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USAMO-2008-notes_5
|
Three nonnegative real numbers $r_1$, $r_2$, $r_3$ are written on a blackboard.
These numbers have the property
that there exist integers $a_1$, $a_2$, $a_3$, not all zero,
satisfying $a_1r_1 + a_2r_2 + a_3r_3 = 0$.
We are permitted to perform the following operation:
find two numbers $x$, $y$ on the blackboard with $x \le y$,
then erase $y$ and write $y - x$ in its place.
Prove that after a finite number of such operations,
we can end up with at least one $0$ on the blackboard.
|
We first show we can decrease the quantity
$\left\lvert a_1 \right\rvert + \left\lvert a_2 \right\rvert + \left\lvert a_3 \right\rvert$
as long as $0 \notin \left\{ a_1,a_2,a_3 \right\}$.
Assume $a_1 > 0$ and $r_1 > r_2 > r_3$
without loss of generality and consider two cases.
\begin{itemize}
\ii Suppose $a_2 > 0$ or $a_3 > 0$; these cases are identical.
(One cannot have both $a_2 > 0$ and $a_3 > 0$.)
If $a_2 > 0$ then $a_3 < 0$ and get
\[ 0 = a_1r_1 + a_2r_2 + a_3r_3 > a_1r_3 + a_3r_3 \implies a_1 + a_3 < 0 \]
so $\left\lvert a_1 + a_3 \right\rvert < \left\lvert a_3 \right\rvert$,
and hence we perform $(r_1,r_2,r_3) \mapsto (r_1-r_3,r_2,r_3)$.
\ii Both $a_2 < 0$ and $a_3 < 0$.
Assume for contradiction that $\left\lvert a_1+a_2 \right\rvert \ge -a_2$
and $\left\lvert a_1+a_3 \right\rvert \ge -a_3$ both hold
(if either fails then we use $(r_1,r_2,r_3) \mapsto (r_1-r_2,r_2,r_3)$
and $(r_1,r_2,r_3) \mapsto (r_1-r_3,r_2,r_3)$, respectively).
Clearly $a_1+a_2$ and $a_1+a_3$ are both positive in this case,
so we get $a_1 + 2a_2$ and $a_1 + 2a_3 \ge 0$; adding gives $a_1+a_2+a_3 \ge 0$.
But
\begin{align*}
0 &= a_1r_1 + a_2r_2 + a_3r_3 \\
&> a_1r_2 + a_2r_2 + a_3r_2 \\
&= r_2(a_1+a_2+a_3) \\
\implies 0 &< a_1+a_2+a_3.
\end{align*}
\end{itemize}
Since this covers all cases, we see that we can always decrease
$\left\lvert a_1 \right\rvert + \left\lvert a_2 \right\rvert + \left\lvert a_3 \right\rvert$
whenever $0 \notin \left\{ a_1,a_2,a_3 \right\}$.
Because the $a_i$ are integers this cannot occur indefinitely,
so eventually one of the $a_i$'s is zero.
At this point we can just apply the Euclidean Algorithm, so we're done.
|
USAMO-2008-notes_6
|
At a certain mathematical conference,
every pair of mathematicians are either friends or strangers.
At mealtime, every participant eats in one of two large dining rooms.
Each mathematician insists upon eating in a room which
contains an even number of his or her friends.
Prove that the number of ways that the mathematicians
may be split between the two rooms is a power of two
(i.e.\ is of the form $2^k$ for some positive integer $k$).
\end{enumerate}
|
Take the obvious graph interpretation
where we are trying to $2$-color a graph.
Let $A$ be the adjacency matrix of the graph over $\FF_2$,
except the diagonal of $A$ has $\deg v \pmod 2$ instead of zero.
Then let $\vec d$ be the main diagonal.
Splittings then correspond to $A \vec v = \vec d$.
It's then immediate that the number of ways is either zero
or a power of two, since if it is nonempty it is a coset of $\ker A$.
Thus we only need to show that:
\begin{claim*}
At least one coloring exists.
\end{claim*}
\begin{proof}
If not, consider a minimal counterexample $G$.
Clearly there is at least one odd vertex $v$.
Consider the graph with vertex set $G-v$,
where all pairs of neighbors of $v$ have their edges complemented.
By minimality, we have a good coloring here.
One can check that this extends to a good coloring on $G$
by simply coloring $v$ with the color matching an even number of its neighbors.
This breaks minimality of $G$, and hence all graphs $G$ have a coloring.
\end{proof}
It's also possible to use linear algebra.
We prove the following lemma:
\begin{lemma*}[grobber]
Let $V$ be a finite dimensional vector space,
$T \colon V \to V$ and $w \in V$.
Then $w$ is in the image of $T$ if and only if
there are no $\xi \in V^\vee$ for which $\xi(w) \neq 0$
and yet $\xi \circ T = 0$.
\end{lemma*}
\begin{proof}
Clearly if $T(v) = w$, then no $\xi$ exists.
Conversely, assume $w$ is not in the image of $T$.
Then the image of $T$ is linearly independent from $w$.
Take a basis $e_1$, \dots, $e_m$ for the image of $T$, add $w$,
and then extend it to a basis for all of $V$.
Then have $\xi$ kill all $e_i$ but not $w$.
\end{proof}
\begin{corollary*}
In a symmetric matrix $A$ mod $2$,
there exists a vector $v$ such that $Av$ is a copy of the diagonal of $A$.
\end{corollary*}
\begin{proof}
Let $\xi$ be such that $\xi \circ T = 0$.
Look at $\xi$ as a column vector $w^\top$, and let $d$ be the diagonal.
Then \[ 0 = w^\top \cdot T \cdot w = \xi(d) \]
because this extracts the sum of coefficients submatrix of $T$,
and all the symmetric entries cancel off.
Thus no $\xi$ as in the previous lemma exists.
\end{proof}
This corollary gives the desired proof.
|
USAMO-2009-notes_1
|
Given circles $\omega_1$ and $\omega_2$ intersecting at points $X$ and $Y$,
let $\ell_1$ be a line through the center of $\omega_1$
intersecting $\omega_2$ at points $P$ and $Q$
and let $\ell_2$ be a line through the center of $\omega_2$
intersecting $\omega_1$ at points $R$ and $S$.
Prove that if $P$, $Q$, $R$, and $S$ lie on a circle
then the center of this circle lies on line $XY$.
|
Let $r_1$, $r_2$, $r_3$ denote the
circumradii of $\omega_1$, $\omega_2$, and $\omega_3$, respectively,
where $\omega_3$ is the circle through $P$, $Q$, $R$, $S$.
\begin{center}
\begin{asy}
pair O_1 = dir(220);
pair O_2 = dir(320);
pair O = dir(110);
pair T = foot(O, O_1, O_2);
pair X = midpoint(O--T);
pair Y = 2*T-X;
filldraw(CP(O_1, X), opacity(0.1)+lightcyan, lightblue);
filldraw(CP(O_2, X), opacity(0.1)+lightcyan, lightblue);
draw(O--Y, red+dashed);
pair K = foot(O_1, O, O_2);
pair P = IP(O_1--K, CP(O_2, X));
pair Q = 2*K-P;
draw(O_1--Q, red);
draw(O--O_2, heavycyan);
pair L = foot(O_2, O, O_1);
pair R = IP(O_2--L, CP(O_1, X));
pair S = 2*L-R;
draw(O_2--S, red);
draw(O--O_1, heavycyan);
draw(arc(O, abs(P-O), 180, 360), heavygreen);
dot("$O_1$", O_1, dir(O_1));
dot("$O_2$", O_2, dir(O_2));
dot("$O$", O, dir(O));
dot("$X$", X, dir(100));
dot("$Y$", Y, dir(Y));
dot("$P$", P, dir(120));
dot("$Q$", Q, dir(Q));
dot("$R$", R, dir(40));
dot("$S$", S, dir(S));
/* TSQ Source:
O_1 = dir 220
O_2 = dir 320
O = dir 110
T := foot O O_1 O_2
X = midpoint O--T R100
Y = 2*T-X
CP O_1 X 0.1 lightcyan / lightblue
CP O_2 X 0.1 lightcyan / lightblue
O--Y red dashed
K := foot O_1 O O_2
P = IP O_1--K CP O_2 X R120
Q = 2*K-P
O_1--Q red
O--O_2 heavycyan
L := foot O_2 O O_1
R = IP O_2--L CP O_1 X R40
S = 2*L-R
O_2--S red
O--O_1 heavycyan
!draw(arc(O, abs(P-O), 180, 360), heavygreen);
*/
\end{asy}
\end{center}
We wish to show that $O_3$ lies on the radical axis of $\omega_1$ and $\omega_2$.
Let us encode the conditions using power of a point.
Because $O_1$ is on the radical axis of $\omega_2$ and $\omega_3$,
\begin{align*}
\opname{Pow}_{\omega_2}(O_1) &= \opname{Pow}_{\omega_3}(O_1) \\
\implies O_1O_2^2 - r_2^2 &= O_1O_3^2 - r_3^2.
\intertext{Similarly, because $O_2$ is on the radical axis of $\omega_1$ and $\omega_3$, we have}
\opname{Pow}_{\omega_1}(O_2) &= \opname{Pow}_{\omega_3}(O_2) \\
\implies O_1O_2^2 - r_1^2 &= O_2O_3^2 - r_3^2. \\
\intertext{Subtracting the two gives}
(O_1O_2^2-r_2^2)-(O_1O_2^2-r_1^2) &= (O_1O_3^2-r_3^2)-(O_2O_3^2-r_3^2) \\
\implies r_1^2-r_2^2 &= O_1O_3^2-O_2O_3^2 \\
\implies O_2O_3^2 - r_2^2 &= O_1O_3^2 - r_1^2 \\
\implies \opname{Pow}_{\omega_2}(O_3) &= \opname{Pow}_{\omega_1} (O_3)
\end{align*}
as desired.
|
USAMO-2009-notes_2
|
Let $n$ be a positive integer.
Determine the size of the largest subset of $\{ -n, -n+1, \dots, n-1, n\}$
which does not contain three elements $a$, $b$, $c$ (not necessarily distinct)
satisfying $a+b+c=0$.
|
The answer is $n$ with $n$ even and $n+1$ with $n$ odd;
the construction is to take all odd numbers.
To prove this is maximal, it suffices to show it for $n$ even;
we do so by induction on even $n \ge 2$ with the base case being trivial.
Letting $A$ be the subset, we consider three cases:
\begin{enumerate}[(i)]
\ii If $|A \cap \{-n,-n+1,n-1,n\}| \le 2$,
then by the hypothesis for $n-2$ we are done.
\ii If both $n \in A$ and $-n \in A$,
then there can be at most $n-2$ elements in $A \setminus \{\pm n\}$,
one from each of the pairs $(1,n-1)$, $(2,n-2)$, $\dots$
and their negations.
\ii If $n, n-1, -n+1 \in A$ and $-n \notin A$,
and at most $n-3$ more can be added,
one from each of $(1, n-2)$, $(2, n-3)$, $\dots$
and $(-2, -n+2)$, $(-3, -n+3)$, $\dots$.
(In particular $-1 \notin A$.
Analogous case for $-A$ if $n \notin A$.)
\end{enumerate}
Thus in all cases, $|A| \le n$ as needed.
\begin{remark*}
A few examples of equality cases:
\begin{itemize}
\ii All odd numbers
\ii All numbers with absolute value at least $n/2$
\ii For $n$ even, the set $\{1, 2, \dots, n\}$
\end{itemize}
This list isn't exhaustive
e.g.\ for $n=4$, the set $\{-3,2,3,4\}$ achieves the optimum.
\end{remark*}
|
USAMO-2009-notes_3
|
We define a \emph{chessboard polygon} to be a simple polygon whose sides are
situated along lines of the form $x=a$ or $y=b$, where $a$ and $b$ are integers.
These lines divide the interior into unit squares, which are shaded alternately
gray and white so that adjacent squares have different colors.
To tile a chessboard polygon by dominoes is to
exactly cover the polygon by non-overlapping $ 1 \times 2$ rectangles.
Finally, a \emph{tasteful tiling} is one which avoids the
two configurations of dominoes and colors shown on the left below.
Two tilings of a $3 \times 4$ rectangle are shown;
the first one is tasteful, while the second is not,
due to the vertical dominoes in the upper right corner.
\begin{center}
\begin{asy}
size(300);
pen rt = linewidth(2.5) + rgb(0.6,0.2,0.2);
void chessboard(int a, int b, int eps, pair P) {
for(int i = 0; i < a; ++i) {
for(int j = 0; j < b; ++j) {
if((i+j+eps)#2 == (i+j+eps)/2)
fill(shift(P.x+i,P.y+j)*unitsquare,rgb(0.6,0.6,0.6));
}
}
draw(P--P+(a,0)--P+(a,b)--P+(0,b)--cycle, rt);
}
chessboard(2,2,0,(0,0));
chessboard(2,2,1,(2.5,0));
chessboard(4,3,1,(6,0));
chessboard(4,3,1,(11,0));
label("Distasteful tilings",(2.25,2.5),fontsize(10));
/* draw lines */
draw((1,0)--(1,2), rt);
draw((2.5,1)--(4.5,1), rt);
draw((7,0)--(7,2)--(6,2)--(10,2)--(9,2)--(9,0)--(9,1)--(7,1), rt);
draw((8,2)--(8,3), rt);
draw((12,0)--(12,2)--(11,2)--(13,2), rt);
draw((13,1)--(15,1)--(14,1)--(14,3), rt);
draw((13,0)--(13,3), rt);
\end{asy}
\end{center}
Prove that (a) if a chessboard polygon can be tiled by dominoes,
then it can be done so tastefully, and (b) such a tasteful tiling is unique.
|
\paragraph{Proof of (a).} This is easier, and by induction.
Let $\mathcal P$ denote the chessboard polygon which can be tiled by dominoes.
Consider a \emph{lower-left} square $s$ of the polygon,
and WLOG is it white (other case similar).
Then we have two cases:
\begin{itemize}
\ii If there exists a domino tiling of $\mathcal P$
where $s$ is covered by a vertical domino,
then delete this domino and apply induction on the rest of $\mathcal P$.
This additional domino will not cause any distasteful tilings.
\ii Otherwise, assume $s$ is covered by a horizontal domino
in \emph{every} tiling.
Again delete this domino and apply induction on the rest of $\mathcal P$.
The resulting tasteful tiling should not have another horizontal domino
adjacent to the one covering $s$, because otherwise
we could have replaced that $2 \times 2$ square
with two vertical dominoes to arrive in the first case.
So this additional domino will not cause any distasteful tilings.
\end{itemize}
\begin{remark*}
The second case can actually arise, for example in the following picture.
\begin{center}
\begin{asy}
unitsize(0.6cm);
fill(shift(1,0)*unitsquare,rgb(0.6,0.6,0.6));
fill(shift(0,1)*unitsquare,rgb(0.6,0.6,0.6));
fill(shift(2,1)*unitsquare,rgb(0.6,0.6,0.6));
draw( (0,0)--(0,3)--(1,3)--(1,2)--(3,2)--(3,1)--(2,1)--(2,0)--cycle,
linewidth(2.5) + rgb(0.6,0.2,0.2));
\end{asy}
\end{center}
Thus one cannot just try to cover $s$ with a vertical
domino and claim the rest of $\mathcal P$ is tile-able.
So the induction is not as easy as one might hope.
One can phrase the solution algorithmically too,
in the following way: any time we see a distasteful tiling,
we rotate it to avoid the bad pattern.
The bottom-left corner eventually becomes stable,
and an induction shows the termination of the algorithm.
\end{remark*}
\paragraph{Proof of (b).} We now turn to proving uniqueness.
Suppose for contradiction there are two distinct tasteful tilings.
Overlaying the two tilings on top of each other induces
several \emph{cycles} of overlapping dominoes at positions where the tilings differ.
Henceforth, it will be convenient to work with the lattice $\ZZ^2$,
treating the squares as black/white points, and we do so.
Let $\gamma$ be any such cycle and let $s$ denote a
lower left point, and again WLOG it is black.
Orient $\gamma$ counterclockwise henceforth.
Restrict attention to the lattice polygon $\mathcal Q$ enclosed by $\gamma$
(we consider points of $\gamma$ as part of $\mathcal Q$).
In one of the two tilings of (lattice points of) $\mathcal Q$,
the point $s$ will be covered by a horizontal domino;
in the other tiling $s$ will be covered by a vertical domino.
From now on we will focus only on the latter one.
Observe that we now have a set of dominoes along $\gamma$,
such that $\gamma$ points from the white point to the black point
within each domino.
Now impose coordinates so that $s = (0,0)$.
Consider the stair-case sequence of points
$p_0 = s = (0,0)$, $p_1 = (1,0)$, $p_2 = (1,1)$, $p_3 = (2,1)$, and so on.
By hypothesis, $p_0$ is covered by a vertical domino.
Then $p_1$ must be covered by a horizontal domino, to avoid a distasteful tiling.
Then if $p_2$ is in $\mathcal Q$, then it must be covered by a vertical domino
to avoid a distasteful tiling, and so on.
We may repeat this argument as long the points $p_i$ lie inside $\mathcal Q$.
(See figure below; the staircase sequence is highlighted by red halos.)
\begin{center}
\begin{asy}
unitsize(1.5cm);
margin m = Margin(3,3);
arrowbar ar = EndArrow(TeXHead);
pen gamma = blue + 1;
draw( (0,0)--(1,0), gamma, ar, m);
draw( (1,0)--(2,0), gamma, ar, m);
draw( (2,0)--(3,0), gamma, ar, m);
draw( (3,0)--(3,-1), gamma, ar, m);
draw( (3,-1)--(4,-1), gamma, ar, m);
draw( (4,-1)--(4,0), gamma, ar, m);
draw( (4,0)--(4,1), gamma, ar, m);
draw( (4,1)--(4,2), gamma, ar, m);
draw( (4,2)--(3,2), gamma, ar, m);
draw( (3,2)--(3,3), gamma, ar, m);
draw( (3,3)--(4,3), gamma, ar, m);
draw( (4,3)--(4,4), gamma, ar, m);
draw( (4,4)--(3,4), gamma, ar, m);
draw( (3,4)--(2,4), gamma, ar, m);
draw( (2,4)--(1,4), gamma, ar, m);
draw( (1,4)--(1,3), gamma, ar, m);
draw( (1,3)--(0,3), gamma, ar, m);
draw( (0,3)--(0,2), gamma, ar, m);
draw( (0,2)--(0,1), gamma, ar, m);
draw( (0,1)--(0,0), gamma, ar, m);
fill( (0,0)--(3,0)--(3,-1)--(4,-1)--(4,2)--(3,2)
--(3,3)--(4,3)--(4,4)--(1,4)--(1,3)--(0,3)--cycle,
opacity(0.1) + lightcyan);
label("$s$", (0,0), 1.5*dir(225));
label("$a$", (3,2), 2*dir(45), red);
label("$b$", (1,0), 2*dir(225), red);
draw( (3,2)--(1,0), lightred);
real r = 0.07;
filldraw(CR( (3,-1), r), gray, black);
filldraw(CR( (0,0), r), gray, black);
filldraw(CR( (2,0), r), gray, black);
filldraw(CR( (4,0), r), gray, black);
filldraw(CR( (1,1), r), gray, black);
filldraw(CR( (3,1), r), gray, black);
filldraw(CR( (0,2), r), gray, black);
filldraw(CR( (2,2), r), gray, black);
filldraw(CR( (4,2), r), gray, black);
filldraw(CR( (1,3), r), gray, black);
filldraw(CR( (3,3), r), gray, black);
filldraw(CR( (2,4), r), gray, black);
filldraw(CR( (4,4), r), gray, black);
filldraw(CR( (4,-1), r), white, black);
filldraw(CR( (1,0), r), white, black);
filldraw(CR( (3,0), r), white, black);
filldraw(CR( (0,1), r), white, black);
filldraw(CR( (2,1), r), white, black);
filldraw(CR( (4,1), r), white, black);
filldraw(CR( (1,2), r), white, black);
filldraw(CR( (3,2), r), white, black);
filldraw(CR( (0,3), r), white, black);
filldraw(CR( (2,3), r), white, black);
filldraw(CR( (4,3), r), white, black);
filldraw(CR( (1,4), r), white, black);
filldraw(CR( (3,4), r), white, black);
real s = 3*r;
fill(CR( (0,0), s), opacity(0.2)+lightred );
fill(CR( (1,0), s), opacity(0.2)+lightred );
fill(CR( (1,1), s), opacity(0.2)+lightred );
fill(CR( (2,1), s), opacity(0.2)+lightred );
fill(CR( (2,2), s), opacity(0.2)+lightred );
fill(CR( (3,2), s), opacity(0.2)+lightred );
pen domino = heavygreen+1.2;
path vd = box( (-0.4, -0.4), (0.4, 1.4) );
path hd = rotate(-90)*vd;
draw(shift(0,0)*vd, domino);
draw(shift(1,1)*vd, domino);
draw(shift(2,2)*vd, domino);
draw(shift(1,0)*hd, domino);
draw(shift(2,1)*hd, domino);
draw(shift(3,2)*hd, domino);
\end{asy}
\end{center}
The curve $\gamma$ by definition should cross $y=x-1$ at the point $b = (1,0)$.
Let $a$ denote the first point of this sequence
after $p_1$ for which $\gamma$ \emph{crosses} $y=x-1$ again.
Now $a$ is tiled by a vertical domino whose black point is to the right of $\ell$.
But the line segment $\ell$ cuts $\mathcal Q$ into two parts,
and the orientation of $\gamma$ has this path also entering from the right.
This contradicts the fact that the orientation of $\gamma$
points only from white to black within dominoes.
This contradiction completes the proof.
\begin{remark*}
Note the problem is false if you allow holes
(consider a $3 \times 3$ with the middle square deleted).
\end{remark*}
|
USAMO-2009-notes_4
|
For $n \ge 2$, let $a_1$, $a_2$, \dots, $a_n$ be positive real numbers such that
\[
\left( a_1 + a_2 + \dots + a_n \right)
\left( \frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_n} \right)
\le \left( n + \half \right)^2.
\]
Prove that $\max\left( a_1, \dots, a_n \right) \le 4 \min\left( a_1, \dots, a_n \right)$.
|
Assume $a_1$ is the largest and $a_2$ is the smallest.
Let $M = a_1/a_2$.
We wish to show $M \le 4$.
In left-hand side of given, write as $a_2+a_1 + \dots + a_n$.
By Cauchy Schwarz, one obtains
\begin{align*}
\left( n+\half \right)^2 &\ge
\left( a_2 + a_1 + a_3 + \dots + a_n \right)
\left( \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3}
+ \dots + \frac{1}{a_n} \right) \\
&\ge \left(\sqrt{\frac{a_2}{a_1}} + \sqrt{\frac{a_1}{a_2}}
+ 1 + \dots + 1 \right)^2 \\
&\ge \left(\sqrt{1/M} + \sqrt{M} + (n-2) \right)^2.
\end{align*}
Expanding and solving for $M$ gives $1/4 \le M \le 4$ as needed.
|
USAMO-2009-notes_5
|
Trapezoid $ABCD$, with $\ol{AB} \parallel \ol{CD}$,
is inscribed in circle $\omega$ and point $G$ lies inside triangle $BCD$.
Rays $AG$ and $BG$ meet $\omega$ again at points $P$ and $Q$, respectively.
Let the line through $G$ parallel to $\ol{AB}$ intersect
$\ol{BD}$ and $\ol{BC}$ at points $R$ and $S$, respectively.
Prove that quadrilateral $PQRS$ is cyclic
if and only if $\ol{BG}$ bisects $\angle CBD$.
|
Perform an inversion around $B$ with arbitrary radius,
and denote the inverse of a point $Z$ with $Z^\ast$.
\begin{center}
\begin{asy}
unitsize(2cm);
pair A = Drawing("A", dir(130), dir(130));
pair B = Drawing("B", dir(50), dir(50));
pair C = Drawing("C", dir(-20), dir(-20));
pair D = Drawing("D", dir(200), dir(200));
pair Q = Drawing("Q", dir(270), dir(270));
pair G = Drawing("G", 0.3*Q+0.7*B, 1.414*dir(290));
pair R = Drawing("R", extension(B,D,G,G-(1,0)), dir(135));
pair S = Drawing("S", extension(B,C,G,G+(1,0)), dir(45));
pair P = Drawing("P", 2*foot(origin,A,G)-A, dir(G-A));
draw(unitcircle);
draw(D--A--B--C--D--B--Q--P--A);
draw(R--S);
draw(circumcircle(P,Q,R), dashed);
\end{asy}
\quad
\begin{asy}
unitsize(2cm);
pair B = Drawing("B", dir(130), dir(110));
pair R = Drawing("R^\ast", dir(210), dir(210));
pair S = Drawing("S^\ast", dir(330), dir(330));
pair G = Drawing("G^\ast", dir(270), dir(290));
pair D = Drawing("D^\ast", .6*R+.4*B, 1.414*dir(230));
pair C = Drawing("C^\ast", .6*S+.4*B, 1.414*dir(-90));
pair Q = Drawing("Q^\ast", extension(B,G,C,D), dir(45));
pair A = Drawing("A^\ast", extension(B,2/(B+conj(B)),C,D), dir(180));
pair P = Drawing("P^\ast", 2*foot(circumcenter(A,B,G),C,D)-A,dir(-25));
draw(unitcircle);
draw(circumcircle(B,C,D));
draw(A--P);
draw(R--B--S--cycle);
draw(G--B);
draw(circumcircle(A,B,G));
draw(A--B);
/*
pair X = Drawing("X", 2*foot(origin,P,G)-G, dir(P-G));
draw(G--X, dashed);
*/
\end{asy}
\end{center}
After inversion, we obtain a cyclic quadrilateral
$BS^\ast G^\ast R^\ast$ and points $C^\ast$, $D^\ast$
on $\ol{BS^\ast}$, $\ol{BR^\ast}$,
such that $(BC^\ast D^\ast)$ is tangent to $(BS^\ast G^\ast R^\ast)$ ---
in other words, so that $\ol{C^\ast D^\ast}$
is parallel to $\ol{S^\ast R^\ast}$.
Point $A^\ast$ lies on line $\ol{C^\ast D^\ast}$
so that $\ol{A^\ast B}$ is tangent to $(BS^\ast G^\ast R^\ast)$.
Points $P^\ast$ and $Q^\ast$ are the
intersections of $(A^\ast BG^\ast)$ and $\ol{BG^\ast}$
with line $C^\ast D^\ast$.
Observe that $P^\ast Q^\ast R^\ast S^\ast$ is a trapezoid,
so it is cyclic if and only if it isosceles.
Let $X$ be the second intersection of line $G^\ast P^\ast$
with $(B S^\ast R^\ast)$. Because
\[ \dang Q^\ast P^\ast G^\ast
= \dang A^\ast B G^\ast = \dang BXG^\ast \]
we find that $BXS^\ast R^\ast$ is an isosceles trapezoid.
If $G^\ast$ is indeed the midpoint of the arc
then everything is clear by symmetry now.
Conversely, if $P^\ast R^\ast = Q^\ast S^\ast$ then
that means $P^\ast Q^\ast R^\ast S^\ast$ is a cyclic trapezoid,
and hence that the perpendicular bisectors of
$\ol{P^\ast Q^\ast}$ and $\ol{R^\ast S^\ast}$ are the same.
Hence $B$, $X$, $P^\ast$, $Q^\ast$ are symmetric around this line.
This forces $G^\ast$ to be the midpoint of arc $R^\ast S^\ast$ as desired.
|
USAMO-2009-notes_6
|
Let $s_1, s_2, s_3, \dots$ be an infinite,
nonconstant sequence of rational numbers,
meaning it is not the case that $s_1 = s_2 = s_3 = \dots$.
Suppose that $t_1, t_2, t_3, \dots$ is also an infinite,
nonconstant sequence of rational numbers
with the property that $(s_i - s_j)(t_i - t_j)$
is an integer for all $i$ and $j$.
Prove that there exists a rational number $r$
such that $(s_i - s_j)r$ and $(t_i - t_j)/r$
are integers for all $i$ and $j$.
\end{enumerate}
|
First we eliminate the silly edge case:
\begin{claim*}
For some $i$ and $j$, we have
$(s_i - s_j)(t_i - t_j) \neq 0$.
\end{claim*}
\begin{proof}
Assume not.
WLOG $s_1 \neq s_2$, then $t_1 = t_2$.
Now select $i$ such that $t_i \neq t_1 = t_2$.
Then either $t_i - s_1 \neq 0$
or $t_i - s_2 \neq 0$, contradiction.
\end{proof}
So, WLOG (by permutation) that $n = (s_1-s_2)(t_1-t_2) \neq 0$.
By shifting and scaling appropriately,
we may assume
\[ s_1 = t_1 = 0, \quad s_2 = 1, \quad t_2 = n. \]
Thus we deduce \[ s_i t_i \in \ZZ, \quad s_i t_j + s_j t_i \in \ZZ
\qquad \forall i,j. \]
\begin{claim*}
For any index $i$,
$t_i \in \ZZ$, $s_i \in \frac 1n \ZZ$.
\end{claim*}
\begin{proof}
We have $s_i t_i \in \ZZ$
and $t_i + n s_i \in \ZZ$ by problem condition.
By looking at $\nu_p$ of this,
we conclude $ns_i, t_i \in \ZZ$
(since if either as negative $p$-adic value,
so does the other, and then $s_i t_i \notin \ZZ$).
\end{proof}
Last claim:
\begin{claim*}
If $d = \gcd t_\bullet$, then $ds_i \in \ZZ$ for all $i$.
\end{claim*}
\begin{proof}
Consider a prime $p \mid n$, and let $e = \nu_p(t_j)$.
We will show $\nu_p(s_i) \ge -e$ for any $i$.
This is already true for $i = j$,
so assume $i \neq j$.
Assume for contradiction $\nu_p(s_i) < -e$.
Then $\nu_p(t_i) > e = \nu_p(t_k)$.
Since $\nu_p(s_k) \ge -e$ we deduce
$\nu_p(s_i t_k) < \nu_p(s_k t_i)$;
so $\nu_p(s_i t_k) \ge 0$ and $\nu_p(s_i) \ge -e$ as desired.
\end{proof}
|
USAMO-2010-notes_1
|
Let $AXYZB$ be a convex pentagon inscribed in a semicircle of diameter $AB$.
Denote by $P$, $Q$, $R$, $S$ the feet of the perpendiculars
from $Y$ onto lines $AX$, $BX$, $AZ$, $BZ$, respectively.
Prove that the acute angle formed by lines $PQ$ and $RS$ is half the size of $\angle XOZ$,
where $O$ is the midpoint of segment $AB$.
|
We present two possible approaches.
The first approach is just ``bare-hands'' angle chasing.
The second approach requires more insight but makes it clearer what is going on;
it shows the intersection point of lines $PQ$ and $RS$ is the foot
from the altitude from $Y$ to $AB$ using Simson lines.
The second approach also has the advantage that it works even if $\ol{AB}$
is not a diameter of the circle.
\paragraph{First approach using angle chasing.}
Define $T = \ol{PQ} \cap \ol{RS}$.
Also, let $2\alpha$, $2\beta$, $2\gamma$, $2\delta$ denote the measures of arcs
$\arc{AX}$, $\arc{XY}$, $\arc{YZ}$, $\arc{ZB}$, respectively,
so that $\alpha + \beta + \gamma + \delta = 90\dg$.
\begin{center}
\begin{asy}
pair A = dir(180);
pair B = dir(0);
pair X = dir(140);
pair Y = dir(100);
pair Z = dir(60);
pair P = foot(Y, A, X);
pair Q = foot(Y, B, X);
pair R = foot(Y, A, Z);
pair S = foot(Y, B, Z);
pair T = extension(P, Q, R, S);
draw(A--X--Y--Z--B, orange);
filldraw(unitcircle, opacity(0.1)+orange, red);
draw(Y--P, red);
draw(Y--Q, red);
draw(Y--R, red);
draw(Y--S, red);
draw(P--X--B, brown);
draw(P--T--S, red);
draw(S--Z--A, brown);
pair O = origin;
draw(arc(O, B, Z), blue+1.3, Margin(2,2));
draw(arc(O, Z, Y), blue+1.3, Margin(2,2));
draw(arc(O, Y, X), blue+1.3, Margin(2,2));
draw(arc(O, X, A), blue+1.3, Margin(2,2));
label("$2\alpha$", dir(160), dir(160), blue);
label("$2\beta$", dir(122), dir(122), blue);
label("$2\gamma$", dir( 82), dir( 82), blue);
label("$2\delta$", dir( 30), dir( 30), blue);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
dot("$Z$", Z, dir(Z));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(210));
dot("$R$", R, dir(340));
dot("$S$", S, dir(S));
dot("$T$", T, dir(-90));
/* TSQ Source:
A = dir 180
B = dir 0
X = dir 140
Y = dir 100
Z = dir 60
P = foot Y A X
Q = foot Y B X
R = foot Y A Z
S = foot Y B Z
T = extension P Q R S
A--X--Y--Z--B orange
unitcircle 0.1 orange / red
Y--P red
Y--Q red
Y--R red
Y--S red
P--X--B brown
P--T--S red
S--Z--A brown
*/
\end{asy}
\end{center}
We now compute the following angles:
\begin{align*}
\angle SRY &= \angle SZY = 90\dg - \angle YZA = 90\dg - (\alpha + \beta) \\
\angle YQP &= \angle YXP = 90\dg - \angle BXY = 90\dg - (\gamma + \delta) \\
\angle QYR &= 180\dg - \angle (\ol{ZR}, \ol{QX})
= 180\dg - \frac{2\beta + 2\gamma + 180\dg}{2} = 90\dg - (\beta + \gamma).
\end{align*}
Hence, we can then compute
\begin{align*}
\angle RTQ &= 360\dg - \left(
\angle QYR + (180\dg - \angle SRY) + (180\dg - \angle YQP) \right) \\
&= \angle SRY + \angle YQP - \angle QYR \\
&= \left( 90\dg - (\alpha + \beta) \right) + \left( 90\dg - (\gamma + \delta) \right)
- \left( 90\dg - (\beta + \gamma) \right) \\
&= 90\dg - (\alpha + \delta) \\
&= \beta + \gamma.
\end{align*}
Since $\angle XOZ = \frac{2\beta+2\gamma}{2} = \beta+\gamma$, the proof is complete.
\paragraph{Second approach using Simson lines, ignoring the diameter condition.}
In this solution, we will ignore the condition that $\ol{AB}$ is a diameter;
the solution works equally well without it,
as long as $O$ is redefined as the center of $(AXYZB)$ instead.
We will again show the angle formed by lines $PQ$ and $RS$ is half the measure of $\arc{XZ}$.
Unlike the previous solution, we instead define $T$ to be the foot from $Y$ to $\ol{AB}$.
Then the Simson line of $Y$ with respect to $\triangle XAB$ passes through $P$, $Q$, $T$.
Similarly, the Simson line of $Y$ with respect to $\triangle ZAB$ passes through $R$, $S$, $T$.
Therefore, point $T$ coincides with $\ol{PQ} \cap \ol{RS}$.
\begin{center}
\begin{asy}
pair A = dir(205);
pair B = -conj(A);
pair X = dir(140);
pair Y = dir(100);
pair Z = dir(60);
pair P = foot(Y, A, X);
pair Q = foot(Y, B, X);
pair R = foot(Y, A, Z);
pair S = foot(Y, B, Z);
pair T = extension(P, Q, R, S);
draw(A--X--Y--Z--B, orange);
filldraw(unitcircle, opacity(0.1)+orange, red);
draw(Y--P, red);
draw(Y--Q, red);
draw(Y--R, red);
draw(Y--S, red);
draw(P--X--B, brown);
draw(P--T--S, red);
draw(S--Z--A, brown);
draw(A--B, deepcyan+1.4);
draw(Y--T, deepcyan+1.4);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
dot("$Z$", Z, dir(Z));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(210));
dot("$R$", R, dir(340));
dot("$S$", S, dir(S));
dot("$T$", T, dir(-90));
/* TSQ Source:
A = dir 180
B = dir 0
X = dir 140
Y = dir 100
Z = dir 60
P = foot Y A X
Q = foot Y B X
R = foot Y A Z
S = foot Y B Z
T = extension P Q R S
A--X--Y--Z--B orange
unitcircle 0.1 orange / red
Y--P red
Y--Q red
Y--R red
Y--S red
P--X--B brown
P--T--S red
S--Z--A brown
*/
\end{asy}
\end{center}
Now it's straightforward to see $APYRT$ is cyclic (in the circle with diameter $\ol{AY}$),
and therefore
\[ \angle RTY = \angle RAY = \angle ZAY. \]
Similarly,
\[ \angle YTQ = \angle YBQ = \angle YBX. \]
Summing these gives $\angle RTQ$ is equal to half the measure of arc $\arc{XZ}$ as needed.
|
USAMO-2010-notes_2
|
There are $n$ students standing in a circle, one behind the other.
The students have heights $h_1<h_2<\dots <h_n$.
If a student with height $h_k$ is standing directly
behind a student with height $h_{k-2}$ or less,
the two students are permitted to switch places.
Prove that it is not possible to make more than
$\binom{n}{3}$ such switches
before reaching a position in which no
further switches are possible.
|
The main claim is the following observation,
which is most motivated in the situation $j-i = 2$.
\begin{claim*}
The students with heights $h_i$ and $h_j$
switch at most $|j-i|-1$ times.
\end{claim*}
\begin{proof}
By induction on $d = |j-i|$, assuming $j > i$.
For $d = 1$ there is nothing to prove.
For $d \ge 2$, look at only students $h_j$, $h_{i+1}$
and $h_{i}$ ignoring all other students.
After $h_j$ and $h_i$ switch the first time,
the relative ordering of the students must be
$h_i \to h_j \to h_{i+1}$.
Thereafter $h_j$ must always switch with $h_{i+1}$
before switching with $h_i$,
so the inductive hypothesis applies to give the bound
$1 + j - (i+1) - 1 = j-i - 1$.
\end{proof}
Hence, the number of switches is at most
\[ \sum_{1 \le i < j \le n} \left( |j-i| - 1 \right) = \binom n3. \]
|
USAMO-2010-notes_3
|
The $2010$ positive real numbers $a_1$, $a_2$, \dots , $a_{2010}$
satisfy the inequality $a_i a_j \le i+j$
for all $1 \le i < j \le 2010$.
Determine, with proof, the largest possible value
of the product $a_1a_2\dots a_{2010}$.
|
The answer is $3 \times 7 \times 11 \times \dots \times 4019$,
which is clearly an upper bound
(and it's not too hard to show this is the lowest
number we may obtain by multiplying $1005$ equalities together;
this is essentially the rearrangement inequality).
The tricky part is the construction.
Intuitively we want $a_i \approx \sqrt{2i}$,
but the details require significant care.
Note that if this is achievable,
we will require $a_n a_{n+1} = 2n+1$ for all odd $n$.
Here are two constructions:
\begin{itemize}
\ii One can take the sequence such that $a_{2008} a_{2010} = 4018$
and $a_n a_{n+1} = 2n+1$ for all $n = 1, 2, \dots, 2009$.
This can be shown to work by some calculation. As an illustrative example,
\[ a_1 a_4 = \frac{a_1a_2 \cdot a_3a_4}{a_2a_3} = \frac{3 \cdot 7}{5} < 5. \]
\ii In fact one can also take $a_n = \sqrt{2n}$ for all even $n$
(and hence $a_{n-1} = \sqrt{2n} - \frac{1}{\sqrt{2n}}$ for such even $n$).
\end{itemize}
\begin{remark*}
This is a chief example of an ``abstract'' restriction-based approach.
One can motivate it in three steps:
\begin{itemize}
\ii The bound $3 \cdot 7 \cdot \dots \cdot 4019$ is provably best possible
upper bound by pairing the inequalities;
also the situation with $2010$ replaced by $4$
is constructible with bound $21$.
\ii We have $a_n \approx \sqrt{2n}$ heuristically;
in fact $a_n = \sqrt{2n}$ satisfies inequalities by AM-GM.
\ii So we are most worried about $a_i a_j \le i+j$
when $|i-j|$ is small, like $|i-j| = 1$.
\end{itemize}
I then proceeded to spend five hours on various constructions,
but it turns out that the right thing to do was just require
$a_k a_{k+1} = 2k+1$, to make sure these pass:
and the problem almost solves itself.
\end{remark*}
\begin{remark*}
When $2010$ is replaced by $4$
it is not too hard to manually write an explicit example:
say $a_1 = \frac{\sqrt3}{1.1}$, $a_2 = 1.1\sqrt3$,
$a_3 = \frac{\sqrt7}{1.1}$ and $a_4 = 1.1\sqrt7$.
So this is a reason one might guess that
$3 \times 7 \times \dots \times 4019$
can actually be achieved in the large case.
\end{remark*}
\begin{remark*}
Victor Wang says:
I believe we can actually prove that WLOG (!) assume $a_i a_{i+1} = 2i+1$
for all $i$ (but there are other ways to motivate that as well,
like linear programming after taking logs),
which makes things a bit simpler to think about.
\end{remark*}
|
USAMO-2010-notes_4
|
Let $ABC$ be a triangle with $\angle A = 90\dg$.
Points $D$ and $E$ lie on sides $AC$ and $AB$, respectively,
such that $\angle ABD = \angle DBC$ and $\angle ACE = \angle ECB$.
Segments $BD$ and $CE$ meet at $I$.
Determine whether or not it is possible for segments $AB$, $AC$, $BI$, $ID$, $CI$, $IE$
to all have integer lengths.
|
The answer is no.
We prove that it is not even possible that $AB$, $AC$, $CI$, $IB$ are all integers.
\begin{center}
\begin{asy}
size(5cm);
pair B = dir(140);
pair A = conj(B);
pair C = -B;
filldraw(A--B--C--cycle, opacity(0.1)+lightblue, blue);
pair I = incenter(A, B, C);
pair D = extension(B, I, A, C);
pair E = extension(C, I, A, B);
draw(B--D);
draw(C--E);
filldraw(incircle(A,B,C), opacity(0.1)+lightred, red+dotted);
dot("$B$", B, dir(B));
dot("$A$", A, dir(A));
dot("$C$", C, dir(C));
dot("$I$", I, dir(45));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
/* Source generated by TSQ */
\end{asy}
\end{center}
First, we claim that $\angle BIC = 135\dg$.
To see why, note that
\[ \angle IBC + \angle ICB = \frac{\angle B}{2} + \frac{\angle C}{2}
= \frac{90\dg}{2} = 45\dg. \]
So, $\angle BIC = 180\dg - (\angle IBC + \angle ICB) = 135\dg$, as desired.
We now proceed by contradiction.
The Pythagorean theorem implies
\[ BC^2 = AB^2 + AC^2 \]
and so $BC^2$ is an integer.
However, the law of cosines gives
\begin{align*}
BC^2 &= BI^2 + CI^2 - 2 BI \cdot CI \cos \angle BIC \\
&= BI^2 + CI^2 + BI \cdot CI \cdot \sqrt 2.
\end{align*}
which is irrational, and this produces the desired contradiction.
|
USAMO-2010-notes_5
|
Let $q = \frac{3p-5}{2}$ where $p$ is an odd prime, and let
\[ S_q = \frac{1}{2\cdot 3 \cdot 4} + \frac{1}{5\cdot 6 \cdot 7}
+ \dots + \frac{1}{q(q+1)(q+2)}. \]
Prove that if $\frac{1}{p}-2S_q = \frac{m}{n}$ for integers $m$ and $n$,
then $m - n$ is divisible by $p$.
|
By partial fractions, we have
\[
\frac{2}{(3k-1)(3k)(3k+1)}
= \frac{1}{3k-1} - \frac{2}{3k} + \frac{1}{3k+1}.
\]
Thus
\begin{align*}
2S_q
&= \left( \frac12 - \frac23 + \frac14 \right)
+ \left( \frac 15 - \frac26 + \frac17 \right)
+ \dots
+ \left( \frac1q - \frac2{q+1} + \frac{1}{q+2} \right) \\
&= \left( \half + \frac13 + \frac14 + \dots + \frac{1}{q+2} \right)
- 3\left( \frac13 + \frac 16 + \dots + \frac{1}{q+1} \right) \\
&= \left( \half + \frac13 + \frac14 + \dots + \frac{1}{q+2} \right)
- \left( \frac11 + \frac 12 + \dots + \frac{1}{\frac{q+1}{3}} \right) \\
\implies 2S_q - \frac1p + 1
&= \left( \frac11 + \frac12 + \dots + \frac{1}{p-1} \right)
+ \left( \frac{1}{p+1} + \frac{1}{p+2} \dots + \frac{1}{q+2} \right)
- \left( \frac11 + \frac 12 + \dots + \frac{1}{\frac{q+1}{3}} \right)
\end{align*}
Now we are ready to take modulo $p$.
The given says that $q-p+2 = \frac{q+1}{3}$, so
\begin{align*}
2S_q - \frac1p + 1
&= \left( \frac11 + \frac12 + \dots + \frac{1}{p-1} \right)
+ \left( \frac{1}{p+1} + \frac{1}{p+2} + \dots + \frac{1}{q+2} \right)
- \left( \frac11 + \frac 12 + \dots + \frac{1}{\frac{q+1}{3}} \right) \\
&\equiv \left( \frac11 + \frac12 + \dots + \frac{1}{p-1} \right)
+ \left( \frac11 + \frac12 + \dots + \frac{1}{q-p+2} \right)
- \left( \frac11 + \frac 12 + \dots + \frac{1}{\frac{q+1}{3}} \right) \\
&= \frac11 + \frac12+ \dots + \frac{1}{p-1} \\
&\equiv 0 \pmod p.
