
|
The answer is all $n$ other than $1,2,3,5$.
\begin{claim*}
The only solution with $n \le 5$ is $n = 4$.
\end{claim*}
\begin{proof}
The case $n=4$ works since $2+2 = 2\cdot2=4$. So assume $n > 4$.
We now contend that $k > 2$.
Indeed, if $a_1 + a_2 = a_1 a_2 = n$ then
$(a_1-a_2)^2 = (a_1+a_2)^2 - 4a_1a_2 = n^2-4n = (n-2)^2 - 4$
is a rational integer square, hence a perfect square.
This happens only when $n = 4$.
Now by AM-GM,
\[ \frac nk = \frac{a_1 + \dots + a_k}{k}
\ge \sqrt[k]{a_1 \dots a_k} = n^{1/k} \]
and so $n \ge k^{\frac{1}{1 - 1/k}} = k^{\frac{k}{k-1}}$.
This last quantity is always greater than $5$, since
\begin{align*}
3^{3/2} &= 3\sqrt3 > 5 \\
4^{4/3} &= 4\sqrt[3]{4} > 5 \\
k^{\frac{k}{k-1}} &> k \ge 5 \qquad \forall k \ge 5.
\end{align*}
This finishes the proof.
\end{proof}
Now, in general:
\begin{itemize}
\ii If $n \ge 6$ is even, we may take
$(a_1, \dots, a_{n/2}) = (n/2, 2, 1, \dots, 1)$.
\ii If $n \ge 9$ is odd, we may take
$(a_1, \dots, a_{(n-3)/2}) = (n/2, 1/2, 4, 1, \dots, 1)$.
\ii A special case $n = 7$: one example is $(4/3, 7/6, 9/2)$.
(Another is $(7/6, 4/3, 3/2, 3)$.)
\end{itemize}
\begin{remark*}
The main hurdle in the problem is the $n=7$ case.
One good reason to believe a construction exists is that
it seems quite difficult to prove that $n=7$ fails.
\end{remark*}
|
USAMO-2006-notes_5
|
A mathematical frog jumps along the number line.
The frog starts at $1$, and jumps according to the following rule:
if the frog is at integer $n$,
then it can jump either to $n+1$ or to $n + 2^{m_n+1}$
where $2^{m_n}$ is the largest power of $2$ that is a factor of $n$.
Show that if $k \ge 2$ is a positive integer
and $i$ is a nonnegative integer,
then the minimum number of jumps needed to reach $2^ik$
is greater than the minimum number of jumps needed to reach $2^i$.
|
We will think about the problem in terms of
finite sequences of jumps $(s_1, s_2, \dots, s_\ell)$,
which we draw as
\[ 1 = x_0
\taking{s_1} x_1
\taking{s_2} x_2
\taking{s_3} \dots
\taking{s_\ell} x_\ell \]
where $s_k = x_k - x_{k-1}$ is the length of some hop.
We say the sequence is \emph{valid} if it has the
property required by the problem:
for each $k$, either $s_k = 1$ or $s_k = 2^{m_{x_{k-1}}+1}$.
An example is shown below.
\begin{lemma*}
Let $(s_1, \dots, s_\ell)$ be a sequence of jumps.
Suppose we delete pick an index $k$ and exponent $e > 0$,
and delete any jumps after the $k$th one
which are divisible by $2^e$.
The resulting sequence is still valid.
\end{lemma*}
\begin{proof}
We only have to look after the $k$th jump.
The launching points of the remaining jumps after the $k$th
one are now shifted by multiples of $2^e$
due to the deletions;
so given a jump $x \taking{s} x+s$
we end up with a jump $x' \taking{s} x'+s$
where $x-x'$ is a multiple of $2^e$.
But since $s < 2^e$, we have $\nu_2(x') < e$
and hence $\nu_2(x) = \nu_2(x')$ so the jump is valid.
\end{proof}
\begin{center}
\begin{asy}
unitsize(0.6cm);
defaultpen(fontsize(8pt));
for (int i=1; i<=24; ++i) {
if (i!=8) label("$"+(string)i+"$", (i,0));
}
label("$\boxed{8}$", (8,0));
void arr(int i, int j, pen p, real eps) {
pair M = ((i+j)/2, eps*(j-i)**0.5/2);
draw((i,0)..M..(j,0), p, EndArrow(TeXHead), Margin(2,2));
label("$"+(string)(j-i)+"$", M, eps*dir(90), p);
}
arr( 1, 3, red, 1);
arr( 3, 4, red, -1);
arr( 4, 12, blue+dashed, 1);
arr(12, 13, red, -1);
arr(13, 14, red, 1);
arr(14, 18, blue+dashed, -1);
arr(18, 19, red, 1);
arr(19, 21, blue+dashed, -1);
arr(21, 23, blue+dashed, 1);
arr(23, 24, red, -1);
add(shift(-8, 3)*CC());
for (int i=1; i<=8; ++i) {
if (i!=8) label("$"+(string)i+"$", (i,0));
}
label("$\boxed{8}$", (8,0));
arr( 1, 3, red, 1);
arr( 3, 4, red, -1);
arr( 4, 5, red, -1);
arr( 5, 6, red, 1);
arr( 6, 7, red, 1);
arr( 7, 8, red, -1);
\end{asy}
\end{center}
Now let's consider a valid path to $2^i k$ with $\ell$ steps, say
\[ 1 = x_0
\taking{s_1} x_1
\taking{s_2} x_2
\taking{s_3} \dots
\taking{s_\ell} x_\ell = 2^i \cdot k \]
where $s_i = x_i - x_{i-1}$ is the distance jumped.