\end{align*}
So $\frac1p - 2S_q \equiv 1 \pmod p$ which is the desired.
|
USAMO-2011-notes_1
|
Let $a$, $b$, $c$ be positive real numbers
such that $a^2+b^2+c^2+(a+b+c)^2 \le 4$. Prove that
\[ \frac{ab+1}{(a+b)^2} + \frac{bc+1}{(b+c)^2} + \frac{ca+1}{(c+a)^2} \ge 3. \]
|
The condition becomes $2 \ge a^2+b^2+c^2 + ab+bc+ca$.
Therefore,
\begin{align*}
\sum_{\text{cyc}} \frac{2ab+2}{(a+b)^2}
&\ge \sum_{\text{cyc}} \frac{2ab+(a^2+b^2+c^2+ab+bc+ca)}{(a+b)^2} \\
&= \sum_{\text{cyc}} \frac{(a+b)^2+(c+a)(c+b)}{(a+b)^2} \\
&= 3 + \sum_{\text{cyc}} \frac{(c+a)(c+b)}{(a+b)^2} \\
&\ge 3 + 3\sqrt[3]{\prod_{\text{cyc}} \frac{(c+a)(c+b)}{(a+b)^2}}
= 3 + 3 = 6
\end{align*}
with the last line by AM-GM.
This completes the proof.
|
USAMO-2011-notes_2
|
An integer is assigned to each vertex of a regular pentagon
so that the sum of the five integers is $2011$.
A turn of a solitaire game consists of subtracting an integer $m$
(not necessarily positive) from each of the integers at two neighboring vertices
and adding $2m$ to the opposite vertex, which is not adjacent
to either of the first two vertices.
(The amount $m$ and the vertices chosen can vary from turn to turn.)
The game is won at a certain vertex if, after some number of turns,
that vertex has the number $2011$ and the other four vertices have the number $0$.
Prove that for any choice of the initial integers,
there is exactly one vertex at which the game can be won.
|
Call the vertices $0$, $1$, $2$, $3$, $4$ in order.
First, notice that the quantity
\[ S \coloneq N_1 + 2N_2 + 3N_3 + 4N_4 \pmod 5 \]
is invariant, where $N_i$ is the amount at vertex $i$.
This immediately implies that at most one vertex can win,
since in a winning situation all $N_i$ are $0$ except for one, which is $2011$.
(For example, if $S \equiv 3 \pmod 5$, any victory must occur at the
vertex $3$, via $N_3 = 2011$, $N_0 = N_1 = N_2 = N_4 = 0$.)
Now we prove we can win on this unique vertex.
Let $a_i$, $x_i$ denote the number initially at $i$
and $x_i$ denote $\sum m$ over all $m$ where vertex $i$ gains $2m$.
WLOG the possible vertex is $0$,
meaning $a_1 + 2a_2 + 3a_3 + 4a_4 \equiv 0 \pmod 5$.
Moreover we want
\begin{align*}
2011 &= a_0 + 2x_0 - x_2 - x_3 \\
0 &= a_1 + 2x_1 - x_3 - x_4 \\
0 &= a_2 + 2x_2 - x_4 - x_0 \\
0 &= a_3 + 2x_3 - x_0 - x_1 \\
0 &= a_4 + 2x_4 - x_1 - x_2.
\end{align*}
We can ignore the first equation since its the sum of the other four,
and we can WLOG shift $x_0 \to 0$ by shifting each $x_i$ by a fixed amount.
We will now solve the resulting system of equations.
First, we have \[ x_4 = 2x_2 + a_2 \text{ and } x_1 = 2x_3 + a_3. \]
Using these to remove all instances of $x_1$ and $x_4$ gives
\[ 2x_2-3x_3 = 2a_3+a_1-a_2 \text{ and } 2x_3-3x_2 = 2a_2+a_4-a_3 \]
whence we have a two-variable system of equations!
To verify its solution is integral, note that
\[ x_2-x_3 = \frac{a_1 - 3a_2 + 3a_3 - a_4}{5} \]
is an integer, since
\[ a_1 - 3a_2 + 3a_3 - a_4 \equiv a_1 + 2a_2 + 3a_3 + 4a_4 \equiv 0 \pmod 5. \]
Abbreviating $\frac{a_1 - 3a_2 + 3a_3 - a_4}{5}$ as $k$, we obtain the desired $x_i$:
\begin{align*}
x_2 &= 2a_3+a_1-a_2 + 2k \\
x_3 &= x_2 + k \\
x_1 &= 2x_3 + a_3 \\
x_4 &= 2x_2 + a_2 \\
x_0 &= 0.
\end{align*}
This is the desired integer solution.
\begin{remark*}
In principle, you could unwind all the definitions above to explicitly
write every $x_i$ as a function of $a_1$, $a_2$, $a_3$, $a_4$.
If you did this, you could get the long equations
\begin{align*}
x_0 &= 0 \\
x_1 &= -\tfrac15(6a_1 + 2a_2 + 3a_3 + 4a_4) \\
x_2 &= -\tfrac15(2a_1 + 4a_2 + a_3 + 3a_4) \\
x_3 &= -\tfrac15(3a_1 + a_2 + 4a_3 + 2a_4) \\
x_4 &= -\tfrac15(4a_1 + 3a_2 + 2a_3 + 6a_4)
\end{align*}
which indeed are all integers whenever $a_1 + 2a_2 + 3a_3 + 4a_4 \equiv 0 \pmod 5$.
However, this is quite tedious and also unnecessary to solve the problem.
That's because we only care that the $x_i$ are integers,
and do not need to actually know the values.
This lets us work more indirectly to avoid long calculation, as we did above.
\end{remark*}
|
USAMO-2011-notes_3
|
In hexagon $ABCDEF$, which is nonconvex but not self-intersecting,
no pair of opposite sides are parallel.
The internal angles satisfy
$\angle A=3\angle D$, $\angle C=3\angle F$, and $\angle E=3\angle B$.
Furthermore $AB=DE$, $BC=EF$, and $CD=FA$.
Prove that diagonals $\ol{AD}$, $\ol{BE}$, and $\ol{CF}$
are concurrent.
|
We present the official solution.
We say a hexagon is \emph{satisfying}
if it obeys the six conditions;
note that intuitively we expect three degrees of freedom
for satisfying hexagons.
Main idea:
\begin{claim*}
In a satisfying hexagon,
$B$, $D$, $F$ are reflections of $A$, $C$, $E$ across
the sides of $\triangle ACE$.
\end{claim*}
(This claim looks plausible because every excellent hexagon is satisfying,
and both configuration spaces are three-dimensional.)
Call a hexagon of this shape ``excellent'';
in a excellent hexagon the diagonals clearly concur (at the orthocenter).
Set $\beta = \angle B$, $\delta = \angle D$, $\varphi = \angle F$.
Now given a satisfying hexagon $ABCDEF$,
construct a ``phantom hexagon'' $A'B'C'D'E'F'$
with the same angles which is excellent (see figure).
This is possible since $\beta + \delta + \varphi = 180\dg$.
\begin{center}
\begin{asy}
size(12cm);
void MA(string s, pair A, pair O, pair B) {
markangle(11.0, A,O,B);
label("$"+s+"$", O, 4*dir((pair)incenter(A,O,B)-O));
}
pair A = dir(110);
pair C = dir(205);
pair E = dir(335);
pair B = -E+2*foot(E, A, C);
pair D = -A+2*foot(A, C, E);
pair F = -C+2*foot(C, E, A);
filldraw(A--B--C--D--E--F--cycle, opacity(0.1)+lightgreen, heavygreen);
draw(A--C--E--cycle, green);
draw(A--D, dashed+green);
draw(B--E, dashed+green);
draw(C--F, dashed+green);
MA("\beta", C,B,A);
MA("\delta", E,D,C);
MA("\varphi", A,F,E);
transform t = shift(-4,0);
add(t*CC());
// Again
pair Ap = A;
pair Cp = C;
pair Ep = E;
pair Bp = B;
pair Dp = D;
pair Fp = F;
filldraw(A--B--C--D--E--F--cycle, opacity(0.1)+lightcyan, heavycyan);
draw(A--C--E--cycle, heavycyan);
draw(A--D, dashed+heavycyan);
draw(B--E, dashed+heavycyan);
draw(C--F, dashed+heavycyan);
MA("\beta", C,B,A);
MA("\delta", E,D,C);
MA("\varphi", A,F,E);
MA("\beta", A,E,C);
MA("\varphi", E,C,A);
MA("\delta", C,A,E);
pair A = t*A;
pair B = t*B;
pair C = t*C;
pair D = t*D;
pair E = t*E;
pair F = t*F;
dot("$A'$", Ap, dir(Ap));
dot("$C'$", Cp, dir(Cp));
dot("$E'$", Ep, dir(Ep));
dot("$B'$", Bp, dir(Bp));
dot("$D'$", Dp, dir(Dp));
dot("$F'$", Fp, dir(Fp));
dot("$A$", A, dir(120));
dot("$B$", B, dir(135));
dot("$C$", C, dir(225));
dot("$D$", D, dir(270));
dot("$E$", E, dir(350));
dot("$F$", F, dir(45));
/* TSQ Source:
!size(12cm);
A := dir 110
C := dir 205
E := dir 335
B := -E+2*foot E A C
D := -A+2*foot A C E
F := -C+2*foot C E A
A--B--C--D--E--F--cycle 0.1 lightgreen / heavygreen
A--C--E--cycle green
A--D dashed green
B--E dashed green
C--F dashed green
!MA("\beta", C,B,A, 0.2);
!MA("\delta", E,D,C, 0.2);
!MA("\varphi", A,F,E, 0.2);
! transform t = shift(-4,0);
!add(t*CC());
// Again
A' = A
C' = C
E' = E
B' = B
D' = D
F' = F
A--B--C--D--E--F--cycle 0.1 lightcyan / heavycyan
A--C--E--cycle heavycyan
A--D dashed heavycyan
B--E dashed heavycyan
C--F dashed heavycyan
! anglepen = heavycyan;
!MA("\beta", C,B,A, 0.2);
!MA("\delta", E,D,C, 0.2);
!MA("\varphi", A,F,E, 0.2);
!MA("\beta", A,E,C, 0.2);
!MA("\varphi", E,C,A, 0.2);
!MA("\delta", C,A,E, 0.2);
A = t*A R120
B = t*B R135
C = t*C R225
D = t*D R270
E = t*E R350
F = t*F R45
*/
\end{asy}
\end{center}
Then it would suffice to prove that:
\begin{lemma*}
A satisfying hexagon is uniquely determined by its angles
up to similarity.
That is, at most one hexagon
(up to similarity) has angles $\beta$, $\delta$, $\gamma$ as above.
\end{lemma*}
\begin{proof}
Consider any two satisfying hexagons $ABCDEF$ and $A'B'C'D'E'F'$
(not necessarily as constructed above!) with the same angles.
We show they are similar.
To do this, consider the unit complex numbers
in the directions $\overrightarrow{BA}$ and $\overrightarrow{DE}$
respectively and let $\vec x$ denote their sum.
Define $\vec y$, $\vec z$ similarly.
Note that the condition $\ol{AB} \nparallel \ol{DE}$
implies $\vec x \neq 0$, and similarly.
Then we have the identities
\[ AB \cdot \vec x + CD \cdot \vec y + EF \cdot \vec z
= A'B' \cdot \vec x + C'D' \cdot \vec y + E'F' \cdot \vec z
= 0. \]
So we would obtain $AB : CD : EF = A'B' : C'D' : E'F'$
if only we could show that $\vec x$, $\vec y$, $\vec z$
are not multiples of each other
(linear dependency reasons).
This is a tiresome computation with arguments,
but here it is.
First note that none of $\beta$, $\delta$, $\varphi$
can be $90\dg$,
since otherwise we get a pair of parallel sides.
Now work in the complex plane,
fix a reference such that $\vec A - \vec B$ has argument $0$,
and assume $ABCDEF$ are labelled counterclockwise.
Then
\begin{itemize}
\ii $\vec B - \vec C$ has argument $\pi-\beta$
\ii $\vec C - \vec D$ has argument $-(\beta+3\varphi)$
\ii $\vec D - \vec E$ has argument $\pi-(\beta+3\varphi+\delta)$
\ii $\vec E - \vec F$ has argument $-(4\beta+3\varphi+\delta)$
% \ii $\vec F - \vec A$ has argument $\pi-(4\beta+4\varphi+\delta)$
\end{itemize}
So the argument of $\vec x$ has argument
$\frac{\pi-(\beta+3\varphi+\delta)}{2} \pmod \pi$.
The argument of $\vec y$ has argument
$\frac{\pi-(5\beta+3\varphi+\delta)}{2} \pmod \pi$.
Their difference is $2\beta \pmod \pi$, and since $\beta \neq 90\dg$,
it follows that $\vec x$ and $\vec y$ are not
multiples of each other; the other cases are similar.
\end{proof}
Then the lemma implies $ABCDEF \sim A'B'C'D'E'F$ and we're done.
\begin{remark*}
This problem turned out to be known already.
It appears in this reference:
\begin{quote}
Nikolai Beluhov, \emph{Matematika}, 2008, issue 6, problem 3.
\end{quote}
It was reprinted as Kvant, 2009, issue 2, problem M2130;
the reprint is available at
\url{http://kvant.ras.ru/pdf/2009/2009-02.pdf}.
\end{remark*}
\begin{remark*}
The vector perspective also shows the condition about parallel
sides cannot be dropped.
Here is a counterexample from Ryan Kim
in the event that it is.
\begin{center}
\begin{asy}
pair A = dir(140);
pair C_0 = conj(A);
pair E = -A;
pair B_0 = 2*C_0-E;
pair D_0 = 2*C_0-A;
pair F = extension(A, E*E/C_0, E, A*A/C_0);
pair C = C_0+2*C_0;
pair B = B_0+2*C_0;
pair D = D_0+2*C_0;
filldraw(A--B--C--D--E--F--cycle, opacity(0.1)+lightcyan, heavycyan);
draw(A--D, heavygreen);
draw(B--E, heavygreen);
draw(C--F, heavygreen);
draw(A--C_0--E--cycle, blue);
draw(B_0--C--D_0--C_0--cycle, orange+dashed);
dot("$A$", A, dir(A));
dot("$C_0$", C_0, dir(45));
dot("$E$", E, dir(E));
dot("$B_0$", B_0, dir(135));
dot("$D_0$", D_0, dir(315));
dot("$F$", F, dir(F));
dot("$C$", C, dir(C));
dot("$B$", B, dir(B));
dot("$D$", D, dir(D));
/* TSQ Source:
A = dir 140
C_0 = conj(A) R45
E = -A
B_0 = 2*C_0-E R135
D_0 = 2*C_0-A R315
F = extension A E*E/C_0 E A*A/C_0
C = C_0+2*C_0
B = B_0+2*C_0
D = D_0+2*C_0
A--B--C--D--E--F--cycle 0.1 lightcyan / heavycyan
A--D heavygreen
B--E heavygreen
C--F heavygreen
A--C_0--E--cycle blue
B_0--C--D_0--C_0--cycle orange dashed
*/
\end{asy}
\end{center}
By adjusting the figure above so that
the triangles are right isosceles (instead of just right),
one also finds an example of a hexagon which is satisfying
and whose diagonals are concurrent,
but which is \emph{not} excellent.
\end{remark*}
|
USAMO-2011-notes_4
|
Consider the assertion that for each positive integer $n\geq2$,
the remainder upon dividing $2^{2^n}$ by $2^n-1$ is a power of $4$.
Either prove the assertion or find (with proof) a counterexample.
|
We claim $n = 25$ is a counterexample.
Since $2^{25} \equiv 2^0 \pmod{2^{25}-1}$, we have
\[ 2^{2^{25}} \equiv 2^{2^{25} \bmod{25}}
\equiv 2^7 \bmod{2^{25}-1} \]
and the right-hand side is actually the remainder,
since $0 < 2^7 < 2^{25}$.
But $2^7$ is not a power of $4$.
\begin{remark*}
Really, the problem is just equivalent
for asking $2^n$ to have odd remainder when divided by $n$.
\end{remark*}
|
USAMO-2011-notes_5
|
Let $P$ be a point inside convex quadrilateral $ABCD$.
Points $Q_1$ and $Q_2$ are located within $ABCD$ such that
\begin{align*}
\angle Q_1BC=\angle ABP, & \qquad \angle Q_1CB=\angle DCP, \\
\angle Q_2AD=\angle BAP, & \qquad \angle Q_2DA=\angle CDP.
\end{align*}
Prove that $\ol{Q_1Q_2} \parallel \ol{AB}$
if and only if $\ol{Q_1Q_2} \parallel \ol{CD}$.
|
If $\ol{AB} \parallel \ol{CD}$ there is nothing to prove.
Otherwise let $X = \ol{AB} \cap \ol{CD}$.
Then the $Q_1$ and $Q_2$ are the isogonal conjugates of $P$
with respect to triangles $XBC$ and $XAD$.
Thus $X$, $Q_1$, $Q_2$ are collinear,
on the isogonal of $\ol{XP}$ with respect to
$\angle DXA = \angle CXB$.
|
USAMO-2011-notes_6
|
Let $A$ be a set with $|A|=225$, meaning that $A$ has $225$ elements.
Suppose further that there are eleven subsets $A_1, \dots, A_{11}$ of $A$
such that $|A_i|=45$ for $1\leq i\leq11$ and $|A_i\cap A_j|=9$
for $1\leq i<j\leq11$.
Prove that $|A_1\cup A_2\cup\dots\cup A_{11}|\geq 165$,
and give an example for which equality holds.
\end{enumerate}
|
Ignore the $225$ --- it is irrelevant.
Denote the elements of $A_1 \cup \dots \cup A_{11}$
by $a_1$, \dots, $a_n$,
and suppose that $a_i$ appears $x_i$ times among $A_i$
for each $1 \le i \le n$ (so $1 \le x_i \le 11$).
Then we have
\[ \sum_{i=1}^{11} x_i = \sum_{i=1}^{11} |A_i| = 45 \cdot 11 \]
and
\[ \sum_{i=1}^{11} \binom{x_i}{2} = \sum_{1 \le i < j \le 11}
\left\lvert A_i \cap A_j \right\rvert
= \binom{11}{2} \cdot 9. \]
Therefore, we deduce that
$\sum x_i = 495$ and $\sum_i x_i^2 = 1485$.
Now, by Cauchy Schwarz
\[ n \left( \sum_i x_i^2 \right) \ge \left( \sum x_i \right)^2 \]
which implies $n \ge \frac{495^2}{1485} = 165$.
Equality occurs if we let $A$ consist of the $165$
three-element subsets of $\left\{ 1, \dots, 11 \right\}$
(plus $60$ of your favorite reptiles
if you really insist $|A|=225$).
Then we let $A_i$ denote those subsets containing $i$,
of which there are $\binom{10}{2} = 45$,
and so that $|A_i \cap A_j| = \binom 91 = 9$.
|
USAMO-2012-notes_1
|
Find all integers $n \ge 3$ such that
among any $n$ positive real numbers
$a_1$, $a_2$, \dots, $a_n$ with
\[ \max(a_1,a_2,\dots,a_n)
\le n \cdot \min(a_1,a_2,\dots,a_n), \]
there exist three that are the side lengths
of an acute triangle.
|
The answer is all $n \ge 13$.
Define $(F_n)$ as the sequence of Fibonacci numbers,
by $F_1 = F_2 = 1$ and $F_{n+1} = F_n + F_{n-1}$.
We will find that Fibonacci numbers show up naturally
when we work through the main proof,
so we will isolate the following calculation now
to make the subsequent solution easier to read.
\begin{claim*}
For positive integers $m$, we have $F_m \le m^2$ if and only if $m \le 12$.
\end{claim*}
\begin{proof}
A table of the first $14$ Fibonacci numbers is given below.
\[
\begin{array}{rrrrr rrrrr rrrr}
F_1 & F_2 & F_3 & F_4 & F_5 & F_6 & F_7 & F_8 & F_9
& F_{10} & F_{11} & F_{12} & F_{13} & F_{14} \\ \hline
1 & 1 & 2 & 3 & 5 & 8 & 13 & 21 & 34 & 55 & 89 & 144 & 233 & 377
\end{array}
\]
By examining the table, we see that $F_m \le m^2$ is true for $m = 1, 2, \dots 12$,
and in fact $F_{12} = 12^2 = 144$.
However, $F_m > m^2$ for $m = 13$ and $m = 14$.
Now it remains to prove that $F_m > m^2$ for $m \ge 15$.
The proof is by induction with base cases
$m = 13$ and $m = 14$ being checked already.
For the inductive step, if $m \ge 15$ then we have
\begin{align*}
F_m &= F_{m-1} + F_{m-2} > (m-1)^2 + (m-2)^2 \\
&= 2m^2 - 6m + 5 = m^2 + (m-1)(m-5) > m^2
\end{align*}
as desired.
\end{proof}
We now proceed to the main problem.
The hypothesis $\max(a_1,a_2,\dots,a_n)
\le n \cdot \min(a_1,a_2,\dots,a_n)$
will be denoted by $(\dagger)$.
\medskip
\textbf{Proof that all $n \ge 13$ have the property.}
We first show now that every $n \ge 13$ has the desired property.
Suppose for contradiction that no three numbers are the sides of an acute triangle.
Assume without loss of generality (by sorting the numbers)
that $a_1 \le a_2 \le \dots \le a_n$.
Then since $a_{i-1}$, $a_i$, $a_{i+1}$ are not the sides of an acute triangle
for each $i \ge 2$, we have that $a_{i+1}^2 \ge a_i^2 + a_{i-1}^2$;
writing this out gives
\begin{align*}
a_3^2 &\ge a_2^2 + a_1^2 \ge 2a_1^2 \\
a_4^2 &\ge a_3^2 + a_2^2 \ge 2a_1^2 + a_1^2 = 3a_1^2 \\
a_5^2 &\ge a_4^2 + a_3^2 \ge 3a_1^2 + 2a_1^2 = 5a_1^2 \\
a_6^2 &\ge a_5^2 + a_4^2 \ge 5a_1^2 + 3a_1^2 = 8a_1^2
\end{align*}
and so on.
The Fibonacci numbers appear naturally and by induction,
we conclude that $a_i^2 \ge F_i a_1^2$.
In particular, $a_n^2 \ge F_n a_1^2$.
However, we know $\max(a_1, \dots, a_n) = a_n$
and $\min(a_1, \dots, a_n) = a_1$,
so $(\dagger)$ reads $a_n \le n \cdot a_1$.
Therefore we have $F_n \le n^2$, and so $n \le 12$, contradiction!
\medskip
\textbf{Proof that no $n \le 12$ have the property.}
Assume that $n \le 12$.
The above calculation also suggests a way to pick the counterexample:
we choose $a_i = \sqrt{F_i}$ for every $i$.
Then $\min(a_1, \dots, a_n) = a_1 = 1$
and $\max(a_1, \dots, a_n) = \sqrt{F_n}$,
so $(\dagger)$ is true as long as $n \le 12$.
And indeed no three numbers form the sides of an acute triangle:
if $i < j < k$,
then $a_k^2 = F_k = F_{k-1} + F_{k-2} \ge F_j + F_i = a_j^2 + a_i^2$.
|
USAMO-2012-notes_2
|
A circle is divided into congruent arcs by $432$ points.
The points are colored in four colors such that
some $108$ points are colored red,
some $108$ points are colored green,
some $108$ points are colored blue,
and the remaining $108$ points are colored yellow.
Prove that one can choose three points of each color in such a way
that the four triangles formed by the chosen points
of the same color are congruent.
|
First, consider the $431$ possible non-identity rotations
of the red points, and count overlaps with green points.
If we select a rotation randomly,
then each red point lies over a green point
with probability $\frac{108}{431}$;
hence the expected number of red-green incidences is
\[ \frac{108}{431} \cdot 108 > 27 \]
and so by pigeonhole, we can find a red $28$-gon
and a green $28$-gon which are rotations of each other.
Now, look at the $430$ rotations of this $28$-gon
(that do not give the all-red or all-green configuration)
and compare it with the blue points.
The same approach gives
\[ \frac{108}{430} \cdot 28 > 7 \]
incidences, so we can find red, green, blue $8$-gons
which are similar under rotation.
Finally, the $429$ nontrivial rotations of this $8$-gon expect
\[ \frac{108}{429} \cdot 8 > 2 \]
incidences with yellow.
So finally we have four monochromatic $3$-gons,
one of each color, which are rotations of each other.
|
USAMO-2012-notes_3
|
Determine which integers $n > 1$ have the property
that there exists an infinite sequence $a_1$, $a_2$, $a_3$, \dots
of nonzero integers such that the equality
\[ a_k + 2a_{2k} + \dots + na_{nk} = 0 \]
holds for every positive integer $k$.
|
Answer: all $n > 2$.
For $n = 2$, we have $a_k + 2a_{2k} = 0$,
which is clearly not possible,
since it implies $a_{2^k} = \frac{a_1}{2^{k-1}}$
for all $k \ge 1$.
For $n \ge 3$ we will construct a \emph{completely multiplicative} sequence
(meaning $a_{ij} = a_i a_j$ for all $i$ and $j$).
Thus $(a_i)$ is determined by its value on primes,
and satisfies the condition as long as
$a_1 + 2a_2 + \dots + na_n = 0$.
The idea is to take two large primes and use Bezout's theorem,
but the details require significant care.
We start by solving the case where $n \ge 9$.
In that case, by Bertrand postulate there exists primes $p$ and $q$ such that
\[
\left\lceil n/2 \right\rceil < q < 2\left\lceil n/2 \right\rceil
\quad\text{and}\quad
\half(q-1) < p < q-1.
\]
Clearly $p \neq q$, and $q \ge 7$, so $p > 3$.
Also, $p < q < n$ but $2q > n$, and $4p \ge 4\left( \half(q+1) \right) > n$.
We now stipulate that $a_r = 1$ for any prime $r \neq p,q$
(in particular including $r=2$ and $r=3$).
There are now three cases, identical in substance.
\begin{itemize}
\ii If $p, 2p, 3p \in [1,n]$ then we would like to
choose nonzero $a_p$ and $a_q$ such that
\[ 6p \cdot a_p + q \cdot a_q = 6p + q - \half n (n+1) \]
which is possible by B\'{e}zout lemma, since $\gcd(6p,q) = 1$.
\ii Else if $p, 2p \in [1,n]$ then we would like to
choose nonzero $a_p$ and $a_q$ such that
\[ 3p \cdot a_p + q \cdot a_q = 3p + q - \half n (n+1) \]
which is possible by B\'{e}zout lemma, since $\gcd(3p, q) = 1$.
\ii Else if $p \in [1,n]$ then we would like to
choose nonzero $a_p$ and $a_q$ such that
\[ p \cdot a_p + q \cdot a_q = p + q - \half n (n+1) \]
which is possible by B\'{e}zout lemma, since $\gcd(p, q) = 1$.
(This case is actually possible in a few edge cases,
for example when $n=9$, $q=7$, $p=5$.)
\end{itemize}
It remains to resolve the cases where $3 \le n \le 8$.
We enumerate these cases manually:
\begin{itemize}
\ii For $n = 3$, let $a_n = (-1)^{\nu_3(n)}$.
\ii For $n = 4$, let $a_n = (-1)^{\nu_2(n) + \nu_3(n)}$.
\ii For $n = 5$, let $a_n = (-2)^{\nu_5(n)}$.
\ii For $n = 6$, let $a_n = 5^{\nu_2(n)} \cdot 3^{\nu_3(n)} \cdot (-42)^{\nu_5(n)}$.
\ii For $n = 7$, let $a_n = (-3)^{\nu_7(n)}$.
\ii For $n = 8$, we can choose $(p,q) = (5,7)$ in the prior construction.
\end{itemize}
This completes the constructions for all $n > 2$.
|
USAMO-2012-notes_4
|
Find all functions $f \colon \NN \to \NN$ such that
$f(n!) = f(n)!$ for all positive integers $n$ and such that
$m-n$ divides $f(m) - f(n)$ for all distinct positive integers $m$, $n$.
|
Answer: $f \equiv 1$, $f \equiv 2$, and $f$ the identity.
As these obviously work, we prove these are the only ones.
By putting $n=1$ and $n=2$ we give $f(1), f(2) \in \{1,2\}$.
Also, we will use the condition
\[ m!-n! \text{ divides } f(m)! - f(n)!. \]
We consider four cases on $f(1)$ and $f(2)$,
and dispense with three of them.
\begin{itemize}
\ii If $f(2) = 1$ then for all $m \ge 3$ we have
$m!-2$ divides $f(m)! - 1$, so $f(m) = 1$
for modulo $2$ reasons.
Then clearly $f(1) = 1$.
\ii If $f(1) = f(2) = 2$ we first obtain
$3!-1 \mid f(3)!-2$, which implies $f(3) = 2$.
Then $m! - 3 \mid f(m)! - 2$ for $m \ge 4$
implies $f(m)=2$ for modulo $3$ reasons.
\end{itemize}
Hence we are left with the case where $f(1) = 1$ and $f(2) = 2$.
Continuing, we have
\[ 3!-1 \mid f(3)! - 1 \quad\text{and}\quad
3!-2 \mid f(3)!-2 \implies f(3) = 3. \]
Continuing by induction, suppose $f(1) = 1$, \dots, $f(k) = k$.
\[ k! \cdot k = (k+1)! - k! \mid f(k+1)! - k! \]
and thus we deduce that $f(k+1) \ge k$, and hence
\[ k \mid \frac{f(k+1)!}{k!} - 1. \]
Then plainly $f(k+1) \le 2k$ for mod $k$ reasons,
but also $f(k+1) \equiv 1 \pmod k$ so we conclude $f(k+1) = k+1$.
\begin{remark*}
Shankar Padmanabhan gives the following way
to finish after verifying that $f(3) = 3$.
Note that if
\[ M = ( ( ( (3!)! )! )! \dots )! \]
for any number of iterated factorials
then $f(M) = M$.
Thus for any $n$, we have
\[ M-n \mid f(M) - f(n) = M - f(n) \implies M - n \mid n - f(n) \]
and so taking $M$ large enough implies $f(n) = n$.
\end{remark*}
|
USAMO-2012-notes_5
|
Let $P$ be a point in the plane of $\triangle ABC$,
and $\gamma$ a line through $P$.
Let $A'$, $B'$, $C'$ be the points where the
reflections of lines $PA$, $PB$, $PC$ with respect to $\gamma$
intersect lines $BC$, $CA$, $AB$ respectively.
Prove that $A'$, $B'$, $C'$ are collinear.
|
We present three solutions.
\paragraph{First solution (complex numbers).}
Let $p=0$ and set $\gamma$ as the real line.
Then $A'$ is the intersection of $bc$ and $p\ol a$.
So, we get
\[ a' = \frac{\ol a(\ol b c - b \ol c)}{(\ol b - \ol c)\ol a-(b-c) a}. \]
\begin{center}
\begin{asy}
size(4cm);
pair A = dir(110), B = dir(210), C = dir(330);
Drawing("A", A, A);
Drawing("B", B, dir(-90));
Drawing("C", C, dir(-90));
draw(A--B--C--cycle);
pair P = Drawing("P", 0.1*dir(40), dir(-35));
pair X1 = P + .5 * dir(10);
pair X2 = P - .5 * dir(10);
Drawing(Line(X1,X2), dashed, Arrows);
pair T = 2*foot(A,X1,X2)-A;
pair A1 = Drawing("A'", extension(P,T,B,C), dir(-40));
draw(A--P--A1);
\end{asy}
\end{center}
Note that
\[ \ol a' = \frac{a (b \ol c - \ol b c)}{(b-c) a - (\ol b - \ol c)\ol a}. \]
Thus it suffices to prove
\[ 0 = \det
\begin{bmatrix}
\frac{\ol a(\ol b c-b \ol c)}{(\ol b - \ol c)\ol a - (b-c) a} & \frac{a (b \ol c - \ol b c)}{(b-c) a - (\ol b - \ol c) \ol a} & 1 \\
\frac{\ol b(\ol c a-c \ol a)}{(\ol c - \ol a)\ol b - (c-a) b} & \frac{b (c \ol a - \ol c a)}{(c-a) b - (\ol c - \ol a) \ol b} & 1 \\
\frac{\ol c(\ol a b-a \ol b)}{(\ol a - \ol b)\ol c - (a-b) c} & \frac{c (a \ol b - \ol a b)}{(a-b) c - (\ol a - \ol b)\ol c} & 1
\end{bmatrix}.
\]
This is equivalent to
\[ 0 = \det
\begin{bmatrix}
\ol a(\ol b c -b \ol c) & a (\ol b c - b \ol c) & (\ol b - \ol c) \ol a - (b-c) a \\
\ol b(\ol c a -c \ol a) & b (\ol c a - c \ol a) & (\ol c - \ol a) \ol b - (c-a) b \\
\ol c(\ol a b -a \ol b) & c (\ol a b - a \ol b) & (\ol a - \ol b) \ol c - (a-b) c \\
\end{bmatrix}. \]
This determinant has the property that the columns sum to zero, and we're done.
\begin{remark*}
Alternatively, if you don't notice that you could just blindly expand:
\begin{align*}
&\phantom{=} \sum_{\text{cyc}} ((\ol b - \ol c) \ol a - (b-c) a) \cdot
- \det
\begin{bmatrix}
b & \ol b \\
c & \ol c
\end{bmatrix}
\left( \ol c a - c \ol a \right)\left( \ol a b - a \ol b \right) \\
&= (\ol b c - c \ol b)(\ol c a - c \ol a)(\ol a b - a \ol b)
\sum_{\text{cyc}} \left( ab - ac + \ol c \ol a - \ol b \ol a \right) = 0.
\end{align*}
\end{remark*}
\paragraph{Second solution (Desargues involution).}
We let $C'' = \ol{A'B'} \cap \ol{AB}$.
Consider complete quadrilateral $ABCA'B'C''C$.
We see that there is an involutive pairing $\tau$ at $P$
swapping $(PA,PA')$, $(PB,PB')$, $(PC,PC'')$.
From the first two, we see $\tau$ coincides with reflection
about $\ell$, hence conclude $C'' = C$.
\paragraph{Third solution (barycentric), by Catherine Xu.}
We will perform barycentric coordinates on the triangle $PCC'$,
with $P=(1,0,0)$, $C'=(0,1,0)$, and $C=(0,0,1)$.
Set $a = CC'$, $b = CP$, $c = C'P$ as usual.
Since $A$, $B$, $C'$ are collinear,
we will define $A = (p : k : q)$ and $B = (p : \ell : q)$.
\begin{claim*}
Line $\gamma$ is the angle bisector of
$\angle APA' $, $\angle BPB'$, and $\angle CPC'$.
\end{claim*}
\begin{proof}
Since $A'P$ is the reflection of $AP$ across $\gamma$, etc.
\end{proof}
Thus $B'$ is the intersection of the isogonal of $B$ with respect to $\angle P$
with the line $CA$; that is,
\[ B' = \left( \frac pk \frac{b^2}{\ell}
: \frac{b^2}{\ell} : \frac{c^2}{q} \right). \]
Analogously, $A'$ is the intersection of the isogonal of $A$ with respect to $\angle P$
with the line $CB$; that is,
\[ A' = \left( \frac{p}{\ell} \frac{b^2}{k}
: \frac{b^2}{k} : \frac{c^2}{q} \right). \]
The ratio of the first to third coordinate in these two points
is both $b^2pq : c^2k\ell$, so it follows $A'$, $B'$, and $C'$ are collinear.
\begin{remark*}
[Problem reference]
The converse of this problem appears as problem 1052
attributed S.\ V.\ Markelov in the book
\emph{Geometriya: 9--11 Klassy: Ot Uchebnoy Zadachi k Tvorcheskoy, 1996},
by I.\ F.\ Sharygin.
\end{remark*}
|
USAMO-2012-notes_6
|
For integer $n \ge 2$,
let $x_1$, $x_2$, \dots, $x_n$ be real numbers satisfying
\[ x_1 + x_2 + \dots + x_n = 0
\quad\text{and}\quad x_1^2 + x_2^2 + \dots + x_n^2 = 1. \]
For each subset $A\subseteq\{1, 2, \dots, n\}$, define $S_A=\sum_{i\in A} x_i$.
(If $A$ is the empty set, then $S_A=0$.)
Prove that for any positive number $\lambda$,
the number of sets $A$ satisfying $S_A\geq\lambda$ is at most $2^{n-3}/\lambda^2$.
For which choices of $x_1$, $x_2$, \dots, $x_n$, $\lambda$ does equality hold?
\end{enumerate}
|
Let $\eps_i$ be a coin flip of $0$ or $1$.
Then we have
\begin{align*} \mathbb E[S_A^2]
&= \mathbb E\left[ \left( \sum \eps_i x_i \right)^2 \right]
= \sum_i \mathbb E[\eps_i^2] x_i^2
+ \sum_{i < j} \mathbb E[\eps_i\eps_j] 2x_i x_j \\
&= \half \sum x_i^2 + \half \sum x_i x_j
= \half + \half \sum_{i<j} x_i x_j
= \half + \half \left( -\half \right) = \frac 14.
\end{align*}
In other words, $\sum_A S_A^2 = 2^{n-2}$.
Since can always pair $A$ with its complement, we conclude
\[ \sum_{S_A > 0} S_A^2 = 2^{n-3}. \]
Equality holds iff $S_A \in \{\pm \lambda, 0\}$ for every $A$.
This occurs when $x_1 = 1/\sqrt2$, $x_2 = -1/\sqrt2$,
$x_3 = \dots = 0$ (or permutations), and $\lambda = 1/\sqrt2$.
|
USAMO-2013-notes_1
|
In triangle $ABC$,
points $P$, $Q$, $R$ lie on sides $BC$, $CA$, $AB$, respectively.
Let $\omega_A$, $\omega_B$, $\omega_C$ denote the
circumcircles of triangles $AQR$, $BRP$, $CPQ$, respectively.
Given the fact that segment $AP$ intersects
$\omega_A$, $\omega_B$, $\omega_C$ again at $X$, $Y$, $Z$ respectively,
prove that $YX/XZ=BP/PC$.
|
Let $M$ be the concurrence point of $\omega_A$, $\omega_B$, $\omega_C$ (by Miquel's theorem).
\begin{center}
\begin{asy}
size(8cm);
defaultpen(fontsize(9pt));
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
pair P = 0.4*B+0.6*C;
pair Q = 0.4*C+0.6*A;
pair R = 0.7*A+0.3*B;
draw(B--A--C);
draw(circumcircle(A, Q, R), blue);
draw(circumcircle(B, R, P), blue);
draw(circumcircle(C, P, Q), blue);
pair O_A = circumcenter(A, Q, R);
pair O_B = circumcenter(B, R, P);
pair O_C = circumcenter(C, P, Q);
pair M = -P+2*foot(P, O_B, O_C);
pair X = -A+2*foot(O_A, A, P);
pair Y = -P+2*foot(O_B, A, P);
pair Z = -P+2*foot(O_C, A, P);
draw(X--M--P, dotted);
draw(R--Q);
draw(A--Y);
draw(Z--P);
draw(B--C, red);
draw(Y--Z, red);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
dot("$R$", R, dir(R));
dot("$M$", M, dir(-90));
dot("$X$", X, dir(225));
dot("$Y$", Y, dir(45));
dot("$Z$", Z, dir(180));
/* Source generated by TSQ
A = dir 110
B = dir 210
C = dir 330
P = 0.4*B+0.6*C
Q = 0.4*C+0.6*A
R = 0.7*A+0.3*B
B--A--C
circumcircle A Q R blue
circumcircle B R P blue
circumcircle C P Q blue
O_A := circumcenter A Q R
O_B := circumcenter B R P
O_C := circumcenter C P Q
M = -P+2*foot P O_B O_C R-90
X = -A+2*foot O_A A P R225
Y = -P+2*foot O_B A P R45
Z = -P+2*foot O_C A P R180
X--M--P dotted
R--Q
A--Y
Z--P
B--C red
Y--Z red
*/
\end{asy}
\end{center}
Then $M$ is the center of a spiral similarity sending
$\ol{YZ}$ to $\ol{BC}$.
So it suffices to show that this spiral similarity
also sends $X$ to $P$, but
\[ \dang MXY = \dang MXA = \dang MRA = \dang MRB = \dang MPB \]
so this follows.
|
USAMO-2013-notes_2
|
For a positive integer $n\geq 3$ plot $n$ equally spaced points around a circle.
Label one of them $A$, and place a marker at $A$.
One may move the marker forward in a clockwise direction
to either the next point or the point after that.
Hence there are a total of $2n$ distinct moves available; two from each point.
Let $a_n$ count the number of ways to advance around the circle exactly twice,
beginning and ending at $A$, without repeating a move.
Prove that $a_{n-1}+a_n=2^n$ for all $n\geq 4$.
|
We present two similar approaches.
\paragraph{First solution.}
Imagine the counter is moving along
the set $S = \{0, 1, \dots, 2n\}$ instead,
starting at $0$ and ending at $2n$, in jumps of length $1$ and $2$.
We can then record the sequence of moves as a matrix of the form
\[
\begin{bmatrix}
p_0 & p_1 & p_2 & \dots & p_{n-1} & p_n \\
p_n & p_{n+1} & p_{n+2} & \dots & p_{2n-1} & p_{2n}
\end{bmatrix}
\]
where $p_i = 1$ if the point $i$ was visited by the counter,
and $p_i = 0$ if the point was not visited by the counter.
Note that $p_0 = p_{2n} = 1$ and the upper-right and lower-left entries are equal.
Then, the problem amounts to finding the number of such matrices
which avoid the contiguous submatrices
\[
\begin{bmatrix} 0 & 0 \end{bmatrix}
\qquad
\begin{bmatrix} 0 \\ 0 \end{bmatrix}
\qquad
\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}
\]
which correspond to forbidding jumps of length greater than $2$,
repeating a length $2$ jump and repeating a length $1$ jump.
We give a nice symmetric phrasing suggested by \texttt{fclvbfm934} at
\url{https://aops.com/community/p27834267}.
If we focus on just the three possible column vectors that appear, say
\[
\mathbf u \coloneq \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \qquad
\mathbf v \coloneq \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \qquad
\mathbf w \coloneq \begin{bmatrix} 1 \\ 1 \end{bmatrix}
\]
then we can instead describe valid matrices as sequences of $n+1$ such column vectors,
where no two column vectors are adjacent, and where the boundary condition is that
\begin{itemize}
\ii either we start with $\mathbf u$ and end with $\mathbf v$, or
\ii either we start with $\mathbf w$ and end with $\mathbf w$.
\end{itemize}
Let $x_n$ and $y_n$ denote the number of such $2 \times (n+1)$ matrices.
(Hence $a_n = x_n + y_n$.)
But owing to the symmetry of the setup with $\mathbf u$, $\mathbf v$, $\mathbf w$,
we could instead view $x_n$ (resp.\ $y_n$) as the number of $2 \times (n+1)$ matrices
for a fixed starting first column whose final column is the same (resp.\ different).
So we have the recursions
\begin{align*}
x_{n+1} &= x_n + y_n \\
y_{n+1} &= 2x_n.
\end{align*}
We also have that
\[ 2x_n + y_n = 2^n \]
which may either be proved directly from the recursions (using $x_1 = 1$ and $y_1 = 0$),
or by noting the left-hand side counts the total number of sequences
of $n+1$ column vectors with no restrictions on the final column at all
(in which case there are simply $2$ choices for each of the $n$ columns
after the first one).
Thus,
\begin{align*}
a_{n+1} + a_n &= (x_{n+1} + y_{n+1}) + (x_n + y_n) \\
&= \left( (x_n + y_n) + 2x_n \right) + (x_n + y_n) \\
&= 2(2x_n + y_n) = 2^{n+1}
\end{align*}
as needed.
\paragraph{Second (longer) solution.}
If one does not notice the nice rephrasing with $\mathbf u$, $\mathbf v$,
$\mathbf w$ above, one may still proceed with the following direct calculation.
Retain the notation of
\[
\begin{bmatrix}
p_0 & p_1 & p_2 & \dots & p_{n-1} & p_n \\
p_n & p_{n+1} & p_{n+2} & \dots & p_{2n-1} & p_{2n}
\end{bmatrix}
\]
described earlier.
We will for now ignore the boundary conditions.
Instead we say a $2 \times m$ matrix is \emph{silver} ($m \ge 2$)
if it avoids the three shapes above.
We consider three types of silver matrices
(essentially doing casework on the last column):
\begin{itemize}
\ii \emph{type B matrices}, of the shape
$\begin{bmatrix}
1 & \dotsb & 1 \\
0 & \dotsb & 0
\end{bmatrix}$
\ii \emph{type C matrices}, of the shape
$\begin{bmatrix}
1 & \dotsb & 0 \\
0 & \dotsb & 1
\end{bmatrix}$.
\ii \emph{type D matrices}, of the shape
$\begin{bmatrix}
1 & \dotsb & 1 \\
0 & \dotsb & 1
\end{bmatrix}$.
\end{itemize}
We let $b_m$, $c_m$, $d_m$
denote matrices of each type, of size $2 \times m$,
and claim the following two recursions for $m \ge 4$:
\begin{align*}
b_m &= c_{m-1} + d_{m-1} \\
c_m &= b_{m-1} + d_{m-1} \\
d_m &= b_{m-1} + c_{m-1}.
\end{align*}
Indeed, if we delete the last column of a type B matrix
and consider what used to be the second-to-last column,
we find that it is either type C or type D.
This establishes the first recursion and the others are analogous.
Note that $b_2 = 0$ and $c_2 = d_2 = 1$.
So using this recursion, the first few values are
\[
\begin{array}{rrrrrrrr}
m & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline
b_m & 0 & 2 & 2 & 6 & 10 & 22 & 42 \\
c_m & 1 & 1 & 3 & 5 & 11 & 21 & 43 \\
d_m & 1 & 1 & 3 & 5 & 11 & 21 & 43 \\
\end{array}
\]
and a calculation gives $b_m = \frac{2^{m-1} + 2(-1)^{m-1}}{3}$,
$c_m = d_m = \frac{2^{m-1} - (-1)^{m-1}}{3}$.
We now relate $a_n$ to $b_m$, $c_m$, $d_m$.
Observe that a matrix as described in the problem is a silver
matrix of one of two forms:
\[
\begin{bmatrix}
1 & p_1 & p_2 & \dots & p_{n-1} & 0 \\
0 & p_{n+1} & p_{n+2} & \dots & p_{2n-1} & 1
\end{bmatrix}
\qquad
\text{or}\qquad
\begin{bmatrix}
1 & p_1 & p_2 & \dots & p_{n-1} & 1 \\
1 & p_{n+1} & p_{n+2} & \dots & p_{2n-1} & 1
\end{bmatrix}.
\]
There are $c_{n+1}$ matrices of the first form.
Moreover, there are $2d_n$ matrices of the second form
(to see this, delete the first column;
we either get a type-D matrix
or an upside-down type-D matrix).
Thus we get
\[ a_n = c_{n+1} + 2d_n
= \frac{2^{n+1} + (-1)^{n+1}}{3}. \]
This implies the result.
\begin{remark*}
The two solutions are closely related.
In fact, $c_n = x_{n-1}$ and $b_n = y_{n-1}$.
So the second solution is really the same as the first solution,
except the symmetry of $\mathbf u$, $\mathbf v$, $\mathbf w$ was not noticed,
thus requiring a third recursion to handle all the cases manually.
\end{remark*}
|
USAMO-2013-notes_3
|
Let $n$ be a positive integer.
There are $\tfrac{n(n+1)}{2}$ tokens,
each with a black side and a white side,
arranged into an equilateral triangle,
with the biggest row containing $n$ tokens.
Initially, each token has the white side up.
An operation is to choose a line parallel to the sides of the triangle,
and flip all the token on that line.
A configuration is called admissible if it can be obtained from the
initial configuration by performing a finite number of operations.
For each admissible configuration $C$,
let $f(C)$ denote the smallest number of operations required to
obtain $C$ from the initial configuration.
Find the maximum value of $f(C)$,
where $C$ varies over all admissible configurations.
|
The answer is
\[
\max_C f(C) =
\begin{cases}
6k & n = 4k \\
6k+1 & n = 4k+1 \\
6k+2 & n = 4k+2 \\
6k+3 & n = 4k+3.
\end{cases}
\]
The main point of the problem is actually to determine
all linear dependencies among the $3n$ possible moves
(since the moves commute and applying a move twice
is the same as doing nothing).
In what follows, assume $n > 1$ for convenience.
To this end, we consider sequences of operations as
additive vectors in $v \in \FF_2^{3n}$, with the
linear map $T \colon \FF_2^{3n} \to \FF_2^{\half n(n+1)}$
denoting the result of applying a vector $v$.
We in particular focus on the following four vectors.
\begin{itemize}
\ii Three vectors $x$, $y$, $z$ are defined by
choosing all $n$ lines parallel to one axis.
Note $T(x) = T(y) = T(z) = \mathbf 1$
(i.e.\ these vectors flip all tokens).
\ii The vector $\theta$ which toggles
all lines with an even number of tokens.
One can check that $T(\theta) = \mathbf 0$.
(Easiest to guess from $n=2$ and $n=3$ case.)
One amusing proof that this works is to use Vivani's theorem:
in an equilateral triangle $ABC$,
the sum of distances from an interior point $P$
to the three sides is equal.
\end{itemize}
The main claim is:
\begin{claim*}
For $n \ge 2$,
the kernel of $T$ has exactly eight elements,
namely $\{\mathbf 0, x+y, y+z, z+x,
\theta, \theta+x+y, \theta+y+z, \theta+z+x \}$.
\end{claim*}
% (This is motivated by the validity of the claim
% when $n = 2$ and $n = 3$.)
\begin{proof}
Suppose $T(v) = 0$.
\begin{itemize}
\ii If $v$ uses the $y$-move of length $n$,
then we replace $v$ with $v+(x+y)$
to obtain a vector in the kernel not using the $y$-move of length $n$.
\ii If $v$ uses the $z$-move of length $n$,
then we replace $v$ with $v+(x+z)$
to obtain a vector in the kernel not using the $z$-move of length $n$.
\ii If $v$ uses the $x$-move of length $2$, then
\begin{itemize}
\ii if $n$ is odd, replace $v$ with $v+\theta$;
\ii if $n$ is even, replace $v$ with $v+(\theta+y+z)$
\end{itemize}
to obtain a vector in the kernel not using the $x$-move of length $2$.
\end{itemize}
A picture is shown below,
with the unused rows being dotted.
\begin{center}
\begin{asy}
pair A = dir(60);
pair B = dir(0);
real r = 0.3;
draw( (-A)--(7*A), dashed );
draw( (7*B-A)--(7*A-B), dashed );
draw( (5*A-B)--(5*A+2*B), dashed );
for (int i=0; i<=6; ++i) {
for (int j=0; i+j<=6; ++j) {
filldraw(CR(i*A+j*B, r), opacity(0.2)+lightcyan, black);
}
}
\end{asy}
\end{center}
Then, it is easy to check inductively that $v$ must now be the
zero vector, after the replacements.
The idea is that for each token $t$,
if two of the moves involving $t$ are unused, so is the third,
and in this way we can show all rows are unused.
Thus the original $v$ was in the kernel we described.
(An alternative proof by induction is feasible too;
as a sequence of movings which does not affect the top $n$ rows
also does not affect the to $n-1$ rows.)
\end{proof}
Then problem is a coordinate bash,
since given any $v$ we now know exactly which vectors $w$
have $T(v) = T(w)$, so given any admissible configuration $C$
one can exactly compute $f(C)$ as the minimum of eight values.
To be explicit, we could represent a vector $v$ as
\[ v \longleftrightarrow (a_1, a_2, b_1, b_2, c_1, c_2) \]
where $a_1$ is the number of $1$'s in odd $x$-indices,
$a_2$ number of $1$'s in even $x$-indices.
Then for example
\begin{align*}
v &\longleftrightarrow (a_1, a_2, b_1, b_2, c_1, c_2) \\
v+x+y &\longleftrightarrow \left( \left\lceil \frac n2 \right\rceil - a_1,
\left\lfloor \frac n2 \right\rfloor - a_2,
\left\lceil \frac n2 \right\rceil - b_1,
\left\lfloor \frac n2 \right\rfloor - b_2, c_1, c_2 \right) \\
v+\theta &\longleftrightarrow \left(a_1, \left\lfloor \frac n2 \right\rfloor - a_2, b_1,
\left\lfloor \frac n2 \right\rfloor -
b_2, c_1, \left\lfloor \frac n2 \right\rfloor - c_2\right)
&\vdotswithin{=}
\end{align*}
and $f(T(v))$ is the smallest sum of the six numbers across all eight $6$-tuples.
So you expect to answer about $\frac32 n$
if all things are about $n/4$.
The details are too annoying to reproduce here, so they are omitted.
|
USAMO-2013-notes_4
|
Find all real numbers $x,y,z \ge 1$ satisfying
\[ \min \left( \sqrt{x+xyz}, \sqrt{y+xyz}, \sqrt{z+xyz} \right)
= \sqrt{x-1} + \sqrt{y-1} + \sqrt{z-1}. \]
|
Set $x = 1+a$, $y = 1+b$, $z = 1+c$
which eliminates the $x,y,z \ge 1$ condition.
Assume without loss of generality that $a \leq b \leq c$.
Then the given equation rewrites as
\[ \sqrt{(1+a)\left( 1+(1+b)(1+c) \right)} = \sqrt a + \sqrt b + \sqrt c. \]
In fact, we are going to prove the left-hand side always exceeds the
right-hand side, and then determine the equality cases.
We have:
\begin{align*}
(1+a)\left( 1 + (1+b)(1+c) \right)
&= (a+1)\left( 1 + (b+1)(1+c) \right) \\
&\ge (a+1) \left( 1 + \left( \sqrt b + \sqrt c \right)^2 \right) \\
&\ge \left( \sqrt a + \left( \sqrt b + \sqrt c \right) \right)^2
\end{align*}
by two applications of Cauchy-Schwarz.
Equality holds if $bc = 1$ and $1/a = \sqrt b + \sqrt c$.
Letting $c = t^2$ for $t \ge 1$,
we recover $b = t^{-2} \le t^2$ and $a = \frac{1}{t+1/t} \le t^2$.
Hence the solution set is
\[ (x,y,z) = \left( 1 + \left( \frac{t}{t^2+1} \right)^2,
1 + \frac{1}{t^2}, 1 + t^2 \right) \]
and permutations, for any $t > 0$.
|
USAMO-2013-notes_5
|
Let $m$ and $n$ be positive integers.
Prove that there exists a positive integer $c$
such that $cm$ and $cn$ have the same nonzero decimal digits.
|
One-line spoiler: $142857$.
More verbosely,
the idea is to look at the decimal representation of $1/D$, $m/D$, $n/D$
for a suitable denominator $D$, which have a ``cyclic shift'' property
in which the digits of $n/D$ are the digits of $m/D$ shifted by $3$.
\begin{remark*}
[An example to follow along]
Here is an example to follow along in the subsequent proof
If $m = 4$ and $n = 23$ then the magic numbers $e = 3$ and $D = 41$ obey
\[ 10^3 \cdot \frac{4}{41} = 97 + \frac{23}{41}. \]
The idea is that
\begin{align*}
\frac{1}{41} &= 0.\ol{02439} \\
\frac{4}{41} &= 0.\ol{09756} \\
\frac{23}{41} &= 0.\ol{56097}
\end{align*}
and so $c = 2349$ works;
we get $4c = 9756$ and $23c = 56097$
which are cyclic shifts of each other by $3$ places
(with some leading zeros appended).
\end{remark*}
Here is the one to use:
\begin{claim*}
There exists positive integers $D$ and $e$
such that $\gcd(D,10)=1$, $D > \max(m,n)$,
and moreover \[ \frac{ 10^e m - n }{ D } \in \ZZ. \]
\end{claim*}
\begin{proof}
Suppose we pick some exponent $e$ and define the number
\[ A = 10^e n - m. \]
Let $r = \nu_2(m)$ and $s = \nu_5(m)$.
As long as $e > \max(r,s)$ we have $\nu_2(A) = r$ and $\nu_5(A) = s$, too.
Now choose any $e > \max(r,s)$ big enough that $A > 2^r 5^s \max(m,n)$ also holds.
Then the number $D = \frac{A}{2^r 5^s}$ works;
the first two properties hold by construction and
\[ 10^e \cdot \frac nD - \frac mD
= \frac{A}{D} = 2^r 5^s \]
is an integer.
\end{proof}
\begin{remark*}
[For people who like obscure theorems]
Kobayashi's theorem implies we can actually pick $D$ to be prime.
\end{remark*}
Now we take $c$ to be the
number under the bar of $1/D$ (leading zeros removed).
Then the decimal representation of $\frac mD$
is the decimal representation of $cm$ repeated
(possibly including leading zeros).
Similarly, $\frac nD$ has the decimal representation of $cm$ repeated (possibly including leading zeros).
Finally, since
\[ 10^e \cdot \frac mD - \frac nD \text{ is an integer} \]
it follows that these repeating decimal representations are rotations of each other by $e$ places,
so in particular they have the same number of nonzero digits.
\begin{remark*}
Many students tried to find a $D$ satisfying the stronger hypothesis
that $1/D$, $2/D$, \dots, $(D-1)/D$
are cyclic shifts of each other.
For example, this holds in the famous $D = 7$ case.
The official USAMO 2013 solutions try to do this by proving
that $10$ is a primitive root modulo $7^e$
for each $e \ge 1$, by Hensel lifting lemma.
I think this argument is actually \emph{incorrect},
because it breaks if either $m$ or $n$ are divisible by $7$.
Put bluntly, $\frac{7}{49}$ and $\frac{8}{49}$
are not shifts of each other.
One may be tempted to resort to using large primes $D$
rather than powers of $7$ to deal with this issue.
However it is an open conjecture
(a special case of Artin's primitive root conjecture)
whether or not $10 \pmod p$ is primitive infinitely often,
which is the necessary conjecture
so this is harder than it seems.
\end{remark*}
|
USAMO-2014-notes_1
|
Let $a$, $b$, $c$, $d$ be real numbers such that $b-d \ge 5$ and all zeros
$x_1$, $x_2$, $x_3$, and $x_4$ of the polynomial $P(x)=x^4+ax^3+bx^2+cx+d$ are real.
Find the smallest value the product $(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1)$ can take.
|
The answer is $\boxed{16}$.
This can be achieved by taking $x_1 = x_2 = x_3 = x_4 = 1$,
whence the product is $2^4 = 16$, and $b-d = 5$.
We now show the quantity is always at least $16$.
We prove:
\begin{claim*}
We always have $(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1) = (b-d-1)^2 + (a-c)^2$.
\end{claim*}
\begin{proof}
Let $i = \sqrt{-1}$.
The key observation is that
\[ \prod_{j=1}^4 \left( x_j^2 + 1 \right)
= \prod_{j=1}^4 (x_j - i)(x_j + i)
= P(i)P(-i) = |P(i)|^2. \]
Since $P(i) = (-1+b-d) + (c-a)i$, the claim follows.
\end{proof}
Since $b-d-1 \ge 4$, we get the desired lower bound of $4^2+0^2=16$.
|
USAMO-2014-notes_2
|
Find all $f \colon \ZZ \to \ZZ$ such that
\[ xf\left( 2f(y)-x \right) + y^2f\left( 2x-f(y) \right)
= \frac{f(x)^2}{x} + f\left( yf(y) \right) \]
for all $x,y \in \ZZ$ such that $x \neq 0$.
|
The answer is $f(x) \equiv 0$ and $f(x) \equiv x^2$. Check that these work.
Now let's prove these are the only solutions.
Put $y=0$ to obtain
\[ x f\left( 2f(0)-x \right) = \frac{f(x)^2}{x} + f(0). \]
The nicest part of the problem is the following step:
\begin{claim*}
We have $f(0)=0$.
\end{claim*}
\begin{proof}
If not, select a prime $p \nmid f(0)$ and put $x=p \neq 0$.
In the above, we find that $p \mid f(p)^2$,
so $p \mid f(p)$ and hence $p \mid \tfrac{f(p)^2}{p}$.
From here we derive $p \mid f(0)$, contradiction.
Hence \[ f(0) = 0. \]
\end{proof}
\begin{claim*}
We have $f(x) \in \{0,x^2\}$ for each individual $x$.
\end{claim*}
\begin{proof}
The above then implies that
\[ x^2f(-x) = f(x)^2 \]
holds for all nonzero $x$, but also for $x=0$.
Let us now check that $f$ is an even function.
In the above, we may also derive $f(-x)^2 = x^2f(x)$.
If $f(x) \neq f(-x)$ (and hence $x \neq 0$),
then subtracting the above and factoring implies that
$f(x) + f(-x) = -x^2$;
we can then obtain by substituting the relation
\[ \left[ f(x) + \frac 12x^2 \right]^2 = -\frac 34 x^4 < 0 \]
which is impossible.
This means $f(x)^2 = x^2f(x)$, thus
\[ f(x) \in \{0, x^2\} \qquad \forall x. \]
\end{proof}
Now suppose there exists a nonzero integer $t$ with $f(t) = 0$.
We will prove that $f(x) \equiv 0$.
Put $y=t$ in the given to obtain that
\[ t^2 f(2x) = 0 \]
for any integer $x \neq 0$, and hence conclude that $f(2 \ZZ) \equiv 0$.
Then selecting $x = 2k \neq 0$ in the given implies that
\[ y^2 f(4k-f(y)) = f(yf(y)). \]
Assume for contradiction that $f(m) = m^2$ now for some odd $m \neq 0$.
Evidently \[ m^2 f(4k-m^2) = f(m^3). \]
If $f(m^3) \neq 0$ this forces $f(4k-m^2) \neq 0$,
and hence $m^2(4k-m^2)^2 = m^6$ for arbitrary $k \neq 0$, which is clearly absurd.
That means \[ f(4k-m^2) = f(m^2-4k) = f(m^3) = 0 \] for each $k \neq 0$.
Since $m$ is odd, $m^2 \equiv 1 \pmod 4$,
and so $f(n) = 0$ for all $n$ other than $\pm m^2$
(since we cannot select $k=0$).
Now $f(m) = m^2$ means that $m = \pm 1$.
Hence either $f(x) \equiv 0$ or
\[ f(x) = \begin{cases} 1 & x = \pm 1 \\ 0 & \text{otherwise}. \end{cases} \]
To show that the latter fails,
we simply take $x=5$ and $y=1$ in the given.
Hence, the only solutions are $f(x) \equiv 0$ and $f(x) \equiv x^2$.
|
USAMO-2014-notes_3
|
Prove that there exists an infinite set of points
\[ \dots, \; P_{-3}, \; P_{-2},\; P_{-1},\; P_0,\; P_1,\; P_2,\; P_3,\; \dots \]
in the plane with the following property:
For any three distinct integers $a$, $b$, and $c$,
points $P_a$, $P_b$, and $P_c$ are collinear if and only if $a+b+c=2014$.
|
The construction
\[ P_n = \left( n-\frac{2014}{3},
\left( n-\frac{2014}{3} \right)^3 \right) \]
works fine, and follows from the following claim:
\begin{claim*}
If $x$, $y$, $z$ are distinct real numbers
then the points $(x,x^3)$, $(y,y^3)$, $(z,z^3)$
are collinear if and only if $x+y+z=0$.
\end{claim*}
\begin{proof}
Note that by the ``shoelace formula'',
the collinearity is equivalent to
\[
0 =
\det \begin{bmatrix}
x & x^3 & 1 \\
y & y^3 & 1 \\
z & z^3 & 1 \\
\end{bmatrix}
\]
But the determinant equals
\[ \sum_{\text{cyc}} x(y^3-z^3)
= (x-y)(y-z)(z-x)(x+y+z). \qedhere \]
\end{proof}
|
USAMO-2014-notes_4
|
Let $k$ be a positive integer.
Two players $A$ and $B$ play a game on an infinite grid of regular hexagons.
Initially all the grid cells are empty.
Then the players alternately take turns with $A$ moving first.
In her move, $A$ may choose two adjacent hexagons in the grid
which are empty and place a counter in both of them.
In his move, $B$ may choose any counter on the board and remove it.
If at any time there are $k$ consecutive grid cells
in a line all of which contain a counter, $A$ wins.
Find the minimum value of $k$ for which $A$ cannot
win in a finite number of moves, or prove that no such minimum value exists.
|
The answer is $k = 6$.
\paragraph{Proof that $A$ cannot win if $k=6$.}
We give a strategy for $B$ to prevent $A$'s victory.
Shade in every third cell, as shown in the figure below.
Then $A$ can never cover two shaded cells simultaneously on her turn.
Now suppose $B$ always removes a counter on a shaded cell
(and otherwise does whatever he wants).
Then he can prevent $A$ from ever getting six consecutive counters,
because any six consecutive cells contain two shaded cells.
\begin{center}
\begin{asy}
path hexagon = scale(0.5)*(dir(30)--dir(90)--dir(150)--dir(210)--dir(270)--dir(330)--cycle);
unitsize(0.6cm);
void mark(int x, int y) {
pair P = dir(0)*x + dir(60)*y;
filldraw(shift(P)*hexagon, purple, black+1);
}
void spot(int x, int y) {
pair P = dir(0)*x + dir(60)*y;
draw(shift(P)*hexagon);
}
int N = 4;
for (int x=-N; x<=N; ++x) {
for (int y=-N; y<=N; ++y) {
if ( (abs(x+y)) > N ) continue;
if (
(x-y) % 3 == 0
) mark(x,y);
else spot(x,y);
}
}
\end{asy}
\end{center}
\paragraph{Example of a strategy for $A$ when $k=5$.}
We describe a winning strategy for $A$ explicitly.
Note that after $B$'s first turn there is one counter,
so then $A$ may create an equilateral triangle,
and hence after $B$'s second turn there are two consecutive counters.
Then, on her third turn,
$A$ places a pair of counters two spaces away on the same line.
Label the two inner cells $x$ and $y$ as shown below.
\begin{center}
\begin{asy}
path hexagon = scale(0.5)*(dir(30)--dir(90)--dir(150)--dir(210)--dir(270)--dir(330)--cycle);
unitsize(0.6cm);
void mark(int x, int y) {
pair P = dir(0)*x + dir(60)*y;
filldraw(shift(P)*hexagon, palecyan, black);
filldraw(CR(P, 0.3), paleblue, blue);
}
void spot(int x, int y) {
pair P = dir(0)*x + dir(60)*y;
draw(shift(P)*hexagon);
}
for (int x=-3; x<=4; ++x) {
for (int y=-2; y<=2; ++y) {
if ( x+y < -3 ) continue;
if ( x+y > 4 ) continue;
spot(x,y);
}
}
mark(-2, 0);
mark(-1, 0);
mark( 2, 0);
mark( 3, 0);
label("$x$", (-1)*dir(0));
label("$y$", (2)*dir(0));
\end{asy}
\end{center}
Now it is $B$'s turn to move;
in order to avoid losing immediately,
he must remove either $x$ or $y$.
Then on any subsequent turn, $A$ can replace $x$ or $y$
(whichever was removed)
and add one more adjacent counter.
This continues until either $x$ or $y$ has all its neighbors filled
(we ask $A$ to do so in such a way that
she avoids filling in the two central cells between $x$ and $y$
as long as possible).
So, let's say without loss of generality (by symmetry) that $x$
is completely surrounded by tokens.
Again, $B$ must choose to remove $x$ (or $A$ wins on her next turn).
After $x$ is removed by $B$, consider the following figure.
\begin{center}
\begin{asy}
path hexagon = scale(0.5)*(dir(30)--dir(90)--dir(150)--dir(210)--dir(270)--dir(330)--cycle);
unitsize(0.6cm);
void mark(int x, int y) {
pair P = dir(0)*x + dir(60)*y;
filldraw(shift(P)*hexagon, palecyan, black);
filldraw(CR(P, 0.3), paleblue, blue);
}
void spot(int x, int y) {
pair P = dir(0)*x + dir(60)*y;
draw(shift(P)*hexagon);
}
for (int x=-4; x<=5; ++x) {
for (int y=-3; y<=3; ++y) {
if ( x+y < -4 ) continue;
if ( x+y > 5 ) continue;
spot(x,y);
}
}
mark(-2, 0);
mark( 0, 0);
mark( 0, -1);
mark(-1, 1);
mark(-1, -1);
mark(-2, 1);
mark( 2, 0);
mark( 2, 1);
mark( 3, 0);
label("$x$", (-1)*dir(0));
label("$y$", (2)*dir(0));
void win(int x, int y, pen p1, pen p2) {
pair P = dir(0)*x + dir(60)*y;
filldraw(shift(P)*hexagon, p1, p2);
}
win(1,0, lightgreen, deepgreen);
win(0,1, lightgreen, deepgreen);
win(0,2, lightred, red);
win(0,3, lightred, red);
win(4,0, lightred, red);
win(5,0, lightred, red);
\end{asy}
\end{center}
We let $A$ play in the two marked green cells.
Then, regardless of what move $B$ plays,
one of the two choices of moves marked in red lets $A$ win.
Thus, we have described a winning strategy when $k=5$ for $A$.
|
USAMO-2014-notes_5
|
Let $ABC$ be a triangle with orthocenter $H$ and
let $P$ be the second intersection of the circumcircle of
triangle $AHC$ with the internal bisector of $\angle BAC$.
Let $X$ be the circumcenter of triangle $APB$
and let $Y$ be the orthocenter of triangle $APC$.
Prove that the length of segment $XY$ is equal to
the circumradius of triangle $ABC$.
|
\begin{center}
\begin{asy}
size(8cm);
pair A = Drawing("A", dir(110), dir(110));
pair B = Drawing("B", dir(210), dir(210));
pair C = Drawing("C", dir(330), dir(330));
draw(A--B--C--cycle);
draw(unitcircle);
pair Q = dir(30); // 330 + 1/2 * (330-210) (mod 360)
Drawing("Q", Q, Q);
pair P = A+C-A*C*conj(Q);
P = Drawing("P", P, dir(Q-P));
pair X = Drawing("X", circumcenter(A,P,B), dir(225));
pair Y = Drawing("Y", orthocenter(A,P,C), dir(225));
pair X1 = Drawing("X'", A+C-A*C*conj(X), dir(45));
pair Y1 = Drawing("Y'", A+C-A*C*conj(Y), dir(45));
Drawing("H", A+B+C, dir(170));
draw(circumcircle(A,P,C));
pair B1 = Drawing("B'", A+C-A*C*conj(B), dir(45));
draw(A--Q--C--cycle);
draw(X--Y);
draw(X1--Y1);
Drawing("O", origin, dir(45));
\end{asy}
\end{center}
We eliminate the floating orthocenter
by reflecting $P$ across $\ol{AC}$ to $Q$.
Then $Q$ lies on $(ABC)$ and moreover $\angle QAC = \tfrac 12 \angle BAC$.
This motivates us to reflect $B$, $X$, $Y$
to $B'$, $X'$, $Y'$ and complex bash with respect to $\triangle AQC$.
Obviously \[ y' = a + q + c. \]
Now we need to compute $x'$.
You can get this using the formula
\[ x' = a + \frac{(b'-a)(q-a)\left( \ol{q-a}-\ol{b'-a} \right)}{(b'-a)\ol{(q-a)}
- \ol{(b'-a)}(q-a)}. \]
Using the angle condition we know
$b = \frac{c^3}{q^2}$, and then that
\[ b' = a+c - ac\ol b = a+c-\frac{aq^2}{c^2}. \]
Therefore
\begin{align*}
x' &= a + \frac{\left( c-\frac{aq^2}{c^2} \right)\left( q-a \right)\left( \frac 1q - \frac 1a - \frac 1c + \frac{c^2}{aq^2} \right)}{\left( c-\frac{aq^2}{c^2} \right)\left( \frac 1q - \frac 1a \right) - \left( \frac 1c - \frac{c^2}{aq^2} \right)\left( q-a \right)} \\
&= a + \frac{\frac{c^3-aq^2}{c^2} \left( q-a \right)\left( \frac 1q - \frac 1a - \frac 1c + \frac{c^2}{aq^2} \right)}{-\frac{c^3-aq^2}{c^2}\frac{q-a}{qa} + \frac{c^3-aq^2}{aq^2c}(q-a)} \\
&= a + \frac{\frac 1q - \frac 1a - \frac 1c + \frac{c^2}{aq^2}}{-\frac{1}{qa} + \frac{c}{aq^2}} \\
&= a + \frac{c^2-q^2 + aq-\frac{aq^2}{c}}{c-q} \\
&= a + c + q + \frac{aq}{c}
\end{align*}
whence
\[ \left\lvert x'-y' \right\rvert
= \left\lvert \frac{aq}{c} \right\rvert
= 1. \]
|
USAMO-2014-notes_6
|
Prove that there is a constant $c>0$ with the following property:
If $a$, $b$, $n$ are positive integers such that $\gcd(a+i, b+j)>1$
for all $i, j \in \{0, 1, \dots, n\}$, then
\[ \min\{a, b\}> (cn)^{n/2}. \]
\end{enumerate}
|
Let $N = n+1$ and assume $N$ is (very) large.
We construct an $N \times N$ with cells $(i,j)$
where $0 \le i, j \le n$ and in each cell
place a prime $p$ dividing $\gcd (a+i, b+j)$.
The central claim is at least $50\%$ of the primes
in this table exceed $0.001n^2$.
We count the maximum number of squares they could occupy:
\[
\sum_p \left\lceil \frac{N}{p} \right\rceil^2
\le \sum_p \left( \frac Np + 1 \right)^2
= N^2 \sum_p \frac{1}{p^2} + 2N \sum_p \frac1p + \sum_p 1. \]
Here the summation runs over primes $p \le 0.001n^2$.
Let $r = \pi(0.001n^2)$ denote the number of such primes.
Now we consider the following three estimates.
First, \[ \sum_p \frac{1}{p^2} < \frac 12 \]
which follows by adding all the primes directly with some computation.
Moreover,
\[ \sum_p \frac 1p < \sum_{k=1}^r \frac 1k = O(\log r) < o(N) \]
using the harmonic series bound,
and \[ \sum_p 1 < r \sim O \left( \frac{N^2}{\ln N} \right) < o(N^2) \]
via Prime Number Theorem.
Hence the sum in question is certainly less than $\half N^2$ for $N$ large
enough, establishing the central claim.
Hence some column $a+i$ has at least one
half of its primes greater than $0.001n^2$.
Because this is greater than $n$ for large $n$,
these primes must all be distinct,
so $a+i$ exceeds their product,
which is larger than
\[ \left( 0.001n^2 \right)^{N/2} > c^n \cdot n^n \]
where $c$ is some constant (better than the requested bound).
|
USAMO-2015-notes_1
|
Solve in integers the equation
\[ x^2+xy+y^2 = \left(\frac{x+y}{3}+1\right)^3. \]
|
We do the trick of setting $a=x+y$ and $b=x-y$.
This rewrites the equation as
\[ \frac14\left( (a+b)^2+(a+b)(a-b)+(a-b)^2 \right) = \left( \frac a3 + 1 \right)^3 \]
where $a,b \in \ZZ$ have the same parity.
This becomes
\[ 3a^2+b^2 = 4\left( \frac a3 + 1 \right)^3 \]
which is enough to imply $3 \mid a$, so let $a = 3c$.
Miraculously, this becomes \[ b^2 = (c-2)^2 (4c+1). \]
So a solution must have $4c+1=m^2$, with $m$ odd.
This gives
\[ x = \frac 18 \left( 3 (m^2-1) \pm (m^3-9m) \right)
\quad\text{and}\quad
y = \frac 18 \left( 3 (m^2-1) \mp (m^3-9m) \right). \]
For mod $8$ reasons,
this always generates a valid integer solution,
so this is the complete curve of solutions.
Actually, putting $m=2n+1$ gives the much nicer curve
\[ \boxed{x = n^3+3n^2-1 \quad\text{and}\quad y = -n^3+3n+1} \]
and permutations.
For $n=0,1,2,3$ this gives the first few solutions
are $(-1,1)$, $(3,3)$, $(19,-1)$, $(53, -17)$, (and permutations).
|
USAMO-2015-notes_2
|
Quadrilateral $APBQ$ is inscribed in circle $\omega$ with
$\angle P = \angle Q = 90^{\circ}$ and $AP = AQ < BP$.
Let $X$ be a variable point on segment $\ol{PQ}$.
Line $AX$ meets $\omega$ again at $S$ (other than $A$).
Point $T$ lies on arc $AQB$ of $\omega$ such that $\ol{XT}$
is perpendicular to $\ol{AX}$.
Let $M$ denote the midpoint of chord $\ol{ST}$.
As $X$ varies on segment $\ol{PQ}$, show that $M$ moves along a circle.
|
We present three solutions,
one by complex numbers, two more synthetic.
(A fourth solution using median formulas is also possible.)
Most solutions will prove that the center of the fixed circle
is the midpoint of $\ol{AO}$
(with $O$ the center of $\omega$);
this can be recovered empirically by letting
\begin{itemize}
\ii $X$ approach $P$ (giving the midpoint of $\ol{BP}$)
\ii $X$ approach $Q$ (giving the point $Q$), and
\ii $X$ at the midpoint of $\ol{PQ}$ (giving the midpoint of $\ol{BQ}$)
\end{itemize}
which determines the circle;
this circle then passes through $P$ by symmetry
and we can find the center by taking the intersection
of two perpendicular bisectors (which two?).
\paragraph{Complex solution (Evan Chen).}
Toss on the complex unit circle with $a = -1$, $b=1$, $z = -\tfrac12$.
Let $s$ and $t$ be on the unit circle. We claim $Z$ is the center.
It follows from standard formulas that
\[ x = \frac 12 \left( s + t - 1 + s/t \right) \]
thus
\[ 4\operatorname{Re} x + 2
= 2\left( x + \frac 1x \right) + 2
= s + t + \frac 1s + \frac 1t + \frac st + \frac ts \]
which depends only on $P$ and $Q$, and not on $X$.
Thus
\[ 4\left\lvert z-\frac{s+t}{2} \right\rvert^2
= \left\lvert s+t+1 \right\rvert^2 = 3 + (4\operatorname{Re}x+2) \]
does not depend on $X$, done.
\paragraph{Homothety solution (Alex Whatley).}
Let $G$, $N$, $O$ denote the centroid, nine-point center,
and circumcenter of triangle $AST$, respectively.
Let $Y$ denote the midpoint of $\ol{AS}$.
Then the three points $X$, $Y$, $M$
lie on the nine-point circle of triangle $AST$,
which is centered at $N$ and has radius $\frac 12 AO$.
\begin{center}
\begin{asy}
size(9cm);
pair A = dir(90);
pair B = dir(-90);
pair S = dir(-50);
pair T = dir(170);
pair O = midpoint(A--B);
pair X = foot(T, A, S);
pair E = dir(0);
pair P = IP(unitcircle, X--(X-2*E));
pair Q = IP(unitcircle, X--(X+2*E));
filldraw(unitcircle, opacity(0.2)+mediumcyan, mediumblue);
pair M = midpoint(S--T);
pair G = centroid(A, S, T);
pair N = 3/2*G;
filldraw(A--P--B--Q--cycle, opacity(0.1)+lightblue, mediumblue);
draw(A--B, mediumblue+dotted);
draw(P--Q, mediumblue+dotted);
filldraw(A--S--T--cycle, opacity(0.4)+mediumgreen, heavygreen);
filldraw(CP(N, M), opacity(0.6)+lightred, red);
draw(T--X, heavygreen);
draw(A--M, heavygreen);
pair Y = midpoint(A--S);
draw(O--N, red);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$S$", S, dir(S));
dot("$T$", T, dir(T));
dot("$O$", O, dir(-45));
dot("$X$", X, dir(45));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
dot("$M$", M, dir(M));
dot("$G$", G, dir(30));
dot("$N$", N, dir(N));
dot("$Y$", Y, dir(Y));
/* Source generated by TSQ
!size(9cm);
A = dir 90
B = dir -90
S = dir -50
T = dir 170
O = midpoint A--B R-45
X = foot T A S R45
E := dir 0
P = IP unitcircle X--(X-2*E)
Q = IP unitcircle X--(X+2*E)
unitcircle 0.2 mediumcyan / mediumblue
M = midpoint S--T
G = centroid A S T R30
N = 3/2*G
A--P--B--Q--cycle 0.1 lightblue / mediumblue
A--B mediumblue dotted
P--Q mediumblue dotted
A--S--T--cycle 0.4 mediumgreen / heavygreen
CP N M 0.6 lightred / red
T--X heavygreen
A--M heavygreen
Y = midpoint A--S
O--N red
*/
\end{asy}
\end{center}
Let $R$ denote the radius of $\omega$.
Note that the nine-point circle of $\triangle AST$
has radius equal to $\half R$, and hence is independent of $S$ and $T$.
Then the power of $A$ with respect to the nine-point circle equals
\[ AN^2 - \left( \half R \right)^2 = AX \cdot AY
= \frac 12 AX \cdot AS = \frac 12 AQ^2 \]
and hence
\[ AN^2 = \left( \half R \right)^2 + \frac 12 AQ^2 \]
which does not depend on the choice of $X$.
So $N$ moves along a circle centered at $A$.
Since the points $O$, $G$, $N$ are collinear on the Euler line of $\triangle AST$
with \[ GO = \frac 23 NO \] it follows by homothety
that $G$ moves along a circle as well,
whose center is situated one-third of the way from $A$ to $O$.
Finally, since $A$, $G$, $M$ are collinear with \[ AM = \frac 32 AG \]
it follows that $M$ moves along a circle
centered at the midpoint of $\ol{AO}$.
\paragraph{Power of a point solution (Zuming Feng, official solution).}
Let $Y$ be the foot of the altitude from $S$ to $\ol{AT}$.
Then $\ol{XY} \perp \ol{AO}$, so $Y$ lies on line $PQ$ too.
We then complete the picture by letting $K$ be the foot of $A$ to $\ol{ST}$.
\begin{center}
\begin{asy}
size(9cm);
pair A = dir(125);
pair B = -A;
pair S = dir(210);
pair T = dir(330);
pair O = midpoint(A--B);
pair X = foot(T, A, S);
pair E = dir(0);
filldraw(unitcircle, opacity(0.2)+mediumcyan, mediumblue);
pair M = midpoint(S--T);
filldraw(A--S--T--cycle, opacity(0.4)+mediumgreen, heavygreen);
draw(T--X, heavygreen);
draw(A--M, heavygreen);
pair Y = foot(S, A, T);
pair K = foot(A, S, T);
filldraw(circumcircle(X, Y, M), opacity(0.1)+yellow, red);
draw(S--Y, heavygreen);
draw(A--K, heavygreen);
pair P = IP(unitcircle, X--(3*Y-2*X));
pair Q = IP(unitcircle, Y--(3*X-2*Y));
pair V = extension(P, Q, S, T);
draw(P--Q, blue);
draw(A--B, blue);
draw(Q--V, blue);
draw(V--S, heavygreen);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$S$", S, dir(S));
dot("$T$", T, dir(T));
dot("$O$", O, dir(45));
dot("$X$", X, dir(135));
dot("$M$", M, dir(M));
dot("$Y$", Y, dir(70));
dot("$K$", K, dir(K));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(150));
dot("$V$", V, dir(V));
/* TSQ Source:
!size(9cm);
A = dir 125
B = -A
S = dir 210
T = dir 330
O = midpoint A--B R45
X = foot T A S R135
E := dir 0
unitcircle 0.2 mediumcyan / mediumblue
M = midpoint S--T
A--S--T--cycle 0.4 mediumgreen / heavygreen
T--X heavygreen
A--M heavygreen
Y = foot S A T R70
K = foot A S T
circumcircle X Y M 0.1 yellow / red
S--Y heavygreen
A--K heavygreen
P = IP unitcircle X--(3*Y-2*X)
Q = IP unitcircle Y--(3*X-2*Y) R150
V = extension P Q S T
P--Q blue
A--B blue
Q--V blue
V--S heavygreen
*/
\end{asy}
\end{center}
The main claim is:
\begin{claim*}
Quadrilateral $PQKM$ is cyclic.
\end{claim*}
\begin{proof}
To see this, we use power of a point: let $V = \ol{QXYP} \cap \ol{SKMT}$.
One approach is that since $(VK;ST) = -1$
we have $VQ \cdot VP = VS \cdot VT = VK \cdot VM$.
A longer approach is more elementary:
\[ VQ \cdot VP = VS \cdot VT = VX \cdot VY = VK \cdot VM \]
using the nine-point circle, and the circle with diameter $\ol{ST}$.
\end{proof}
But the circumcenter of $PQKM$,
is the midpoint of $\ol{AO}$,
since it lies on the perpendicular bisectors of $\ol{KM}$ and $\ol{PQ}$.
So it is fixed, the end.
|
USAMO-2015-notes_4
|
Steve is piling $m\geq 1$ indistinguishable stones
on the squares of an $n\times n$ grid.
Each square can have an arbitrarily high pile of stones.
After he finished piling his stones in some manner,
he can then perform \emph{stone moves}, defined as follows.
Consider any four grid squares, which are corners of a rectangle,
i.e.\ in positions $(i, k)$, $(i, l)$, $(j, k)$, $(j, l)$
for some $1\leq i, j, k, l\leq n$, such that $i<j$ and $k<l$.
A stone move consists of either removing one stone from each of
$(i, k)$ and $(j, l)$ and moving them to $(i, l)$ and $(j, k)$ respectively,
or removing one stone from each of $(i, l)$ and $(j, k)$
and moving them to $(i, k)$ and $(j, l)$ respectively.
Two ways of piling the stones are equivalent if they can be obtained
from one another by a sequence of stone moves.
How many different non-equivalent ways can Steve pile the stones on the grid?
|
The answer is $\binom{m+n-1}{n-1}^2$.
The main observation is that the ordered sequence of column counts
(i.e.\ the number of stones in the first, second, etc.\ column)
is invariant under stone moves, as does the analogous sequence of row counts.
\paragraph{Definitions.}
Call these numbers $(c_1, c_2, \dots, c_n)$
and $(r_1, r_2, \dots, r_n)$ respectively,
with $\sum c_i = \sum r_i = m$.
We say that the sequence $(c_1, \dots, c_n, r_1, \dots, r_n)$
is the \emph{signature} of the configuration.
These are the $2m$ blue and red numbers shown in the example below
(in this example we have $m=8$ and $n=3$).
\begin{center}
\begin{asy}
unitsize(1.5cm);
for (int i=0; i<=3; ++i) {
draw( (0,i)--(3,i), gray );
draw( (i,0)--(i,3), gray );
}
label("$c_1=\boxed{5}$", (0.5,3), dir(90), blue);
label("$c_2=\boxed{2}$", (1.5,3), dir(90), blue);
label("$c_3=\boxed{1}$", (2.5,3), dir(90), blue);
label("$r_1=\boxed{3}$", (0,2.5), dir(180), red);
label("$r_2=\boxed{3}$", (0,1.5), dir(180), red);
label("$r_3=\boxed{2}$", (0,0.5), dir(180), red);
real r = 0.1;
filldraw(circle((0.3,2.7), r), gray, black);
filldraw(circle((0.6,2.4), r), gray, black);
filldraw(circle((1.4,2.3), r), gray, black);
filldraw(circle((0.6,1.7), r), gray, black);
filldraw(circle((0.4,1.4), r), gray, black);
filldraw(circle((0.7,0.5), r), gray, black);
filldraw(circle((2.4,1.6), r), gray, black);
filldraw(circle((1.6,0.4), r), gray, black);
label("Signature: $(5,2,1;3,3,2)$", (1.5,0), dir(-90));
\end{asy}
\end{center}
By stars-and-bars, the number of possible values $(c_1, \dots, c_n)$
is $\binom{m+n-1}{n-1}$.
The same is true for $(r_1, \dots, r_n)$.
So if we're just counting \emph{signatures},
the total number of possible signatures is $\binom{m+n-1}{n-1}^2$.
\paragraph{Outline and setup.}
We are far from done.
To show that the number of non-equivalent ways is also this number,
we need to show that signatures correspond to pilings.
In other words, we need to prove:
\begin{enumerate}
\ii Check that signatures are invariant around moves
(trivial; we did this already);
\ii Check conversely that two configurations
are equivalent if they have the same signatures
(the hard part of the problem); and
\ii Show that each signature is realized by at least
one configuration (not immediate, but pretty easy).
\end{enumerate}
Most procedures to the second step are algorithmic in nature,
but Ankan Bhattacharya gives the following far cleaner approach.
Rather than having a grid of stones,
we simply consider the multiset of ordered pairs $(x,y)$
corresponding to the stones.
Then:
\begin{itemize}
\ii a stone move corresponds to switching two $y$-coordinates
in two different pairs.
\ii we \emph{redefine} the signature to be the multiset $(X,Y)$
of $x$ and $y$ coordinates which appear.
Explicitly, $X$ is the multiset that contains $c_i$ copies
of the number $i$ for each $i$.
\end{itemize}
For example, consider the earlier example which had
\begin{itemize}
\ii Two stones each at $(1,1)$, $(1,2)$.
\ii One stone each at $(1,3)$, $(2,1)$, $(2,3)$, $(3,2)$.
\end{itemize}
Its signature can then be reinterpreted as
\[
(5,2,1; 3,3,2)
\longleftrightarrow
\begin{cases}
X = \{1,1,1,1,1,2,2,3\} \\
Y = \{1,1,1,2,2,2,3,3\}.
\end{cases}
\]
In that sense, the entire grid is quite misleading!
\paragraph{Proof that two configurations with the same signature are equivalent.}
The second part is completed just because
transpositions generate any permutation.
To be explicit, given two sets of stones,
we can permute the labels so that the first set is
$(x_1, y_1)$, \dots, $(x_m, y_m)$ and
the second set of stones is
$(x_1, y_1')$, \dots, $(x_m, y_m')$.
Then we just induce the correct permutation on $(y_i)$ to get $(y_i')$.
\paragraph{Proof that any signature has at least one configuration.}
Sort the elements of $X$ and $Y$ arbitrarily (say, in non-decreasing order).
Put a stone whose $x$-coordinate is the $i$th element of $X$,
and whose $y$-coordinate is the $i$th element of $Y$,
for each $i = 1, 2, \dots, m$.
Then this gives a stone placement of $m$ stones with signature $(X,Y)$.
For example, if
\begin{align*}
X &= \{1,1,1,1,1,2,2,3\} \\
Y &= \{1,1,1,2,2,2,3,3\}
\end{align*}
then placing stones at
$(1,1)$, $(1,1)$, $(1,1)$, $(1,2)$, $(1,2)$, $(2,2)$, $(2,3)$, $(3,3)$
gives a valid piling with this signature.
|
USAMO-2015-notes_5
|
Let $a$, $b$, $c$, $d$, $e$ be distinct positive integers
such that $a^4+b^4=c^4+d^4=e^5$.
Show that $ac+bd$ is a composite number.
|
Assume to the contrary that $p = ac+bd$, so that
\begin{align*}
ac &\equiv -bd \pmod p \\
\implies a^4c^4 &\equiv b^4d^4 \pmod p \\
\implies a^4 (e^5 - d^4) &\equiv (e^5 - a^4) d^4 \pmod p \\
\implies a^4 e^5 &\equiv d^4 e^5 \pmod p \\
\implies e^5(a^4-d^4) &\equiv 0 \pmod p
\end{align*}
and hence \[ p \mid e^5(a-d)(a+d)(a^2+d^2). \]
\begin{claim*}
We should have $p > e$.
\end{claim*}
\begin{proof}
We have $e^5 = a^4 + b^4 \le a^5 + b^5 < (ac+bd)^5 = p^5$.
\end{proof}
Thus the above equation implies $p \le \max(a-d, a+d, a^2+d^2) = a^2+d^2$.
Similarly, $p \le b^2+c^2$.