We delete jumps in the following way:
starting from the largest $e$ and going downwards until $e=0$,
we delete all the jumps of length $2^e$
which end at a point exceeding the target $2^i$.
By the lemma, at each stage, the path remains valid.
We claim more:
\begin{claim*}
Let $e \ge 0$.
After the jumps of length greater than $2^e$ are deleted,
the resulting end-point is at least $2^i$,
and divisible by $2^{\min(i,e)}$.
\end{claim*}
\begin{proof}
By downwards induction.
Consider any step where \emph{some} jump is deleted.
Then, the end-point must be strictly greater than $x = 2^i - 2^e$
(i.e.\ we must be within $2^e$ of the target $2^i$).
It is also divisible by $2^{\min(i,e)}$ by induction hypothesis,
since we are changing the end-point by multiples of $2^e$.
And the smallest multiple of $2^{\min(i,e)}$ exceeding $x$ is $2^i$.
\end{proof}
On the other hand by construction when the process ends
the reduced path ends at a point at most $2^i$,
so it is $2^i$ as desired.
Therefore we have taken a path to $2^i k$
and reduced it to one to $2^i$ by deleting some jumps.
This proves the result.
|
USAMO-2006-notes_6
|
Let $ABCD$ be a quadrilateral,
and let $E$ and $F$ be points on sides $AD$ and $BC$,
respectively, such that $\frac{AE}{ED} = \frac{BF}{FC}$.
Ray $FE$ meets rays $BA$ and $CD$ at $S$ and $T$, respectively.
Prove that the circumcircles of triangles $SAE$, $SBF$, $TCF$, and $TDE$
pass through a common point.
\end{enumerate}
|
\begin{center}
\begin{asy}
size(8cm);
defaultpen(fontsize(9pt));
pair A = (-1,3);
pair B = (-3,-2);
pair C = (6,-2);
pair D = (3,5);
pair M = (A*C-B*D)/(A+C-B-D);
pair E = 0.7*A+0.3*D;
pair F = 0.7*B+0.3*C;
pair S = extension(E, F, A, B);
pair T = extension(C, D, E, F);
draw(T--F);
draw(A--B--C--D--cycle);
draw(A--S);
draw(T--D);
draw(circumcircle(S, A, E), dotted+blue);
draw(circumcircle(S, B, F), dotted+blue);
draw(circumcircle(T, C, F), dotted+blue);
draw(circumcircle(T, D, E), dotted+blue);
pair P = extension(A, B, C, D);
pair Q = extension(A, D, B, C);
draw(A--Q--B, dashed);
draw(S--P, dashed);
draw(circumcircle(P, A, D), dotted+red);
draw(circumcircle(P, B, C), dotted+red);
draw(circumcircle(Q, A, B), dotted+red);
draw(circumcircle(Q, C, D), dotted+red);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$M$", M, dir(M));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$S$", S, dir(135));
dot("$T$", T, dir(T));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
/* Source generated by TSQ */
\end{asy}
\end{center}
Let $M$ be the Miquel point of $ABCD$.
Then $M$ is the center of a spiral similarity taking $AD$ to $BC$.
The condition guarantees that it also takes $E$ to $F$.
Hence, we see that $M$ is the center of a spiral similarity taking $\ol{AB}$ to $\ol{EF}$,
and consequently the circumcircles of $QAB$, $QEF$, $SAE$, $SBF$ concur at point $M$.
In other words, the Miquel point of $ABCD$ is also the Miquel point of $ABFE$.
Similarly, $M$ is also the Miquel point of $EDCF$, so all four circles concur at $M$.
|
USAMO-2007-notes_1
|
Let $n$ be a positive integer.
Define a sequence by setting $a_1 = n$ and, for each $k > 1$,
letting $a_k$ be the unique integer in the range $0 \leq a_k \leq k-1$
for which $a_1 + a_2 + \dots + a_k$ is divisible by $k$.
(For instance, when $n = 9$
the obtained sequence is $9,1,2,0,3,3,3,\dots$.)
Prove that for any $n$ the sequence $a_1$, $a_2$, \dots
eventually becomes constant.
|
For each $k$, the number
\[ b_k \coloneq \frac 1k (a_1 + \dots + a_k) \]
is a nonnegative integer.
\begin{claim*}
The sequence $(b_k)$ is eventually constant.
\end{claim*}
\begin{proof}
Since
\[ b_{k+1}
= \frac{a_1 + \dots + a_k + a_{k+1}}{k+1}
\le \frac{k b_k + k}{k+1} < b_k + 1 \]
and therefore $b_{k+1} \le b_k$ for all $k$.
Hence the sequence $b_k$ must eventually be constant.
\end{proof}
This can only happen once the sequence is constant:
indeed if $N$ is an index such that $b_N = b_{N+1} = \dots = b$ then
\[ a_k = (k+1) b_{k+1} - k \cdot b_k
= (k+1) \cdot b - k \cdot b = b \]
for any $k \ge N$.