So \[ ac+bd = p \le \min \left\{ a^2+d^2, b^2+c^2 \right\} \]
or by subtraction
\[ 0 \le \min \left\{ a(a-c) + d(d-b),
b(b-d) + c(c-a) \right\}. \]
But since $a^4+b^4 = c^4+d^4$ the numbers $a-c$ and $d-b$
should have the same sign, and so this is an obvious contradiction.
|
USAMO-2015-notes_6
|
Fix $0 < \lambda < 1$, and let $A$ be a multiset of positive integers.
Let $A_n = \{a\in A: a\leq n\}$.
Assume that for every $n \in \NN$, the multiset $A_n$ contains at most $n\lambda$ numbers.
Show that there are infinitely many $n\in\NN$ for
which the sum of the elements in $A_n$ is at most $\frac{n(n+1)}{2}\lambda$.
\end{enumerate}
|
For brevity, $\#S$ denotes $|S|$.
Let $x_n = n\lambda - \#A_n \ge 0$.
We now proceed by contradiction by assuming
the conclusion fails for $n$ large enough; that is,
\begin{align*}
\frac{n(n+1)}{2}\lambda
&< \sum_{a \in A_n} a \\
&= 1(\#A_1-\#A_0)
+ 2(\#A_2 - \#A_1)
+ \dots + n(\#A_n - \#A_{n-1}) \\
&= n \# A_n - (\# A_1 + \dots + \# A_{n-1}) \\
&= n(n \lambda - x_n) - \left[ (\lambda - x_1)
+ (2\lambda - x_2) + \dots + ((n-1)\lambda - x_{n-1}) \right] \\
&= \frac{n(n+1)}{2} \lambda - n x_n
+ (x_1 + \dots + x_{n-1}).
\end{align*}
This means that for all sufficiently large $n$, say $n \ge N_0$, we have
\[ x_{n} < \frac{x_1 + \dots + x_{n-1}}{n}
\qquad \forall n \ge N_0. \]
In particular, each $x_n$ is the less
than the average of all preceding terms.
Intuitively this means $x_n$ should become close to each other,
since they are also nonnegative.
However, we have a second condition we haven't used yet:
the ``integer'' condition implies
\[
\left\lvert x_{n+1} - x_n \right\rvert
= \left\lvert \lambda - \#\{n \in A \} \right\rvert
> \eps
\]
for some fixed $\eps > 0$,
namely $\eps = \min \left\{ \lambda, 1 - \lambda \right\}$.
Using the fact that consecutive terms differ by some fixed $\eps$,
we will derive a contradiction.
If we let $M$ be the average of $x_1$, \dots, $x_{N_0}$,
then we ought to have
\[ x_n < M \qquad \forall n > N_0. \]
Hence for $n > N_0$ we have $x_n + x_{n+1} < 2M - \eps$,
and so for large enough $n$ the average
must drop to just above $M - \half \eps$.
Thus for some large $N_1 > N_0$, we will have
\[ x_n < M - \frac13 \eps \qquad \forall n > N_1. \]
If we repeat this argument then with a large $N_2 > N_1$, we obtain
\[ x_n < M - \frac23 \eps \qquad \forall n > N_2 \]
and so on and so forth.
This is a clear contradiction.
\begin{remark*}
Note that if $A = \{2,2,3,4,5,\dots\}$ and $\lambda = 1$ then contradiction.
So the condition that $0 < \lambda < 1$ cannot be dropped,
and (by scaling) neither can the condition that $A \subseteq \ZZ$.
\end{remark*}
\begin{remark*}
[Suggested by Zhao Ting-wei]
Despite the relation
\[ x_{n} < \frac{x_1 + \dots + x_{n-1}}{n}
\qquad \forall n \ge N_0 \]
implying that $x_n$ is bounded,
it does not alone imply that $x_n$ converges, not even to some nonzero value.
Zhao Ting-Wei showed me that one can have a sequence which is zero
``every so often'' yet where the average is nonzero.
A counterexample is given explicitly by
\[
x_n
= \begin{cases}
1000 & n = 1 \\
0 & n \text{ is a power of $10$} \\
1 + \frac 1n & \text{otherwise}
\end{cases}
\]
which does not have a limit.
For completeness, let's check this ---
let $H_n$ denote the $n$'th harmonic number, and compute
\begin{align*}
\sum_1^{n-1} x_n
&= 1000 + (n-1) + H_{n-1} - \sum_{k=1}^{\left\lfloor \log_{10} n \right\rfloor} \left( 1 + \frac{1}{10^k} \right) \\
&> n + 999 + H_{n-1} - \log_{10} n - \left( 1 + \frac{1}{10} + \dots \right) \\
&> n + 997 + H_{n-1} - \log_{10} n > n + 1
\end{align*}
so $1 + \frac 1n < \frac 1n \sum_1^{n-1} x_n$ as needed.
\end{remark*}
|
USAMO-2016-notes_1
|
Let $X_1$, $X_2$, \dots, $X_{100}$ be a sequence of
mutually distinct nonempty subsets of a set $S$.
Any two sets $X_i$ and $X_{i+1}$ are disjoint and their union is not the whole set $S$,
that is, $X_i \cap X_{i+1} = \emptyset$ and $X_i \cup X_{i+1} \neq S$,
for all $i \in \{1, \dots, 99\}$.
Find the smallest possible number of elements in $S$.
|
Solution with Danielle Wang: the answer is that $|S| \ge 8$.
\paragraph{Proof that $|S| \ge 8$ is necessary.}
Since we must have $2^{|S|} \geq 100$, we must have $|S| \geq 7$.
To see that $|S| = 8$ is the minimum possible size,
consider a chain on the set $S = \{1, 2, \dots, 7\}$
satisfying $X_i \cap X_{i+1} = \emptyset$ and $X_i \cup X_{i+1} \neq S$.
Because of these requirements any subset of size $4$ or more
can only be neighbored by sets of size $2$ or less,
of which there are $\binom 71 + \binom 72 = 28$ available.
Thus, the chain can contain no more than $29$ sets of size $4$ or more
and no more than $28$ sets of size $2$ or less.
Finally, since there are only $\binom 73 = 35$ sets of size $3$ available,
the total number of sets in such a chain
can be at most $29 + 28 + 35 = 92 < 100$, contradiction.
\paragraph{Construction.}
We will provide an inductive construction for a chain of subsets
$X_1, X_2, \dots, X_{2^{n-1} + 1}$ of $S = \left\{ 1, \dots, n \right\}$
satisfying $X_i \cap X_{i+1} = \varnothing$
and $X_i \cup X_{i+1} \neq S$ for each $n \geq 4$.
For $S = \{1, 2, 3, 4\}$,
the following chain of length $2^3 + 1 = 9$ will work:
\[
\begin{array}{ccccccccc}
34 & 1 & 23 & 4 & 12 & 3 & 14 & 2 & 13
\end{array}.
\]
Now, given a chain of subsets of $\{1, 2, \dots, n\}$
the following procedure produces a
chain of subsets of $\{1, 2, \dots, n+1\}$:
\begin{enumerate}
\item take the original chain, delete any element,
and make two copies of this chain, which now has even length;
\item glue the two copies together,
joined by $\varnothing$ in between; and then
\item insert the element $n+1$ into the sets in alternating positions
of the chain starting with the first.
\end{enumerate}
For example, the first iteration of this construction gives:
\[
\begin{array}{ccccccccc}
345 & 1 & 235 & 4 & 125 & 3 & 145 & 2 & 5 \\
34 & 15 & 23 & 45 & 12 & 35 & 14 & 25 &
\end{array}
\]
It can be easily checked that if the original chain satisfies the requirements,
then so does the new chain, and if the original chain has length $2^{n-1}+1$,
then the new chain has length $2^{n}+1$, as desired.
This construction yields a chain of length $129$
when $S = \{1, 2, \dots, 8\}$.
\begin{remark*}
Here is the construction for $n=8$ in its full glory.
\[
\begin{array}{ccccccccc}
345678 & 1 & 235678 & 4 & 125678 & 3 & 145678 & 2 & 5678 \\
34 & 15678 & 23 & 45678 & 12 & 35678 & 14 & 678 & \\
345 & 1678 & 235 & 4678 & 125 & 3678 & 145 & 2678 & 5 \\
34678 & 15 & 23678 & 45 & 12678 & 35 & 78 & & \\\hline
3456 & 178 & 2356 & 478 & 1256 & 378 & 1456 & 278 & 56 \\
3478 & 156 & 2378 & 456 & 1278 & 356 & 1478 & 6 & \\
34578 & 16 & 23578 & 46 & 12578 & 36 & 14578 & 26 & 578 \\
346 & 1578 & 236 & 4578 & 126 & 8 & & & \\ \hline\hline
34567 & 18 & 23567 & 48 & 12567 & 38 & 14567 & 28 & 567 \\
348 & 1567 & 238 & 4567 & 128 & 3567 & 148 & 67 & \\
3458 & 167 & 2358 & 467 & 1258 & 367 & 1458 & 267 & 58 \\
3467 & 158 & 2367 & 458 & 1267 & 358 & 7 & & \\\hline
34568 & 17 & 23568 & 47 & 12568 & 37 & 14568 & 27 & 568 \\
347 & 1568 & 237 & 4568 & 127 & 3568 & 147 & 68 & \\
3457 & 168 & 2357 & 468 & 1257 & 368 & 1457 & 268 & 57 \\
3468 & 157 & 2368 & 457 & 1268 & & & & \\
\end{array}
\]
\end{remark*}
|
USAMO-2016-notes_2
|
Prove that for any positive integer $k$,
\[ (k^2)!\cdot\displaystyle\prod_{j=0}^{k-1}\frac{j!}{(j+k)!} \]
is an integer.
|
We show the exponent of any given prime $p$ is nonnegative in the expression.
Recall that the exponent of $p$ in $n!$ is equal to
$\sum_{i \ge 1} \left\lfloor n/p^i \right\rfloor$.
In light of this, it suffices to show
that for any prime power $q$, we have
\[
\left\lfloor \frac{k^2}{q} \right\rfloor
+ \sum_{j=0}^{k-1} \left\lfloor \frac{j}{q} \right\rfloor
\ge \sum_{j=0}^{k-1} \left\lfloor \frac{j+k}{q} \right\rfloor
\]
Since both sides are integers, we show
\[
\left\lfloor \frac{k^2}{q} \right\rfloor
+ \sum_{j=0}^{k-1} \left\lfloor \frac{j}{q} \right\rfloor
> -1 + \sum_{j=0}^{k-1} \left\lfloor \frac{j+k}{q} \right\rfloor.
\]
If we denote by $\left\{ x \right\}$ the fractional part of $x$,
then $\left\lfloor x \right\rfloor = x - \left\{ x \right\}$
so it's equivalent to
\[
\left\{ \frac{k^2}{q} \right\}
+ \sum_{j=0}^{k-1} \left\{ \frac{j}{q} \right\}
< 1 + \sum_{j=0}^{k-1} \left\{ \frac{j+k}{q} \right\}.
\]
However, the sum of remainders when $0, 1, \dots, k-1$ are taken modulo $q$
is easily seen to be less than the sum of remainders
when $k, k+1, \dots, 2k-1$ are taken modulo $q$.
So \[ \sum_{j=0}^{k-1} \left\{ \frac{j}{q} \right\}
\le \sum_{j=0}^{k-1} \left\{ \frac{j+k}{q} \right\} \] follows,
and we are done upon noting $\left\{ k^2/q \right\} < 1$.
|
USAMO-2016-notes_3
|
Let $ABC$ be an acute triangle
and let $I_B$, $I_C$, and $O$ denote its
$B$-excenter, $C$-excenter, and circumcenter, respectively.
Points $E$ and $Y$ are selected on $\ol{AC}$ such that
$\angle ABY = \angle CBY$ and $\ol{BE} \perp \ol{AC}$.
Similarly, points $F$ and $Z$ are selected on $\ol{AB}$ such that
$\angle ACZ = \angle BCZ$ and $\ol{CF} \perp \ol{AB}$.
Lines $I_B F$ and $I_C E$ meet at $P$.
Prove that $\ol{PO}$ and $\ol{YZ}$ are perpendicular.
|
We present two solutions.
\paragraph{First solution.}
Let $I_A$ denote the $A$-excenter and $I$ the incenter.
Then let $D$ denote the foot of the altitude from $A$.
Suppose the $A$-excircle is tangent to $\ol{BC}$, $\ol{AB}$, $\ol{AC}$
at $A_1$, $B_1$, $C_1$ and let
$A_2$, $B_2$, $C_2$ denote the reflections of $I_A$ across these points.
Let $S$ denote the circumcenter of $\triangle I I_B I_C$.
\begin{center}
\begin{asy}
size(14cm);
pair A = dir(110);
pair B = dir(208);
pair C = dir(332);
draw(A--B--C--cycle, blue+1);
pair I = incenter(A, B, C);
pair O = circumcenter(A, B, C);
pair E = foot(B, C, A);
pair F = foot(C, A, B);
pair M_A = extension(A, I, O, B+C);
pair I_A = 2*M_A-I;
pair M_B = extension(B, I, O, C+A);
pair I_B = 2*M_B-I;
pair M_C = extension(C, I, O, A+B);
pair I_C = 2*M_C-I;
pair P = extension(E, I_C, F, I_B);
draw(A--I_A, blue+dashed);
pair D = foot(A, B, C);
filldraw(I--I_B--I_C--cycle, opacity(0.2)+lightred, red);
filldraw(unitcircle, opacity(0.1)+lightblue, blue);
pair A_1 = foot(I_A, B, C);
pair B_1 = foot(I_A, A, B);
pair C_1 = foot(I_A, C, A);
draw(arc(I_A, abs(I_A-A_1), 15, 165), blue+dotted);
draw(C--C_1, blue+dashed);
draw(B--B_1, blue+dashed);
pair A_2 = 2*A_1-I_A;
pair B_2 = 2*B_1-I_A;
pair C_2 = 2*C_1-I_A;
draw(D--A_2, heavygreen);
draw(I_B--B_2, heavygreen);
draw(I_C--C_2, heavygreen);
draw(I_A--A_2, heavygreen+dashed);
draw(I_A--B_2, heavygreen+dashed);
draw(I_A--C_2, heavygreen+dashed);
draw(I_B--I_A--I_C, red+dashed);
draw(B--I--C, red+dashed);
filldraw(A_2--B_2--C_2--cycle, opacity(0.2)+lightgreen, deepgreen+1);
pair S = circumcenter(I, I_B, I_C);
draw(S--I_A, orange);
draw(arc(S, abs(S-I), 185, 355), red+dotted);
pair Y = extension(B, I, A, C);
pair Z = extension(C, I, A, B);
draw(Y--Z, red);
dot("$A$", A, dir(A));
dot("$B$", B, dir(180));
dot("$C$", C, dir(0));
dot("$I$", I, dir(270));
dot("$O$", O, dir(-45));
dot("$E$", E, dir(30));
dot("$F$", F, dir(F));
dot("$I_A$", I_A, dir(I_A));
dot("$I_B$", I_B, dir(I_B));
dot("$I_C$", I_C, dir(I_C));
dot("$P$", P, dir(135));
dot("$D$", D, dir(-90));
dot("$A_1$", A_1, dir(A_1));
dot("$B_1$", B_1, dir(200));
dot("$C_1$", C_1, dir(20));
dot("$A_2$", A_2, dir(A_2));
dot("$B_2$", B_2, dir(B_2));
dot("$C_2$", C_2, dir(C_2));
dot("$S$", S, dir(S));
dot("$Y$", Y, dir(350));
dot("$Z$", Z, dir(110));
/* Source generated by TSQ
!size(14cm);
A = dir 110
B = dir 208 R180
C = dir 332 R0
A--B--C--cycle blue+1
I = incenter A B C R270
O = circumcenter A B C R-45
E = foot B C A R30
F = foot C A B
M_A := extension A I O B+C
I_A = 2*M_A-I
M_B := extension B I O C+A
I_B = 2*M_B-I
M_C := extension C I O A+B
I_C = 2*M_C-I
P = extension E I_C F I_B R140
A--I_A blue dashed
D = foot A B C R-90
I--I_B--I_C--cycle 0.2 lightred / red
unitcircle 0.1 lightblue / blue
A_1 = foot I_A B C
B_1 = foot I_A A B R200
C_1 = foot I_A C A R20
!draw(arc(I_A, abs(I_A-A_1), 15, 165), blue+dotted);
C--C_1 blue dashed
B--B_1 blue dashed
A_2 = 2*A_1-I_A
B_2 = 2*B_1-I_A
C_2 = 2*C_1-I_A
D--A_2 heavygreen
I_B--B_2 heavygreen
I_C--C_2 heavygreen
I_A--A_2 heavygreen dashed
I_A--B_2 heavygreen dashed
I_A--C_2 heavygreen dashed
I_B--I_A--I_C red dashed
B--I--C red dashed
A_2--B_2--C_2--cycle 0.2 lightgreen / deepgreen+1
S = circumcenter I I_B I_C
S--I_A orange
!draw(arc(S, abs(S-I), 185, 355), red+dotted);
Y = extension B I A C R350
Z = extension C I A B R110
Y--Z red
*/
\end{asy}
\end{center}
We begin with the following observation:
\begin{claim*}
Points $D$, $I$, $A_2$ are collinear.
Similarly, points $E$, $I_C$, $C_2$ are collinear
and points $F$, $I_B$, $B_2$ are collinear.
\end{claim*}
\begin{proof}
This basically follows from the ``midpoints of altitudes'' lemma.
To see $D$, $I$, $A_2$ are collinear,
recall first that $\ol{IA_1}$ passes
through the midpoint $M$ of $\ol{AD}$.
\begin{center}
\begin{asy}
pair A = dir(125);
pair B = dir(220);
pair C = dir(320);
draw(A--B--C--cycle, blue);
pair I = incenter(A, B, C);
pair O = circumcenter(A, B, C);
pair M_A = extension(A, I, O, B+C);
pair I_A = 2*M_A-I;
filldraw(incircle(A, B, C), opacity(0.2)+lightblue, blue);
pair D = foot(A, B, C);
filldraw(unitcircle, opacity(0.1)+palecyan, blue);
draw(A--D, heavygreen);
pair A_1 = foot(I_A, B, C);
draw(arc(I_A, abs(I_A-A_1), 0, 180), blue+dashed);
pair B_1 = foot(I_A, A, B);
pair C_1 = foot(I_A, C, A);
draw(C--C_1, blue+dotted);
draw(B--B_1, blue+dotted);
pair A_2 = 2*A_1-I_A;
draw(A--I_A, blue);
draw(D--A_2, red);
draw(I_A--A_2, blue+dotted);
pair P = foot(I, B, C);
pair Q = 2*I-P;
draw(A--A_1, heavygreen);
draw(P--Q, heavygreen);
pair U = extension(Q, Q+B-C, A, B);
pair V = extension(Q, Q+C-B, A, C);
draw(U--V, heavygreen);
pair M = midpoint(A--D);
draw(M--A_1, dotted);
dot(P); dot(Q);
dot("$A$", A, dir(A));
dot("$B$", B, dir(180));
dot("$C$", C, dir(0));
dot("$I$", I, dir(10));
dot("$I_A$", I_A, dir(I_A));
dot("$D$", D, dir(-90));
dot("$A_1$", A_1, dir(-45));
dot("$B_1$", B_1, dir(200));
dot("$C_1$", C_1, dir(20));
dot("$A_2$", A_2, dir(A_2));
dot("$M$", M, dir(0));
/* TSQ Source:
A = dir 125
B = dir 220 R180
C = dir 320 R0
A--B--C--cycle blue
I = incenter A B C R10
O := circumcenter A B C R45
M_A := extension A I O B+C
I_A = 2*M_A-I
incircle A B C 0.2 lightblue / blue
D = foot A B C R-90
unitcircle 0.1 palecyan / blue
A--D heavygreen
A_1 = foot I_A B C R-45
!draw(arc(I_A, abs(I_A-A_1), 0, 180), blue+dashed);
B_1 = foot I_A A B R200
C_1 = foot I_A C A R20
C--C_1 blue dotted
B--B_1 blue dotted
A_2 = 2*A_1-I_A
A--I_A blue
D--A_2 red
I_A--A_2 blue dotted
P := foot I B C
Q := 2*I-P
A--A_1 heavygreen
P--Q heavygreen
U := extension Q Q+B-C A B
V := extension Q Q+C-B A C
U--V heavygreen
M = midpoint A--D R0
M--A_1 dotted
!dot(P); dot(Q);
*/
\end{asy}
\end{center}
Now since $\ol{AD} \parallel \ol{I_A A_2}$,
and $M$ and $A_1$ are the midpoints of $\ol{AD}$ and $\ol{I_A A_2}$,
it follows from the collinearity of $A$, $I$, $I_A$
that $D$, $I$, $A_2$ are collinear as well.
The other two claims follow in a dual fashion.
For example, using the homothety taking the $A$ to $C$-excircle,
we find that $\ol{C_1I_C}$ bisects the altitude $\ol{BE}$.
(The homothety at $B$ will send $C_1$ to the point on the $C$-excircle which is
diametrically opposite the foot from $I_C$ to $AC$. This diameter of the
$C$-excircle is then mapped through homothety through $C_1$ onto the $B$-altitude.)
Then since $I_C$, $B$, $I_A$ are collinear the same argument
now gives $I_C$, $E$, $C_2$ are collinear.
The fact that $I_B$, $F$, $B_2$ are collinear is symmetric.
\end{proof}
Observe that $\ol{B_2C_2} \parallel \ol{B_1C_1} \parallel \ol{I_B I_C}$.
Proceeding similarly on the other sides,
we discover $\triangle I I_B I_C$ and $\triangle A_2 B_2 C_2$ are homothetic.
Hence $P$ is the center of this homothety (in particular, $D$, $I$, $P$, $A_2$ are collinear).
Moreover, $P$ lies on the line joining $I_A$ to $S$,
which is the Euler line of $\triangle I I_B I_C$,
so it passes through the nine-point center of $\triangle I I_B I_C$,
which is $O$. Consequently, $P$, $O$, $I_A$ are collinear as well.
To finish, we need only prove that $\ol{OS} \perp \ol{YZ}$.
In fact, we claim that $\ol{YZ}$ is the radical axis
of the circumcircles of $\triangle ABC$ and $\triangle I I_B I_C$.
Actually, $Y$ is the radical center of these two circumcircles
and the circle with diameter $\ol{II_B}$
(which passes through $A$ and $C$).
Analogously $Z$ is the radical center of the circumcircles
and the circle with diameter $\ol{II_C}$, and the proof is complete.
\paragraph{Second solution (barycentric, outline, Colin Tang).}
we are going to use barycentric coordinates
to show that the line through $O$ perpendicular to $\ol{YZ}$
is concurrent with $\ol{I_B F}$ and $\ol{I_C E}$.
The displacement vector $\overrightarrow{YZ}$
is proportional to $(a(b-c):-b(a+c):c(a+b))$,
and so by strong perpendicularity criterion
and doing a calculation gives the line
\[ x(b-c)bc(a+b+c) + y(a+c)ac(a+b-c) + z(a+b)ab(-a+b-c) = 0. \]
On the other hand, line $I_C E$ has equation
\[
0 =
\det \begin{bmatrix}
a & b & -c \\
S_C & 0 & S_A \\
x & y & z
\end{bmatrix}
= bS_a \cdot x + (-cS_C-aS_A) \cdot y + (-bS_C) \cdot z
\]
and similarly for $I_B F$.
Consequently, concurrence of these lines is equivalent to
\[
\det \begin{bmatrix}
bS_A & -cS_C - aS_A & -bS_C \\
cS_A & -cS_B & -aS_A - bS_B \\
(b-c)bc(a+b+c) & (a+c)ac(a+b-c) & (a+b)ab(-a+b-c)
\end{bmatrix}
= 0
\]
which is a computation.
\paragraph{Authorship comments.}
I was intrigued by a Taiwan TST problem which implied that,
in the configuration above, $\angle I_B D I_C$ was bisected by $\ol{DA}$.
This motivated me to draw all three properties above where $I_A$ and $P$
were isogonal conjugates with respect to $DEF$.
After playing around with this picture for a long time,
I finally noticed that $O$ was on line $PI_A$.
(So the original was to show that $I_B F$, $I_C E$, $DA_2$ concurrent).
Eventually I finally noticed in the picture that $PI_A$
actually passed through the circumcenter of $ABC$ as well.
This took me many hours to prove.
The final restatement (which follows quickly from $P$, $O$, $I_A$ collinear)
was discovered by Telv Cohl when I showed him the problem.
|
USAMO-2016-notes_4
|
Find all functions $f \colon \RR \to \RR$ such that
for all real numbers $x$ and $y$,
\[ (f(x)+xy) \cdot f(x-3y) + (f(y)+xy) \cdot f(3x-y) = (f(x+y))^2. \]
|
We claim that the only two functions satisfying
the requirements are $f(x) \equiv 0$ and $f(x) \equiv x^2$.
These work.
First, taking $x=y=0$ in the given yields $f(0) = 0$,
and then taking $x=0$ gives $f(y)f(-y) = f(y)^2$.
So also $f(-y)^2 = f(y)f(-y)$, from which we conclude $f$ is even.
Then taking $x = -y$ gives
\[ \forall x \in \RR : \qquad
f(x) = x^2 \qquad\text{or}\qquad f(4x) = 0 \qquad(\bigstar) \]
for all $x$.
\begin{remark*}
Note that an example of a function satisfying $(\bigstar)$ is
\[
f(x)
=
\begin{cases}
x^2 & \text{if } |x| < 1 \\
1 - \cos \left( \frac{\pi}{2} \cdot x^{1337} \right) & \text{if } 1 \le |x| < 4 \\
0 & \text{if } |x| \ge 4.
\end{cases}
\]
So, yes, we are currently in a world of trouble, still.
(This function is even continuous; I bring this up to emphasize that
``continuity'' is completely unrelated to the issue at hand.)
\end{remark*}
Now we claim
\begin{claim*}
$f(z) = 0 \iff f(2z) = 0 \qquad (\spadesuit)$.
\end{claim*}
\begin{proof}
Let $(x,y)=(3t,t)$ in the given to get
\[ \left( f(t)+3t^2 \right)f(8t) = f(4t)^2. \]
Now if $f(4t) \neq 0$ (in particular, $t \neq 0$),
then $f(8t) \neq 0$.
Thus we have $(\spadesuit)$ in the reverse direction.
Then $f(4t) \neq 0 \overset{(\bigstar)}{\implies} f(t) = t^2 \neq 0
\overset{(\spadesuit)}{\implies} f(2t) \neq 0$
implies the forwards direction,
the last step being the reverse direction $(\spadesuit)$.
\end{proof}
%so by $(\bigstar)$ we have $f(t) = t^2$, and moreover $f(8t) \neq 0$.
%But we also have
%\[ \left( f(t/4)+\frac{3}{16}t^2 \right)f(2t) = f(t)^2 = t^4 \neq 0 \]
%and hence $f(2t) \neq 0$ as well.
%Thus we have seen that $f(4t) \neq 0 \implies f(8t), f(2t) \neq 0$
%which is logically equivalent to $(\spadesuit)$.
By putting together $(\bigstar)$ and $(\spadesuit)$ we finally get
\[ \forall x \in \RR : \qquad
f(x) = x^2 \qquad\text{or}\qquad f(x) = 0 \qquad(\heartsuit) \]
We are now ready to approach the main problem.
Assume there's an $a \neq 0$ for which $f(a) = 0$;
we show that $f \equiv 0$.
Let $b \in \RR$ be given.
Since $f$ is even, we can assume without loss of generality that $a, b > 0$.
Also, note that $f(x) \ge 0$ for all $x$ by $(\heartsuit)$.
By using $(\spadesuit)$ we can generate $c > b$
such that $f(c) = 0$ by taking $c = 2^n a$ for a large enough integer $n$.
Now, select $x, y > 0$ such that
$x-3y=b$ and $x+y=c$. That is,
\[
(x,y) = \left( \frac{3c+b}{4}, \frac{c-b}{4} \right).
\]
Substitution into the original equation gives
\[
0 = \left( f(x) + xy \right) f(b)
+ \left( f(y) + xy \right) f(3x-y)
\ge \left( f(x) + xy \right) f(b).
\]
But since $f(b) \ge 0$, it follows $f(b) = 0$, as desired.
|
USAMO-2016-notes_5
|
An equilateral pentagon $AMNPQ$ is inscribed in triangle $ABC$
such that $M \in \ol{AB}$, $Q \in \ol {AC}$, and $N,P \in \ol{BC}$.
Let $S$ be the intersection of $\ol{MN}$ and $\ol{PQ}$.
Denote by $\ell$ the angle bisector of $\angle MSQ$.
Prove that $\ol{OI}$ is parallel to $\ell$,
where $O$ is the circumcenter of triangle $ABC$,
and $I$ is the incenter of triangle $ABC$.
|
\paragraph{First solution (complex).}
In fact, we only need $AM = AQ = NP$ and $MN = QP$.
We use complex numbers with $ABC$ the unit circle,
assuming WLOG that $A$, $B$, $C$ are labeled counterclockwise.
Let $x$, $y$, $z$ be the complex numbers corresponding to the arc midpoints
of $BC$, $CA$, $AB$, respectively; thus $x+y+z$ is the incenter of $\triangle ABC$.
Finally, let $s > 0$ be the side length of $AM = AQ = NP$.
Then, since $MA = s$ and $MA \perp OZ$, it follows that
\[ m - a = i \cdot sz. \]
Similarly, $p-n = i \cdot sy$ and $a-q = i \cdot sx$, so summing these up gives
\[ i \cdot s(x+y+z) = (p-q) + (m-n) = (m-n) - (q-p). \]
Since $MN = PQ$, the argument of $(m-n) - (q-p)$ is along
the external angle bisector of the angle formed, which is perpendicular to $\ell$.
On the other hand, $x+y+z$ is oriented in the same direction as $OI$, as desired.
\paragraph{Second solution (trig, Danielle Wang).}
Let $\delta$ and $\epsilon$ denote $\angle MNB$ and $\angle CPQ$.
Also, assume $AMNPQ$ has side length $1$.
In what follows, assume $AB < AC$.
First, we note that
\begin{align*}
BN &= (c-1) \cos B + \cos \delta \textrm{,} \\
CP &= (b-1) \cos C + \cos \epsilon \textrm{, and} \\
a &= 1 + BN + CP
\end{align*}
from which it follows that
\[
\cos \delta + \cos \epsilon = \cos B + \cos C - 1
\]
Also, by the Law of Sines, we have $\frac{c-1}{\sin\delta} = \frac{1}{\sin B}$
and similarly on triangle $CPQ$, and from this we deduce
\[ \sin \epsilon - \sin \delta = \sin B - \sin C. \]
The sum-to-product formulas
\begin{align*}
\sin{\epsilon} - \sin{\delta} &= 2 \cos\left( \frac{\epsilon + \delta}{2} \right) \sin\left( \frac{\epsilon - \delta}{2} \right) \\
\cos{\epsilon} - \cos{\delta} &= 2 \cos\left( \frac{\epsilon + \delta}{2} \right) \cos\left( \frac{\epsilon - \delta}{2} \right)
\end{align*}
give us
\[
\tan \left( \frac{\epsilon-\delta}{2} \right) = \frac{\sin{\epsilon} - \sin{\delta}}{\cos{\epsilon} - \cos{\delta}}
= \frac{\sin B - \sin C}{\cos B + \cos C - 1}. \]
Now note that $\ell$ makes an angle of
$\frac{1}{2}(\pi+\epsilon-\delta)$ with line $BC$.
Moreover, if line $OI$ intersects line $BC$ with angle $\varphi$ then
\[ \tan\varphi = \frac{r - R \cos A}{\frac{1}{2}(b-c)}. \]
So in order to prove the result, we only need to check that
\[ \frac{r - R \cos A}{\frac{1}{2}(b-c)}
= \frac{\cos B + \cos C + 1}{\sin B - \sin C}. \]
Using the fact that $b = 2R\sin B$, $c = 2R\sin C$, this reduces to the fact that
$r/R + 1 = \cos A + \cos B + \cos C$, which is the so-called Carnot theorem.
|
USAMO-2016-notes_6
|
Integers $n$ and $k$ are given, with $n \ge k \ge 2$.
You play the following game against an evil wizard.
The wizard has $2n$ cards; for each $i=1,\dots,n$, there are two cards labeled $i$.
Initially, the wizard places all cards face down in a row, in unknown order.
You may repeatedly make moves of the following form: you point to any $k$ of the cards.
The wizard then turns those cards face up.
If any two of the cards match, the game is over and you win.
Otherwise, you must look away, while the wizard arbitrarily permutes
the $k$ chosen cards and then turns them back face-down. Then, it is your turn again.
We say this game is \emph{winnable} if there exist some positive integer $m$ and
some strategy that is guaranteed to win in at most $m$ moves,
no matter how the wizard responds.
For which values of $n$ and $k$ is the game winnable?
\end{enumerate}
|
The game is winnable if and only if $k < n$.
First, suppose $2 \le k < n$. Query the cards in positions $\left\{ 1, \dots, k \right\}$,
then $\left\{ 2, \dots, k+1 \right\}$, and so on, up to $\left\{ 2n-k+1, 2n \right\}$.
Indeed, by taking the difference of the $i$th and $(i+1)$st query,
we can deduce the value of the $i$th card, for $1 \le i \le 2n-k$.
(This is possible because the cards are flipped face up before they are re-shuffled,
so even if two adjacent queries return the same set,
one can still determine the $i$th card.
It is possible to solve the problem even without the flipped information, though.)
If $k \le n$, this is more than $n$ cards, so we can find a matching pair.
For $k = n$ we remark the following:
at each turn after the first, assuming one has not won,
there are $n$ cards representing each of the $n$ values exactly once,
such that the player has no information about the order of those $n$ cards.
We claim that consequently the player cannot guarantee victory.
Indeed, let $S$ denote this set of $n$ cards, and $\ol{S}$ the other $n$ cards.
The player will never win by picking only cards in $S$ or $\ol{S}$.
Also, if the player selects some cards in $S$ and some cards in $\ol{S}$,
then it is possible that the choice of cards in $S$ is exactly the complement
of those selected from $\ol{S}$; the strategy cannot prevent this since
the player has no information on $S$. This implies the result.
|
USAMO-2017-notes_1
|
Prove that there exist infinitely many pairs of
relatively prime positive integers $a,b > 1$
for which $a+b$ divides $a^b+b^a$.
|
One construction: let $d \equiv 1 \pmod 4$, $d > 1$.
Let $x = \frac{d^d+2^d}{d+2}$. Then set
\[ a = \frac{x+d}{2}, \qquad
b = \frac{x-d}{2}. \]
To see this works, first check that $b$ is odd and $a$ is even.
Let $d = a-b$ be odd.
Then:
\begin{align*}
a+b \mid a^b+b^a &\iff
(-b)^b + b^a \equiv 0 \pmod{a+b} \\
&\iff b^{a-b} \equiv 1 \pmod{a+b} \\
&\iff b^d \equiv 1 \pmod{d+2b} \\
&\iff (-2)^d \equiv d^d \pmod{d+2b} \\
&\iff d+2b \mid d^d + 2^d.
\end{align*}
So it would be enough that
\[ d+2b = \frac{d^d+2^d}{d+2}
\implies b = \half \left( \frac{d^d+2^d}{d+2} - d \right) \]
which is what we constructed.
Also, since $\gcd(x,d) = 1$ it follows $\gcd(a,b) = \gcd(d,b) = 1$.
\begin{remark*}
Ryan Kim points out that in fact,
$(a,b) = (2n-1,2n+1)$ is always a solution.
\end{remark*}
|
USAMO-2017-notes_2
|
Let $m_1$, $m_2$, \dots, $m_n$ be a collection of $n$ positive integers,
not necessarily distinct.
For any sequence of integers $A = (a_1, \dots, a_n)$
and any permutation $w = w_1, \dots, w_n$ of $m_1, \dots, m_n$,
define an $A$-inversion of $w$ to be a pair of
entries $w_i, w_j$ with $i < j$ for which
one of the following conditions holds:
\begin{itemize}
\ii $a_i \ge w_i > w_j$,
\ii $w_j > a_i \ge w_i$, or
\ii $w_i > w_j > a_i$.
\end{itemize}
Show that, for any two sequences of integers
$A = (a_1, \dots, a_n)$ and $B = (b_1, \dots, b_n)$,
and for any positive integer $k$, the number of permutations
of $m_1, \dots, m_n$ having exactly $k$ $A$-inversions
is equal to the number of permutations of $m_1, \dots, m_n$
having exactly $k$ $B$-inversions.
|
The following solution was posted by Michael Ren,
and I think it is the most natural one
(since it captures all the combinatorial ideas
using a $q$-generating function that is easier to think about,
and thus makes the problem essentially a long computation).
Denote by $M$ our multiset of $n$ positive integers.
Define an \emph{inversion} of a permutation to be pair $i < j$
with $w_i < w_j$ (which is a $(0,\dots,0)$-inversion in the problem statement);
this is the usual definition
(see \url{https://en.wikipedia.org/wiki/Inversion_(discrete_mathematics)}).
So we want to show the number of $A$-inversions
is equal to the number of usual inversions.
In what follows we count permutations on $M$ with multiplicity:
so $M = \{1,1,2\}$ still has $3!=6$ permutations.
We are going to do what is essentially recursion,
but using generating functions in a variable $q$ to do our book-keeping.
(Motivation: there's no good closed form for the number of inversions,
but there's a great generating function known --- which is even better
for us, since we're only trying to show two numbers are equal!)
First, we prove two claims.
\begin{claim*}
For any positive integer $n$,
the generating function for the number of permutations
of $(1, 2, \dots, n)$ with exactly $k$ inversions is
\[ n!_q \coloneq 1 \cdot (1+q) \cdot (1+q+q^2) \cdot \dots (1+q+\dots+q^{n-1}). \]
\end{claim*}
Here we mean that the coefficient of $q^s$ above
gives the number of permutations with exactly $s$ inversions.
\begin{proof}
This is an induction on $n$, with $n=1$ being trivial.
Suppose we choose the first element to be $i$, with $1 \le i \le n$.
Then there will always be exactly $i-1$ inversions
using the first element,
so this contributes $q^i \cdot (n-1)!_q$.
Summing $1 \le i \le n$ gives the result.
\end{proof}
Unfortunately, the main difficulty of the problem
is that there are repeated elements,
which makes our notation much more horrific.
Let us define the following.
We take our given multiset $M$ of $n$ positive integers,
we suppose the distinct numbers are
$\theta_1 < \theta_2 < \dots < \theta_m$.
We let $e_i$ be the number of times $\theta_i$ appears.
Therefore the multiplicities $e_i$ should have sums
\[ e_1 + \dots + e_m = n \]
and $m$ denotes the number of distinct elements.
Finally, we let
\[ F(e_1, \dots, e_m)
= \sum_{\text{permutations } \sigma}
q^{\text{number inversions of } \sigma} \]
be the associated generating function for the number of inversions.
For example, the first claim we proved says that $F(1, \dots, 1) = n!_q$.
\begin{claim*}
We have the explicit formula
\[ F(e_1, \dots, e_m) = n!_q \cdot \prod_{i=1}^m \frac{e_i!}{e_i!_q}. \]
\end{claim*}
\begin{proof}
First suppose we perturb all the elements slightly,
so that they are no longer equal.
Then the generating function would just be $n!_q$.
Then, we undo the perturbations for each group,
one at a time, and claim that we get the above $e_i!_q$ factor each time.
Indeed, put the permutations into classes of $e_1!$ each
where permutations in the same classes differ
only in the order of the perturbed $\theta_1$'s
(with the other $n-e_1$ elements being fixed).
Then there is a factor of $e_1!_q$ from each class,
owing to the slightly perturbed inversions we added within each class.
So we remove that factor and add $e_1! \cdot q^0$ instead.
This accounts for the first term of the product.
Repeating this now with each term of the product implies the claim.
\end{proof}
Thus we have the formula for the number of inversions in general.
We wish to show this also equals the generating function
the number of $A$-inversions, for any fixed choice of $A$.
This will be an induction by $n$, with the base case being immediate.
For the inductive step, fix $A$,
and assume the first element satisfies
$\theta_k \le a_1 < \theta_{k+1}$ (so $0 \le k \le m$;
we for convenience set $\theta_0 = -\infty$ and $\theta_m = +\infty$).
We count the permutations based on what the first element
$\theta_i$ of the permutation is.
Then:
\begin{itemize}
\ii Consider permutations starting with
$\theta_i \in \left\{ \theta_1, \dots, \theta_k \right\}$.
Then the number of inversions which will
use this first term is $(e_1 + \dots + e_{i-1})
+ (e_{k+1} + \dots + e_m)$.
Also, there are $e_i$ ways to pick which
$\theta_i$ gets used as the first term.
So we get a contribution of
\[ q^{e_1 + \dots + e_{i-1} + (e_{k+1} + \dots + e_m)}
\cdot e_i \cdot F(e_1, \dots, e_i-1, \dots, e_m) \]
in this case
(with inductive hypothesis to get the last $F$-term).
\ii Now suppose $\theta_i \in \left\{ \theta_{k+1}, \dots, \theta_m \right\}$.
Then the number of inversions which will
use this first term is $e_{k+1} + \dots + e_{i-1}$.
Thus by a similar argument the contribution is
\[ q^{e_{k+1} + \dots + e_{i-1}}
\cdot e_i \cdot F(e_1, \dots, e_i-1, \dots, e_m). \]
\end{itemize}
Therefore, to complete the problem it suffices to prove
\begin{align*}
&\phantom+ \sum_{i=1}^k q^{(e_1 + \dots + e_{i-1}) + (e_{k+1} + \dots + e_m)}
\cdot e_i \cdot F(e_1, \dots, e_i-1, \dots, e_m) \\
&+ \sum_{i=k+1}^m q^{e_{k+1} + \dots + e_{i-1}}
\cdot e_i \cdot F(e_1, \dots, e_i-1, \dots, e_m) \\
&= F(e_1, \dots, e_m).
\end{align*}
Now, we see that
\[ \frac{e_i \cdot F(e_1, \dots, e_i-1, \dots, e_m)}{F(e_1, \dots, e_m)}
= \frac{1+\dots+q^{e_i-1}}{1+q+\dots+q^{n-1}}
= \frac{1-q^{e_i}}{1-q^n} \]
so it's equivalent to show
\[ 1-q^n
= q^{e_{k+1} + \dots + e_m} \sum_{i=1}^k q^{e_1 + \dots + e_{i-1}}(1-q^{e_i})
+ \sum_{i=k+1}^m q^{e_{k+1} + \dots + e_{i-1}} (1-q^{e_i})
\]
which is clear,
since the left summand telescopes to
$q^{e_{k+1}+\dots+e_m} - q^n$
and the right summand telescopes to
$1 - q^{e_{k+1}+\dots+e_m}$.
\begin{remark*}
Technically, we could have skipped straight to the induction,
without proving the first two claims.
However I think the solution reads more naturally this way.
\end{remark*}
|
USAMO-2017-notes_3
|
Let $ABC$ be a scalene triangle with circumcircle $\Omega$ and incenter $I$.
Ray $AI$ meets $\ol{BC}$ at $D$ and $\Omega$ again at $M$;
the circle with diameter $\ol{DM}$ cuts $\Omega$ again at $K$.
Lines $MK$ and $BC$ meet at $S$, and $N$ is the midpoint of $\ol{IS}$.
The circumcircles of $\triangle KID$ and $\triangle MAN$
intersect at points $L_1$ and $L_2$.
Prove that $\Omega$ passes through the midpoint of either $\ol{IL_1}$ or $\ol{IL_2}$.
|
Let $W$ be the midpoint of $\ol{BC}$,
let $X$ be the point on $\Omega$ opposite $M$.
Observe that $\ol{KD}$ passes through $X$,
and thus lines $BC$, $MK$, $XA$ concur at
the orthocenter of $\triangle DMX$, which we call $S$.
Denote by $I_A$ the $A$-excenter of $ABC$.
Next, let $E$ be the foot of the altitude from $I$ to $\ol{XI_A}$;
observe that $E$ lies on the circle centered at $M$
through $I$, $B$, $C$, $I_A$.
Then, $S$ is the radical center of $\Omega$
and the circles with diameter $\ol{IX}$ and $\ol{II_A}$;
hence line $SI$ passes through $E$;
accordingly $I$ is the orthocenter of $\triangle XSI_A$;
denote by $L$ the foot from $X$ to $\ol{SI_A}$.
\begin{center}
\begin{asy}
size(10cm);
pair A = dir(160);
pair B = dir(210);
pair C = dir(330);
pair I = incenter(A, B, C);
pair D = extension(A, I, B, C);
pair M = circumcenter(B, I, C);
pair I_A = 2*M-I;
pair X = -M;
pair K = foot(M, D, X);
pair S = extension(M, K, B, C);
pair N = midpoint(I--S);
pair W = midpoint(B--C);
pair E = foot(I, X, I_A);
draw(A--B--C--cycle, orange);
filldraw(unitcircle, opacity(0.1)+lightred, orange);
filldraw(CP(M, I), opacity(0.1)+yellow, olive);
draw(A--I_A, orange);
draw(M--X, orange);
draw(S--E, olive);
draw(X--I_A, olive);
draw(B--S, orange);
pair L = foot(X, I_A, S);
draw(M--S--X--cycle, heavyred+1);
draw(X--K, heavyred+1);
draw(A--M, heavyred+1);
draw(S--W, heavyred+1);
draw(S--I_A, orange);
draw(circumcircle(K, I, D), palered+1);
filldraw(circumcircle(M, A, N), opacity(0.1)+yellow, palered+1);
draw(X--L, orange+1);
pair T = midpoint(I--L);
draw(T--M, orange);
dot("$A$", A, dir(150));
dot("$B$", B, dir(B));
dot("$C$", C, dir(-10));
dot("$I$", I, dir(I));
dot("$D$", D, dir(D));
dot("$M$", M, dir(M));
dot("$I_A$", I_A, dir(I_A));
dot("$X$", X, dir(X));
dot("$K$", K, dir(K));
dot("$S$", S, dir(S));
dot("$N$", N, dir(135));
dot("$W$", W, dir(-45));
dot("$E$", E, dir(45));
dot("$L$", L, dir(L));
dot("$T$", T, dir(T));
/* Source generated by TSQ:
!size(10cm);
A = dir 160 R150
B = dir 210
C = dir 330 R-10
I = incenter A B C
D = extension A I B C
M = circumcenter B I C
I_A = 2*M-I
X = -M
K = foot M D X
S = extension M K B C
N = midpoint I--S R135
W = midpoint B--C R-45
E = foot I X I_A R45
A--B--C--cycle orange
unitcircle 0.1 lightred / orange
CP M I 0.1 yellow / olive
A--I_A orange
M--X orange
S--E olive
X--I_A olive
B--S orange
L = foot X I_A S
M--S--X--cycle heavyred+1
X--K heavyred+1
A--M heavyred+1
S--W heavyred+1
S--I_A orange
circumcircle K I D palered+1
circumcircle M A N 0.1 yellow / palered+1
X--L orange+1
T = midpoint I--L
T--M orange
*/
\end{asy}
\end{center}
We claim that this $L$ lies on
both the circumcircle of $\triangle KID$ and $\triangle MAN$.