In other words, $a_N = a_{N+1} = \dots = b$ as well.
|
USAMO-2007-notes_2
|
Decide whether it possible to cover all lattice points in $\RR^2$
by an (infinite) family of disks whose interiors are disjoint
such that the radius of each disk is at least $5$.
|
The answer is no.
Assume not.
Take a disk $\odot O$ not touching any member of the family,
and then enlarge it until it is maximal.
Then, it must be tangent to at least three other disks,
say $\odot A$, $\odot B$, $\odot C$.
Suppose WLOG that $\angle AOB \le 120\dg$.
Denote the radii of $\odot O$, $\odot A$, $\odot B$ by $r$, $a$, $b$.
But the Law of Cosines gives
\[ (a+b)^2 \le (a+r)^2 + (b+r)^2 + (a+r)(b+r) \]
which rewrites as
\[ 12r^2 \ge (a-3r)(b-3r) \ge (5-3r)^2 \]
which one can check is impossible for $r \le 1/\sqrt2$.
Thus $r > 1/\sqrt2$.
In particular $(\odot O)$ must contain a lattice point
as it contains a unit square.
\begin{remark*}
The order of the argument here matters in subtle ways.
A common approach is to try and reduce to the ``optimal'' case
where we have three mutually tangent circles,
and then apply the Descarte circle theorem.
There are ways in which this approach can fail
if the execution is not done with care.
(In particular, one cannot simply say to reduce
to this case, without some justification.)
For example: it is not true that, given an infinite family of disks,
we can enlarge disks until we get three mutually tangent ones.
As a counterexample consider the ``square grid''
in which a circle is centered at $(10m, 10n)$ for each $m,n \in \ZZ$
and has radius $5$.
Thus it is also not possible to simply pick three nearby circles
and construct a circle tangent to all three:
that newly constructed circle might intersect a fourth disk
not in the picture.
Thus, when constructing the small disk $\odot O$
in the above solution,
it seems easiest to start with a point not covered and grow $\odot O$
until it is tangent to \emph{some} three circles,
and then argue by cosine law.
Otherwise it not easy to determine which three
circles to start with.
In all solutions it seems easier to prove that a disjoint circle
of radius $1/\sqrt2$ exists,
and then \emph{finally} deduce it has a lattice point,
rather than trying to work the lattice point into the existence proof.
\end{remark*}
|
USAMO-2007-notes_3
|
Let $S$ be a set containing $n^2+n-1$ elements.
Suppose that the $n$-element subsets of $S$ are partitioned into two classes.
Prove that there are at least $n$ pairwise disjoint sets in the same class.
|
We present two solutions which are really equivalent, but phrased differently.
We refer to the two classes as ``red'' and ``blue'', respectively.
\paragraph{First solution (Grant Yu).}
We define a set of $n+1$ elements to be \emph{useful}
if it has $n$-element subsets in each class.
Consider a \textbf{maximal collection of disjoint useful sets}
and assume there are $p$ such sets.
Then, let $T$ be the set of elements remaining (i.e.\ not in one of chosen useful sets).
\begin{claim*}
All subsets of $T$ of size $n$ are the same color.
\end{claim*}
\begin{proof}
Assume there was a red set $R$ in $T$.
Replace the elements of $R$ one by one
until we get to any other subset $R'$ of $T$.
At each step, because no sets of $T$ form a useful set,
the set remains red --- so $R'$ is red too.
Since $R'$ is arbitrary, this proves the claim.
\end{proof}
We have $|T| = n^2+n-1 - p(n+1)$, and in particular $p < n$.
WLOG all sets in $T$ are red.
We can extract another red set from each of our chosen useful sets.
So we can get at least
\[ p + \left\lfloor \frac{|T|}{n} \right\rfloor
= p + \left\lfloor n+1-p - \frac{1+p}{n} \right\rfloor
\ge p + (n-p) = n. \]
\paragraph{Second solution (by induction).}
We prove more strongly that:
\begin{claim*}
Let $S$ be a set containing $k \cdot (n+1)-1$ elements.
Then we can find $k$ pairwise disjoint sets of the same color.
\end{claim*}
The proof is by induction on $k \ge 1$.
The base case $k=1$ this is immediate; $\binom{S}{n}$ is a single set.
For the inductive step, assume for contradiction the problem fails.
Let $T$ be any subset of $S$ of size $(k-1)(n+1)-1$.
By the induction hypothesis, among the subsets of $T$ alone,
we can already find $k-1$ pairwise disjoint sets of the same color.
Now $S \setminus T$ has size $n+1$, and so we would have to have
that all $\binom{n+1}{n}$ subsets of $S \setminus T$ are the same color.
By varying $T$, the set $S \setminus T$ ranges over all of $\binom{S}{k+1}$.
This causes all sets to be the same color, contradiction.
\begin{remark*}
Victor Wang writes the following:
\begin{quote}
I don't really like this problem,
but I think the main motivation for generalizing the problem
is that the original problem doesn't allow you to look at small cases.
(Also, it's not initially clear where the $n^2+n-1$ comes from.)
And pretty much the simplest way to get lots of similarly-flavored
small cases is to start with $k=2,3$ in
``find the smallest $N(n,k)$ such that when we partition the $n$-subsets
of a $\ge N(n,k)$-set into $2$ classes,
we can find some $k$ pairwise disjoint sets in the same class''.
\end{quote}
\end{remark*}
|
USAMO-2007-notes_4
|
An \emph{animal} with $n$ \emph{cells}
is a connected figure consisting of $n$ equal-sized square cells
(equivalently, a polyomino with $n$ cells).
A \emph{dinosaur} is an animal with at least $2007$ cells.
It is said to be \emph{primitive}
it its cells cannot be partitioned into two or more dinosaurs.
Find with proof the maximum number of cells in a primitive dinosaur.
|
In fact it's true for any tree with maximum degree $\le 4$.
Here is the solution of Andrew Geng.
Let $T$ be such a tree (a spanning tree of the dinosaur graph)
which corresponds to a primitive dinosaur.
\begin{claim*}
There exists a vertex $v$ such that when $v$ is deleted, no dinosaurs result.
\end{claim*}
\begin{proof}
Assume for contradiction that all vertices are bad
(leave a dinosaur when deleted).
Consider two adjacent vertices $v$, $w$ in $T$.
By checking possibilities, one sees that, say,
the dinosaur in $T-v$ contains $w$ and the dinosaur of $T-w$.
We can repeat in this way; since $T$ is acyclic,
this eventually becomes a contradiction.
\end{proof}
When this vertex is deleted,
we get at most $4$ components, each with $\le 2006$ vertices,
giving the answer of $4 \cdot 2006 + 1 = 8025$.
The construction is easy (take a ``cross'', for example).
|
USAMO-2007-notes_5
|
Prove that for every nonnegative integer $n$,
the number $7^{7^n}+1$ is the product
of at least $2n+3$ (not necessarily distinct) primes.
|
We prove this by induction on $n$ by showing that
\[ \frac{X^7+1}{X+1} = X^6 - X^5 + \dots + 1 \]
is never prime for $X = 7^{7^n}$,
hence we gain at least two additional prime factors
whenever we increase $n$ by one.
Indeed, the quotient may be written as
\[ \left( X+1 \right)^6 - 7X \cdot (X^2+X+1)^2 \]
which becomes a difference of squares, hence composite.
|
USAMO-2007-notes_6
|
Let $ABC$ be an acute triangle
with $\omega$, $S$, and $R$ being its incircle,
circumcircle, and circumradius, respectively.
Circle $\omega_{A}$ is tangent internally to $S$
at $A$ and tangent externally to $\omega$.
Circle $S_{A}$ is tangent internally to $S$
at $A$ and tangent internally to $\omega$.
Let $P_A$ and $Q_A$ denote
the centers of $\omega_A$ and $S_A$, respectively.
Define points $P_B$, $Q_B$, $P_C$, $Q_C$ analogously.
Prove that
\[ 8 P_A Q_A \cdot P_B Q_B \cdot P_C Q_C \le R^3 \]
with equality if and only if triangle $ABC$ is equilateral.
\end{enumerate}
|
It turns out we can compute $P_AQ_A$ explicitly.
Let us invert around $A$ with radius $s-a$
(hence fixing the incircle) and then compose this
with a reflection around the angle bisector of $\angle BAC$.
We denote the image of the composed map via
\[ \bullet \mapsto \bullet^\ast \mapsto \bullet^+. \]
We overlay this inversion with the original diagram.
Let $P_AQ_A$ meet $\omega_A$ again at $P$ and $S_A$ again at $Q$.
Now observe that $\omega_A^\ast$ is a line parallel to $S^\ast$;
that is, it is perpendicular to $\ol{PQ}$.
Moreover, it is tangent to $\omega^\ast = \omega$.
Now upon the reflection,
we find that $\omega^+ = \omega^\ast = \omega$,
but line $\ol{PQ}$ gets mapped to the altitude from $A$ to $\ol{BC}$,
since $\ol{PQ}$ originally contained the circumcenter $O$
(isogonal to the orthocenter).
But this means that $\omega_A^\ast$ is none other than the $\ol{BC}$!
Hence $P^+$ is actually the foot of the altitude from $A$
onto $\ol{BC}$.
By similar work, we find that $Q^+$
is the point on $\ol{AP^+}$ such that $P^+Q^+ = 2r$.