It lies on the circumcircle of $\triangle MAN$
since this circle is the nine-point circle of $\triangle XSI_A$.
Also, $XD \cdot XK = XW \cdot XM = XA \cdot XS = XI \cdot XL$,
so $KDIL$ are concyclic.
%For the other, note that $\triangle MWI \sim \triangle MIX$,
%according to which $\angle IWM = \angle MIX = 180^{\circ} - \angle LIM = 180^{\circ} - \angle MLI$,
%enough to imply that quadrilateral $MWIL$ is cyclic.
%But lines $IL$, $DK$, and $WM$ meet at $X$, implying the conclusion.
All that remains to show is that the
midpoint $T$ of $\ol{IL}$ lies on $\Omega$.
But this follows from the fact that
$\ol{TM} \parallel \ol{LI_A} \implies \angle MTX = 90^{\circ}$,
thus the problem is solved.
\begin{remark*}
Some additional facts about this picture:
the point $T$ is the contact point of the $A$-mixtilinear incircle
(since it is collinear with $X$ and $I$),
while the point $K$ is such that $\ol{AK}$ is an $A$-symmedian
(since $\ol{KD}$ and $\ol{AD}$ bisect $\angle A$ and $\angle K$, say).
\end{remark*}
\begin{center}
\begin{asy}
pair I_A = dir(115);
pair I_B = dir(215);
pair I_C = dir(325);
pair A = foot(I_A, I_B, I_C);
pair B = foot(I_B, I_C, I_A);
pair C = foot(I_C, I_A, I_B);
pair X = midpoint(I_B--I_C);
pair S = extension(B, C, I_B, I_C);
pair I = incenter(A, B, C);
pair L = extension(X, I, S, I_A);
pair D = extension(A, I, B, C);
pair M = midpoint(I--I_A);
pair K = foot(M, D, X);
pair N = midpoint(I--S);
draw(I_A--I_B--I_C--cycle, heavycyan);
filldraw(circumcircle(I_A, I_B, I_C), opacity(0.1)+palered, heavycyan);
filldraw(circumcircle(A, B, C), opacity(0.1)+orange, orange);
draw(A--B--C--cycle, orange);
draw(S--I_A, orange);
draw(S--I, olive);
draw(S--I_B, heavycyan);
draw(A--I_A, heavycyan);
filldraw(circumcircle(K, I, D), opacity(0.07)+yellow, palered+1);
filldraw(circumcircle(M, A, N), opacity(0.07)+yellow, palered+1);
draw(X--L, orange+1);
pair T = midpoint(I--L);
draw(M--X, olive);
draw(K--X, olive);
draw(M--S, olive);
draw(X--I_A, heavycyan);
draw(S--C, orange);
filldraw(circumcircle(I, B, C), opacity(0.1)+palecyan, heavycyan);
dot("$I_A$", I_A, dir(I_A));
dot("$I_B$", I_B, dir(I_B));
dot("$I_C$", I_C, dir(I_C));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$X$", X, dir(X));
dot("$S$", S, dir(S));
dot("$I$", I, dir(I));
dot("$L$", L, dir(L));
dot("$D$", D, dir(D));
dot("$M$", M, dir(M));
dot("$K$", K, dir(K));
dot("$N$", N, dir(80));
dot("$T$", T, dir(T));
/* TSQ Source:
I_A = dir 115
I_B = dir 215
I_C = dir 325
A = foot I_A I_B I_C
B = foot I_B I_C I_A
C = foot I_C I_A I_B
X = midpoint I_B--I_C
S = extension B C I_B I_C
I = incenter A B C
L = extension X I S I_A
D = extension A I B C
M = midpoint I--I_A
K = foot M D X
N = midpoint I--S R80
I_A--I_B--I_C--cycle heavycyan
circumcircle I_A I_B I_C 0.1 palered / heavycyan
circumcircle A B C 0.1 orange / orange
A--B--C--cycle orange
S--I_A orange
S--I olive
S--I_B heavycyan
A--I_A heavycyan
circumcircle K I D 0.07 yellow / palered+1
circumcircle M A N 0.07 yellow / palered+1
X--L orange+1
T = midpoint I--L
M--X olive
K--X olive
M--S olive
X--I_A heavycyan
S--C orange
circumcircle I B C 0.1 palecyan / heavycyan
*/
\end{asy}
\end{center}
\begin{remark*}
In fact, the point $L$ is the Miquel point of cyclic quadrilateral
$I_B I_C B C$ (inscribed in the circle with diameter $\ol{I_B I_C}$).
This implies many of the properties that $L$ has above.
For example, it directly implies that $L$ lies on the circumcircles
of triangles $I_A I_B I_C$ and $BCI_A$,
and that the point $L$ lies on $\ol{SI_A}$
(since $S = \ol{BC} \cap \ol{I_B I_C}$).
For this reason, many students found it easier to think about
the problem in terms of $\triangle I_A I_B I_C$ rather than $\triangle ABC$.
\end{remark*}
|
USAMO-2017-notes_4
|
Let $P_1$, $P_2$, \dots, $P_{2n}$ be $2n$ distinct points on the
unit circle $x^2+y^2=1$, other than $(1,0)$.
Each point is colored either red or blue,
with exactly $n$ red points and $n$ blue points.
Let $R_1$, $R_2$, \dots, $R_n$ be any ordering of the red points.
Let $B_1$ be the nearest blue point to $R_1$ traveling
counterclockwise around the circle starting from $R_1$.
Then let $B_2$ be the nearest of the remaining blue points to $R_2$
traveling counterclockwise around the circle from $R_2$, and so on,
until we have labeled all of the blue points $B_1$, \dots, $B_n$.
Show that the number of counterclockwise arcs of the form $R_i \to B_i$
that contain the point $(1,0)$ is independent of the way we chose the
ordering $R_1$, \dots, $R_n$ of the red points.
|
We present two solutions, one based on
swapping and one based on an invariant.
\paragraph{First ``local'' solution by swapping two points.}
Let $1 \le i < n$ be any index and consider the two red points
$R_i$ and $R_{i+1}$.
There are two blue points $B_i$ and $B_{i+1}$ associated with them.
\begin{claim*}
If we swap the locations of points $R_i$ and $R_{i+1}$ then
the new arcs $R_i \to B_i$ and $R_{i+1} \to B_{i+1}$
will cover the same points.
\end{claim*}
\begin{proof}
Delete all the points $R_1$, \dots, $R_{i-1}$
and $B_1$, \dots, $B_{i-1}$;
instead focus on the positions of $R_i$ and $R_{i+1}$.
The two blue points can then be located in three possible ways:
either $0$, $1$, or $2$ of them lie on the arc $R_i \to R_{i+1}$.
For each of the cases below, we illustrate on the left
the locations of $B_i$ and $B_{i+1}$
and the corresponding arcs in green;
then on the right we show the modified picture
where $R_i$ and $R_{i+1}$ have swapped.
(Note that by hypothesis there are no other blue points in the green arcs).
\begin{center}
\begin{asy}
unitsize(1cm);
pair O = (0,0);
picture init(bool flip) {
picture pic;
draw(pic, unitcircle);
if (flip) {
dot(pic, "$R_{i}$", dir(0), dir(0), red);
dot(pic, "$R_{i+1}$", dir(180), dir(180), red);
}
else {
dot(pic, "$R_{i+1}$", dir(0), dir(0), red);
dot(pic, "$R_{i}$", dir(180), dir(180), red);
}
return pic;
}
picture L1 = init(true);
picture L2 = init(true);
picture L3 = init(true);
picture R1 = init(false);
picture R2 = init(false);
picture R3 = init(false);
real r = 0.8;
real s = 0.9;
// Case 1
dot(L1, "$B_i$", dir(60), dir(60), blue);
dot(L1, "$B_{i+1}$", dir(120), dir(120), blue);
dot(R1, "$B_i$", dir(60), dir(60), blue);
dot(R1, "$B_{i+1}$", dir(120), dir(120), blue);
draw(L1, arc(O, r, 0, 60), deepgreen, EndArrow(TeXHead));
draw(L1, arc(O, s, 180, 480), deepgreen, EndArrow(TeXHead));
draw(R1, arc(O, r, 180, 420), deepgreen, EndArrow(TeXHead));
draw(R1, arc(O, s, 0, 120), deepgreen, EndArrow(TeXHead));
dot(L1, dir(140), blue);
dot(L1, dir(150), blue);
dot(R1, dir(140), blue);
dot(R1, dir(150), blue);
// Case 2
dot(L2, "$B_i$", dir(90), dir(90), blue);
dot(L2, "$B_{i+1}$", dir(270), dir(270), blue);
dot(R2, "$B_i$", dir(270), dir(270), blue);
dot(R2, "$B_{i+1}$", dir(90), dir(90), blue);
draw(L2, arc(O, s, 0, 90), deepgreen, EndArrow(TeXHead));
draw(L2, arc(O, s, 180, 270), deepgreen, EndArrow(TeXHead));
draw(R2, arc(O, s, 180, 270), deepgreen, EndArrow(TeXHead));
draw(R2, arc(O, s, 0, 90), deepgreen, EndArrow(TeXHead));
dot(L2, dir(128), blue);
dot(L2, dir(155), blue);
dot(R2, dir(128), blue);
dot(R2, dir(155), blue);
dot(L2, dir(297), blue);
dot(L2, dir(335), blue);
dot(R2, dir(297), blue);
dot(R2, dir(335), blue);
// Case 3
dot(L3, "$B_i$", dir(240), dir(240), blue);
dot(L3, "$B_{i+1}$", dir(300), dir(300), blue);
dot(R3, "$B_i$", dir(240), dir(240), blue);
dot(R3, "$B_{i+1}$", dir(300), dir(300), blue);
draw(L3, arc(O, r, 0, 240), deepgreen, EndArrow(TeXHead));
draw(L3, arc(O, s, 180, 300), deepgreen, EndArrow(TeXHead));
draw(R3, arc(O, r, 180, 240), deepgreen, EndArrow(TeXHead));
draw(R3, arc(O, s, 0, 300), deepgreen, EndArrow(TeXHead));
dot(L3, dir(328), blue);
dot(L3, dir(342), blue);
dot(R3, dir(328), blue);
dot(R3, dir(342), blue);
real t = 3.5;
add(shift(0,0)*L1);
add(shift(t,0)*R1);
add(shift(0,-t)*L2);
add(shift(t,-t)*R2);
add(shift(0,-2*t)*L3);
add(shift(t,-2*t)*R3);
label("Case 1", (-2.5,0));
label("Case 2", (-2.5,-t));
label("Case 3", (-2.5,-2*t));
\end{asy}
\end{center}
Observe that in all cases, the number of arcs covering
any given point on the circumference is not changed.
Consequently, this proves the claim.
\end{proof}
Finally, it is enough to recall that any permutation
of the red points can be achieved by swapping consecutive points
(put another way: $(i \; i+1)$ generates the permutation group $S_n$).
This solves the problem.
\begin{remark*}
This proof does \emph{not} work if one tries to swap
$R_i$ and $R_j$ if $|i-j| \neq 1$.
For example if we swapped $R_i$ and $R_{i+2}$
then there are some issues caused by the
possible presence of the blue point $B_{i+1}$
in the green arc $R_{i+2} \to B_{i+2}$.
\end{remark*}
\paragraph{Second longer solution using an invariant.}
Visually, if we draw all the segments $R_i \to B_i$
then we obtain a set of $n$ chords.
Say a chord is \emph{inverted} if satisfies the problem condition, and \emph{stable} otherwise.
The problem contends that the number of stable/inverted chords
depends only on the layout of the points and not on the choice of chords.
\begin{center}
\begin{asy}
size(6cm);
pair A(int i) { return dir(22.5+45*i); }
draw(unitcircle, gray);
dot("$(1,0)$", dir(0), dir(0));
dotfactor *= 2;
draw(A(7)--A(0), EndArrow, Margins);
draw(A(1)--A(2), EndArrow, Margins);
draw(A(3)--A(5), EndArrow, Margins);
draw(A(4)--A(6), EndArrow, Margins);
dot("$-1$", A(0), A(0), blue);
dot("$0$", A(1), A(1), red);
dot("$-1$", A(2), A(2), blue);
dot("$0$", A(3), A(3), red);
dot("$+1$", A(4), A(4), red);
dot("$0$", A(5), A(5), blue);
dot("$-1$", A(6), A(6), blue);
dot("$0$", A(7), A(7), red);
\end{asy}
\end{center}
In fact we'll describe the number of inverted chords explicitly.
Starting from $(1,0)$ we keep a running tally of $R-B$;
in other words we start the counter at $0$ and decrement by $1$ at each
blue point and increment by $1$ at each red point.
Let $x \le 0$ be the lowest number ever recorded. Then:
\begin{claim*}
The number of inverted chords is $-x$
(and hence independent of the choice of chords).
\end{claim*}
This is by induction on $n$.
I think the easiest thing is to delete chord $R_1 B_1$;
note that the arc cut out by this chord contains no blue points.
So if the chord was stable certainly no change to $x$.
On the other hand, if the chord is inverted,
then in particular the last point before $(1,0)$ was red, and so $x < 0$.
In this situation one sees that deleting the chord changes $x$ to $x+1$, as desired.
|
USAMO-2017-notes_5
|
Find all real numbers $c > 0$ such that there exists a labeling
of the lattice points in $\ZZ^2$ with positive integers for which:
\begin{itemize}
\ii only finitely many distinct labels occur, and
\ii for each label $i$, the distance between any two points
labeled $i$ is at least $c^i$.
\end{itemize}
|
The answer is $c < \sqrt2$. Here is a solution with Calvin Deng.
The construction for any $c < \sqrt 2$ can be done as follows.
Checkerboard color the lattice points and label the black ones with $1$.
The white points then form a copy of $\ZZ^2$ again
scaled up by $\sqrt 2$ so we can repeat the procedure with $2$
on half the resulting points.
Continue this dyadic construction until a large $N$
for which $c^N < 2^{\half (N-1)}$,
at which point we can just label all the points with $N$.
I'll now prove that $c = \sqrt 2$ (and hence $c \ge \sqrt 2$) can't be done.
\begin{claim*}
It is impossible to fill a $2^n \times 2^n$ square
with labels not exceeding $2n$.
\end{claim*}
The case $n = 1$ is clear.
So now assume it's true up to $n-1$;
and assume for contradiction a $2^n \times 2^n$ square
$S$ only contains labels up to $2n$.
(Of course every $2^{n-1} \times 2^{n-1}$ square
contains an instance of a label at least $2n-1$.)
\begin{center}
\begin{asy}
unitsize(0.75cm);
for (int i=0; i<=8; ++i) {
if ((i!=0) && (i!=4) && (i!=8)) {
draw( (0,i)--(8,i), gray );
draw( (i,0)--(i,8), gray );
}
else {
draw( (0,i)--(8,i), black+1 );
draw( (i,0)--(i,8), black+1 );
}
}
filldraw( box((2.1,4.1),(5.9,7.9)), opacity(0.2)+palered, red+1 );
filldraw( box((1.1,2.1),(4.9,5.9)), opacity(0.2)+lightgreen, blue+1 );
fill(box((0,4),(4,8)), opacity(0.1)+lightcyan);
label("$A$", (3.5,8), dir(90), red);
label("$B$", (2.5,2), dir(-90), blue);
label("$2$", (0.5,7.5));
label("$1$", (1.5,7.5));
label("$2$", (2.5,7.5));
label("$1$", (3.5,7.5));
label("$1$", (0.5,6.5));
label("$5$", (1.5,6.5), red);
label("$1$", (2.5,6.5));
label("$3$", (3.5,6.5));
label("$2$", (0.5,5.5));
label("$1$", (1.5,5.5));
label("$2$", (2.5,5.5));
label("$1$", (3.5,5.5));
label("$1$", (0.5,4.5));
label("$3$", (1.5,4.5));
label("$1$", (2.5,4.5));
label("$4$", (3.5,4.5));
label("$6$", (5.5,6.5), blue);
\end{asy}
\end{center}
Now, we contend there are fewer than four copies of $2n$:
\begin{lemma*}
In a unit square, among any four points,
two of these points have distance $\le 1$ apart.
\end{lemma*}
\begin{proof}
Look at the four rays emanating from the origin and note
that two of them have included angle $\le 90\dg$.
\end{proof}
So WLOG the northwest quadrant has no $2n$'s.
Take a $2n-1$ in the northwest and draw a square of size $2^{n-1} \times 2^{n-1}$
directly right of it (with its top edge coinciding with the top of $S$).
Then $A$ can't contain $2n-1$, so it must contain a $2n$ label;
that $2n$ label must be in the northeast quadrant.
Then we define a square $B$ of size $2^{n-1} \times 2^{n-1}$ as follows.
If $2n-1$ is at least as high $2n$, let $B$ be a $2^{n-1} \times 2^{n-1}$
square which touches $2n-1$ north and is bounded east by $2n$.
Otherwise let $B$ be the square that touches $2n-1$ west
and is bounded north by $2n$.
We then observe $B$ can neither have $2n-1$ nor $2n$ in it, contradiction.
\begin{remark*}
To my knowledge, essentially all density arguments fail
because of hexagonal lattice packing.
\end{remark*}
|
USAMO-2017-notes_6
|
Find the minimum possible value of
\[ \frac{a}{b^3+4} + \frac{b}{c^3+4} + \frac{c}{d^3+4} + \frac{d}{a^3+4} \]
given that $a$, $b$, $c$, $d$ are nonnegative real numbers
such that $a+b+c+d=4$.
\end{enumerate}
|
The minimum $\frac23$ is achieved
at $(a,b,c,d) = (2,2,0,0)$ and cyclic permutations.
The problem is an application of the tangent line trick:
we observe the miraculous identity
\[ \frac{1}{b^3+4} \ge \frac14 - \frac{b}{12} \]
since $12-(3-b)(b^3+4) = b(b+1)(b-2)^2 \ge 0$.
Moreover,
\[ ab+bc+cd+da = (a+c)(b+d)
\le \left( \frac{(a+c)+(b+d)}{2} \right)^2 = 4. \]
Thus
\[ \sum_{\text{cyc}} \frac{a}{b^3+4}
\ge \frac{a+b+c+d}{4} - \frac{ab+bc+cd+da}{12}
\ge 1 - \frac13 = \frac23. \]
\begin{remark*}
The main interesting bit is the equality at $(a,b,c,d) = (2,2,0,0)$.
This is the main motivation for trying tangent line trick,
since a lower bound of the form $\sum a(1-\lambda b)$
preserves the unusual equality case above.
Thus one takes the tangent at $b=2$ which miraculously
passes through the point $(0,1/4)$ as well.
\end{remark*}
|
USAMO-2018-notes_1
|
Let $a$, $b$, $c$ be positive real numbers such that $a+b+c = 4\sqrt[3]{abc}$.
Prove that
\[ 2(ab+bc+ca) + 4 \min (a^2, b^2, c^2) \ge a^2 + b^2 + c^2. \]
|
WLOG let $c = \min(a,b,c) = 1$ by scaling.
The given inequality becomes equivalent to
\[ 4ab + 2a + 2b + 3 \ge (a+b)^2 \qquad \forall a+b = 4(ab)^{1/3}-1. \]
Now, let $t = (ab)^{1/3}$ and eliminate $a+b$ using the condition, to get
\[ 4t^3 + 2(4t-1) + 3 \ge (4t-1)^2
\iff 0 \le 4t^3 - 16t^2 + 16t = 4t(t-2)^2 \]
which solves the problem.
Equality occurs only if $t=2$,
meaning $ab = 8$ and $a+b=7$, which gives
\[ \{a,b\} = \left\{ \frac{7 \pm \sqrt{17}}{2} \right\} \]
with the assumption $c = 1$.
Scaling gives the curve of equality cases.
|
USAMO-2018-notes_2
|
Find all functions $f \colon (0,\infty) \to (0,\infty)$ such that
\[
f\left( x+\frac 1y \right)
+ f\left( y+\frac 1z \right)
+ f\left( z+\frac 1x \right)
= 1
\]
for all $x,y,z > 0$ with $xyz = 1$.
|
The main part of the problem is to show all solutions are linear.
As always, let $x=b/c$, $y=c/a$, $z=a/b$
(classical inequality trick).
Then the problem becomes
\[ \sum_{\text{cyc}} f\left( \frac{b+c}{a} \right) = 1. \]
Let $f(t) = g(\frac{1}{t+1})$,
equivalently $g(s) = f(1/s-1)$.
Thus $g \colon (0,1) \to (0,1)$
which satisfies $\sum_{\text{cyc}} g\left( \frac{a}{a+b+c} \right) = 1$,
or equivalently
\[ \boxed{g(a) + g(b) + g(c) = 1} \qquad \forall a+b+c=1. \]
The rest of the solution is dedicated to solving this
equivalent functional equation in $g$.
It is a lot of technical details and I will only outline them
(with apologies to the contestants who didn't have that luxury).
\begin{claim*}
The function $g$ is linear.
\end{claim*}
\begin{proof}
This takes several steps, all of which are technical.
We begin by proving $g$ is linear over $[1/8, 3/8]$.
\begin{itemize}
\ii First, whenever $a+b \le 1$ we have
\[ 1 - g(1-(a+b)) = g(a) + g(b) = 2 g\left( \frac{a+b}{2} \right). \]
Hence $g$ obeys Jensen's functional equation over $(0,1/2)$.
\ii Define $h \colon [0,1] \to \RR$
by $h(t) = g(\frac{2t+1}{8}) - (1-t) \cdot g(1/8) - t \cdot g(3/8)$,
then $h$ satisfies Jensen's functional equation too over $[0,1]$.
We have also arranged that $h(0) = h(1) = 0$, hence $h(1/2) = 0$ as well.
\ii Since
\[ h(t) = h(t) + h(1/2) = 2h(t/2+1/4)
= h(t+1/2) + h(0) = h(t+1/2) \]
for any $t < 1/2$, we find $h$ is periodic modulo $1/2$.
It follows one can extend $\wt h$ by
\[ \wt h \colon \RR \to \RR \qquad\text{by}\qquad
\wt h(t) = h(t - \left\lfloor t \right\rfloor) \]
and still satisfy Jensen's functional equation.
Because $\wt h(0) = 0$, it's well-known this implies $\wt h$ is additive
(because $\wt h(x+y) = 2\wt h\left( (x+y)/2 \right) = \wt h(x) + \wt h(y)$
for any real numbers $x$ to $y$).
\end{itemize}
But $\wt h$ is bounded below on $[0,1]$ since $g \ge 0$,
and since $\wt h$ is also additive,
it follows (well-known) that $\wt h$ is linear.
Thus $h$ is the zero function.
So, the function $g$ is linear over $[1/8,3/8]$;
thus we may write $g(x) = kx + \ell$, valid for $1/8 \le x \le 3/8$.
Since $3g(1/3) = 1$, it follows $k + 3\ell = 1$.
For $0 < x < 1/8$ we have
$g(x) = 2g(0.15) - g(0.3-x) = 2(0.15k+\ell) - (k(0.3-x)+\ell) = kx+\ell$,
so $g$ is linear over $(0,3/8)$ as well.
Finally, for $3/8 < x < 1$, we use the given equation
\[
1 = g\left( \frac{1-x}{2} \right)
+ g\left( \frac{1-x}{2} \right)
+ g(x)
\implies g(x) = 1 - 2\left( k \cdot \frac{1-x}{2} + \ell \right)
= kx+\ell
\]
since $\frac{1-x}{2} < \frac{5}{16} < \frac{3}{8}$.
Thus $g$ is linear over all.
\end{proof}
Putting this back in,
we deduce that $g(x) = kx + \frac{1-k}{3}$ for some $k \in [-1/2,1]$,
and so \[ f(x) = \frac{k}{x+1} + \frac{1-k}{3} \]
for some $k \in [-1/2,1]$.
All such functions work.
|
USAMO-2018-notes_3
|
Let $n \ge 2$ be an integer, and let $\{a_1, \dots, a_m\}$ denote
the $m = \varphi(n)$ integers less than $n$ and relatively prime to $n$.
Assume that every prime divisor of $m$ also divides $n$.
Prove that $m$ divides $a_1^k + \dots + a_m^k$ for every positive
integer $k$.
|
For brevity, given any $n$, we let $A(n) = \left\{ 1 \le x \le n, \gcd(x,n) = 1 \right\}$
(thus $|A(n)| = \varphi(n)$).
Also, let $S(n,k) = \sum_{a \in A(n)} a^k$.
We will prove the stronger statement (which eliminates the hypothesis on $n$).
\begin{claim*}
Let $n \ge 2$ be arbitrary (and $k \ge 0$).
If $p \mid n$, then \[ \nu_p (\varphi(n)) \le \nu_p(S(n,k)). \]
\end{claim*}
We start with the special case where $n$ is a prime power.
\begin{lemma*}
Let $p$ be prime, $e \ge 1$, $k \ge 0$.
We always have
\[ S(p^e, k) = \sum_{x \in A(p^e)} x^k \equiv 0 \pmod{p^{e-1}}. \]
\end{lemma*}
\begin{proof}
For $p$ odd, this follows by taking a primitive root $g$ modulo $p^{e}$.
We will have
\[ S(p^e, k) \equiv 1 + g^k + g^{2k} + \dots + g^{(\varphi(p^e)-1)k}
\equiv \frac{g^{\varphi(p^e) k} - 1}{g^k - 1}. \]
If $p-1 \nmid k$, then the denominator is not divisible by $p$
and hence the entire expression is $0 \pmod{p^e}$.
In the other case where $p-1 \mid k$, since $\nu_p(\varphi(p^e)) = e-1$,
the exponent lifting lemma implies
\[ \nu_p\left( (g^k)^{\varphi(p^e)} - 1 \right) = \nu_p(g^k-1) + (e-1) \]
and so the conclusion is true here too.
In the annoying case $p = 2$, the proof is broken into two cases:
for $k$ odd it follows by pairing $x$ with $2^e-x$
and when $k$ is even one can take $5$ as a generator
of all the quadratic residues as in the $p > 2$ case.
\end{proof}
\begin{corollary*}
We have $\nu_p(1^k + \dots + t^k) \ge \nu_p(t) - 1$ for any $k$, $t$, $p$.
\end{corollary*}
\begin{proof}
Assume $p \mid t$.
Handle the terms in that sum divisible by $p$ (by induction)
and apply the lemma a bunch of times.
\end{proof}
Now the idea is to add primes $q$ one at a time to $n$, starting from the base case $n = p^e$.
So, formally we proceed by induction on the number of prime divisors of $n$.
We'll also assume $k \ge 1$ in what follows since the base case $k=0$ is easy.
\begin{itemize}
\ii First, suppose we want to go from $n$ to $nq$ where $q \nmid n$.
In that case $\varphi(nq)$ gained $\nu_p(q-1)$ factors of $p$
and then we need to show $\nu_p(S(nq,k)) \ge \nu_p(\varphi(n)) + \nu_p(q-1)$.
The trick is to write
\[ A(nq) = \left\{ a + nh \mid a \in A(n)
\text{ and } h = 0,\dots,q-1 \right\}
\setminus qA(n) \]
and then expand using binomial theorem:
\begin{align*}
S(nq, k) &= \sum_{a \in A(n)} \sum_{h=0}^{q-1} (a+nh)^k - \sum_{a \in A(n)} (qa)^k \\
&= -q^k S(n,k) + \sum_{a \in A(n)} \sum_{h=0}^{q-1} \sum_{j=0}^k
\left[ \binom kj a^{k-j} n^j h^j \right] \\
&= -q^k S(n,k) +
\sum_{j=0}^k \left[ \binom kj n^j \left( \sum_{a \in A(n)} a^{k-j} \right)
\left( \sum_{h=0}^{q-1} h^j \right) \right] \\
&= -q^k S(n,k) + \sum_{j=0}^k \left[ \binom kj n^j S(n,k-j)
\left( \sum_{h=1}^{q-1} h^j \right) \right] \\
&= (q-q^k) S(n,k)
+ \sum_{j=1}^k \left[ \binom kj n^j S(n,k-j)
\left( \sum_{h=1}^{q-1} h^j \right) \right].
\end{align*}
We claim every term here has enough powers of $p$.
For the first term, $S(n,k)$ has at least $\nu_p(\varphi(n))$ factors of $p$;
and we have the $q-q^k$ multiplier out there.
For the other terms, we apply induction to $S(n,k-j)$;
moreover $\sum_{h=1}^{q-1} h^j$ has at least $\nu_p(q-1)-1$ factors of $p$ by corollary,
and we get one more factor of $p$ (at least) from $n^j$.
\ii On the other hand, if $q$ already divides $n$, then this time
\[ A(nq) = \left\{ a + nh \mid a \in A(n)
\text{ and } h = 0,\dots,q-1 \right\}. \]
and we have no additional burden of $p$ to deal with; the same calculation gives
\[ S(nq,k) = qS(n,k)
+ \sum_{j=1}^k \left[ \binom kj n^j S(n,k-j)
\left( \sum_{h=1}^{q-1} h^j \right) \right] \]
which certainly has enough factors of $p$ already.
\end{itemize}
\begin{remark*}
A curious bit about the problem is that $\nu_p(\varphi(n))$ can exceed $\nu_p(n)$,
and so it is not true that the residues of $A(n)$ are well-behaved modulo $\varphi(n)$.
As an example, let $n = 2 \cdot 3 \cdot 7 \cdot 13 = 546$,
so $m = \varphi(n) = 1 \cdot 2 \cdot 6 \cdot 12 = 144$.
Then $A(n)$ contains $26$ elements which are $1 \bmod 9$
and $23$ elements which are $4 \bmod 9$.
\end{remark*}
\begin{remark*}
The converse of the problem is true too
(but asking both parts would make this too long for exam).
\end{remark*}
|
USAMO-2018-notes_4
|
Let $p$ be a prime, and let $a_1$, \dots, $a_p$ be integers.
Show that there exists an integer $k$ such that the numbers
\[ a_1 + k, \; a_2 + 2k, \; \dots, \; a_p + pk \]
produce at least $\half p$ distinct remainders upon division by $p$.
|
For each $k = 0, \dots, p-1$ let $G_k$ be the graph
on $\{1, \dots, p\}$ where we join $\{i,j\}$ if and only if
\[ a_i + ik \equiv a_j + jk \pmod p
\iff k \equiv - \frac{a_i - a_j}{i-j} \pmod p. \]
So we want a graph $G_k$ with at least $\half p$ connected components.
However, each $\{i,j\}$ appears in exactly one graph $G_k$,
so some graph has at most $\frac 1p \binom p2 = \half(p-1)$ edges
(by ``pigeonhole'').
This graph has at least $\half(p+1)$ connected components, as desired.
\begin{remark*}
Here is an example for $p=5$ showing equality can occur:
\[
\begin{bmatrix}
0 & 0 & 3 & 4 & 3 \\
0 & 1 & 0 & 2 & 2 \\
0 & 2 & 2 & 0 & 1 \\
0 & 3 & 4 & 3 & 0 \\
0 & 4 & 1 & 1 & 4
\end{bmatrix}.
\]
Ankan Bhattacharya points out more generally
that $a_i = i^2$ is sharp in general.
\end{remark*}
|
USAMO-2018-notes_5
|
Let $ABCD$ be a convex cyclic quadrilateral with
$E = \ol{AC} \cap \ol{BD}$, $F = \ol{AB} \cap \ol{CD}$,
$G = \ol{DA} \cap \ol{BC}$.
The circumcircle of $\triangle ABE$
intersects line $CB$ at $B$ and $P$,
and the circumcircle of $\triangle ADE$
intersects line $CD$ at $D$ and $Q$.
Assume $C$, $B$, $P$, $G$
and $C$, $Q$, $D$, $F$ are collinear in that order.
Let $M = \ol{FP} \cap \ol{GQ}$.
Prove that $\angle MAC = 90\dg$.
|
We present three general routes.
(The second route, using the fact that $\ol{AC}$
is an angle bisector, has many possible variations.)
\paragraph{First solution (Miquel points).}
This is indeed a Miquel point problem,
but the main idea is to focus on the self-intersecting
cyclic quadrilateral $PBQD$ as the key player,
rather than on the given $ABCD$.
Indeed, we will prove that $A$ is its Miquel point;
this follows from the following two claims.
\begin{claim*}
The self-intersecting quadrilateral $PQDB$ is cyclic.
\end{claim*}
\begin{proof}
By power of a point from $C$:
$CQ \cdot CD = CA \cdot CE = CB \cdot CP$.
\end{proof}
\begin{claim*}
Point $E$ lies on line $PQ$.
\end{claim*}
\begin{proof}
$\dang AEP = \dang ABP = \dang ABC = \dang ADC
= \dang ADQ = \dang AEQ$.
\end{proof}
\begin{center}
\begin{asy}
pair P = dir(240);
pair B = dir(300);
pair Q = dir(20);
pair D = dir(50);
pair E = extension(Q, P, B, D);
pair C = extension(D, Q, B, P);
pair H = extension(D, P, B, Q);
pair A = foot(H, E, C);
filldraw(unitcircle, opacity(0.1)+lightcyan, lightblue);
pair G = extension(D, A, B, C);
pair F = extension(B, A, C, D);
filldraw(circumcircle(P, A, E), opacity(0.1)+green, heavygreen);
filldraw(circumcircle(Q, A, E), opacity(0.1)+green, heavygreen);
filldraw(circumcircle(A, B, C), opacity(0.05)+yellow, heavygreen+dotted);
draw(B--F--C, gray);
draw(D--G--C, gray);
draw(P--H--B, blue);
draw(P--B--Q--D--cycle, blue);
draw(A--C, gray);
draw(P--Q, blue);
draw(B--D, blue);
pair M = extension(P, F, G, Q);
draw(F--P, heavycyan);
draw(G--Q, heavycyan);
dot("$P$", P, dir(P));
dot("$B$", B, dir(B));
dot("$Q$", Q, dir(Q));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$C$", C, dir(C));
dot("$H$", H, dir(H));
dot("$A$", A, dir(E-C));
dot("$G$", G, dir(G));
dot("$F$", F, dir(F));
dot("$M$", M, dir(B-M));
/* TSQ Source:
P = dir 240
B = dir 300
Q = dir 20
D = dir 50
E = extension Q P B D
C = extension D Q B P
H = extension D P B Q
A = foot H E C RE-C
unitcircle 0.1 lightcyan / lightblue
G = extension D A B C
F = extension B A C D
circumcircle P A E 0.1 green / heavygreen
circumcircle Q A E 0.1 green / heavygreen
circumcircle A B C 0.05 yellow / heavygreen dotted
B--F--C gray
D--G--C gray
P--H--B blue
P--B--Q--D--cycle blue
A--C gray
P--Q blue
B--D blue
M = extension P F G Q
F--P heavycyan
G--Q heavycyan
*/
\end{asy}
\end{center}
To finish, let $H = \ol{PD} \cap \ol{BQ}$.
By properties of the Miquel point, we have $A$ is the
foot from $H$ to $\ol{CE}$.
But also, points $M$, $A$, $H$ are collinear
by Pappus theorem on $\ol{BPG}$ and $\ol{DQF}$, as desired.
\paragraph{Second solution (projective).}
We start with a synthetic observation.
\begin{claim*}
The line $\ol{AC}$ bisects $\angle PAD$ and $\angle BAQ$.
\end{claim*}
\begin{proof}
Angle chase:
$\dang PAC = \dang PAE = \dang PBE = \dang CBD = \dang CAD$.
\end{proof}
There are three ways to finish from here:
\begin{itemize}
\ii (Michael Kural) Suppose the external bisector
of $\angle PAD$ and $\angle BAQ$
meet lines $BC$ and $DC$ at $X$ and $Y$.
Then \[ -1 = (GP;XC) = (FD;YC) \]
which is enough to imply that $\ol{XY}$, $\ol{GQ}$, $\ol{PF}$
are concurrent (by so-called prism lemma).
\ii (Daniel Liu) Alternatively,
apply the dual Desargues involution theorem
to complete quadrilateral $GQFPCM$, through the point $A$.
This gives that an involutive pairing of
\[ (AC,AM) \; (AP,AQ) \; (AG, AF). \]
This is easier to see if we project it onto the
line $\ell$ through $C$ perpendicular to $\ol{AC}$:
let $P'$, $Q'$, $G'$, $F'$ be the images of the last four lines,
so $C$ is the common midpoint of $\ol{F'Q'}$ and $\ol{G'P'}$.
Hence the involution swapping $G' \leftrightarrow F'$ and $P' \leftrightarrow Q'$
coincides with negative inversion through $C$ with power $\sqrt{CP' \cdot CQ'}$
which implies that $\ol{AM} \cap \ell$ is an infinity point, as desired.
\ii (Kada Williams) The official solution instead shows the external angle
bisector by a long trig calculation.
\end{itemize}
\paragraph{Third solution (inversion, Andrew Wu).}
Noting that $CE \cdot CA = CP \cdot CB = CQ \cdot CD$,
we perform an inversion at $C$ swapping these pairs of points.
The point $G$ is mapped to a point $G^\ast$ ray $CB$
for which $QEG^\ast C$ is cyclic, but then
\[ \dang CG^\ast E = \dang CQE = \dang CQP = \dang DBC = \dang CBE \]
and so we conclude $EB = EG^\ast$.
Similarly, $ED = EF^\ast$.
Finally, $M^\ast = (CG^\ast D) \cap (CF^\ast B) \neq C$,
and we wish to show that $\angle EM^\ast C = 90\dg$.
\begin{center}
\begin{asy}
pair D = dir(100);
pair B = dir(220);
pair C = dir(320);
pair E = 0.3*D+0.7*B;
pair K = foot(E, B, C);
pair L = foot(E, D, C);
pair Gs = 2*K-B;
pair Fs = 2*L-D;
pair Ms = (D*B-Fs*Gs)/(D+B-Fs-Gs);
filldraw(circumcircle(Gs, D, C), opacity(0.05)+yellow, orange);
filldraw(circumcircle(Fs, B, C), opacity(0.05)+yellow, orange);
fill(B--C--D--cycle, opacity(0.1)+yellow);
draw(B--C--D, orange);
draw(B--D, red);
draw(Gs--E--K, red);
draw(Fs--E--L, red);
filldraw(circumcircle(E, L, K), opacity(0.1)+red, lightred+dashed);
dot("$D$", D, dir(D));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$E$", E, dir(E));
dot("$K$", K, dir(K));
dot("$L$", L, dir(L));
dot("$G^\ast$", Gs, dir(Gs));
dot("$F^\ast$", Fs, dir(30));
dot("$M^\ast$", Ms, dir(Ms));
/* TSQ Source:
D = dir 100
B = dir 220
C = dir 320
E = 0.3*D+0.7*B
K = foot E B C
L = foot E D C
G* = 2*K-B
F* = 2*L-D R30
M* = (D*B-Fs*Gs)/(D+B-Fs-Gs)
circumcircle Gs D C 0.05 yellow / orange
circumcircle Fs B C 0.05 yellow / orange
!fill(B--C--D--cycle, opacity(0.1)+yellow);
B--C--D orange
B--D red
Gs--E--K red
Fs--E--L red
circumcircle E L K 0.1 red / lightred dashed
*/
\end{asy}
\end{center}
Note that $M^\ast$ is the center of the
spiral similarity sending $\ol{BG^\ast}$ to $\ol{F^\ast D}$.
Hence it also maps the midpoint $K$ of $BG^\ast$
to the midpoint $L$ of $\ol{F^\ast E}$.
Consequently, $M^\ast$ lies on the circumcircle $KLC$ as well.
In other words, $ELCKM^\ast$ is a cyclic pentagon
with circumdiameter $\ol{CE}$, as desired.
|
USAMO-2018-notes_6
|
Let $a_n$ be the number of permutations $(x_1, \dots, x_n)$ of $(1, \dots, n)$
such that the ratios $x_k / k$ are all distinct.
Prove that $a_n$ is odd for all $n \ge 1$.
\end{enumerate}
|
This is the official solution; the proof has two main insights.
The first idea:
\begin{lemma*}
If a permutation $x$ works, so does the inverse permutation.
\end{lemma*}
Thus it suffices to consider permutations $x$
in which all cycles have length at most $2$.
Of course, there can be at most one fixed point (since that gives the ratio $1$),
and hence exactly one if $n$ is odd, none if $n$ is even.
We consider the graph $K_n$ such that the edge
$\{i,j\}$ is labeled with $i / j$ (for $i < j$).
The permutations we're considering are then equivalent to maximal
matchings of this $K_n$.
We call such a matching \emph{fantastic} if it has all distinct edge labels.
\bigskip
Now the second insight is that if edges $ab$ and $cd$
have the same label for $a<b$ and $c<d$, then so do edges $ac$ and $bd$.
Thus:
\begin{definition*}
Given a matching $\mathcal M$ as above we say the \emph{neighbors}
of $\mathcal M$ are those other matchings obtained as follows:
for each label $\ell$, we take some disjoint pairs of edges (possibly none)
with label $\ell$ and apply the above switching operation
(in which we replace $ab$ and $cd$ with $ac$ and $bd$).
\end{definition*}
This neighborship relation is reflexive, and most importantly it is \emph{symmetric}
(because one can simply reverse the moves).
But it is not transitive.
The second observation is that:
\begin{claim*}
The matching $\mathcal M$ has an odd number of neighbors
(including itself) if and only if it is fantastic.
\end{claim*}
\begin{proof}
Consider the label $\ell$, and assume it appears $n_\ell \ge 1$ times.
If we pick $k$ disjoint pairs and swap them, the number of ways to do this is
$\binom{n_\ell}{2k} (2k-1)!!$, and so the total
number of ways to perform operations on the edges labeled $\ell$ is
\[ \sum_k \binom{n_\ell}{2k} (2k-1)!!
\equiv \sum_k \binom{n_\ell}{2k}
= 2^{n_\ell-1} \pmod 2. \]
This is even if and only if $n_\ell > 1$.
Finally, note that the number of neighbors
of $\mathcal M$ is the product across all $\ell$ of the above.
So it is odd if and only if each factor is odd,
if and only if $n_\ell = 1$ for every $\ell$.
\end{proof}
To finish, consider a huge simple graph $\Gamma$ on all the
maximal matchings, with edge relations given by neighbor relation
(we don't consider vertices to be connected to themselves).
Observe that:
\begin{itemize}
\ii The fantastic matchings of the vertices of $\Gamma$ have odd degrees
(one less than the neighbor count).
\ii The other matchings correspond to isolated vertices
(of degree zero, with no other neighbors) of $\Gamma$.
\ii The graph $\Gamma$ has an even number of vertices of odd degree
(this is true for any simple graph, see ``handshake lemma'').
\ii The number of vertices of $\Gamma$ is odd,
namely $\left( 2\left\lceil n/2 \right\rceil - 1 \right)!!$.
\end{itemize}
This concludes the proof.
|
USAMO-2019-notes_1
|
A function $f \colon \NN \to \NN$ satisfies
\[ \underbrace{f(f(\dots f}_{f(n)\text{ times}} (n)\dots)) %chktex 9
= \frac{n^2}{f(f(n))} \]
for all positive integers $n$.
What are all possible values of $f(1000)$?
|
Actually, we classify all such functions: $f$ can be any function which fixes odd integers
and acts as an involution on the even integers.
In particular, $f(1000)$ may be any even integer.
It's easy to check that these all work, so now we check they are the only solutions.
\begin{claim*}
$f$ is injective.
\end{claim*}
\begin{proof}
If $f(a) = f(b)$, then $a^2 = f^{f(a)}(a) f(f(a)) = f^{f(b)}(b) f(f(b)) = b^2$,
so $a = b$.
\end{proof}
\begin{claim*}
We have $f(n) = n$ for every odd integer $n \ge 1$.
\end{claim*}
\begin{proof}
We prove this by induction on odd $n \ge 1$.
Assume $f$ fixes $S = \{1,3,\dots,n-2\}$ now (allowing $S = \varnothing$ for $n=1$).
Now we have that
\[ f^{f(n)}(n) \cdot f^2(n) = n^2. \]
However, neither of the two factors on the left-hand side can be in $S$ since $f$ was injective.
Therefore they must both be $n$, and we have $f^2(n) = n$.
Now let $y = f(n)$, so $f(y) = n$.
Substituting $y$ into the given yields
\[ y^2 = f^n(y) \cdot y = f^{n+1}(n) \cdot y = ny \]
since $n+1$ is even.
We conclude $n=y$, as desired.
\end{proof}
\begin{remark*}
[Motivation]
After obtaining $f(1) = 1$ and $f$ injective,
here is one way to motivate where the above proof comes from.