\begin{center}
\begin{asy}
size(8.5cm);
pair A = dir(140);
pair B = dir(210);
pair C = dir (330);
filldraw(A--B--C--cycle, opacity(0.1)+yellow, gray);
dot("$A$", A, A);
dot("$B$", B, B);
dot("$C$", C, C);
draw(incircle(A,B,C), black+1);
pair D = foot(A,B,C);
dot("$P^+$", D, dir(-45), brown);
draw(A--D, brown);
pair I = incenter(A,B,C);
pair O = circumcenter(A,B,C);
real a = abs(B-C);
real b = abs(C-A);
real c = abs(A-B);
real s = 0.5*(a+b+c);
real ha = abs(A-D);
real k = ((s-a)**3) / (s*ha*abs(A-I)**2);
pair PA = k*b*c/2*dir(O-A)+A;
dot("$P_A$", PA, dir(255), blue);
draw(CP(PA,A), lightblue);
pair P = Drawing("P", 2*PA-A, dir(180));
pair QA = PA + dir(PA-A)*( (s-a)*a**2 / (a*b*c) );
dot("$Q_A$", QA, dir(-80)*1.4, blue);
draw(CP(QA,A), lightblue);
pair Q = Drawing("Q", 2*QA-A, dir(-70));
draw(unitcircle, gray);
draw(A--Q);
real r = abs(I-foot(I,B,C));
Drawing("I", I, dir(90));
pair X = I + dir(O-A)*r;
pair P1 = foot(X,P,Q);
draw(Line(X,P1,2.5), red, Arrows(TeXHead));
dot("$P^\ast$", P1, dir(5), red);
draw(CP(A, foot(I,A,B)), dashed+deepgreen);
pair Q2 = D+dir(A-D)*2*r;
dot("$Q^+$", Q2, dir(145), brown);
draw(Q2--(I+r*dir(90)), brown);
\end{asy}
\end{center}
Now we can compute all the lengths directly. We have that
\[ AP_A = \half AP = \frac{(s-a)^2}{2AP^+}
= \half (s-a)^2 \cdot \frac{1}{h_a} \]
and
\[ AQ_A = \half AQ = \frac{(s-a)^2}{2AQ^+}
= \half (s-a)^2 \cdot \frac{1}{h_a-2r} \]
where $h_a = \frac{2K}{a}$ is the length of the $A$-altitude, with $K$ the area of $ABC$ as usual. Now it follows that
\[ P_AQ_A = \half (s-a)^2 \left( \frac{2r}{h_a(h_a-2r)} \right). \]
This can be simplified, as
\[ h_a - 2r = \frac{2K}{a} - \frac{2K}{s} = 2K \cdot \frac{s-a}{as}. \]
Hence
\[ P_AQ_A = \frac{a^2rs(s-a)}{4K^2} = \frac{a^2(s-a)}{4K}. \]
Hence, the problem is just asking us to show that
\[ a^2b^2c^2(s-a)(s-b)(s-c) \le 8 (RK)^3. \]
Using $abc=4RK$ and $(s-a)(s-b)(s-c) = \frac 1s K^2 = rK$, we find that this becomes
\[ 2(s-a)(s-b)(s-c) \le RK \iff 2r \le R \]
which follows immediately from $IO^2=R(R-2r)$.
Alternatively, one may rewrite this as Schur's Inequality in the form
\[ abc \ge (-a+b+c)(a-b+c)(a+b-c). \]
|
USAMO-2008-notes_1
|
Prove that for each positive integer $n$,
there are pairwise relatively prime integers $k_0$, \dots, $k_n$,
all strictly greater than $1$, such that
$k_0k_1 \dots k_n - 1$ is the product of two consecutive integers.
|
In other words, if we let
\[ P(x) = x(x+1) + 1 \]
then we would like there to be infinitely many primes
dividing some $P(t)$ for some integer $t$.
In fact, this result is true in much greater generality.
We first state:
\begin{theorem}
[Schur's theorem]
If $P(x) \in \ZZ[x]$ is nonconstant and $P(0) = 1$,
then there are infinitely many primes
which divide $P(t)$ for some integer $t$.
\end{theorem}
\begin{proof}
If $P(0) = 0$, this is clear.
So assume $P(0) = c \neq 0$.
Let $S$ be any finite set of prime numbers.
Consider then the value
\[ P\left(k \prod_{p \in S} p\right) \]
for some integer $k$.
It is $1 \pmod p$ for each prime $p$,
and if $k$ is large enough it should not be equal to $1$
(because $P$ is not constant).
Therefore, it has a prime divisor not in $S$.
\end{proof}
\begin{remark*}
In fact the result holds without the assumption $P(0) \neq 1$.
The proof requires only small modifications,
and a good exercise would be to write down a similar
proof that works first for $P(0) = 20$,
and then for any $P(0) \neq 0$.
(The $P(0) = 0$ case is vacuous,
since then $P(x)$ is divisible by $x$.)
\end{remark*}
To finish the proof, let $p_1$, \dots, $p_n$ be primes
and $x_i$ be integers such that
\begin{align*}
P(x_1) &\equiv 0 \pmod{p_1} \\
P(x_2) &\equiv 0 \pmod{p_2} \\
&\vdotswithin\equiv \\
P(x_n) &\equiv 0 \pmod{p_n}
\end{align*}
as promised by Schur's theorem.
Then, by Chinese remainder theorem,
we can find $x$ such that $x \equiv x_i \pmod{p_i}$
for each $i$, whence $P(x)$ has at least $n$ prime factor.
|
USAMO-2008-notes_2
|
Let $ABC$ be an acute, scalene triangle,
and let $M$, $N$, and $P$ be the midpoints of
$\ol{BC}$, $\ol{CA}$, and $\ol{AB}$, respectively.
Let the perpendicular bisectors of $\ol{AB}$ and $\ol{AC}$
intersect ray $AM$ in points $D$ and $E$ respectively,
and let lines $BD$ and $CE$ intersect in point $F$, inside triangle $ABC$.
Prove that points $A$, $N$, $F$, and $P$ all lie on one circle.
|
We present four solutions.
\paragraph{Barycentric solution.}
First, we find the coordinates of $D$.
As $D$ lies on $\ol{AM}$, we know $D=(t:1:1)$ for some $t$.
Now by perpendicular bisector formula, we find
\[ 0 = b^2(t-1) + (a^2-c^2) \implies t = \frac{c^2+b^2-a^2}{b^2}. \]
Thus we obtain \[ D = \left( 2S_A : c^2 : c^2 \right). \]
Analogously $E = (2S_A : b^2 : b^2)$,
and it follows that \[ F = \left( 2S_A : b^2 : c^2 \right). \]
The sum of the coordinates of $F$ is
\[ (b^2+c^2-a^2)+b^2+c^2= 2b^2+2c^2-a^2. \]
Hence the reflection of $A$ over $F$ is simply
\[ 2F-A = \left( 2(b^2+c^2-a^2)-(2b^2+2c^2-a^2) : 2b^2 : 2c^2 \right)
= \left(-a^2 : 2b^2 : 2c^2 \right). \]
It is evident that $F'$ lies on $(ABC) : -a^2yz - b^2zx - c^2xy = 0$,
and we are done.
\paragraph{Synthetic solution (harmonic).}
Here is a synthetic solution.
Let $X$ be the point so that $APXN$ is a cyclic harmonic quadrilateral.
We contend that $X = F$.
To see this it suffices to prove $B$, $X$, $D$ collinear
(and hence $C$, $X$, $E$ collinear by symmetry).
\begin{center}
\begin{asy}
pair A = dir(130);
pair P = dir(210);
pair N = dir(330);
pair B = 2*P-A;
pair C = 2*N-A;
pair O = -A;
pair T = midpoint(P--N);
pair D = extension(A, T, P, O);
pair E = extension(A, T, N, O);
pair X = extension(B, D, E, C);
pair M = extension(A, T, B, C);
draw(N--T, heavygreen);
draw(T--P, lightblue);
draw(X--P, heavygreen);
draw(B--C--N, lightblue);
draw(P--O--N, lightblue);
draw(N--E, lightblue);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
draw(D--X--C, red+dashed);
filldraw(A--B--X--cycle, opacity(0.1)+lightgreen, heavygreen);
filldraw(A--M--N--cycle, opacity(0.1)+lightgreen, heavygreen);
dot("$A$", A, dir(A));
dot("$P$", P, dir(P));
dot("$N$", N, dir(10));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$O$", O, dir(O));
dot("$T$", T, dir(45));
dot("$D$", D, dir(60));
dot("$E$", E, dir(240));
dot("$X$", X, dir(X));
dot("$M$", M, dir(M));
/* TSQ Source:
A = dir 130
P = dir 210
N = dir 330 R10
B = 2*P-A
C = 2*N-A
O = -A
T = midpoint P--N R45
D = extension A T P O R60
E = extension A T N O R240
X = extension B D E C
M = extension A T B C
N--T heavygreen
T--P lightblue
X--P heavygreen
B--C--N lightblue
P--O--N lightblue
N--E lightblue
unitcircle 0.1 lightcyan / blue
D--X--C red dashed
A--B--X--cycle 0.1 lightgreen / heavygreen
A--M--N--cycle 0.1 lightgreen / heavygreen
*/
\end{asy}
\end{center}
Let $T$ be the midpoint of $\ol{PN}$,
so $\triangle APX \sim \triangle ATN$.
So $\triangle ABX \sim \triangle AMN$, ergo
\[ \dang XBA = \dang NMA = \dang BAM = \dang BAD = \dang DBA \]
as desired.
\paragraph{Angle chasing solution (Mason Fang).}
Obviously $ANOP$ is concyclic.
\begin{claim*}
Quadrilateral $BFOC$ is cyclic.
\end{claim*}
\begin{proof}
Write
\begin{align*}
\dang BFC = \dang FBC+\dang BCF
&=\dang FBA +\dang ABC+\dang BCA+\dang ACF\\
&=\dang DBA +\dang ABC+\dang BCA+\dang ACE\\
&=\dang BAD +\dang ABC+\dang BCA+\dang EAC\\
&=2\angle BAC=\angle BOC. \qedhere
\end{align*}
\end{proof}
Define $Q = \ol{AA} \cap \ol{BC}$.
\begin{claim*}
Point $Q$ lies on $\ol{FO}$.
\end{claim*}
\begin{proof}
Write
\begin{align*}
\dang BOQ = \dang BOA+\dang AOQ
&=2\dang BCA+90^{\dg}+\dang AQO\\
&=2\dang BCA+90^{\dg}+\dang AMO\\
&=2\dang BCA+90^{\dg}+\dang AMC+90^{\dg}\\
&=\dang BCA+\dang MAC =\dang BCA+\dang ACE \\
&=\dang BCE =\dang BOF. \qedhere
\end{align*}
\end{proof}
As $Q$ is the radical center of $(ANOP)$, $(ABC)$ and $(BFOC)$,
this implies the result.
\paragraph{Inversive solution (Kelin Zhu).}
Invert about $A$ with radius $\sqrt{bc}$
followed by a reflection over the angle bisector of $\angle A$,
and denote the image of a point $X$ by $X'$.
The inverted problem now states the following:
\begin{quote}
In triangle $AP^{\ast}N^{\ast}$,
let $B^{\ast}$, $C^{\ast}$ be the midpoints of $AP^{\ast}$, $AN^{\ast}$
and $D^{\ast}$, $E^{\ast}$ be the intersection of the $A$ symmedian
with $(AP^{\ast})$, $(AN^{\ast})$, respectively.