From the equation
\[ f^{f(n)}(n) \cdot f^2(n) = n^2 \]
it would be natural to consider the case where $n$ is \emph{prime},
because $p^2$ only has a few possible factorizations.
In fact, actually because of injectivity and $f(1)=1$, we would need to have
\[ f^{f(p)}(p) = f^2(p) = p \]
in order for the equation to be true.
Continuing on as in the proof above, one then gets $f(p) = p$ for odd primes $p$
(but no control over $f(2)$).
The special case of prime $n$ then serves as a foothold by which one can continue
the induction towards all numbers $n$, finding the induction works out exactly
when the prime $2$ never appears.
As a general point, in mathematical problem-solving,
one often needs to be willing to try out a proof idea or strategy
and then retroactively determine what hypothesis is needed,
rather than hoping one will always happen to guess exactly the right claim first.
In other words, it may happen that one begins working out a proof of a claim
\emph{before} knowing exactly what the claim will turn out to say, and this is the case here
(despite the fact the proof strategy uses induction).
\end{remark*}
Thus, $f$ maps even integers to even integers.
In light of this, we may let $g \coloneq f(f(n))$ (which is also injective),
so we conclude that
\[ g^{f(n)/2} (n) g(n) = n^2 \qquad \text{ for } n = 2, 4, \dots. \]
\begin{claim*}
The function $g$ is the identity function.
\end{claim*}
\begin{proof}
The proof is similar to the earlier proof of the claim.
Note that $g$ fixes the odd integers already.
We proceed by induction to show $g$ fixes the even integers;
so assume $g$ fixes the set $S = \{1, 2, \dots, n-1\}$, for some even integer $n \ge 2$.
In the equation
\[ g^{f(n)/2}(n) \cdot g(n) = n^2 \]
neither of the two factors may be less than $n$.
So they must both be $n$.
\end{proof}
These three claims imply that the solutions we claimed earlier are the only ones.
\begin{remark*}
The last claim is not necessary to solve the problem;
after realizing $f$ and injective fixes the odd integers,
this answers the question about the values of $f(1000)$.
However, we chose to present the ``full'' solution anyways.
\end{remark*}
\begin{remark*}
After noting $f$ is injective, another approach is outlined below.
Starting from any $n$, consider the sequence
\[ n, \; f(n), \; f(f(n)), \; \]
and so on.
We may let $m$ be the smallest term of the sequence;
then $m^2 = f(f(m)) \cdot f^{f(m)}(m)$
which forces $f^{f(m)}(m) = f(f(m)) = m$ by minimality.
Thus the sequence is $2$-periodic.
Therefore, $f(f(n)) = n$ always holds,
which is enough to finish.
\end{remark*}
\paragraph{Authorship comments.}
I will tell you a great story about this problem.
Two days before the start of grading of USAMO 2017,
I had a dream that I was grading a functional equation.
When I woke up, I wrote it down, and it was
\[ f^{f(n)}(n) = \frac{n^2}{f(f(n))}. \]
You can guess the rest of the story
(and imagine how surprised I was the solution set was interesting).
%Much to my surprise, when I woke up and remembered my dream
%I was actually able to solve the problem
%(and found that the solution set was nontrivial).
I guess some dreams do come true, huh?
|
USAMO-2019-notes_2
|
Let $ABCD$ be a cyclic quadrilateral satisfying $AD^2 + BC^2 = AB^2$.
The diagonals of $ABCD$ intersect at $E$.
Let $P$ be a point on side $\ol{AB}$ satisfying $\angle APD = \angle BPC$.
Show that line $PE$ bisects $\ol{CD}$.
|
Here are three solutions.
The first two are similar although the first one makes use of symmedians.
The last solution by inversion is more advanced.
\paragraph{First solution using symmedians.}
We define point $P$ to obey
\[ \frac{AP}{BP} = \frac{AD^2}{BC^2} = \frac{AE^2}{BE^2} \]
so that $\ol{PE}$ is the $E$-symmedian of $\triangle EAB$,
therefore the $E$-median of $\triangle ECD$.
Now, note that
\[ AD^2 = AP \cdot AB \quad\text{and}\quad BC^2 = BP \cdot BA. \]
This implies $\triangle APD \sim \triangle ADB$ and $\triangle BPC \sim \triangle BCA$.
Thus
\[ \dang DPA = \dang ADB = \dang ACB = \dang BCP \]
and so $P$ satisfies the condition as in the statement
(and is the unique point to do so), as needed.
\paragraph{Second solution using only angle chasing (by proposer).}
We again re-define $P$ to obey $AD^2 = AP \cdot AB$ and $BC^2 = BP \cdot BA$.
As before, this gives $\triangle APD \sim \triangle ABD$
and $\triangle BPC \sim \triangle BDP$ and so we let
\[ \theta \coloneq \dang DPA = \dang ADB = \dang ACB = \dang BCP. \]
Our goal is to now show $\ol{PE}$ bisects $\ol{CD}$.
Let $K = \ol{AC} \cap \ol{PD}$ and $L = \ol{BD} \cap \ol{PC}$.
Since $\dang KPA = \theta = \dang ACB$, quadrilateral $BPKC$ is cyclic.
Similarly, so is $APLD$.
\begin{center}
\begin{asy}
size(8cm);
pair A = dir(200);
pair B = -conj(A);
pair D = dir(140);
pair O = origin;
pair P = extension(D, foot(D, A, O), A, B);
pair Z = 100*foot(P,O,B)-99*P;
pair C = IP(P--Z, unitcircle);
markangle(13.0, A, D, B, red);
markangle(13.0, A, C, B, red);
markangle(13.0, D, P, A, red);
markangle(13.0, B, P, C, red);
filldraw(unitcircle, opacity(0.1)+palecyan, deepcyan);
draw(A--B--C--D--cycle, deepcyan);
draw(A--C, deepgreen);
draw(B--D, deepgreen);
pair K = extension(D, P, A, C);
pair L = extension(C, P, D, B);
draw(D--P--C, blue);
pair E = extension(A, C, B, D);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$D$", D, dir(D));
dot("$P$", P, dir(270));
dot("$C$", C, dir(C));
dot("$K$", K, dir(K));
dot("$L$", L, dir(20));
dot("$E$", E, dir(90));
/* TSQ Source:
!size(8cm);
A = dir 200
B = -conj(A)
D = dir 140
O := origin
P 270 = extension D foot D A O A B
Z := 100*foot(P,O,B)-99*P
C = IP P--Z unitcircle
!markangle(13.0, A, D, B, red);
!markangle(13.0, A, C, B, red);
!markangle(13.0, D, P, A, red);
!markangle(13.0, B, P, C, red);
unitcircle 0.1 palecyan / deepcyan
A--B--C--D--cycle deepcyan
A--C deepgreen
B--D deepgreen
K = extension D P A C
L 20 = extension C P D B
D--P--C blue
E = extension A C B D R90
*/
\end{asy}
\end{center}
Finally $AKLB$ is cyclic since
\[ \dang BKA = \dang BKC = \dang BPC = \theta = \dang DPA = \dang DLA = \dang BLA. \]
This implies $\dang CKL = \dang LBA = \dang DCK$, so $\ol{KL} \parallel \ol{CD}$.
Then $PE$ bisects $\ol{BC}$ by Ceva's theorem on $\triangle PCD$.
\paragraph{Third solution (using inversion).}
By hypothesis, the circle $\omega_a$ centered at $A$ with radius $AD$
is orthogonal to the circle $\omega_b$ centered at $B$ with radius $BC$.
For brevity, we let $\mathbf{I}_a$ and $\mathbf{I}_b$
denote inversion with respect to $\omega_a$ and $\omega_b$.
We let $P$ denote the intersection of $\ol{AB}$ with the radical axis of $\omega_a$ and $\omega_b$;
hence $P = \mathbf{I}_a(B) = \mathbf{I}_b(A)$.
This already implies that
\[ \dang DPA \overset{\mathbf{I}_a}{=} \dang ADB = \dang ACB \overset{\mathbf{I}_b}{=} \dang BPC \]
so $P$ satisfies the angle condition.
\begin{center}
\begin{asy}
size(10cm);
pair A = dir(200);
pair B = -conj(A);
pair D = dir(140);
pair O = origin;
pair P = extension(D, foot(D, A, O), A, B);
pair Z = 100*foot(P,O,B)-99*P;
pair C = IP(P--Z, unitcircle);
filldraw(unitcircle, opacity(0.1)+palecyan, deepcyan);
draw(A--B--C--D--cycle, deepcyan);
filldraw(CP(A, D), opacity(0.1)+lightred, red);
filldraw(CP(B, C), opacity(0.1)+orange, orange);
pair X = IP(CP(A, D), CP(B, C));
pair Y = OP(CP(A, D), CP(B, C));
draw(X--Y, yellow);
draw(A--C, deepcyan+dotted);
draw(B--D, deepcyan+dotted);
pair K = extension(D, P, A, C);
pair L = extension(C, P, D, B);
draw(D--P, red+dashed);
draw(C--P, orange+dashed);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$D$", D, dir(D));
dot("$P$", P, dir(270));
dot("$C$", C, dir(C));
dot(X);
dot(Y);
dot("$K$", K, dir(K));
dot("$L$", L, dir(20));
/* TSQ Source:
!size(12cm);
A = dir 200
B = -conj(A)
D = dir 140
O := origin
P 270 = extension D foot D A O A B
Z := 100*foot(P,O,B)-99*P
C = IP P--Z unitcircle
unitcircle 0.1 palecyan / deepcyan
A--B--C--D--cycle deepcyan
CP A D 0.1 lightred / red
CP B C 0.1 orange / orange
X .= IP CP A D CP B C
Y .= OP CP A D CP B C
X--Y yellow
A--C deepcyan dotted
B--D deepcyan dotted
K = extension D P A C
L 20 = extension C P D B
D--P red dashed
C--P orange dashed
*/
\end{asy}
\end{center}
\begin{claim*}
The point $K = \mathbf{I}_a(C)$ lies on $\omega_b$ and $\ol{DP}$.
Similarly $L = \mathbf{I}_b(D)$ lies on $\omega_a$ and $\ol{CP}$.
\end{claim*}
\begin{proof}
The first assertion follows from the fact that $\omega_b$ is orthogonal to $\omega_a$.
For the other, since $(BCD)$ passes through $A$, it follows $P = \mathbf{I}_a(B)$, $K = \mathbf{I}_a(C)$,
and $D = \mathbf{I}_a(D)$ are collinear.
\end{proof}
Finally, since $C$, $L$, $P$ are collinear,
we get $A$ is concyclic with $K = \mathbf{I}_a(C)$,
$L = \mathbf{I}_a(L)$, $B = \mathbf{I}_a(P)$, i.e.\ that $AKLB$ is cyclic.
So $\ol{KL} \parallel \ol{CD}$ by Reim's theorem,
and hence $\ol{PE}$ bisects $\ol{CD}$ by Ceva's theorem.
|
USAMO-2019-notes_3
|
Let $K$ be the set of positive integers not containing the decimal digit $7$.
Determine all polynomials $f(x)$ with nonnegative coefficients
such that $f(x) \in K$ for all $x \in K$.
|
The answer is only the obvious ones:
$f(x) = 10^e x$, $f(x) = k$,
and $f(x) = 10^e x + k$, for any choice of $k \in K$
and $e > \log_{10} k$ (with $e \ge 0$).
Now assume $f$ satisfies $f(K) \subseteq K$;
such polynomials will be called \emph{stable}.
We first prove the following claim
which reduces the problem to the study of monomials.
\begin{lemma*}
[Reduction to monomials]
If $f(x) = a_0 + a_1 x + a_2 x^2 + \dots$ is stable,
then each monomial $a_0$, $a_1 x$, $a_2 x^2$, \dots is stable.
\end{lemma*}
\begin{proof}
For any $x \in K$, plug in $f(10^e x)$ for large enough $e$:
the decimal representation of $f$ will contain
$a_0$, $a_1 x$, $a_2 x^2$ with some zeros padded in between.
\end{proof}
Let's tackle the linear case next.
Here is an ugly but economical proof.
\begin{claim*}
[Linear classification]
If $f(x) = cx$ is stable, then $c = 10^e$
for some nonnegative integer $e$.
\end{claim*}
\begin{proof}
We will show when $c \neq 10^e$ then we can find $x \in K$
such that $cx$ starts with the digit $7$.
This can actually be done with the following explicit cases
in terms of how $c$ starts in decimal notation:
\begin{itemize}
\ii For $9 \cdot 10^e \le c < 10 \cdot 10^e$, pick $x = 8$.
\ii For $8 \cdot 10^e \le c < 9 \cdot 10^e$, pick $x = 88$.
\ii For $7 \cdot 10^e \le c < 8 \cdot 10^e$, pick $x = 1$.
\ii For $4.4 \cdot 10^e \le c < 7 \cdot 10^e$, pick $11 \le x \le 16$.
\ii For $2.7 \cdot 10^e \le c < 4.4 \cdot 10^e$, pick $18 \le x \le 26$.
\ii For $2 \cdot 10^e \le c < 2.7 \cdot 10^e$, pick $28 \le x \le 36$.
\ii For $1.6 \cdot 10^e \le c < 2 \cdot 10^e$, pick $38 \le x \le 46$.
\ii For $1.3 \cdot 10^e \le c < 1.6 \cdot 10^e$, pick $48 \le x \le 56$.
\ii For $1.1 \cdot 10^e \le c < 1.3 \cdot 10^e$, pick $58 \le x \le 66$.
\ii For $1 \cdot 10^e \le c < 1.1 \cdot 10^e$, pick $x = 699\dots9$
for suitably many $9$'s. \qedhere
\end{itemize}
\end{proof}
The hardest part of the problem is the case where $\deg f > 1$.
We claim that no solutions exist then:
\begin{claim*}
[Higher-degree classification]
No monomial of the form $f(x) = cx^d$ is stable
for any $d > 1$.
\end{claim*}
\begin{proof}
Note that $f(10x+3)$ is stable too.
Thus
\[ f(10x+3) = 3^d + 10d \cdot 3^{d-1} x
+ 100 \binom d2 \cdot 3^{d-1} x^2 + \dots \]
is stable.
By applying the lemma the linear monomial
$10d \cdot 3^{d-1} x$ is stable,
so $10d \cdot 3^{d-1}$ is a power of $10$,
which can only happen if $d = 1$.
\end{proof}
Thus the only nonconstant stable polynomials with nonnegative coefficients
must be of the form $f(x) = 10^e x + k$ for $e \ge 0$.
It is straightforward to show we then need $k < 10^e$
and this finishes the proof.
\begin{remark*}
The official solution replaces the proof for $f(x) = cx$ with Kronecker density.
Here is how it goes.
From $f(1) = c \in K$, we get $f(c) = c^2 \in K$,
et cetera and hence $c^n \in K$.
However, I claim that when $c$ is not a power of $10$,
some power of $c$ starts with \emph{any} specified prefix.
For example, there should be an integer $n$
such that $c^n$ starts with the digits $98765$.
And this is obviously a contradiction to $c^n \in K$.
To see why, note this is equivalent to requiring that
\[ \log_{10}(9.8765) \le \{ n \log_{10} c \} < \log_{10}(9.8766) \]
where $\{x\}$ is the fractional part of $x$.
However, when $c$ is not a power of $10$, the number $\log_{10}(c)$ is irrational.
And Kronecker density theorem asserts that whenever $\alpha$ is irrational,
the fractional parts $\{ n \alpha \}$ for $n \ge 0$
are actually equally distributed across the entire open interval $(0,1)$.
In particular, there are $n$ for which the fractional part lie in any
interval of positive length.
\end{remark*}
|
USAMO-2019-notes_4
|
Let $n$ be a nonnegative integer.
Determine the number of ways to choose sets
$S_{ij} \subseteq \{1, 2, \dots, 2n\}$,
for all $0 \le i \le n$ and $0 \le j \le n$
(not necessarily distinct), such that
\begin{itemize}
\ii $|S_{ij}| = i+j$, and
\ii $S_{ij} \subseteq S_{kl}$ if $0 \le i \le k \le n$
and $0 \le j \le l \le n$.
\end{itemize}
|
The answer is $(2n)! \cdot 2^{n^2}$.
First, we note that
$\varnothing = S_{00} \subsetneq S_{01} \subsetneq \dots \subsetneq S_{nn}
= \left\{ 1, \dots, 2n \right\}$
and thus multiplying by $(2n)!$
we may as well assume $S_{0i} = \left\{ 1, \dots, i \right\}$
and $S_{in} = \left\{ 1, \dots, n+i \right\}$.
We illustrate this situation by placing the sets in a grid,
as below for $n = 4$;
our goal is to fill in the rest of the grid.
\[
\begin{bmatrix}
1234 & 12345 & 123456 & 1234567 & 12345678\\
123 \\
12 \\
1 \\
\varnothing
\end{bmatrix}
\]
We claim the number of ways to do so is $2^{n^2}$.
In fact, more strongly even the partial fillings
are given exactly by powers of $2$.
\begin{claim*}
Fix a choice $T$ of cells we wish to fill in,
such that whenever a cell is in $T$,
so are all the cells above and left of it.
(In other words, $T$ is a Young tableau.)
The number of ways to fill in these cells with sets
satisfying the inclusion conditions is $2^{|T|}$.
\end{claim*}
An example is shown below, with an indeterminate set marked in red
(and the rest of $T$ marked in blue).
\[
\begin{bmatrix}
1234 & 12345 & 123456 & 1234567 & 12345678\\
123 & {\color{blue}1234} & {\color{blue}12346} & {\color{blue}123467} \\
12 & {\color{blue}124} & {\color{red}1234 \text{ or } 1246} \\
1 & {\color{blue}12} \\
\varnothing & {\color{blue}2}
\end{bmatrix}
\]
\begin{proof}
The proof is by induction on $|T|$,
with $|T| = 0$ being vacuous.
Now suppose we have a corner $\begin{bmatrix}
B & C \\ A & {\color{red}S} \end{bmatrix}$
where $A$, $B$, $C$ are fixed and $S$ is to be chosen.
Then we may write $B = A \cup \{x\}$ and $C = A \cup \{x,y\}$
for $x,y \notin A$.
Then the two choices of $S$ are $A \cup \{x\}$ (i.e.\ $B$)
and $A \cup \{y\}$, and both of them are seen to be valid.
In this way, we gain a factor of $2$
any time we add one cell as above to $T$.
Since we can achieve any Young tableau in this way,
the induction is complete.
\end{proof}
|
USAMO-2019-notes_6
|
Find all polynomials $P$ with real coefficients such that
\[ \frac{P(x)}{yz} + \frac{P(y)}{zx} + \frac{P(z)}{xy} = P(x-y) + P(y-z) + P(z-x) \]
for all nonzero real numbers $x$, $y$, $z$ obeying $2xyz = x+y+z$.
\end{enumerate}
|
The given can be rewritten as saying that
\begin{align*}
Q(x,y,z) &\coloneq xP(x) + yP(y) + zP(z) \\ &- xyz \left( P(x-y) + P(y-z) + P(z-x) \right)
\end{align*}
is a polynomial vanishing whenever $xyz \neq 0$ and $2xyz = x+y+z$, for real numbers $x$, $y$, $z$.
\begin{claim*}
This means $Q(x,y,z)$ vanishes also for any complex numbers $x$, $y$, $z$ obeying $2xyz=x+y+z$.
\end{claim*}
\begin{proof}
Indeed, this means that the rational function
\[ R(x,y) \coloneq Q\left( x,y,\frac{x+y}{2xy-1} \right) \]
vanishes for any real numbers $x$ and $y$
such that $xy \neq \half$, $x \neq 0$, $y \neq 0$, $x+y \neq 0$.
This can only occur if $R$ is identically zero as a rational function with real coefficients.
If we then regard $R$ as having complex coefficients, the conclusion then follows.
\end{proof}
\begin{remark*}
[Algebraic geometry digression on real dimension]
Note here we use in an essential way that $z$
can be solved for in terms of $x$ and $y$.
If $s(x,y,z) = 2xyz-(x+y+z)$ is replaced with some general condition,
the result may become false; e.g.\ we would certainly not expect the result to hold when
$s(x,y,z) = x^2+y^2+z^2-(xy+yz+zx)$ since for real numbers $s = 0$ only when $x=y=z$!
The general condition we need here is that $s(x,y,z) = 0$ should have ``real dimension two''.
Here is a proof using this language, in our situation.
Let $M \subset \RR^3$ be the surface $s=0$.
We first contend $M$ is two-dimensional manifold.
Indeed, the gradient $\nabla s = \left< 2yz-1, 2zx-1, 2xy-1 \right>$
vanishes only at the points $(\pm 1/\sqrt2, \pm 1/\sqrt2, \pm 1/\sqrt2)$
where the $\pm$ signs are all taken to be the same.
These points do not lie on $M$, so the result follows by the \emph{regular value theorem}.
In particular the topological closure of points on $M$
with $xyz \neq 0$ is all of $M$ itself; so $Q$ vanishes on all of $M$.
If we now identify $M$ with the semi-algebraic set consisting of
maximal ideals $(x-a,y-b,z-c)$ in $\opname{Spec} \RR[x,y,z]$
satisfying $2abc = a+b+c$, then we have
\href{https://en.wikipedia.org/wiki/Dimension_of_an_algebraic_variety#Real_dimension}{real dimension} two,
and thus the Zariski closure of $M$
is a two-dimensional closed subset of $\opname{Spec} \RR[x,y,z]$.
Thus it must be $Z = \mathcal V(2xyz-(x+y+z))$,
since this $Z$ is an irreducible two-dimensional closed subset
(say, by \emph{Krull's principal ideal theorem}) containing $M$.
Now $Q$ is a global section vanishing on all of $Z$,
therefore $Q$ is contained in the (radical, principal) ideal $(2xyz-(x+y+z))$ as needed.
So it is actually divisible by $2xyz-(x+y+z)$ as desired.
\end{remark*}
Now we regard $P$ and $Q$ as complex polynomials instead.
First, note that substituting $(x,y,z) = (t,-t,0)$ implies $P$ is even.
We then substitute
\[ (x,y,z) = \left(x, \frac{i}{\sqrt2}, \frac{-i}{\sqrt2}\right) \]
to get
\begin{align*}
&\phantom= xP(x) + \frac{i}{\sqrt2}
\left( P\left( \frac{i}{\sqrt2} \right)
- P\left( \frac{-i}{\sqrt2} \right) \right) \\
&= \half x\left( P(x-i/\sqrt2) + P(x+i/\sqrt2) + P(\sqrt2i) \right)
\end{align*}
which in particular implies that
\[ P\left( x + \frac{i}{\sqrt2} \right) + P\left( x - \frac{i}{\sqrt2} \right)
- 2P(x) \equiv P(\sqrt 2i) \]
identically in $x$.
The left-hand side is a second-order finite difference in $x$ (up to scaling the argument),
and the right-hand side is constant, so this implies $\deg P \le 2$.
Since $P$ is even and $\deg P \le 2$, we must have $P(x) = cx^2 + d$ for some real numbers $c$ and $d$.
A quick check now gives the answer $P(x) = c(x^2+3)$ which all work.
|
USAMO-2020-notes_1
|
Let $ABC$ be a fixed acute triangle
inscribed in a circle $\omega$ with center $O$.
A variable point $X$ is chosen on minor arc $AB$ of $\omega$,
and segments $CX$ and $AB$ meet at $D$.
Denote by $O_1$ and $O_2$ the circumcenters of
triangles $ADX$ and $BDX$, respectively.
Determine all points $X$ for which
the area of triangle $OO_1O_2$ is minimized.
|
We prove $[OO_1O_2] \ge \frac14 [ABC]$,
with equality if and only if $\ol{CX} \perp \ol{AB}$.
\paragraph{First approach (Bobby Shen).}
We use two simultaneous inequalities:
\begin{itemize}
\ii Let $M$ and $N$ be the midpoints of $CX$ and $DX$.
Then $MN$ equals the length of the $O$-altitude of $\triangle OO_1O_2$,
since $\ol{O_1O_2}$ and $\ol{DX}$ meet at $N$ at a right angle.
Moreover, we have \[ MN = \half CD \ge \half h_a \]
where $h_a$ denotes the $A$-altitude.
\ii The projection of $O_1 O_2$ onto line $AB$ has length exactly $AB/2$.
Thus \[ O_1 O_2 \ge \half AB. \]
\end{itemize}
So, we find
\[ [OO_1O_2] = \half \cdot MN \cdot O_1 O_2
\ge \frac 18 h_a \cdot AB = \frac14 [ABC]. \]
Note that equality occurs in both cases if and only if $\ol{CX} \perp \ol{AB}$.
So the area is minimized exactly when this occurs.
\paragraph{Second approach (Evan's solution).}
We need two claims.
\begin{claim*}
We have $\triangle O O_1 O_2 \sim \triangle CBA$, with opposite orientation.
\end{claim*}
\begin{proof}
Notice that $\ol{OO_1} \perp \ol{AX}$
and $\ol{O_1 O_2} \perp \ol{CX}$,
so $\dang OO_1O_2 = \dang AXC = \dang ABC$.
Similarly $\dang OO_2O_1 = \dang BAC$.
\end{proof}
Therefore, the problem is equivalent to minimizing $O_1 O_2$.
\begin{center}
\begin{asy}
pair C = dir(130);
pair A = dir(200);
pair B = dir(340);
pair X = dir(280);
pair D = extension(C, X, A, B);
pair O_1 = circumcenter(A, D, X);
pair O_2 = circumcenter(B, D, X);
pair O = origin;
filldraw(O--O_1--O_2--cycle, opacity(0.1)+lightcyan, deepcyan);
filldraw(A--X--B--cycle, opacity(0.1)+lightred, red);
filldraw(A--B--C--cycle, opacity(0.1)+lightcyan, blue);
draw(X--C, red);
filldraw(O_1--X--O_2--cycle, opacity(0.1)+yellow, orange);
draw(unitcircle, gray);
dot("$C$", C, dir(C));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$X$", X, dir(X));
dot("$D$", D, dir(245));
dot("$O_1$", O_1, dir(O_1));
dot("$O_2$", O_2, dir(20));
dot("$O$", O, dir(45));
/* TSQ Source:
C = dir 130
A = dir 200
B = dir 340
X = dir 280
D = extension C X A B R245
O_1 = circumcenter A D X
O_2 = circumcenter B D X R20
O = origin R45
O--O_1--O_2--cycle 0.1 lightcyan / deepcyan
A--X--B--cycle 0.1 lightred / red
A--B--C--cycle 0.1 lightcyan / blue
X--C red
O_1--X--O_2--cycle 0.1 yellow / orange
unitcircle gray
*/
\end{asy}
\end{center}
\begin{claim*}
[Salmon theorem]
We have $\triangle X O_1 O_2 \sim \triangle XAB$.
\end{claim*}
\begin{proof}
It follows from the fact that $\triangle AO_1 X \sim \triangle BO_2 X$
(since $\dang ADX = \dang XDB \implies \dang XO_1A = \dang XO_2B$)
and that spiral similarities come in pairs.
\end{proof}
Let $\theta = \angle ADX$.
The ratio of similarity in the previous claim is equal to
$\frac{XO_1}{XA} = \frac{1}{2\sin \theta}$.
In other words,
\[ O_1 O_2 = \frac{AB}{2 \sin \theta}. \]
This is minimized when $\theta = 90\dg$,
in which case $O_1 O_2 = AB/2$
and $[OO_1O_2] = \frac14 [ABC]$.
This completes the solution.
|
USAMO-2020-notes_2
|
An empty $2020 \times 2020 \times 2020$ cube is given,
and a $2020 \times 2020$ grid of square unit cells is drawn on each of its six faces.
A \emph{beam} is a $1 \times 1 \times 2020$ rectangular prism.
Several beams are placed inside the cube subject to the following conditions:
\begin{itemize}
\item The two $1 \times 1$ faces of each beam coincide
with unit cells lying on opposite faces of the cube.
(Hence, there are $3 \cdot 2020^2$ possible positions for a beam.)
\item No two beams have intersecting interiors.
\item The interiors of each of the four $1 \times 2020$ faces of each beam touch
either a face of the cube or the interior of the face of another beam.
\end{itemize}
What is the smallest positive number of beams that can be placed to satisfy these conditions?
|
\paragraph{Answer.} $3030$ beams.
\paragraph{Construction.}
We first give a construction with $3n/2$ beams for any $n \times n \times n$ box,
where $n$ is an even integer.
Shown below is the construction for $n=6$, which generalizes.
(The left figure shows the cube in 3d;
the right figure shows a direct view of the three visible faces.)
\begin{center}
\begin{asy}
import three;
import bsp;
size(8cm);
settings.prc = false;
settings.render = 0;
settings.tex="pdflatex";
settings.outformat="pdf";
triple O = (9,7,5);
currentprojection = orthographic(O);
currentlight = light(gray(2),specular=gray(0.7), specularfactor=3, (7,7,5));
pen finalpen = yellow+2;
pen meshpenlight = rgb(0.95,0.95,0.99);
int M = 6;
int s = 0;
for (int i=0; i<=M; ++i) {
draw( (s,i,0)--(s,i,M), meshpenlight);
draw( (s,0,i)--(s,M,i), meshpenlight);
draw( (i,s,0)--(i,s,M), meshpenlight);
draw( (0,s,i)--(M,s,i), meshpenlight);
draw( (i,0,s)--(i,M,s), meshpenlight);
draw( (0,i,s)--(M,i,s), meshpenlight);
}
draw(box((0,0,0), (M,M,M)), finalpen);
pen penA = red;
pen penB = cyan;
pen penC = green;
draw(shift(0,0,0)*unitcube, penA);
draw(shift(0,0,1)*unitcube, penA);
draw(shift(0,0,2)*unitcube, penA);
draw(shift(0,0,3)*unitcube, penA);
draw(shift(0,0,4)*unitcube, penA);
draw(shift(0,0,5)*unitcube, penA);
draw(shift(1,0,0)*unitcube, penB);
draw(shift(1,1,0)*unitcube, penB);
draw(shift(1,2,0)*unitcube, penB);
draw(shift(1,3,0)*unitcube, penB);
draw(shift(1,4,0)*unitcube, penB);
draw(shift(1,5,0)*unitcube, penB);
draw(shift(0,1,1)*unitcube, penC);
draw(shift(1,1,1)*unitcube, penC);
draw(shift(2,1,1)*unitcube, penC);
draw(shift(3,1,1)*unitcube, penC);
draw(shift(4,1,1)*unitcube, penC);
draw(shift(5,1,1)*unitcube, penC);
draw(shift(2,2,0)*unitcube, penA);
draw(shift(2,2,1)*unitcube, penA);
draw(shift(2,2,2)*unitcube, penA);
draw(shift(2,2,3)*unitcube, penA);
draw(shift(2,2,4)*unitcube, penA);
draw(shift(2,2,5)*unitcube, penA);
draw(shift(3,0,2)*unitcube, penB);
draw(shift(3,1,2)*unitcube, penB);
draw(shift(3,2,2)*unitcube, penB);
draw(shift(3,3,2)*unitcube, penB);
draw(shift(3,4,2)*unitcube, penB);
draw(shift(3,5,2)*unitcube, penB);
draw(shift(0,3,3)*unitcube, penC);
draw(shift(1,3,3)*unitcube, penC);
draw(shift(2,3,3)*unitcube, penC);
draw(shift(3,3,3)*unitcube, penC);
draw(shift(4,3,3)*unitcube, penC);
draw(shift(5,3,3)*unitcube, penC);
draw(shift(4,4,0)*unitcube, penA);
draw(shift(4,4,1)*unitcube, penA);
draw(shift(4,4,2)*unitcube, penA);
draw(shift(4,4,3)*unitcube, penA);
draw(shift(4,4,4)*unitcube, penA);
draw(shift(4,4,5)*unitcube, penA);
draw(shift(5,0,4)*unitcube, penB);
draw(shift(5,1,4)*unitcube, penB);
draw(shift(5,2,4)*unitcube, penB);
draw(shift(5,3,4)*unitcube, penB);
draw(shift(5,4,4)*unitcube, penB);
draw(shift(5,5,4)*unitcube, penB);
draw(shift(0,5,5)*unitcube, penC);
draw(shift(1,5,5)*unitcube, penC);
draw(shift(2,5,5)*unitcube, penC);
draw(shift(3,5,5)*unitcube, penC);
draw(shift(4,5,5)*unitcube, penC);
draw(shift(5,5,5)*unitcube, penC);
pen meshpen = gray;
int s = M;
for (int i=1; i<=M-1; ++i) {
draw( (s,i,0)--(s,i,M), meshpen);
draw( (s,0,i)--(s,M,i), meshpen);
draw( (i,s,0)--(i,s,M), meshpen);
draw( (0,s,i)--(M,s,i), meshpen);
draw( (i,0,s)--(i,M,s), meshpen);
draw( (0,i,s)--(M,i,s), meshpen);
}
draw((M,M,M)--(0,M,M), finalpen);
draw((M,M,M)--(M,0,M), finalpen);
draw((M,M,M)--(M,M,0), finalpen);
\end{asy}
\quad
\begin{asy}
size(8cm);
picture redface, greenface, blueface;
void draw_grid(picture pic) {
for (int i=1; i<6; ++i) {
draw(pic, (i,0)--(i,6), gray);
draw(pic, (0,i)--(6,i), gray);
}
draw(pic, scale(6)*unitsquare, yellow+2);
}
fill(redface, shift(0,5)*unitsquare, red);
fill(redface, shift(2,3)*unitsquare, red);
fill(redface, shift(4,1)*unitsquare, red);
draw_grid(redface);
fill(greenface, shift(5,5)*unitsquare, green);
fill(greenface, shift(3,3)*unitsquare, green);
fill(greenface, shift(1,1)*unitsquare, green);
draw_grid(greenface);
fill(blueface, shift(0,4)*unitsquare, cyan);
fill(blueface, shift(2,2)*unitsquare, cyan);
fill(blueface, shift(4,0)*unitsquare, cyan);
draw_grid(blueface);
add(shift(0,8.5)*shift(3,3)*rotate(-45)*shift(-3,-3)*redface);
add(shift(-4,0)*greenface);
add(shift( 4,0)*blueface);
label("Left face", (-1,0), dir(-90));
label("Right face", (7,0), dir(-90));
label("Top face", (3,16), dir(90));
\end{asy}
\end{center}
To be explicit,
impose coordinate axes such that one corner of the cube is the origin.
We specify a beam by two opposite corners.
The $3n/2$ beams come in three directions, $n/2$ in each direction:
\begin{itemize}
\ii $(0,0,0) \to (1,1,n)$, $(2,2,0) \to (3,3,n)$, $(4,4,0) \to (5,5,n)$, and so on;
\ii $(1,0,0) \to (2,n,1)$, $(3,0,2) \to (4,n,3)$, $(5,0,4) \to (6,n,5)$, and so on;
\ii $(0,1,1) \to (n,2,2)$, $(0,3,3) \to (n,4,4)$, $(0,5,5) \to (n,6,6)$, and so on.
\end{itemize}
This gives the figure we drew earlier and shows $3030$ beams is possible.
\paragraph{Necessity.}
We now show at least $3n/2$ beams are necessary.
Maintain coordinates, and call the beams $x$-beams, $y$-beams, $z$-beams
according to which plane their long edges are perpendicular too.
Let $N_x$, $N_y$, $N_z$ be the number of these.
\begin{claim*}
If $\min(N_x, N_y, N_z) = 0$, then at least $n^2$ beams are needed.
\end{claim*}
\begin{proof}
Assume WLOG that $N_z = 0$.
Orient the cube so the $z$-plane touches the ground.
Then each of the $n$ layers of the cube (from top to bottom)
must be completely filled, and so at least $n^2$ beams are necessary,
\end{proof}
We henceforth assume $\min(N_x, N_y, N_z) > 0$.
\begin{claim*}
If $N_z > 0$, then we have $N_x + N_y \ge n$.
\end{claim*}
\begin{proof}
Again orient the cube so the $z$-plane touches the ground.
We see that for each of the $n$ layers of the cube (from top to bottom),
there is at least one $x$-beam or $y$-beam.
(Pictorially, some of the $x$ and $y$ beams form a ``staircase''.)
This completes the proof.
\end{proof}
Proceeding in a similar fashion, we arrive at the three relations
\begin{align*}
N_x + N_y &\ge n \\
N_y + N_z &\ge n \\
N_z + N_x &\ge n.
\end{align*}
Summing gives $N_x + N_y + N_z \ge 3n/2$ too.
\begin{remark*}
The problem condition has the following ``physics'' interpretation.
Imagine the cube is a metal box which is sturdy enough that
all beams must remain orthogonal to the faces of the box
(i.e.\ the beams cannot spin).
Then the condition of the problem is exactly what is needed so that,
if the box is shaken or rotated, the beams will not move.
\end{remark*}
\begin{remark*}
Walter Stromquist points out that the number of constructions
with $3030$ beams is actually enormous:
not dividing out by isometries,
the number is $(2 \cdot 1010!)^3$.
\end{remark*}
|
USAMO-2020-notes_3
|
Let $p$ be an odd prime.
An integer $x$ is called a \emph{quadratic non-residue}
if $p$ does not divide $x-t^2$ for any integer $t$.
Denote by $A$ the set of all integers $a$
such that $1 \le a < p$,
and both $a$ and $4-a$ are quadratic non-residues.
Calculate the remainder
when the product of the elements of $A$ is divided by $p$.
|
The answer is that $\prod_{a \in A} a \equiv 2 \pmod p$
regardless of the value of $p$.
In the following solution,
we work in $\FF_p$ always
and abbreviate ``quadratic residue'' and ``non-quadratic residue''
to ``QR'' and ``non-QR'', respectively.
We define
\begin{align*}
A &= \left\{ a \in \FF_p \mid a, 4-a \text{ non-QR} \right\} \\
B &= \left\{ b \in \FF_p \mid b, 4-b \text{ QR}, b \neq 0, b \neq 4 \right\}.
\end{align*}
Notice that
\[ A \cup B = \left\{ n \in \FF_p \mid
n(4-n) \text{ is QR} \;
n \neq 0, 4 \right\}. \]
We now present two approaches both based on the set $B$.
\paragraph{First approach (based on Holden Mui's presentation in Mathematics Magazine).}
The idea behind this approach is that $n(4-n)$
is itself an element of $B$ for $n \in A \cup B$,
because $4 - n(4-n) = (n-2)^2$.
This motivates the following claim.
\begin{claim*}
The map
\[ A \cup B \setminus \{2\} \to B
\qquad \text{by}\quad n \mapsto n(4-n) \]
is a well-defined two-to-one map,
i.e.\ every $b \in B$ has exactly two pre-images.
\end{claim*}
\begin{proof}
Since $n \notin \{0,2,4\}$,
we have $n(4-n) \notin \{0,4\}$,
so as discussed previously, $n(4-n) \in B$.
Thus this map is well-defined.
Choose $b \in B$.
The quadratic equation $n(4-n) = b$ in $n$
rewrites as $n^2-4n+b=0$,
and has discriminant $4(4-b)$,
which is a nonzero QR.
Hence there are exactly two values of $n$, as desired.
\end{proof}
Therefore, it follows that
\[ \prod_{n \in A \cup B \setminus \{2\}} n
= \prod_{b \in B} b \]
by pairing $n$ with $4-n$ on the left-hand side.
So, $\prod_{a \in A} a = 2$.
\paragraph{Second calculation approach (along the lines of official solution).}
We now do the following magical calculation in $\FF_p$:
\begin{align*}
\prod_{b \in B} b
= \prod_{b \in B} (4-b)
&= \prod_{\substack{1 \le y \le (p-1)/2 \\ y \neq 2 \\ 4-y^2 \text{ is QR}}} (4-y^2) \\
&= \prod_{\substack{1 \le y \le (p-1)/2 \\ y \neq 2 \\ 4-y^2 \text{ is QR}}} (2+y)
\prod_{\substack{1 \le y \le (p-1)/2 \\ y \neq 2 \\ 4-y^2 \text{ is QR}}} (2-y) \\
&= \prod_{\substack{1 \le y \le (p-1)/2 \\ y \neq 2 \\ 4-y^2 \text{ is QR}}} (2+y)
\prod_{\substack{(p+1)/2 \le y \le p-1 \\ y \neq p-2 \\ 4-y^2 \text{ is QR}}} (2+y) \\
&= \prod_{\substack{1 \le y \le p-1 \\ y \neq 2, p-2 \\ 4-y^2 \text{ is QR}}} (2+y) \\
&= \prod_{\substack{3 \le z \le p+1 \\ z \neq 4,p \\ z(4-z) \text{ is QR}}} z \\
&= \prod_{\substack{0 \le z \le p-1 \\ z \neq 0,4,2 \\ z(4-z) \text{ is QR}}} z.
\end{align*}
Note $z(4-z)$ is a nonzero QR if and only if $z \in A \cup B$.
So the right-hand side is almost the product over $z \in A \cup B$,
except it is missing the $z=2$ term.
Adding it in gives
\[ \prod_{b \in B} b
= \half \prod_{\substack{0 \le z \le p-1 \\ z \neq 0,4 \\ z(4-z) \text{ is QR}}} z
= \half \prod_{a \in A} a \prod_{b \in B} b. \]
This gives $\prod_{a \in A} a = 2$ as desired.
|
USAMO-2020-notes_4
|
Suppose that $(a_1, b_1)$, $(a_2, b_2)$, \dots, $(a_{100}, b_{100})$
are distinct ordered pairs of nonnegative integers.
Let $N$ denote the number of pairs of integers $(i,j)$ satisfying
$1 \le i < j \le 100$ and $\left\lvert a_ib_j - a_jb_i \right\rvert = 1$.
Determine the largest possible value of $N$
over all possible choices of the $100$ ordered pairs.
|
The answer is $197$.
In general, if $100$ is replaced by $n \ge 2$ the answer is $2n-3$.
The idea is that if we let $P_i = (a_i, b_i)$ be a point
in the coordinate plane, and let $O = (0,0)$
then we wish to maximize the number of triangles
$\triangle O P_i P_j$ which have area $1/2$.
Call such a triangle \emph{good}.
\paragraph{Construction of $197$ points.}
It suffices to use the points
$(1,0)$, $(1,1)$, $(2,1)$, $(3,1)$, \dots, $(99,1)$ as shown.
Notice that:
\begin{itemize}
\ii There are $98$ good triangles
with vertices $(0,0)$, $(k,1)$ and $(k+1,1)$ for $k=1, \dots, 98$.
\ii There are $99$ good triangles
with vertices $(0,0)$, $(1,0)$ and $(k,1)$ for $k=1, \dots, 99$.
\end{itemize}
This is a total of $98 + 99 = 197$ triangles.
\begin{center}
\begin{asy}
draw( (6,0)--(0,0)--(0,2), gray, Arrows );
dot("$O$", (0,0), dir(225), red );
dot("$(1,0)$", (1,0), dir(-90), blue);
dot("$(1,1)$", (1,1), dir(90), blue);
dot("$(2,1)$", (2,1), dir(90), blue);
dot("$(3,1)$", (3,1), dir(90), blue);
dot("$(4,1)$", (4,1), dir(90), blue);
label("$\dotsb$", (5,1), blue);
\end{asy}
\end{center}
\paragraph{Proof that $197$ points is optimal.}
We proceed by induction on $n$ to show the bound of $2n-3$.
The base case $n=2$ is evident.
For the inductive step,
suppose (without loss of generality) that the point $P = P_n = (a,b)$
is the farthest away from the point $O$ among all points.
\begin{claim*}
This farthest point $P = P_n$ is part of at most two good triangles.
\end{claim*}
\begin{proof}
We must have $\gcd(a,b) = 1$ for $P$ to be in any good triangles at all,
since otherwise any divisor of $\gcd(a,b)$ also divides $2[OPQ]$.
Now, we consider the locus of all points $Q$ for which $[OPQ] = 1/2$.
It consists of two parallel lines passing with slope $OP$, as shown.
\begin{center}
\begin{asy}
size(10cm);
int M = 6;
pair P = (3,2);
pair O = (0,0);
draw(O--P, blue);
draw(arc(O,abs(P),0,90), deepcyan);
for (int i=-M#2; i<=M; ++i) {
draw( (-M#2,i)--(M,i), gray);
draw( (i,-M#2)--(i,M), gray);
}
draw( (-M#2,0)--(M,0), black+1.4, Arrows(TeXHead) );
draw( (0,-M#2)--(0,M), black+1.4, Arrows(TeXHead) );
pair X = (1,1);
pair Y = O+P-X;
dot("$(u,v)$", X, dir(135), red);
dot("$(u',v')$", Y, dir(-45), red);
pair X1 = X+P;
pair X2 = X-P;
draw(X1--X2, red+dashed, Arrows);
pair Y1 = Y+P;
pair Y2 = Y-P;
draw(Y1--Y2, red+dashed, Arrows);
dot("$O$", O, dir(225), blue);
dot("$P=(a,b)$", P, dir(30), blue);
\end{asy}
\end{center}
Since $\gcd(a,b)=1$, see that only two lattice points on this locus
actually lie inside the quarter-circle centered at $O$ with radius $OP$.
Indeed if one of the points is $(u,v)$
then the others on the line are $(u \pm a, v \pm b)$ where the signs match.
This proves the claim.
\end{proof}
This claim allows us to complete the induction by simply deleting $P_n$.
|
USAMO-2020-notes_5
|
A finite set $S$ of points in the coordinate plane
is called \emph{overdetermined} if $|S| \ge 2$
and there exists a nonzero polynomial $P(t)$,
with real coefficients and of degree at most $|S|-2$,
satisfying $P(x)=y$ for every point $(x,y) \in S$.
For each integer $n \ge 2$,
find the largest integer $k$ (in terms of $n$) such that
there exists a set of $n$ distinct points
that is \emph{not} overdetermined,
but has $k$ overdetermined subsets.
|
We claim the answer is $k = 2^{n-1}-n$.