$(AB^{\ast}D^{\ast})$, $(AC^{\ast}E^{\ast})$
intersect at a point $F^{\ast}$; prove that it lies on $P^{\ast}N^{\ast}$.
\end{quote}
I claim that, in fact, the midpoint of $P^{\ast}N^{\ast}$ is the desired intersection.
Redefine that point as $F^{\ast}$ and I will prove that
$(AB^{\ast}D^{\ast})$, $(AC^{\ast}E^{\ast})$ pass through it.
Note that
\[ \angle AD^{\ast}B^{\ast}=\angle D^{\ast}AB^{\ast}=\angle F^{\ast}AN^{\ast}=\angle AF^{\ast}B^{\ast}, \]
where the first equality is due to $B^{\ast}$ being the circumcircle of $AD^{\ast}P^{\ast}$,
the second equality is due to the definition of the symmedian,
and the third equality is due to the parallelogram $AB^{\ast}F^{\ast}C^{\ast}$.
A symmetric argument for $C$ finishes.
|
USAMO-2008-notes_3
|
Let $n$ be a positive integer. Denote by $S_n$ the set of points $(x, y)$
with integer coordinates such that
\[ \left\lvert x\right\rvert + \left\lvert y + \half \right\rvert < n. \]
A path is a sequence of distinct points
$(x_1 , y_1), (x_2, y_2), \dots, (x_\ell, y_\ell)$ in $S_n$
such that, for $i = 2, \dots, \ell$,
the distance between $(x_i , y_i)$ and $(x_{i-1} , y_{i-1} )$ is $1$.
Prove that the points in $S_n$ cannot be partitioned into fewer than $n$ paths.
|
\paragraph{First solution (local).}
We proceed by induction on $n$.
The base case $n=1$ is clear, so suppose $n > 1$.
Let $S$ denote the set of points
\[ S = \left\{ (x,y) : x + \left\lvert y+\frac12 \right\rvert \ge n - 2 \right\}. \]
An example when $n=4$ is displayed below.
\begin{center}
\begin{asy}
int n = 4;
for (int a=-4; a<=4; ++a) {
for (int b=-4; b<=4; ++b) {
if (abs(a) + abs(b+0.5) < n) {
if (a + abs(b+0.5) > n-2) dot( (a,b), red+4 );
else dot( (a,b), black+3 );
}
}
}
draw((0,3)--(0,2)--(1,2)--(1,1)--(2,1)--(2,0)--(3,0)--(3,-1)--(2,-1)--(2,-2)--(1,-2)--(1,-3)--(0,-3)--(0,-4), red);
draw(CR((3,0), 0.3), deepgreen);
label("$a$", (3.3,0), dir(0), deepgreen);
\end{asy}
\end{center}
For any minimal partition $\mathcal P$ of $S_n$,
let $P$ denote the path passing through the point $a = (n-1,0)$.
Then the intersection of $P$ with $S$ consists of several disconnected paths;
let $N$ be the number of nodes in the component containing $a$, and
pick $\mathcal P$ such that $N$ is maximal.
We claim that in this case $P = S$.
Assume not.
First, note $a = (n-1,0)$ must be connected to $b = (n-1, -1)$
(otherwise join them to decrease the number of paths).
Now, starting from $a = (n-1,0)$ walk along $P$ away from $b$
until one of the following three conditions is met:
\begin{itemize}
\ii We reach a point $v$ not in $S$.
Let $w$ be the point before $v$,
and $x$ the point in $S$ adjacent to $w$.
Then delete $vw$ and add $wx$.
This increases $N$ while leaving the number of edges unchanged:
so this case can't happen.
\ii We reach an endpoint $v$ of $P$ (which may be $a$),
lying inside the set $S$, which is not the topmost point $(0,n-1)$.
Let $w$ be the next point of $S$.
Delete any edge touching $w$ and add edge $vw$.
This increases $N$ while leaving the number of edges unchanged:
so this case can't happen.
\ii We reach the topmost point $(0,n-1)$.
\end{itemize}
Thus we see that $P$ must follow $S$ until reaching the topmost point $(0,n-1)$.
Similarly it must reach the bottom-most point $(0,-n)$.
Hence $P=S$.
The remainder of $S_n$ is just $S_{n-1}$,
and hence this requires at least $n-1$ paths to
cover by the inductive hypothesis.
So $S_n$ requires at least $n$ paths, as desired.
%A bit easy for a \#3, seeing that I got this during while trying to sleep on a train ride to visit my grandmother (and I normally suck at combo). This is more or less equivalent to MellowMelon's solution, but here's how I phrased it in my head.
\begin{remark*}
[Motivational comments from Evan]
Basically the idea is that I wanted to peel away the
right path $S$ highlighted in red in the figure, so that one could induct.
But the problem is that the red path might not actually exist,
e.g.\ the set of paths might contain the mirror of $S$ instead.
Nonetheless, in those equality cases I found I could perturb some edges
(e.g.\ change from $(-1,n-2)$--$(0,n-2)$ to $(0,n-2)$--$(1,n-2)$).
So the idea then was to do little changes and try to convert
the given partition into one where the red path $S$ exists,
(and then peel it away for induction)
without decreasing the total number of paths.
To make this work, you actually want the incisions to begin
ear the points $a$ and $b$,
because that's the point of $S$ that is most constrained
(e.g.\ you get $a$--$b$ right away for free),
and assemble the path from there.
(If you try to do it from the top, it's much less clear what's happening.)
That's why the algorithm starts the mutations from around a.
\end{remark*}
\paragraph{Second solution (global).}
Here is a much shorter official solution,
which is much trickier to find, and ``global'' in nature.
Color the upper half of the diagram with a blue/red checkerboard pattern
such that the uppermost point $(n-1,0)$ is blue.
Reflect it over to the bottom, as shown.
\begin{center}
\begin{asy}
bool even(int x) {
return (x % 2 == 0
);
}
int n = 4;
for (int a=-4; a<=4; ++a) {
for (int b=-4; b<=4; ++b) {
if (abs(a) + abs(b+0.5) < n) {
if ( (b >= 0) && even(a+b+n+1)) dot( (a,b), blue+4 );
else if ( (b < 0) && even(a+b+n)) dot( (a,b), blue+4 );
else dot( (a,b), red+3 );
}
}
}
\end{asy}
\end{center}
Assume there are $m$ paths.
Cut in two any paths with two adjacent blue points;
this occurs only along the horizontal symmetry axis.
Thus:
\begin{itemize}
\ii After cutting there are at most $m+n$ paths,
since at most $n$ cuts occur.
\ii On the other hand,
there are $2n$ more blue points than red points.
Hence after cutting there must be at least $2n$ paths
(since each path alternates colors, except possibly for double-red pairs).
\end{itemize}
So $m+n \ge 2n$, hence $m \ge n$.
\begin{remark*}
This problem turned out to be known already.
It appears in this reference:
\begin{quote}
Nikolai Beluhov, Nyakolko Zadachi po Shahmatna
Kombinatorika, \emph{Matematika Plyus},
2006, issue 4, pages 61--64.
\end{quote}
Section 1 of 2 was reprinted with revisions as
Nikolai Beluhov, Dolgii Put Korolya, \emph{Kvant},
2010, issue 4, pages 39--41.
The reprint is available at
\url{http://kvant.ras.ru/pdf/2010/2010-04.pdf}.
\end{remark*}
\begin{remark*}
[Nikolai Beluhov]
As pointed out in the reference above,
this problem arises naturally when we try to
estimate the greatest possible length of a closed king tour
on the chessboard of size $n \times n$ with $n$ even,
a question posed by Igor Akulich in Progulki
Korolya, \emph{Kvant}, 2000, issue 3, pages 47--48.
Each one of the two references above contains a proof
that the answer is $n + \sqrt{2}(n^2 - n)$.
\end{remark*}
|
USAMO-2008-notes_4
|
For which integers $n \ge 3$ can one find a triangulation
of regular $n$-gon consisting only of isosceles triangles?
|
The answer is $n$ of the form $2^a(2^b+1)$
where $a$ and $b$ are nonnegative integers not both zero.
Call the polygon $A_1 \dots A_n$
with indices taken modulo $n$.
We refer to segments $A_1 A_2$, $A_2 A_3$, \dots, $A_n A_1$
as \emph{short sides}.
Each of these must be in the triangulation.
Note that
\begin{itemize}
\ii when $n$ is even,
the isosceles triangles triangle using a short side $A_1 A_2$
are $\triangle A_n A_1 A_2$ and $\triangle A_1 A_2 A_3$ only,
which we call \emph{small}.
\ii when $n$ is odd, in addition to the small triangles,
we have $\triangle A_{\half(n+3)} A_1 A_2$,
which we call \emph{big}.
\end{itemize}
This leads to the following two claims.
\begin{claim*}
If $n > 4$ is even, then $n$ works iff $n/2$ does.
\end{claim*}
\begin{proof}
All short sides must be part of a small triangle;
after drawing these in, we obtain an $n/2$-gon.
\begin{center}
\begin{asy}
size(3cm);
draw(unitcircle, dotted);
picture p;
dot(p, dir(0));
dot(p, dir(30));
dot(p, dir(60));
fill(p, dir(0)--dir(30)--dir(60)--cycle, opacity(0.1)+lightred);
draw(p, dir(0)--dir(30)--dir(60), blue);
draw(p, dir(0)--dir(60), red);
for (int i=0; i<6; ++i) {
add(rotate(60*i)*p);
}
\end{asy}
\end{center}
Thus the sides of $\mathcal P$ must pair off,
and when we finish drawing we have an $n/2$-gon.
\end{proof}
Since $n = 4$ works, this implies all powers of $2$ work
and it remains to study the case when $n$ is odd.
\begin{claim*}
If $n > 1$ is odd, then $n$ works if and only if $n = 2^b+1$
for some positive integer $b$.
\end{claim*}
\begin{proof}
We cannot have all short sides part of small triangles
for parity reasons, so some side, must be part of a big triangle.
Since big triangles contain the center $O$,
there can be at most one big triangle too.
Then we get $\half(n-1)$ small triangles,
pairing up the remaining sides.
Now repeating the argument with the triangles on each half
shows that the number $n-1$ must be a power of $2$, as needed.
\end{proof}
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.