We denote the $n$ points by $A$.
Throughout the solution we will repeatedly use the following fact:
\begin{lemma*}
If $S$ is a finite set of points in the plane
there is at most one polynomial with real coefficients
and of degree at most $|S|-1$
whose graph passes through all points of $S$.
\end{lemma*}
\begin{proof}
If any two of the points have the same $x$-coordinate
then obviously no such polynomial may exist at all.
Otherwise, suppose $f$ and $g$ are two such polynomials.
Then $f-g$ has degree at most $|S|-1$,
but it has $|S|$ roots, so is the zero polynomial.
\end{proof}
\begin{remark*}
Actually Lagrange interpolation implies
that such a polynomial exists
as long as all the $x$-coordinates are different!
\end{remark*}
\medskip
\textbf{Construction}:
Consider the set $A = \left\{ (1,a), (2,b), (3,b), (4,b), \dots, (n,b) \right\}$,
where $a$ and $b$ are two distinct nonzero real numbers.
Any subset of the latter $n-1$ points with at least one element
is overdetermined, and there are $2^{n-1}-n$ such sets.
\medskip
\textbf{Bound}:
Say a subset $S$ of $A$ is \emph{flooded}
if it is not overdetermined.
For brevity, an \emph{$m$-set}
refers simply to a subset of $A$ with $m$ elements.
\begin{claim*}
If $S$ is an flooded $m$-set for $m \ge 3$,
then at most one $(m-1)$-subset of $S$ is not flooded.
\end{claim*}
\begin{proof}
Let $S = \left\{ p_1, \dots, p_m \right\}$ be flooded.
Assume for contradiction that $S - \{p_i\}$
and $S - \{p_j\}$ are both overdetermined.
Then we can find polynomials $f$ and $g$ of degree at most $m-3$
passing through $S - \{p_i\}$ and $S - \{p_j\}$, respectively.
Since $f$ and $g$ agree on $S - \{p_i, p_j\}$,
which has $m-2$ elements, by the lemma we have $f = g$.
Thus this common polynomial (actually of degree at most $m-3$)
witnesses that $S$ is overdetermined,
which is a contradiction.
\end{proof}
\begin{claim*}
For all $m = 2, 3, \dots, n$ there are at least $\binom{n-1}{m-1}$
flooded $m$-sets of $A$.
\end{claim*}
\begin{proof}
The proof is by downwards induction on $m$.
The base case $m = n$ is by assumption.
For the inductive step, suppose it's true for $m$.
Each of the $\binom{n-1}{m-1}$ flooded $m$-sets
has at least $m-1$ flooded $(m-1)$-subsets.
Meanwhile, each $(m-1)$-set has exactly $n-(m-1)$ parent $m$-sets.
We conclude the number of flooded sets of size $m-1$ is at least
\[ \frac{m-1}{n-(m-1)} \binom{n-1}{m-1} = \binom{n-1}{m-2} \]
as desired.
\end{proof}
This final claim completes the proof,
since it shows there are at most
\[ \sum_{m=2}^n \left( \binom nm - \binom{n-1}{m-1} \right)
= 2^{n-1} - n \]
overdetermined sets, as desired.
\begin{remark*}
[On repeated $x$-coordinates]
Suppose $A$ has two points $p$ and $q$ with repeated $x$-coordinates.
We can argue directly that $A$ satisfies the bound.
Indeed, any overdetermined set contains at most one of $p$ and $q$.
Moreover, given $S \subseteq A - \{p,q\}$,
if $S \cup \{p\}$ is overdetermined
then $S \cup \{q\}$ is not, and vice-versa.
(Recall that overdetermined sets always
have distinct $x$-coordinates.)
This gives a bound $\left[ 2^{n-2}-(n-2)-1 \right]
+ \left[ 2^{n-2}-1 \right] = 2^{n-1}-n$ already.
\end{remark*}
\begin{remark*}
[Alex Zhai]
An alternative approach to the double-counting
argument is to show that any overdetermined $m$-set
has an flooded $m$-superset.
Together with the first claim,
this lets us pair overdetermined sets in a way
that implies the bound.
\end{remark*}
|
USAMO-2020-notes_6
|
Let $n \geq 2$ be an integer.
Let $x_1 \ge x_2 \ge \dots \ge x_n$ and $y_1 \ge y_2 \ge \dots \ge y_n$
be $2n$ real numbers such that
\begin{align*}
0 &= x_1 + x_2 + \dots + x_n = y_1 + y_2 + \dots + y_n, \\
\text{and}\quad
1 &= x_1^2 + x_2^2 + \dots + x_n^2 = y_1^2 + y_2^2 + \dots + y_n^2.
\end{align*}
Prove that
\[ \sum_{i=1}^n (x_i y_i - x_i y_{n+1-i}) \geq \frac{2}{\sqrt{n-1}}. \]
\end{enumerate}
|
We present two approaches.
In both approaches, it's helpful motivation that for even $n$, equality occurs at
\begin{align*}
(x_i) &= \Big(
\underbrace{\frac{1}{\sqrt n}, \dots, \frac{1}{\sqrt n}}_{n/2},
\underbrace{-\frac{1}{\sqrt n}, \dots, -\frac{1}{\sqrt n}}_{n/2} \Big) \\
(y_i) &= \Big( \frac{n-1}{\sqrt{n(n-1)}},
\underbrace{- \frac{1}{\sqrt{n(n-1)}},
\dots,
- \frac{1}{\sqrt{n(n-1)}}}_{n-1} \Big)
\end{align*}
\paragraph{First approach (expected value).}
For a permutation $\sigma$ on $\{1, 2, \dots, n\}$ we define
\[ S_\sigma = \sum_{i=1}^n x_i y_{\sigma(i)}. \]
\begin{claim*}
For random permutations $\sigma$,
$\mathbb E[S_\sigma] = 0$
and $\mathbb E[S_\sigma^2] = \frac{1}{n-1}$.
\end{claim*}
\begin{proof}
The first one is clear.
Since $\sum_{i < j} 2x_i x_j = -1$, it follows that (for fixed $i$ and $j$),
$\mathbb E[y_{\sigma(i)} y_{\sigma(j)}] = -\frac{1}{n(n-1)}$,
Thus
\begin{align*}
\sum_{i} x_i^2 \cdot \mathbb E \left[ y_{\sigma(i)}^2 \right]
&= \frac 1n \\
2\sum_{i < j} x_i x_j \cdot \mathbb E \left[ y_{\sigma(i)} y_{\sigma(j)} \right]
&= (-1) \cdot \frac{1}{n(n-1)}
\end{align*}
the conclusion follows.
\end{proof}
%\begin{claim*}
% For two independent random permutations $\sigma$ and $\tau$,
% \[ \mathbb E \left[ \left( S_\sigma - S_\tau \right)^2
% \right] = \frac{2}{n-1}. \]
%\end{claim*}
%\begin{proof}
% Since
% \[ \mathbb E[S_\sigma S_\tau]
% = \sum_i \sum_{i'} x_i x_{i'}
% \mathbb E \left[ y_{\sigma(i)} \right]
% \mathbb E \left[ y_{\tau(i')} \right] = 0 \]
% the conclusion follows.
%\end{proof}
\begin{claim*}
[A random variable in {$[0,1]$} has variance at most $1/4$]
If $A$ is a random variable with mean $\mu$
taking values in the closed interval $[m, M]$ then
\[ \mathbb E[(A-\mu)^2] \le \frac14 (M-m)^2. \]
\end{claim*}
\begin{proof}
By shifting and scaling, we may assume $m = 0$ and $M = 1$,
so $A \in [0,1]$ and hence $A^2 \le A$.
Then
\[
\mathbb E[(A-\mu)^2]
= \mathbb E[A^2] - \mu^2
\le \mathbb E[A] - \mu^2 = \mu-\mu^2 \le \frac14.
\]
This concludes the proof.
\end{proof}
Thus the previous two claims together give
\[ \max_\sigma S_\sigma - \min_\sigma S_\sigma
\ge \sqrt{\frac{4}{n-1}} = \frac{2}{\sqrt{n-1}}.
\]
But $\sum_i x_i y_i = \max_\sigma S_\sigma$ and $\sum_i x_i y_{n+1-i} = \min_\sigma S_\sigma$
by rearrangement inequality and therefore we are done.
\paragraph{Outline of second approach (by convexity, due to Alex Zhai).}
We will instead prove a converse result: given the hypotheses
\begin{itemize}
\ii $x_1 \ge \dots \ge x_n$
\ii $y_1 \ge \dots \ge y_n$
\ii $\sum_i x_i = \sum_i y_i = 0$
\ii $\sum_i x_i y_i - \sum_i x_i y_{n+1-i} = \frac{2}{\sqrt{n-1}}$
\end{itemize}
we will prove that $\sum x_i^2 \sum y_i^2 \le 1$.
Fix the choice of $y$'s.
We see that we are trying to maximize a convex function in $n$
variables $(x_1, \dots, x_n)$ over a convex domain
(actually the intersection of two planes with several half planes).
So a maximum can only happen at the boundaries:
when at most two of the $x$'s are different.
An analogous argument applies to $y$.
In this way we find that it suffices to consider situations
where $x_\bullet$ takes on at most two different values.
The same argument applies to $y_\bullet$.
At this point the problem can be checked directly.
|
USAMO-2021-notes_1
|
Rectangles $BCC_1B_2$, $CAA_1C_2$, and $ABB_1A_2$ are erected
outside an acute triangle $ABC$. Suppose that
\[ \angle BC_1C + \angle CA_1A + \angle AB_1B = 180^\circ. \]
Prove that lines $B_1C_2$, $C_1A_2$, and $A_1B_2$ are concurrent.
|
The angle condition implies the circumcircles of the three
rectangles concur at a single point $P$.
\begin{center}
\begin{asy}
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
pair P = 0.2*dir(190);
filldraw(A--B--C--cycle, opacity(0.2)+lightcyan, blue);
pair X = circumcenter(P, B, C);
pair Y = circumcenter(P, C, A);
pair Z = circumcenter(P, A, B);
pair C_1 = 2*X-B;
pair B_2 = 2*X-C;
pair A_1 = 2*Y-C;
pair C_2 = 2*Y-A;
pair B_1 = 2*Z-A;
pair A_2 = 2*Z-B;
filldraw(circumcircle(A, B, P), opacity(0.05)+yellow, red);
filldraw(circumcircle(B, C, P), opacity(0.05)+yellow, red);
filldraw(circumcircle(C, A, P), opacity(0.05)+yellow, red);
draw(B_1--C_2, deepgreen+dashed);
draw(C_1--A_2, deepgreen+dashed);
draw(A_1--B_2, deepgreen+dashed);
draw(C--C_1--B_2--B, red);
draw(A--A_1--C_2--C, red);
draw(B--B_1--A_2--A, red);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$P$", P, dir(280));
dot("$C_1$", C_1, dir(C_1));
dot("$B_2$", B_2, dir(B_2));
dot("$A_1$", A_1, dir(A_1));
dot("$C_2$", C_2, dir(C_2));
dot("$B_1$", B_1, dir(B_1));
dot("$A_2$", A_2, dir(A_2));
/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
A = dir 110
B = dir 210
C = dir 330
P 280 = 0.2*dir(190)
A--B--C--cycle / 0.2 lightcyan / blue
X := circumcenter P B C
Y := circumcenter P C A
Z := circumcenter P A B
C_1 = 2*X-B
B_2 = 2*X-C
A_1 = 2*Y-C
C_2 = 2*Y-A
B_1 = 2*Z-A
A_2 = 2*Z-B
circumcircle A B P / 0.1 yellow / red
circumcircle B C P / 0.1 yellow / red
circumcircle C A P / 0.1 yellow / red
B_1--C_2 / deepgreen
C_1--A_2 / deepgreen
A_1--B_2 / deepgreen
C--C_1--B_2--B / red
A--A_1--C_2--C / red
B--B_1--A_2--A / red
*/
\end{asy}
\end{center}
Then $\dang C P B_2 = \dang C P A_1 = 90\dg$,
hence $P$ lies on $A_1 B_2$ etc., so we're done.
\begin{remark*}
As one might guess from the two-sentence solution,
the entire difficulty of the problem
is getting the characterization of the concurrence point.
\end{remark*}
|
USAMO-2021-notes_2
|
The Planar National Park is a undirected 3-regular planar graph
(i.e.\ all vertices have degree $3$).
A visitor walks through the park as follows:
she begins at a vertex and starts walking along an edge.
When she reaches the other endpoint, she turns left.
On the next vertex she turns right, and so on,
alternating left and right turns at each vertex.
She does this until she gets back to the vertex where she started.
What is the largest possible number of times she could have entered
any vertex during her walk, over all possible layouts of the park?
|
The answer is $3$.
We consider the trajectory of the visitor as an ordered sequence of \emph{turns}.
A turn is defined by specifying a vertex, the incoming edge, and the outgoing edge.
Hence there are six possible turns for each vertex.
\begin{claim*}
Given one turn in the sequence, one can reconstruct the entire sequence of turns.
\end{claim*}
\begin{proof}
This is clear from the process's definition: given a turn $t$,
one can compute the turn after it and the turn before it.
\end{proof}
This implies already that the trajectory of the visitor,
when extended to an infinite sequence, is totally periodic (not just eventually periodic),
because there are finitely many possible turns, so some turn must be repeated.
So, any turn appears at most once in the period of the sequence,
giving a na\"{\i}ve bound of $6$ for the original problem.
However, the following claim improves the bound to $3$.
\begin{claim*}
It is impossible for both of the turns $a \to b \to c$ and $c \to b \to a$ to occur
in the same trajectory.
\end{claim*}
\begin{proof}
If so, then extending the path, we get $a \to b \to c \to d \to e \to \dotsb$
and $\dots \to e \to d \to c \to b \to a$, as illustrated below in red and blue respectively.
\begin{center}
\begin{asy}
defaultpen(fontsize(11pt));
dotfactor *= 1.8;
pair A = (0,0);
pair B = (1,1);
pair C = (0,2);
pair D = (1,3);
pair E = (0,4);
draw(A--B--C--D--E--(0.5,4.5), black+1.3);
draw( (0.5,4.5)--(0.8,4.8), black+1.3+dotted );
draw((1,-1)--A--(-1,0), gray);
draw(B--(2,1), gray);
draw(C--(-1,2), gray);
draw(D--(2,3), gray);
draw(E--(-1,4), gray);
dot("$a$", A, -dir(B-A), deepgreen);
dot("$b$", B, -dir(C-B), deepgreen);
dot("$c$", C, -dir(D-C), deepgreen);
dot("$d$", D, -dir(E-D), deepgreen);
dot("$e$", E, dir(90), deepgreen);
draw( (1,0.2)..(1.4,1)..(1,1.8), red+1.8, EndArrow(TeXHead));
draw( (1,2.2)..(1.4,3)..(1,3.8), lightred+1, EndArrow(TeXHead));
draw( (0.8,1.6)..(0.4,2)..(0.8,2.4), lightred+1, EndArrow(TeXHead));
draw( (0.8,3.6)..(0.4,4)..(0.8,4.4), lightred+1, EndArrow(TeXHead));
draw( (0.2,1.4)..(0.6,1)..(0.2,0.6), blue+1.8, EndArrow(TeXHead));
draw( (0.2,3.4)..(0.6,3)..(0.2,2.6), lightblue+1, EndArrow(TeXHead));
draw( (0,2.8)..(-0.4,2)..(0,1.2), lightblue+1, EndArrow(TeXHead));
draw( (0,4.8)..(-0.4,4)..(0,3.2), lightblue+1, EndArrow(TeXHead));
\end{asy}
\end{center}
However, we assumed for contradiction the red and blue paths
were part of the same trajectory, yet they clearly never meet.
\end{proof}
It remains to give a construction showing $3$ can be achieved.
There are many, many valid constructions.
One construction due to Danielle Wang is given here, who provided the following motivation:
``I was lying in bed and drew the first thing I could think of''.
The path is $CAHIFGDBAHEFGJBAC$ which visits $A$ three times.
\begin{center}
\begin{asy}
size(10cm);
defaultpen(fontsize(11pt));
pair A = (-2,2);
pair B = (2,2);
pair C = (-1,1);
pair D = (1,1);
pair E = (-1,-1);
pair F = (0,-1);
pair G = (1,-1);
pair H = (-2,-2);
pair I = (0,-2);
pair J = (2,-2);
pen gr = fontsize(14pt) + deepgreen;
draw(A--H, blue);
label("$2,9$", A--H, gr);
draw(A--C, blue);
label("$16$", A--C, gr);
draw(A--B, blue);
draw(B--D, blue);
draw(B--J, blue);
draw(B--A, blue);
label("$8,15$", B--A, gr);
draw(C--E, blue);
draw(C--D, blue);
draw(C--A, blue);
label("$1$", C--A, gr);
draw(D--C, blue);
draw(D--G, blue);
draw(D--B, blue);
label("$7$", D--B, gr);
draw(E--H, blue);
draw(E--F, blue);
label("$11$", E--F, gr);
draw(E--C, blue);
draw(F--I, blue);
draw(F--G, blue);
label("$5,12$", F--G, gr);
draw(F--E, blue);
draw(G--F, blue);
draw(G--J, blue);
label("$13$", G--J, gr);
draw(G--D, blue);
label("$6$", G--D, gr);
draw(H--I, blue);
label("$3$", H--I, gr);
draw(H--E, blue);
label("$10$", H--E, gr);
draw(H--A, blue);
draw(I--H, blue);
draw(I--J, blue);
draw(I--F, blue);
label("$4$", I--F, gr);
draw(J--B, blue);
label("$14$", J--B, gr);
draw(J--G, blue);
draw(J--I, blue);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, -dir(C));
dot("$D$", D, -dir(D));
dot("$E$", E, -dir(E));
dot("$F$", F, -dir(F));
dot("$G$", G, -dir(G));
dot("$H$", H, dir(H));
dot("$I$", I, dir(I));
dot("$J$", J, dir(J));
margin m = Margin(3,3);
pair[] trajectory = {C,A,H,I,F,G,D,B,A,H,E,F,G,J,B,A,C};
for (int i=0; i<trajectory.length-1; ++i) {
draw(trajectory[i]--trajectory[i+1], red+1.4, m);
draw(trajectory[i]--trajectory[i+1], red, EndArrow, m);
}
\end{asy}
\end{center}
\begin{remark*}
As the above example shows it is possible to transverse
an edge more than once even in the same direction, as in edge $AH$ above.
\end{remark*}
|
USAMO-2021-notes_4
|
A finite set $S$ of positive integers has the property that,
for each $s\in S$, and each positive integer divisor $d$ of $s$,
there exists a unique element $t\in S$ satisfying $\gcd(s,t) = d$.
(The elements $s$ and $t$ could be equal.)
Given this information, find all possible values for the
number of elements of $S$.
|
The answer is that $|S|$ must be a power of $2$ (including $1$),
or $|S| = 0$ (a trivial case we do not discuss further).
\paragraph{Construction.}
For any nonnegative integer $k$, a construction for $|S| = 2^k$ is given by
\[ S = \left\{
(p_1 \text{ or } q_1)
\times
(p_2 \text{ or } q_2)
\times
\dots
\times
(p_k \text{ or } q_k)
\right\}
\]
for $2k$ distinct primes $p_1$, \dots, $p_k$, $q_1$, \dots, $q_k$.
\paragraph{Converse.}
The main claim is as follows.
\begin{claim*}
In any valid set $S$, for any prime $p$ and $x \in S$, $\nu_p(x) \le 1$.
\end{claim*}
\begin{proof}
Assume for contradiction $e = \nu_p(x) \ge 2$.
\begin{itemize}
\ii On the one hand, let $s = x$ in the statement.
Vary $t \in S$ across all the elements in $S$ to get each divisor of $x$ once.
Since $\frac{e}{e+1}$ of the divisors of $x$ are divisible by $p$,
it thus follows that $\frac{e}{e+1}$ of the elements are divisible by $p$.
\ii On the other hand, consider a $y \in S$ such that $\nu_p(y)=1$,
which must exist (say if $\gcd(x,y) = p$).
Taking $s = y$ in the statement and repeating the same argument,
we see $\half$ of the elements of $S$ are divisible by $p$.
\end{itemize}
So $e=1$, contradiction.
\end{proof}
Now since $|S|$ equals the number of divisors of any element of $S$, we are done.
|
USAMO-2021-notes_5
|
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*}
|
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*}
|
USAMO-2021-notes_6
|
Let $ABCDEF$ be a convex hexagon satisfying
$\ol{AB} \parallel \ol{DE}$,
$\ol{BC} \parallel \ol{EF}$,
$\ol{CD} \parallel \ol{FA}$, and
\[ AB \cdot DE = BC \cdot EF = CD \cdot FA. \]
Let $X$, $Y$, and $Z$ be the midpoints
of $\ol{AD}$, $\ol{BE}$, and $\ol{CF}$.
Prove that
the circumcenter of $\triangle ACE$,
the circumcenter of $\triangle BDF$, and
the orthocenter of $\triangle XYZ$ are collinear.
\end{enumerate}
|
We present two solutions.
\paragraph{Parallelogram solution found by contestants.}
Note that the following figure is intentionally
\emph{not} drawn to scale, to aid legibility.
We construct parallelograms $ABCE'$, etc as shown.
Note that this gives two congruent triangles $A'C'E'$ and $B'D'F'$.
(Assuming that triangle $XYZ$ is non-degenerate,
the triangles $A'C'E'$ and $B'D'F'$ will also be non-degenerate.)
\begin{center}
\begin{asy}
pair B = dir(123);
pair C = dir(150);
pair D = dir(250);
pair E = -conj(D);
pair F = B*C/E;
pair A = D*C/F;
pair Ap = C+E-D;
pair Cp = E+A-F;
pair Ep = A+C-B;
pair Bp = D+F-E;
pair Dp = F+B-A;
pair Fp = B+D-C;
filldraw(A--B--C--D--E--F--cycle, opacity(0.1)+lightcyan, blue);
draw(A--Cp, red);
draw(C--Ep, red);
draw(E--Ap, red);
draw(B--Fp, deepgreen);
draw(D--Bp, deepgreen);
draw(F--Dp, deepgreen);
draw(A--D, gray);
pair X = midpoint(A--D);
pair M = midpoint(Cp--Ep);
pair N = midpoint(Bp--Fp);
draw(M--N, gray);
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$A$", A, dir(A));
dot("$A'$", Ap, dir(90));
dot("$C'$", Cp, dir(160));
dot("$E'$", Ep, dir(310));
dot("$B'$", Bp, dir(90));
dot("$D'$", Dp, dir(180));
dot("$F'$", Fp, dir(315));
dot("$X$", X, dir(X));
dot("$M$", M, dir(350));
dot("$N$", N, dir(270));
/* TSQ Source:
B = dir 123
C = dir 150
D = dir 250
E = -conj(D)
F = B*C/E
A = D*C/F
A' = C+E-D R90
C' = E+A-F R160
E' = A+C-B R310
B' = D+F-E R90
D' = F+B-A R180
F' = B+D-C R315
A--B--C--D--E--F--cycle 0.1 lightcyan / blue
A--Cp red
C--Ep red
E--Ap red
B--Fp deepgreen
D--Bp deepgreen
F--Dp deepgreen
A--D gray
X = midpoint A--D
M = midpoint Cp--Ep R350
N = midpoint Bp--Fp R270
M--N gray
*/
\end{asy}
\end{center}
\begin{claim*}
If $AB \cdot DE = BC \cdot EF = CD \cdot FA = k$,
then the circumcenters of $ACE$ and $A'C'E'$ coincide.
\end{claim*}
\begin{proof}
The power of $A$ to $(A'C'E')$ is
$AE' \cdot AC' = BC \cdot EF = k$;
same for $C$ and $E$.
\end{proof}
\begin{center}
\begin{asy}
size(10cm);
pair B = dir(123);
pair C = dir(150);
pair D = dir(250);
pair E = -conj(D);
pair F = B*C/E;
pair A = D*C/F;
pair Ap = C+E-D;
pair Cp = E+A-F;
pair Ep = A+C-B;
pair Bp = D+F-E;
pair Dp = F+B-A;
pair Fp = B+D-C;
pair Ma = midpoint(Cp--Ep);
pair Mc = midpoint(Ep--Ap);
pair Me = midpoint(Ap--Cp);
pair Mb = midpoint(Dp--Fp);
pair Md = midpoint(Fp--Bp);
pair Mf = midpoint(Bp--Dp);
pair X = midpoint(A--D);
pair Y = midpoint(B--E);
pair Z = midpoint(C--F);
draw(Ap--Cp--Ep--cycle, red);
draw(Bp--Dp--Fp--cycle, deepgreen);
draw(X--Y--Z--cycle, blue);
draw(Ma--Mc--Me--cycle, lightred);
draw(Md--Mf--Mb--cycle, lightgreen);
draw(Mc--Mf, gray);
draw(Me--Mb, gray);
draw(Ma--Md, gray);
dot("$A'$", Ap, dir(180));
dot("$C'$", Cp, dir(275));
dot("$E'$", Ep, dir(0));
dot("$B'$", Bp, dir(0));
dot("$D'$", Dp, dir(180));
dot("$F'$", Fp, dir(275));
dot("$X$", X, dir(340), blue);
dot("$Y$", Y, dir(225), blue);
dot("$Z$", Z, dir(130), blue);
dot(Ma, red);
dot(Mc, red);
dot(Me, red);
dot(Mb, deepgreen);
dot(Md, deepgreen);
dot(Mf, deepgreen);
dot(orthocentercenter(Ma, Mc, Me), red);
dot(orthocentercenter(Mb, Md, Mf), deepgreen);
dot(orthocentercenter(X, Y, Z), blue);
\end{asy}
\end{center}
\begin{claim*}
Triangle $XYZ$ is the vector average
of the (congruent) medial triangles of
triangles $A'C'E'$ and $B'D'F'$.
\end{claim*}
\begin{proof}
If $M$ and $N$ are the midpoints
of $\ol{C'E'}$ and $\ol{B'F'}$,
then $X$ is the midpoint of $\ol{MN}$ by vector calculation:
\begin{align*}
\frac{\vec M + \vec N}{2}
&= \frac{\frac{\vec C' + \vec E'}{2} + \frac{\vec B' + \vec F'}{2}}{2} \\
&= \frac{\vec C' + \vec E' + \vec B' + \vec F'}{4} \\
&= \frac{(\vec A + \vec E - \vec F) + (\vec C + \vec A - \vec B)
+ (\vec D + \vec F - \vec E) + (\vec B + \vec D - \vec C)}{4} \\
&= \frac{\vec A + \vec D}{2} = \vec X. \qedhere
\end{align*}
\end{proof}
Hence the orthocenter of $XYZ$
is the midpoint of the orthocenters
of the medial triangles of $A'C'E'$ and $B'D'F'$ --- that is,
their circumcenters.
\paragraph{Author's solution.}
Call $MNP$ and $UVW$ the medial triangles
of $ACE$ and $BDF$.
\begin{center}
\begin{asy}
size(11cm);
pair M = dir(253.96);
pair N = dir(14.7);
pair P = dir(133.84);
pair U = dir(283.18);
pair V = dir(44.37);
pair W = dir(163.96);
pair A = N+P-M;
pair C = P+M-N;
pair E = M+N-P;
pair B = V+W-U;
pair D = W+U-V;
pair F = U+V-W;
draw(A--C--E--cycle, lightred);
draw(M--N--P--cycle, lightred);
draw(B--D--F--cycle, lightgreen);
draw(U--V--W--cycle, lightgreen);
pair X = midpoint(A--D);
pair Y = midpoint(B--E);
pair Z = midpoint(C--F);
draw(A--D, gray);
draw(B--E, gray);
draw(C--F, gray);
draw(unitcircle, dotted);
draw(M--V, deepcyan+1);
draw(B--C, deepcyan+1);
draw(E--F, deepcyan+1);
draw(A--B, blue+1);
draw(N--W, blue+1);
draw(D--E, blue+1);
draw(A--F, purple+1);
draw(U--P, purple+1);
draw(D--C, purple+1);
dot("$M$", M, dir(M));
dot("$N$", N, dir(N));
dot("$P$", P, dir(P));
dot("$U$", U, dir(U));
dot("$V$", V, dir(V));
dot("$W$", W, dir(W));
dot("$A$", A, dir(A));
dot("$C$", C, dir(C));
dot("$E$", E, dir(E));
dot("$B$", B, dir(B));
dot("$D$", D, dir(D));
dot("$F$", F, dir(F));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
dot("$Z$", Z, dir(Z));
/* TSQ Source:
!size(11cm);
M = dir 253.96
N = dir 14.7
P = dir 133.84
U = dir 283.18
V = dir 44.37
W = dir 163.96
A = N+P-M
C = P+M-N
E = M+N-P
B = V+W-U
D = W+U-V
F = U+V-W
A--C--E--cycle lightred
M--N--P--cycle lightred
B--D--F--cycle lightgreen
U--V--W--cycle lightgreen
X = midpoint A--D
Y = midpoint B--E
Z = midpoint C--F
A--D gray
B--E gray
C--F gray
unitcircle dotted
M--V deepcyan+1
B--C deepcyan+1
E--F deepcyan+1
A--B blue+1
N--W blue+1
D--E blue+1
A--F purple+1
U--P purple+1
D--C purple+1
*/
\end{asy}
\end{center}
\begin{claim*}
In trapezoid $ABDE$, the perpendicular
bisector of $\ol{XY}$ is the same as the
perpendicular bisector of the midline $\ol{WN}$.
\end{claim*}
\begin{proof}
This is true for any trapezoid:
because $WX = \half AB = YN$.
\end{proof}
\begin{claim*}
The points $V$, $W$, $M$, $N$ are cyclic.
\end{claim*}
\begin{proof}
By power of a point from $Y$, since
\[ WY \cdot YN = \half DE \cdot \half AB
= \half EF \cdot \half BC
= VY \cdot YM. \qedhere \]
\end{proof}
Applying all the cyclic variations of the above two claims,
it follows that all six points $U$, $V$, $W$, $M$, $N$, $P$
are concyclic, and the center of that circle
coincides with the circumcenter of $\triangle XYZ$.
\begin{remark*}
It is also possible to implement ideas from the first solution here,
by showing all six midpoints have equal power to $(XYZ)$.
\end{remark*}
\begin{claim*}
The orthocenter of $XYZ$ is the midpoint
of the circumcenters of $\triangle ACE$ and $\triangle BDF$.
\end{claim*}
\begin{proof}
Apply complex numbers with the unit circle coinciding
with the circumcircle of $NVPWMU$.
Then
\begin{align*}
\opname{orthocenter}(XYZ) &= x+y+z = \frac{a+b+c+d+e+f}{2} \\
\opname{circumcenter}(ACE) &= \opname{orthocenter}(MNP) \\
&= m+n+p = \frac{c+e}{2} + \frac{e+a}{2} + \frac{a+c}{2} = a+c+e \\
\opname{circumcenter}(BDF) &= \opname{orthocenter}(UVW) \\
&= u+v+w = \frac{d+f}{2} + \frac{f+b}{2} + \frac{b+d}{2} = b+d+f.
\qedhere
\end{align*}
\end{proof}
|
USAMO-2022-notes_1
|
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.
|
USAMO-2022-notes_2
|
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}
|
USAMO-2022-notes_3
|
Solve over positive real numbers the functional equation
\[ f(x) = f(f(f(x)) + y) + f(xf(y)) f(x+y). \]
|
The answer is $f(x) \equiv c/x$ for any $c > 0$.
This works, so we'll prove this is the only solution.
The following is based on the solution posted by \texttt{pad} on AoPS.
In what follows, $f^n$ as usual denotes $f$ iterated $n$ times,
and $P(x,y)$ is the given statement.
Also, we introduce the notation $Q$ for the statement
\[ Q(a,b) : \qquad f(a) \ge f(b) \implies f(f(b)) \ge a. \]
To see why this statement $Q$ is true,
assume for contradiction that $a > f(f(b))$;
then consider $P(b, a-f(f(b)))$ to get a contradiction.
The main idea of the problem is the following:
\begin{claim*}
Any function $f \colon \RR_{>0} \to \RR_{>0}$ obeying
statement $Q$ satisfies $f^2(x) = f^4(x)$.
\end{claim*}
\begin{proof}
From $Q(t,t)$ we get
\[ f^2(t) \ge t \qquad \text{ for all } t > 0. \]
So this already implies $f^4(x) \ge f^2(x)$ by choosing $t = f^2(x)$.
It also gives $f(x) \le f^3(x) \le f^5(x)$
by choosing $t = f(x)$, $t = f^3(x)$.
Then $Q(f^4(x), x)$ is valid and gives $f^2(x) \ge f^4(x)$, as needed.
\end{proof}
\begin{claim*}
The function $f$ is injective.
\end{claim*}
\begin{proof}
Suppose $f(u) = f(v)$ for some $u > v$.
From $Q(u,v)$ and $Q(v,u)$ we have $f^2(v) \ge u$ and $f^2(u) \ge v$.
Note that for all $x > 0$ we have statements
\begin{align*}
P(f^2(x), u) &\implies f^3(x) = f(x+u) + f(xf(u)) f(x+u) = (1+f(xf(u)))f(x+u) \\
P(f^2(x), v) &\implies f^3(x) = f(x+v) + f(xf(v)) f(x+v) = (1+f(xf(v)))f(x+v).
\end{align*}
It follows that $f(x+u) = f(x+v)$ for all $x > 0$.
This means that $f$ is periodic with period $T = u-v > 0$.
However, this is incompatible with $Q$,
because we would have $Q(1+nT, 1)$ for all positive integers $n$,
which is obviously absurd.
\end{proof}
Since $f$ is injective, we obtain that $f^2(x) = x$.
Thus $P(x,y)$ now becomes the statement
\[ P(x,y) : \qquad f(x) = f(x+y) \cdot \Big[ 1+f(xf(y)) \Big]. \]
In particular
\[ P(1,y) \implies f(1+y) = \frac{f(1)}{1+y} \]
so $f$ is determined on inputs greater than $1$.
Finally, if $a,b > 1$ we get
\[ P(a,b) \implies \frac 1a = \frac{1}{a+b} \cdot
\left[ 1 + f\left( \frac ab f(1) \right) \right] \]
which is enough to determine $f$ on all inputs, by varying $(a,b)$.
|
USAMO-2022-notes_4
|
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}
|
USAMO-2022-notes_5
|
A function $f \colon \RR \to \RR$ is
\emph{essentially increasing}
if $f(s) \leq f(t)$ holds
whenever $s\leq t$ are real numbers such that $f(s)\neq 0$ and $f(t)\neq 0$.
Find the smallest integer $k$ such that
for any $2022$ real numbers $x_1$, $x_2$, \dots, $x_{2022}$,
there exist $k$ essentially increasing functions $f_1, \dots, f_k$ such that
\[ f_1(n) + f_2(n) + \dotsb + f_k(n) = x_n
\qquad \hbox{ for every } n = 1, 2, \dots, 2022. \]
|
The answer is $11$ and, more generally, if $2022$ is replaced by $N$
then the answer is $\left\lfloor \log_2 N \right\rfloor + 1$.
\paragraph{Bound.}
Suppose for contradiction that $2^k-1 > N$
and choose $x_n = -n$ for each $n = 1, \dots, N$.
Now for each index $1 \le n \le N$, define
\[ S(n) = \left\{
\text{indices $i$ for which $f_i(n) \neq 0$}
\right\} \subseteq \{1,\dots,k\}.
\]
As each $S(nt)$ is nonempty, by pigeonhole, two $S(n)$'s coincide,
say $S(n) = S(n')$ for $n < n'$.
But it's plainly impossible that $x_n > x_{n'}$ in that case
due to the essentially increasing condition.
\paragraph{Construction.}
It suffices to do $N = 2^k-1$.
Rather than drown the reader in notation, we'll just illustrate
an example of the (inductive) construction for $k=4$.
Empty cells are zero.
\[
\begin{array}{r|rrrr}
& f_1 & f_2 & f_3 & f_4 \\ \hline
x_1 = 3 & \mathbf{\color{red}3} &&& \\ \hline
x_2 = 1 & 10 & \mathbf{\color{red}-9} && \\
x_3 = 4 & & \mathbf{\color{red}4} && \\ \hline
x_4 = 1 & 100 & 200 & \mathbf{\color{red}-299} & \\
x_5 = 5 & & 200 & \mathbf{\color{red}-195} & \\
x_6 = 9 & 100 & & \mathbf{\color{red}-91} & \\
x_7 = 2 & & & \mathbf{\color{red}2} & \\ \hline
x_8 = 6 & 1000 & 2000 & 4000 & \mathbf{\color{red}-6994} \\
x_{9} = 5 & & 2000 & 4000 & \mathbf{\color{red}-5995} \\
x_{10} = 3 & 1000 & & 4000 & \mathbf{\color{red}-4997} \\
x_{11} = 5 & & & 4000 & \mathbf{\color{red}-3995} \\
x_{12} = 8 & 1000 & 2000 & & \mathbf{\color{red}-2992} \\
x_{13} = 9 & & 2000 & & \mathbf{\color{red}-1991} \\
x_{14} = 7 & 1000 & & & \mathbf{\color{red}-993} \\
x_{15} = 9 & & & & \mathbf{\color{red}9}
\end{array}
\]
The general case is handled in the same way with powers of $10$
replaced by powers of $B$, for a sufficiently large number $B$.
|
USAMO-2022-notes_6
|
There are $2022$ users on a social network called Mathbook,
and some of them are Mathbook-friends.
(On Mathbook, friendship is always mutual and permanent.)
Starting now, Mathbook will only allow a new friendship to be formed
between two users if they have at least two friends in common.
What is the minimum number of friendships that must already exist
so that every user could eventually become friends with every other user?
\end{enumerate}
|
With $2022$ replaced by $n$, the answer is
$\left\lceil \frac 32 n \right\rceil - 2$.
\paragraph{Terminology.}
Standard graph theory terms:
starting from a graph $G$ on $n$ vertices,
we're allowed to take any $C_4$ in the graph and complete it to a $K_4$.
The problem asks the minimum number of edges needed
so that this operation lets us transform $G$ to $K_n$.
\paragraph{Construction.}
For even $n$, start with an edge $ab$,
and then create $n/2-1$ copies of $C_4$ that use $ab$ as an edge, as shown
below for $n=14$ (six copies of $C_4$).
\begin{center}
\begin{asy}
dotfactor *= 2;
pair A = (-1,0);
pair B = (1,0);
pair X, Y;
for (int i=1; i<=6; ++i) {
X = (-(i+3)/10, i/7 + 0.4);
Y = ((i+3)/10, i/7 + 0.4);
dot(X, gray);
dot(Y, gray);
draw(A--X--Y--B, gray);
}
draw(A--B, red+2);
dot("$a$", A, 2*dir(-90), red);
dot("$b$", B, 2*dir(-90), red);
\end{asy}
\end{center}
This can be completed into $K_n$ by first completing the $n/2-1$
$C_4$'s into $K_4$, then connecting red vertices to every gray vertex,
and then finishing up.
The construction for odd $n$ is the same except with one extra vertex $c$
which is connected to both $a$ and $b$.
\paragraph{Bound.}
Notice that additional operations or connections can never hurt.
So we will describe a \emph{specific} algorithm that performs operations
on the graph until no more operations are possible.
This means that if this algorithm terminates with anything other $G = K_n$,
the graph was never completable to $K_n$ to begin with.
The algorithm uses the following data:
it keeps a list $\mathcal C$ of cliques of $G$,
and a labeling $\mathcal{L} \colon E(G) \to \mathcal C$ which assigns to every
edge one of the cliques that contains it.
\begin{itemize}
\ii Initially, $\mathcal C$ consists of one $K_2$ for every edge of $G$,
and each edge is labeled in the obvious way.
\ii At each step, the algorithm arbitrarily takes any $C_4 = abcd$
whose four edges $ab$, $bc$, $cd$, $da$ do not all have the same label.
Consider these labels that appear (at least two, and up to four),
and let $V$ be the union of all vertices in any of these 2-4 cliques.
\ii Do the following graph operations:
connect $ac$ and $bd$, then connect every vertex in $V - \{a,b,c,d\}$
to each of $\{a,b,c,d\}$.
Finally, complete this to a clique on $V$.
\ii Update $\mathcal C$ by merging these 2-4 cliques into a single clique $K_V$.
\ii Update $\mathcal{L}$ by replacing every edge
that was labeled with one of these 2-4 cliques with the label $K_V$.
Also, update every \emph{newly} created edge to have label $K_V$.
However, if there were existing edges
not labeled with one of the 2-4 cliques, then we do \emph{not} update these!
\ii Stop once every $C_4$ has only one label appearing among its edges.
When this occurs, no operations are possible at all on the graph.
\end{itemize}
A few steps of the process are illustrated below for a graph on six vertices
with nine initial edges.
There are initially nine $K_2$'s labeled A, B, \dots, I.
Original edges are always bolder than added edges.
The relabeled edges in each step are highlighted in color.
Notice how we need an entirely separate operation to get G to become L,
even though no new edges are drawn in the graph.
\begin{center}
\begin{asy}
size(12cm);
picture base;
dotfactor *= 2;
pair P1 = (0,1);
pair P2 = (1,1);
pair P3 = (2,1);
pair P4 = (2,0);
pair P5 = (1,0);
pair P6 = (0,0);
dot(base, "$1$", P1, 2*dir(90));
dot(base, "$2$", P2, 2*dir(90));
dot(base, "$3$", P3, 2*dir(90));
dot(base, "$4$", P4, 2*dir(-90));
dot(base, "$5$", P5, 2*dir(-90));
dot(base, "$6$", P6, 2*dir(-90));
real r = 7;
transform t = shift(0,0);
draw("A", t*(P6--P1), dir(180), black+2);
draw("B", t*(P1--P2), dir(90), black+2);
draw("C", t*(P2--P3), dir(90), black+2);
draw("D", t*(P3--P4), dir(0), black+2);
draw("E", t*(P4--P5), dir(-90), black+2);
draw("F", t*(P5--P6), dir(-90), black+2);
draw("G", t*(P6--P2), dir(135), black+2);
draw("H", t*(P2--P5), dir(180), black+2);
draw("I", t*(P5--P3), dir(135), black+2);
add(t*base);
label("Initial setup", P5, r*dir(270));
t = shift(3,0);
draw("J", t*(P6--P1), dir(180), red+2);
draw("J", t*(P1--P2), dir(90), red+2);
draw("C", t*(P2--P3), dir(90), black+2);
draw("D", t*(P3--P4), dir(0), black+2);
draw("E", t*(P4--P5), dir(-90), black+2);
draw("J", t*(P5--P6), dir(-90), red+2);
draw("G", t*(P6--P2), dir(180), black+2);
draw("J", t*(P2--P5), dir(180), red+2);
draw("I", t*(P5--P3), dir(135), black+2);
draw("J", t*(P1--P5), dir(0), red);
add(t*base);
label(minipage("Step 1: Operate on $1256$. \\ Merges ABFH into J. \\
$\theta(\text{J}) = 4$", 120pt), t*P5, r*dir(270));
t = shift(0,-2.5);
draw("K", t*(P6--P1), dir(180), deepgreen+2);
draw("K", t*(P1--P2), dir(90), deepgreen+2);
draw("K", t*(P2--P3), dir(90), deepgreen+2);
draw("D", t*(P3--P4), dir(0), black+2);
draw("E", t*(P4--P5), dir(-90), black+2);
draw("K", t*(P5--P6), dir(-90), deepgreen+2);
draw("G", t*(P6--P2), dir(180), black+2);
draw("K", t*(P2--P5), dir(180), deepgreen+2);
draw("K", t*(P5--P3), dir(135), deepgreen+2);
draw("K", t*(P1--P5), dir(0), deepgreen);
draw("K", t*(P6--P3), dir(0), deepgreen);
draw("K", t*(P1..(1,1.3)..P3), dir(90), deepgreen);
add(t*base);
label(minipage("Step 2: Operate on $1235$. \\ Merges CIJ into K. \\
$\theta(\text{K}) = 6$", 120pt), t*P5, r*dir(270));
t = shift(3,-2.5);
draw("L", t*(P6--P1), dir(180), blue+2);
draw("L", t*(P1--P2), dir(90), blue+2);
draw("L", t*(P2--P3), dir(90), blue+2);
draw("D", t*(P3--P4), dir(0), black+2);
draw("E", t*(P4--P5), dir(-90), black+2);
draw("L", t*(P5--P6), dir(-90), blue+2);
draw("L", t*(P6--P2), dir(180), blue+2);
draw("L", t*(P2--P5), dir(180), blue+2);
draw("L", t*(P5--P3), dir(135), blue+2);
draw("L", t*(P1--P5), dir(0), blue);
draw("L", t*(P6--P3), dir(0), blue);
draw("L", t*(P1..(1,1.3)..P3), dir(90), blue);
add(t*base);
label(minipage("Step 3: Operate on $2356$. \\ Merges GK into L. \\
$\theta(\text{L}) = 7$", 120pt), t*P5, r*dir(270));
\end{asy}
\end{center}
As we remarked, if the graph is going to be completable to $K_n$ at all,
then this algorithm must terminate with $\mathcal C = \{K_n\}$.
We will use this to prove our bound.
We proceed by induction in the following way.
For a clique $K$, let $\theta(K)$ denote the number of edges
of the \emph{original} graph $G$ which are labeled by $K$
(this does \emph{not} include new edges added by the algorithm);
hence the problem amounts to estimating how small $\theta(K_n)$ can be.
We are trying to prove:
\begin{claim*}
At any point in the operation, if $K$ is a clique in
the cover $\mathcal C$, then
\[ \theta(K) \ge \frac{3 |K|}{2} - 2. \]
where $|K|$ is the number of vertices in $K$.
\end{claim*}
\begin{proof}
By induction on the time step of the algorithm.
The base case is clear, because then $K$ is just a single edge of $G$,
so $\theta(K) = 1$ and $|K| = 2$.
The inductive step is annoying casework based on the how the merge occurred.
Let $C_4 = abcd$ be the $4$-cycle operated on.
In general, the $\theta$ value of a newly created $K$ is
exactly the sum of the $\theta$ values of the merged cliques, by definition.
Meanwhile, $|K|$ is the number of vertices in the union of the merged cliques;
so it's the sum of the sizes of these cliques
minus some error due to overcounting of vertices appearing more than once.
To be explicit:
\begin{itemize}
\ii Suppose we merged four cliques $W$, $X$, $Y$, $Z$.
By definition,
\begin{align*}
\theta(K) &= \theta(W)+\theta(X)+\theta(Y)+\theta(Z) \\
&\ge \frac32(|W|+|X|+|Y|+|Z|) - 8 = \frac32(|W|+|X|+|Y|+|Z|-4) - 2.
\end{align*}
On the other hand $|K| \le |W|+|X|+|Y|+|Z|-4$;
the $-4$ term comes from each of $\{a,b,c,d\}$ being in
two (or more) of $\{W,X,Y,Z\}$. So this case is OK.
\ii Suppose we merged three cliques $X$, $Y$, $Z$.
By definition,
\begin{align*}
\theta(K) &= \theta(X)+\theta(Y)+\theta(Z) \\
&\ge \frac32(|X|+|Y|+|Z|) - 6 = \frac32\left(|X|+|Y|+|Z|-\frac83\right) - 2.
\end{align*}
On the other hand, $|K| \le |X|+|Y|+|Z| - 3$,
since at least $3$ of $\{a,b,c,d\}$ are repeated among $X$, $Y$, $Z$.
Note in this case the desired inequality is actually strict.
\ii Suppose we merged two cliques $Y$, $Z$.
By definition,
\begin{align*}
\theta(K) &= \theta(Y)+\theta(Z) \\
&\ge \frac32(|Y|+|Z|) - 4 = \frac32\left(|Y|+|Z|-\frac43\right) - 2.
\end{align*}
On the other hand, $|K| \le |Y|+|Z| - 2$,
since at least $2$ of $\{a,b,c,d\}$ are repeated among $Y$, $Z$.
Note in this case the desired inequality is actually strict. \qedhere
\end{itemize}
\end{proof}
\begin{remark*}
Several subtle variations of this method do not seem to work.
\begin{itemize}
\ii It does not seem possible to require the cliques in $\mathcal C$
to be disjoint, which is why it's necessary to introduce
a label function $\mathcal L$ as well.
\ii It seems you do have to label the newly created edges,
even though they do not count towards any $\theta$ value.
Otherwise the termination of the algorithm doesn't tell you enough.
\ii Despite this, relabeling existing edges,
like G in step 1 of the example, 1 seems to cause a lot of issues.
The induction becomes convoluted if $\theta(K)$ is not exactly
the sum of $\theta$-values of the subparts,
while the disappearance of an edge from a clique
will also break induction.
\end{itemize}
\end{remark*}
|
USAMO-2023-notes_1
|
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.
|
USAMO-2023-notes_2
|
Solve over the positive real numbers the functional equation
\[ f(xy+f(x)) = xf(y) + 2. \]
|
The answer is $f(x) \equiv x+1$,
which is easily verified to be the only linear solution.
We show conversely that $f$ is linear. Let $P(x,y)$ be the assertion.
\begin{claim*}
$f$ is weakly increasing.
\end{claim*}
\begin{proof}
Assume for contradiction $a>b$ but $f(a)<f(b)$.
Choose $y$ such that $ay+f(a)=by+f(b)$, that is $y=\frac{f(b)-f(a)}{a-b}$.
Then $P(a,y)$ and $P(b,y)$ gives $af(y)+2=bf(y)+2$, which is impossible.
\end{proof}
\begin{claim*}
[Up to an error of $2$, $f$ is linear]
We have
\[ \left\lvert f(x) - (K x + C) \right\rvert \le 2 \]
where $K \coloneq \frac{2}{f(1)}$ and $C \coloneq f(2f(1))-2$ are constants.
\end{claim*}
\begin{proof}
Note $P(1,y)$ gives $\boxed{f(y+f(1)) = f(y)+2}$.
Hence, $f(nf(1)) = 2(n-1)+f(f(1))$ for $n \ge 1$.
Combined with weakly increasing, this gives
\[ 2\left\lfloor \frac{x}{f(1)} \right\rfloor + C
\le f(x) \le 2\left\lceil \frac{x}{f(1)} \right\rceil + C \]
which implies the result.
\end{proof}
However, we will only need to remember from the earlier claim that
\[ f(x) = Kx+O(1). \]
Then for any $x$ and $y$, the above claim gives
\[ K\left( xy + Kx + O(1) \right) + O(1) = x f(y) + 2 \]
which means that
\[ x \cdot \left( Ky + K^2 - f(y) \right) = O(1). \]
If we fix $y$ and consider large $x$,
we see this can only happen if $Ky+K^2-f(y)=0$, i.e.\ $f$ is linear.
|
USAMO-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 all possible values of $k(C)$ as a function of $n$.
|
The answer is that
\[ k(C) \in \left\{ 1, 2, \dots, \left( \frac{n-1}{2} \right)^2 \right\}
\cup \left\{ \left( \frac{n+1}{2} \right)^2 \right\}. \]
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.
We now proceed in two cases:
\paragraph{The special squares have both odd coordinates.}
We 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.
Thus all special squares have an outgoing edge except the empty cell.
\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}
\begin{claim*}
Any undirected connected component of $G$ is acyclic unless the cycle
contains the empty square inside it.
\end{claim*}
\begin{proof}
Consider a cycle of $G$; we are going to prove
that the number of chessboard cells enclosed is always odd.
This can be proven directly by induction,
but for theatrical effect, we use Pick's theorem.
Mark the center of every chessboard cell on or inside the cycle
to get a lattice.
The dominoes of the cycle then enclose a polyominoe which
actually consists of $2 \times 2$ squares,
meaning its area is a multiple of $4$.
\begin{center}
\begin{asy}
unitsize(0.9cm);
filldraw(box((-0.5,1.5), (0.5,3.5)), paleyellow, gray+1.4);
filldraw(box((-0.5,3.5), (1.5,4.5)), paleyellow, gray+1.4);
filldraw(box((1.5,3.5), (3.5,4.5)), paleyellow, gray+1.4);
filldraw(box((3.5,2.5), (4.5,4.5)), paleyellow, gray+1.4);
filldraw(box((3.5,0.5), (4.5,2.5)), paleyellow, gray+1.4);
filldraw(box((2.5,-0.5), (4.5,0.5)), paleyellow, gray+1.4);
filldraw(box((1.5,-0.5), (2.5,1.5)), paleyellow, gray+1.4);
filldraw(box((0.5,1.5), (2.5,2.5)), paleyellow, gray+1.4);
draw((0,2)--(0,4), blue+1.2, EndArrow(TeXHead), Margins);
draw((0,4)--(2,4), blue+1.2, EndArrow(TeXHead), Margins);
draw((2,4)--(4,4), blue+1.2, EndArrow(TeXHead), Margins);
draw((4,4)--(4,2), blue+1.2, EndArrow(TeXHead), Margins);
draw((4,2)--(4,0), blue+1.2, EndArrow(TeXHead), Margins);
draw((4,0)--(2,0), blue+1.2, EndArrow(TeXHead), Margins);
draw((2,0)--(2,2), blue+1.2, EndArrow(TeXHead), Margins);
draw((2,2)--(0,2), blue+1.2, EndArrow(TeXHead), Margins);
draw((2,2)--(2,4), blue+dashed, Margins);
draw((2,2)--(4,2), blue+dashed, Margins);
dotfactor *= 1.8;
dot((2,0), blue);
dot((4,0), blue);
dot((0,2), blue);
dot((2,2), blue);
dot((4,2), blue);
dot((0,4), blue);
dot((2,4), blue);
dot((4,4), blue);
dot((0,3), black);
dot((1,4), black);
dot((3,4), black);
dot((4,3), black);
dot((4,1), black);
dot((3,0), black);
dot((2,1), black);
dot((1,2), black);
dot((1,3), red);
dot((2,3), red);
dot((3,3), red);
dot((3,2), red);
dot((3,1), red);
\end{asy}
\end{center}
Hence $B/2+I-1$ is a multiple of $4$, in the notation of Pick's theorem.
As $B$ is twice the number of dominoes,
and a parity argument on the special squares shows that number is even,
it follows that $B$ is also a multiple of $4$
(these correspond to blue and black in the figure above).
This means $I$ is odd (the red dots in the figure above), as desired.
\end{proof}
Let $T$ be the connected component containing the empty cell.
By the claim, $T$ is acyclic, so it's a tree.
Now, notice that all the arrows point along $T$ 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 $k(C) \le \left( \frac{n+1}{2} \right)^2$ in this case.
Equality is achieved if $T$ is a spanning tree of $G$.
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}
\begin{remark*}
In Russia 1997/11.8 it's shown that as long as the missing square is a corner,
we have $G = T$.
The proof is given implicitly from our work here:
when the empty cell is in a corner, it cannot be surrounded,
ergo the resulting graph has no cycles at all.
And since the overall graph has one fewer edge than vertex, it's a tree.
\end{remark*}
Conversely, suppose $T$ was \emph{not} a spanning tree, i.e.\ $T \neq G$.
Since in this odd-odd case, $G$ has one fewer edge than vertex,
if $G$ is not a tree, then it must contain at least one cycle.
That cycle encloses every special square of $T$.
In particular, this means that $T$ can't contain any special squares
from the outermost row or column of the $n \times n$ grid.
In this situation, we therefore have $k(C) \le \left( \frac{n-3}{2} \right)^2$.
\paragraph{The special squares have both even coordinates.}
We construct the analogous graph $G$ on the same special squares.
However, in this case, some of the points may not have outgoing edges,
because their domino may ``point'' outside the grid.
\begin{center}
\begin{asy}
size(8cm);
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((0, 0), (1, 1)), gray, border);
filldraw(box((0, -1), (2, 0)), yellow, border);
filldraw(box((2, -1), (4, 0)), brown, border);
filldraw(box((4, -1), (6, 0)), yellow, border);
filldraw(box((5, 0), (6, 2)), brown, border);
filldraw(box((5, 4), (6, 6)), brown, border);
filldraw(box((1, 5), (3, 6)), brown, border);
filldraw(box((-1, 5), (1, 6)), yellow, border);
filldraw(box((-1, -1), (0, 1)), brown, border);
filldraw(box((-1, 1), (0, 3)), yellow, border);
filldraw(box((-1, 3), (0, 5)), brown, border);
filldraw(box((3, 4), (4, 6)), purple, border);
filldraw(box((4, 4), (5, 6)), magenta, border);
filldraw(box((4, 3), (6, 4)), lightblue, border);
filldraw(box((4, 2), (6, 3)), palered, border);
filldraw(box((3, 2), (4, 4)), yellow, 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(B--E, EndArrow, Margin(3,3));
draw(E--D, 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}
As before, the connected component $T$ containing the empty square is a tree,
and $k(C)$ is exactly the number of vertices of $T$.
Thus to finish the problem we need to give,
for each $k \in \{1,2,\dots,\left( \frac{n-1}{2} \right)^2\}$,
an example of a configuration where $G$ has exactly $k$ vertices.
The construction starts with a ``snake'' picture for
$k = \left( \frac{n-1}{2} \right)^2$,
then decreases $k$ by one by perturbing a suitable set of dominoes.
Rather than write out the procedure in words,
we show the sequence of nine pictures for $n=7$ (where $k=9,8,\dots,1$);
the generalization to larger $n$ is straightforward.
\begin{center}
\begin{asy}
unitsize(0.6cm);
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);
/* Picture 0 */
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);
filldraw(box((0, -1), (2, 0)), yellow, border);
filldraw(box((2, -1), (4, 0)), brown, border);
filldraw(box((4, -1), (6, 0)), yellow, border);
filldraw(box((5, 0), (6, 2)), brown, border);
filldraw(box((5, 2), (6, 4)), yellow, border);
filldraw(box((5, 4), (6, 6)), brown, border);
filldraw(box((3, 5), (5, 6)), yellow, border);
filldraw(box((1, 5), (3, 6)), brown, border);
filldraw(box((-1, 5), (1, 6)), yellow, border);
filldraw(box((-1, -1), (0, 1)), brown, border);
filldraw(box((-1, 1), (0, 3)), yellow, border);
filldraw(box((-1, 3), (0, 5)), brown, 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));
picture p0 = rotate(0)*currentpicture;
currentpicture.erase();
/* Picture 1 */
filldraw(box((4, 4), (6, 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);
filldraw(box((0, -1), (2, 0)), yellow, border);
filldraw(box((2, -1), (4, 0)), brown, border);
filldraw(box((4, -1), (6, 0)), yellow, border);
filldraw(box((5, 0), (6, 2)), brown, border);
filldraw(box((5, 2), (6, 4)), yellow, border);
filldraw(box((3, 4), (4, 6)), brown, border);
filldraw(box((4, 5), (6, 6)), yellow, border);
filldraw(box((1, 5), (3, 6)), brown, border);
filldraw(box((-1, 5), (1, 6)), yellow, border);
filldraw(box((-1, -1), (0, 1)), brown, border);
filldraw(box((-1, 1), (0, 3)), yellow, border);
filldraw(box((-1, 3), (0, 5)), brown, border);
dot(A); dot(B); dot(C);
dot(D); dot(E); dot(F);
dot(X); dot(Y); dot(Z);
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));
picture p1 = rotate(0)*currentpicture;
currentpicture.erase();
/* Picture 2 */
filldraw(box((4, 4), (6, 5)), palecyan, border);
filldraw(box((2, 4), (4, 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);
filldraw(box((0, -1), (2, 0)), yellow, border);
filldraw(box((2, -1), (4, 0)), brown, border);
filldraw(box((4, -1), (6, 0)), yellow, border);
filldraw(box((5, 0), (6, 2)), brown, border);
filldraw(box((5, 2), (6, 4)), yellow, border);
filldraw(box((1, 4), (2, 6)), brown, border);
filldraw(box((4, 5), (6, 6)), yellow, border);
filldraw(box((2, 5), (4, 6)), brown, border);
filldraw(box((-1, 5), (1, 6)), yellow, border);
filldraw(box((-1, -1), (0, 1)), brown, border);
filldraw(box((-1, 1), (0, 3)), yellow, border);
filldraw(box((-1, 3), (0, 5)), brown, border);
dot(A); dot(B); dot(C);
dot(D); dot(E); dot(F);
dot(X); dot(Y); dot(Z);
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));
picture p2 = rotate(0)*currentpicture;
currentpicture.erase();
/* Picture 3 */
filldraw(box((4, 4), (6, 5)), palecyan, border);
filldraw(box((2, 4), (4, 5)), palecyan, border);
filldraw(box((-1, 4), (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);
filldraw(box((0, -1), (2, 0)), yellow, border);
filldraw(box((2, -1), (4, 0)), brown, border);
filldraw(box((4, -1), (6, 0)), yellow, border);
filldraw(box((5, 0), (6, 2)), brown, border);
filldraw(box((5, 2), (6, 4)), yellow, border);
filldraw(box((1, 4), (2, 6)), brown, border);
filldraw(box((4, 5), (6, 6)), yellow, border);
filldraw(box((2, 5), (4, 6)), brown, border);
filldraw(box((-1, 5), (1, 6)), yellow, border);
filldraw(box((-1, -1), (0, 1)), brown, border);
filldraw(box((-1, 1), (0, 3)), yellow, border);
filldraw(box((-1, 3), (1, 4)), brown, border);
dot(A); dot(B); dot(C);
dot(D); dot(E); dot(F);
dot(X); dot(Y); dot(Z);
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));
picture p3 = rotate(0)*currentpicture;
currentpicture.erase();
/* Picture 4 */
filldraw(box((4, 4), (6, 5)), palecyan, border);
filldraw(box((2, 4), (4, 5)), palecyan, border);
filldraw(box((-1, 4), (1, 5)), palecyan, border);
filldraw(box((-1, 2), (1, 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((-1, 1), (1, 2)), lightred, border);
filldraw(box((2, 1), (4, 2)), lightred, border);
filldraw(box((0, -1), (2, 0)), yellow, border);
filldraw(box((2, -1), (4, 0)), brown, border);
filldraw(box((4, -1), (6, 0)), yellow, border);
filldraw(box((5, 0), (6, 2)), brown, border);
filldraw(box((5, 2), (6, 4)), yellow, border);
filldraw(box((1, 4), (2, 6)), brown, border);
filldraw(box((4, 5), (6, 6)), yellow, border);
filldraw(box((2, 5), (4, 6)), brown, border);
filldraw(box((-1, 5), (1, 6)), yellow, border);
filldraw(box((-1, -1), (0, 1)), brown, border);
filldraw(box((1, 1), (2, 3)), yellow, border);
filldraw(box((-1, 3), (1, 4)), brown, border);
dot(A); dot(B); dot(C);
dot(D); dot(E); dot(F);
dot(X); dot(Y); dot(Z);
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));
picture p4 = rotate(0)*currentpicture;
currentpicture.erase();
/* Picture 5 */
filldraw(box((4, 4), (6, 5)), palecyan, border);
filldraw(box((2, 4), (4, 5)), palecyan, border);
filldraw(box((-1, 4), (1, 5)), palecyan, border);
filldraw(box((-1, 2), (1, 3)), palecyan, border);
filldraw(box((1, 2), (3, 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((-1, 1), (1, 2)), lightred, border);
filldraw(box((1, 1), (3, 2)), lightred, border);
filldraw(box((0, -1), (2, 0)), yellow, border);
filldraw(box((2, -1), (4, 0)), brown, border);
filldraw(box((4, -1), (6, 0)), yellow, border);
filldraw(box((5, 0), (6, 2)), brown, border);
filldraw(box((5, 2), (6, 4)), yellow, border);
filldraw(box((1, 4), (2, 6)), brown, border);
filldraw(box((4, 5), (6, 6)), yellow, border);
filldraw(box((2, 5), (4, 6)), brown, border);
filldraw(box((-1, 5), (1, 6)), yellow, border);
filldraw(box((-1, -1), (0, 1)), brown, border);
filldraw(box((3, 1), (4, 3)), yellow, border);
filldraw(box((-1, 3), (1, 4)), brown, border);
dot(A); dot(B); dot(C);
dot(D); dot(E); dot(F);
dot(X); dot(Y); dot(Z);
draw(F--Z, EndArrow, Margin(3,3));
draw(Z--Y, EndArrow, Margin(3,3));
draw(Y--X, EndArrow, Margin(3,3));
picture p5 = rotate(0)*currentpicture;
currentpicture.erase();
/* Picture 6 */
filldraw(box((4, 4), (6, 5)), palecyan, border);
filldraw(box((2, 4), (4, 5)), palecyan, border);
filldraw(box((-1, 4), (1, 5)), palecyan, border);
filldraw(box((-1, 2), (1, 3)), palecyan, border);
filldraw(box((1, 2), (3, 3)), palecyan, border);
filldraw(box((3, 2), (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((-1, 1), (1, 2)), lightred, border);
filldraw(box((1, 1), (3, 2)), lightred, border);
filldraw(box((0, -1), (2, 0)), yellow, border);
filldraw(box((2, -1), (4, 0)), brown, border);
filldraw(box((4, -1), (6, 0)), yellow, border);
filldraw(box((5, 0), (6, 2)), brown, border);
filldraw(box((5, 2), (6, 4)), yellow, border);
filldraw(box((1, 4), (2, 6)), brown, border);
filldraw(box((4, 5), (6, 6)), yellow, border);
filldraw(box((2, 5), (4, 6)), brown, border);
filldraw(box((-1, 5), (1, 6)), yellow, border);
filldraw(box((-1, -1), (0, 1)), brown, border);
filldraw(box((3, 1), (5, 2)), yellow, border);
filldraw(box((-1, 3), (1, 4)), brown, border);
dot(A); dot(B); dot(C);
dot(D); dot(E); dot(F);
dot(X); dot(Y); dot(Z);
draw(Z--Y, EndArrow, Margin(3,3));
draw(Y--X, EndArrow, Margin(3,3));
picture p6 = rotate(0)*currentpicture;
currentpicture.erase();
/* Picture 7 */
filldraw(box((4, 4), (6, 5)), palecyan, border);
filldraw(box((2, 4), (4, 5)), palecyan, border);
filldraw(box((-1, 4), (1, 5)), palecyan, border);
filldraw(box((-1, 2), (1, 3)), palecyan, border);
filldraw(box((1, 2), (3, 3)), palecyan, border);
filldraw(box((3, 2), (5, 3)), palecyan, border);
filldraw(box((4, 0), (6, 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((-1, 1), (1, 2)), lightred, border);
filldraw(box((1, 1), (3, 2)), lightred, border);
filldraw(box((0, -1), (2, 0)), yellow, border);
filldraw(box((2, -1), (4, 0)), brown, border);
filldraw(box((4, -1), (6, 0)), yellow, border);
filldraw(box((3, 0), (4, 2)), brown, border);
filldraw(box((5, 2), (6, 4)), yellow, border);
filldraw(box((1, 4), (2, 6)), brown, border);
filldraw(box((4, 5), (6, 6)), yellow, border);
filldraw(box((2, 5), (4, 6)), brown, border);
filldraw(box((-1, 5), (1, 6)), yellow, border);
filldraw(box((-1, -1), (0, 1)), brown, border);
filldraw(box((4, 1), (6, 2)), yellow, border);
filldraw(box((-1, 3), (1, 4)), brown, border);
dot(A); dot(B); dot(C);
dot(D); dot(E); dot(F);
dot(X); dot(Y); dot(Z);
draw(Y--X, EndArrow, Margin(3,3));
picture p7 = rotate(0)*currentpicture;
currentpicture.erase();
/* Picture 8 */
filldraw(box((4, 4), (6, 5)), palecyan, border);
filldraw(box((2, 4), (4, 5)), palecyan, border);
filldraw(box((-1, 4), (1, 5)), palecyan, border);
filldraw(box((-1, 2), (1, 3)), palecyan, border);
filldraw(box((1, 2), (3, 3)), palecyan, border);
filldraw(box((3, 2), (5, 3)), palecyan, border);
filldraw(box((4, 0), (6, 1)), palecyan, border);
filldraw(box((2, 0), (4, 1)), palecyan, border);
fill(unitsquare, mediumgray);
filldraw(box((3, 3), (5, 4)), lightred, border);
filldraw(box((1, 3), (3, 4)), lightred, border);
filldraw(box((-1, 1), (1, 2)), lightred, border);
filldraw(box((2, 1), (4, 2)), lightred, border);
filldraw(box((0, -1), (2, 0)), yellow, border);
filldraw(box((2, -1), (4, 0)), brown, border);
filldraw(box((4, -1), (6, 0)), yellow, border);
filldraw(box((1, 0), (2, 2)), brown, border);
filldraw(box((5, 2), (6, 4)), yellow, border);
filldraw(box((1, 4), (2, 6)), brown, border);
filldraw(box((4, 5), (6, 6)), yellow, border);
filldraw(box((2, 5), (4, 6)), brown, border);
filldraw(box((-1, 5), (1, 6)), yellow, border);
filldraw(box((-1, -1), (0, 1)), brown, border);
filldraw(box((4, 1), (6, 2)), yellow, border);
filldraw(box((-1, 3), (1, 4)), brown, border);
dot(A); dot(B); dot(C);
dot(D); dot(E); dot(F);
dot(X); dot(Y); dot(Z);
picture p8 = rotate(0)*currentpicture;
currentpicture.erase();
add(shift(0,0)*p0);
add(shift(8,0)*p1);
add(shift(16,0)*p2);
add(shift(0,-8)*p3);
add(shift(8,-8)*p4);
add(shift(16,-8)*p5);
add(shift(0,-16)*p6);
add(shift(8,-16)*p7);
add(shift(16,-16)*p8);
\end{asy}
\end{center}
|
USAMO-2023-notes_4
|
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}
|
USAMO-2023-notes_5
|
Let $n\geq3$ be an integer. We say that an arrangement of the numbers
$1$, $2$, \dots, $n^2$ in an $n \times n$ table is \emph{row-valid}
if the numbers in each row can be permuted to form an arithmetic progression,
and \emph{column-valid} if the numbers in each column
can be permuted to form an arithmetic progression.
For what values of $n$ is it possible to transform any row-valid arrangement
into a column-valid arrangement by permuting the numbers in each row?
|
Answer: $n$ prime only.
\paragraph{Proof for $n$ prime.}
Suppose $n = p$.
In an arithmetic progression with $p$ terms, it's easy to see that either
every term has a different residue modulo $p$ (if the common difference
is not a multiple of $p$), or all of the residues coincide
(when the common difference is a multiple of $p$).
So, look at the multiples of $p$ in a row-valid table;
there is either $1$ or $p$ per row.
As there are $p$ such numbers total, there are two cases:
\begin{itemize}
\ii If all the multiples of $p$ are in the same row,
then the common difference in each row is a multiple of $p$.
In fact, it must be exactly $p$ for size reasons.
In other words, up to permutation the rows are just
the $k \pmod p$ numbers in some order, and this is obviously column-valid
because we can now permute such that the $k$th column
contains exactly $\{(k-1)p+1, (k-1)p+2, \dots, kp\}$.
\ii If all the multiples of $p$ are in different rows,
then it follows each row contains every residue modulo $p$ exactly once.
So we can permute to a column-valid arrangement by ensuring
the $k$th column contains all the $k \pmod p$ numbers.
\end{itemize}
\paragraph{Counterexample for $n$ composite (due to Anton Trygub).}
Let $p$ be any prime divisor of $n$.
Construct the table as follows:
\begin{itemize}
\ii Row $1$ contains $1$ through $n$.
\ii Rows $2$ through $p+1$ contain the numbers from $p+1$ to $np+n$
partitioned into arithmetic progressions with common difference $p$.
\ii The rest of the rows contain the remaining numbers in reading order.
\end{itemize}
For example, when $p=2$ and $n=10$, we get the following table:
\[
\begin{bmatrix}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\mathbf{\color{red} 11} &
\mathbf{\color{red} 13} &
\mathbf{\color{red} 15} &
\mathbf{\color{red} 17} &
\mathbf{\color{red} 19} &
\mathbf{\color{red} 21} &
\mathbf{\color{red} 23} &
\mathbf{\color{red} 25} &
\mathbf{\color{red} 27} &
\mathbf{\color{red} 29} \\
\mathbf{\color{red} 12} &
\mathbf{\color{red} 14} &
\mathbf{\color{red} 16} &
\mathbf{\color{red} 18} &
\mathbf{\color{red} 20} &
\mathbf{\color{red} 22} &
\mathbf{\color{red} 24} &
\mathbf{\color{red} 26} &
\mathbf{\color{red} 28} &
\mathbf{\color{red} 30} \\
31 & 32 & 33 & 34 & 35 & 36 & 37 & 38 & 39 & 40 \\
41 & 42 & 43 & 44 & 45 & 46 & 47 & 48 & 49 & 50 \\
51 & 52 & 53 & 54 & 55 & 56 & 57 & 58 & 59 & 60 \\
61 & 62 & 63 & 64 & 65 & 66 & 67 & 68 & 69 & 70 \\
71 & 72 & 73 & 74 & 75 & 76 & 77 & 78 & 79 & 80 \\
81 & 82 & 83 & 84 & 85 & 86 & 87 & 88 & 89 & 90 \\
91 & 92 & 93 & 94 & 95 & 96 & 97 & 98 & 99 & 100
\end{bmatrix}
\]
We claim this works fine.
Assume for contradiction the rows may be permuted to obtain
a column-valid arrangement.
Then the $n$ columns should be arithmetic progressions
whose smallest element is in $[1,n]$
and whose largest element is in $[n^2-n+1, n^2]$.
These two elements must be congruent modulo $n-1$,
so in particular the column containing $2$ must end with $n^2-n+2$.
Hence in that column, the common difference must in fact be exactly $n$.
And yet $n+2$ and $2n+2$ are in the same row, contradiction.
|
USAMO-2023-notes_6
|
Let $ABC$ be a triangle with incenter $I$
and excenters $I_a$, $I_b$, $I_c$ opposite $A$, $B$, and $C$, respectively.
Given an arbitrary point $D$ on the circumcircle of $\triangle ABC$
that does not lie on any of the lines $II_a$, $I_bI_c$, or $BC$,
suppose the circumcircles of $\triangle DII_a$ and $\triangle DI_bI_c$
intersect at two distinct points $D$ and $F$.
If $E$ is the intersection of lines $DF$ and $BC$,
prove that $\angle BAD = \angle EAC$.
\end{enumerate}
|
Here are two approaches.
\begin{center}
\begin{asy}
size(9cm);
pair A = dir(115);
pair B = dir(210);
pair C = dir(330);
pair D = dir(240);
pair E = extension(B, C, A, B*C/D);
pair I = incenter(A, B, C);
pair M = dir(270);
pair I_a = 2*M-I;
pair I_b = extension(B, I, C, I_a);
pair I_c = extension(C, I, B, I_a);
pair F = -D+2*foot(circumcenter(D, I, I_a), D, E);
draw(D--F, deepgreen);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
filldraw(A--B--C--cycle, opacity(0.1)+lightcyan, blue);
draw(circumcircle(D, I, I_a), lightred);
draw(circumcircle(D, I_b, I_c), lightred);
draw(A--I_a, blue);
draw(I_a--I_b--I_c--cycle, gray);
draw(D--A--E, deepcyan+dashed);
dot("$A$", A, dir(A));
dot("$B$", B, dir(170));
dot("$C$", C, dir(350));
dot("$D$", D, dir(D));
dot("$E$", E, dir(280));
dot("$I$", I, dir(160));
dot("$M$", M, dir(225));
dot("$I_a$", I_a, dir(I_a));
dot("$I_b$", I_b, dir(I_b));
dot("$I_c$", I_c, dir(I_c));
dot("$F$", F, dir(F));
/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
!size(9cm);
A = dir 115
B 170 = dir 210
C 350 = dir 330
D = dir 240
E 280 = extension B C A B*C/D
I 160 = incenter A B C
M 225 = dir 270
I_a = 2*M-I
I_b = extension B I C I_a
I_c = extension C I B I_a
F = -D+2*foot (circumcenter D I I_a) D E
D--F deepgreen
unitcircle / 0.1 lightcyan / blue
A--B--C--cycle / 0.1 lightcyan / blue
circumcircle D I I_a / lightred
circumcircle D I_b I_c / lightred
A--I_a / blue
I_a--I_b--I_c--cycle / gray
D--A--E / deepcyan dashed
*/
\end{asy}
\end{center}
\paragraph{Barycentric coordinates (Carl Schildkraut).}
With reference triangle $\triangle ABC$, set $D = (r:s:t)$.
\begin{claim*}
The equations of $(DII_a)$ and $(DI_bI_c)$ are, respectively,
\begin{align*}
0 &= -a^2yz-b^2zx-c^2xy + (x+y+z)
\cdot \left( bc x - \frac{bcr}{cs-bt} (cy-bz) \right) \\
0 &= -a^2yz-b^2zx-c^2xy + (x+y+z)
\cdot \left( -bc x + \frac{bcr}{cs+bt} (cy+bz) \right).
\end{align*}
\end{claim*}
\begin{proof}
Since $D \in (ABC)$, we have $a^2st+b^2tr+c^2rs=0$.
Now each equation can be verified by direct substitution of three points.
\end{proof}
By EGMO Lemma 7.24, the radical axis is then given by
\[ \ol{DF}: bc x - \frac{bcr}{cs-bt} (cy-bz)
= -bc x + \frac{bcr}{cs+bt} (cy+bz). \]
Now the point
\[ \left( 0 : \frac{b^2}{s} : \frac{c^2}{t} \right)
= \left( 0 : b^2t : c^2s \right) \]
lies on line $DF$ by inspection, and is obviously on line $BC$,
hence it coincides with $E$.
This lies on the isogonal of $\ol{AD}$ (by EGMO Lemma 7.6), as needed.
\paragraph{Synthetic approach (Anant Mudgal).}
Focus on just $(DII_a)$.
Let $P$ be the second intersection of $(DII_a)$ with $(ABC)$,
and let $M$ be the midpoint of minor arc $\arc{BC}$.
Then by radical axis, lines $AM$, $DP$, and $BC$ are concurrent at a point $K$.
Let $E' = \ol{PM} \cap \ol{BC}$.
\begin{center}
\begin{asy}
size(9cm);
pair A = dir(140);
pair B = dir(210);
pair C = dir(330);
pair D = dir(240);
pair E_prime = extension(B, C, A, B*C/D);
pair I = incenter(A, B, C);
pair M = dir(270);
pair I_a = 2*M-I;
pair K = extension(A, I, B, C);
pair P = extension(D, K, M, E_prime);
pair X = 2*M-E_prime;
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
filldraw(A--B--C--cycle, opacity(0.1)+lightcyan, blue);
filldraw(circumcircle(D, I, I_a), opacity(0.1)+yellow, blue);
draw(A--I_a, blue);
draw(D--P--X, orange);
filldraw(X--I--E_prime--I_a--cycle, opacity(0.1)+magenta, red);
filldraw(circumcircle(B, I, C), opacity(0.1)+gray, gray);
draw(D--A--E_prime, deepcyan+dashed);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(350));
dot("$D$", D, dir(D));
dot("$E'$", E_prime, dir(325));
dot("$I$", I, dir(160));
dot("$M$", M, dir(M));
dot("$I_a$", I_a, dir(I_a));
dot("$K$", K, dir(325));
dot("$P$", P, dir(P));
dot("$X$", X, dir(X));
/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
!size(9cm);
A = dir 140
B = dir 210
C 350 = dir 330
D = dir 240
E' 325 = extension B C A B*C/D
I 160 = incenter A B C
M = dir 270
I_a = 2*M-I
K 325 = extension A I B C
P = extension D K M E'
X = 2*M-E'
unitcircle / 0.1 lightcyan / blue
A--B--C--cycle / 0.1 lightcyan / blue
circumcircle D I I_a / 0.1 yellow / blue
A--I_a / blue
D--P--X / orange
X--I--E'--I_a--cycle / 0.1 magenta / red
circumcircle B I C / 0.1 gray / gray
D--A--E' / deepcyan dashed
*/
\end{asy}
\end{center}
\begin{claim*}
We have $\dang BAD = \dang E'AC$.
\end{claim*}
\begin{proof}
By shooting lemma, $AKE'P$ is cyclic, so
\[ \dang KAE' = \dang KPE' = \dang DPM = \dang DAM. \qedhere \]
\end{proof}
\begin{claim*}
The power of point $E'$ with respect to $(DII_a)$ is $2 E'B \cdot E'C$.
\end{claim*}
\begin{proof}
Construct parallelogram $IE'I_aX$.
Since $MI^2 = ME' \cdot MP$, we can get
\[ \dang XI_aI = \dang I_aIE' = \dang MIE' = \dang MPI = \dang XPI. \]
Hence $X$ lies on $(DII_a)$,
and $E'X \cdot E'P = 2 E'M \cdot E'P = 2 E'B \cdot E'C$.
\end{proof}
Repeat the argument on $(DI_bI_c)$;
the same point $E'$ (because of the first claim)
then has power $2 E'B \cdot E'C$ with respect to $(DI_bI_c)$.
Hence $E'$ lies on the radical axis of $(DII_a)$ and $(DI_bI_c)$, ergo $E'=E$.
The first claim then solves the problem.
|
USAMO-2024-notes_1
|
Find all integers $n \geq 3$ such that the following property holds:
if we list the divisors of $n!$ in increasing order
as $1 = d_1 < d_2 < \dots < d_k = n!$, then we have
\[ d_2 - d_1 \leq d_3 - d_2 \leq \dots \leq d_k - d_{k-1}. \]
|
The answer is $n \in \{3,4\}$. These can be checked by listing all the divisors:
\begin{itemize}
\ii For $n=3$ we have $(1,2,3,6)$.
\ii For $n=4$ we have $(1,2,3,4,6,8,12,24)$.
\end{itemize}
We prove these are the only ones.
The numbers $5 \le n \le 12$ all fail because:
\begin{itemize}
\ii For $n = 5$ we have $20 - 15 > 24 - 20$.
\ii For $n = 6$ we have $120 - 90 > 144 - 20$.
\ii For $7 \le n \le 12$ we have because $14-12 > 15-14$ (and $13 \nmid n!$).
\end{itemize}
Now assume $n \ge 13$.
In that case, we have
\[ \left\lfloor \frac n2 \right\rfloor^2 - 1 \ge 2n. \]
So by Bertrand postulate, we can find a prime $p$ such that
\[ n < p < \left\lfloor \frac n2 \right\rfloor^2 - 1. \]
However, note that
\[ \left\lfloor \frac n2 \right\rfloor^2 - 1
= \left( \left\lfloor \frac n2 \right\rfloor - 1 \right) \left( \left\lfloor \frac n2 \right\rfloor + 1 \right),
\qquad \left\lfloor \frac n2 \right\rfloor^2
\]
are consecutive integers both dividing $n!$ (the latter number divides
$\left\lfloor \frac n2 \right\rfloor \cdot \left( 2 \left\lfloor \frac n2 \right\rfloor \right)$).
So the prime $p$ causes the desired contradiction.
|
USAMO-2024-notes_2
|
Let $S_1$, $S_2$, \dots, $S_{100}$ be finite sets of integers whose intersection is not empty.
For each non-empty $T \subseteq \{S_1, S_2, \ldots, S_{100}\}$,
the size of the intersection of the sets in $T$ is a multiple of $|T|$.
What is the smallest possible number of elements which are in at least $50$ sets?
|
The answer is $50 \binom{100}{50}$.
\paragraph{Rephrasing (cosmetic translation only, nothing happens yet).}
We encode with binary strings $v \in \FF_2^{100}$ of length $100$.
Write $v \subseteq w$ if $w$ has $1$'s in every component $v$ does,
and let $|v|$ denote the number of $1$'s in $v$.
Then for each $v$, we let $f(v)$ denote the number of elements $x \in \bigcup S_i$
such that $x \in S_i \iff v_i = 1$.
For example,
\begin{itemize}
\ii $f(1\dots1)$ denotes $|\bigcap_1^{100} S_i|$,
so we know $f(1 \dots 1) \equiv 0 \pmod{100}$.
\ii $f(1\dots10)$ denotes the number of elements in $S_1$ through $S_{99}$
but not $S_{100}$
so we know that $f(1 \dots 1) + f(1 \dots 10) \equiv 0 \pmod{99}$.
\ii \dots And so on.
\end{itemize}
So the problem condition means that $f(v)$ translates to the statement
\[ P(u): \qquad |u| \text{ divides } \sum_{v \supseteq u} f(v) \]
for any $u \neq 0\dots0$, plus one extra condition $f(1\dots1) > 0$.
And the objective function is to minimize the quantity
\[ A \coloneq \sum_{|v| \ge 50} f(v). \]
So the problem is transformed into an system of equations over $\ZZ_{\ge 0}$
(it's clear any assignment of values of $f(v)$
can be translated to a sequence $(S_1, \dots, S_{100})$ in the
original notation).
Note already that:
\begin{claim*}
It suffices to assign $f(v)$ for $|v| \ge 50$.
\end{claim*}
\begin{proof}
If we have found a valid assignment of values to $f(v)$ for $|v| \ge 50$,
then we can always arbitrarily assign values of $f(v)$ for $|v| < 50$ by downwards
induction on $|v|$ to satisfy the divisibility condition (without changing $M$).
\end{proof}
Thus, for the rest of the solution, we altogether ignore $f(v)$ for $|v| < 50$
and only consider $P(u)$ for $|u| \ge 50$.
\paragraph{Construction.}
Consider the construction
\[ f_0(v) = 2 |v| - 100. \]
This construction is valid since if $|u| = 100-k$ for $k \le 50$ then
\begin{align*}
\sum_{v \supseteq u} f_0(v)
&= \binom{k}{0} \cdot 100 + \binom{k}{1} \cdot 98 + \binom{k}{2} \cdot 96
+ \dots + \binom{k}{k} \cdot (100-2k) \\
&= (100-k) \cdot 2^k = |u| \cdot 2^k
\end{align*}
is indeed a multiple of $|u|$, hence $P(u)$ is true.
In that case, the objective function is
\[ A = \sum_{i=50}^{100} \binom{100}{i} (2i-100) = 50\binom{100}{50} \]
as needed.
\begin{remark*}
This construction is the ``easy'' half of the problem because it
coincides with what you get from a greedy algorithm by downwards induction on $|u|$
(equivalently, induction on $k = 100-|u| \ge 0$).
To spell out the first three steps,
\begin{itemize}
\ii We know $f(1\dots1)$ is a nonzero multiple of $100$,
so it makes sense to guess $f(1\dots1) = 100$.
\ii Then we have $f(1\dots10) + 100 \equiv 0 \pmod{99}$,
and the smallest multiple of $99$ which is at least $100$ is $198$.
So it makes sense to guess $f(1\dots10) = 98$, and similarly guess $f(v) = 98$
whenever $|v| = 99$.
\ii Now when we consider, say $v= 1 \dots 100$ with $|v| = 98$, we get
\[ f(1\dots100) + \underbrace{f(1\dots101)}_{=98} + \underbrace{f(1\dots110)}_{=98}
+ \underbrace{f(1\dots111)}_{=100} \equiv 0 \pmod{98}. \]
we obtain $f(1\dots100) \equiv 96 \pmod{98}$.
That makes $f(1\dots100) = 96$ a reasonable guess.
\end{itemize}
Continuing in this way gives the construction above.
\end{remark*}
\paragraph{Proof of bound.}
We are going to use a smoothing argument:
if we have a general working assignment $f$, we will mold it into $f_0$.
We define a \textbf{push-down on $v$} as the following operation:
\begin{itemize}
\ii Pick any $v$ such that $|v| \ge 50$ and $f(v) \ge |v|$.
\ii Decrease $f(v)$ by $|v|$.
\ii For every $w$ such that $w \subseteq v$ and $|w| = |v| - 1$,
increase $f(w)$ by $1$.
\end{itemize}
\begin{claim*}
Apply a push-down preserves the main divisibility condition.
Moreover, it doesn't change $A$ unless $|v| = 50$, where it decreases $A$ by $50$ instead.
\end{claim*}
\begin{proof}
The statement $P(u)$ is only affected when $u \subseteq v$:
to be precise, one term on the right-hand side of $P(u)$ decreases by $|v|$,
while $|v|-|u|$ terms increase by $1$, for a net change of $-|u|$.
So $P(u)$ still holds.
To see $A$ doesn't change for $|v| > 50$,
note $|v|$ terms increase by $1$ while one term decreases by $-|v|$.
When $|v| = 50$, only $f(v)$ decreases by $50$.
\end{proof}
Now, given a valid assignment, we can modify it as follows:
\begin{itemize}
\ii First apply pushdowns on $1\dots1$ until $f(1\dots1) = 100$;
\ii Then we may apply pushdowns on each $v$ with $|v| = 99$ until $f(v) < 99$;
\ii Then we may apply pushdowns on each $v$ with $|v| = 98$ until $f(v) < 98$;
\ii \dots and so on, until we have $f(v) < 50$ for $|v| = 50$.
\end{itemize}
Hence we get $f(1\dots1) = 100$ and $0 \le f(v) < |v|$ for all $50 \le |v| \le 100$.
However, by downwards induction on $|v| = 99, 98, \dots, 50$, we also have
\[ f(v) \equiv f_0(v) \pmod{|v|} \implies f(v) = f_0(v) \]
since $f_0(v)$ and $f(v)$ are both strictly less than $|v|$.
So in fact $f = f_0$, and we're done.
\begin{remark*}
The fact that push-downs actually don't change $A$
shows that the equality case we described is far from unique:
in fact, we could have made nearly arbitrary sub-optimal
decisions during the greedy algorithm and still ended up with an equality case.
For a concrete example, the construction
\[
f(v) = \begin{cases}
500 & |v| = 100 \\
94 & |v| = 99 \\
100-2|v| & 50 \le |v| \le 98
\end{cases}
\]
works fine as well (where we arbitrarily chose $500$ at the start,
then used the greedy algorithm thereafter).
\end{remark*}
|
USAMO-2024-notes_3
|
Let $(m,n)$ be positive integers with $n \ge 3$ and draw a regular $n$-gon.
We wish to triangulate this $n$-gon into $n-2$ triangles,
each colored one of $m$ colors, so that each color has an equal sum of areas.
For which $(m,n)$ is such a triangulation and coloring possible?
|
The answer is if and only if $m$ is a proper divisor of $n$.
Throughout this solution, we let $\omega = \exp\left( 2 \pi i / n \right)$
and let the regular $n$-gon have vertices $1$, $\omega$, \dots, $\omega^{n-1}$.
We cache the following frequent calculation:
\begin{lemma*}
The triangle with vertices $\omega^k$, $\omega^{k+a}$, $\omega^{k+b}$ has signed area
\[ T(a,b) \coloneq \frac{(\omega^a-1)(\omega^b-1)(\omega^{-b}-\omega^{-a})}{4i}. \]
\end{lemma*}
\begin{proof}
Rotate by $\omega^{-k}$ to assume WLOG that $k=0$.
Apply complex shoelace to the triangles with vertices $1$, $\omega^a$, $\omega^b$ to get
\[
\frac{i}{4}
\det \begin{bmatrix}
1 & 1 & 1 \\
\omega^a & \omega^{-a} & 1 \\
\omega^b & \omega^{-b} & 1 \\
\end{bmatrix}
=
\frac{i}{4}
\det \begin{bmatrix}
0 & 0 & 1 \\
\omega^a-1 & \omega^{-a}-1 & 1 \\
\omega^b-1 & \omega^{-b}-1 & 1 \\
\end{bmatrix}
\]
which equals the above.
\end{proof}
\paragraph{Construction.}
It suffices to actually just take all the diagonals from the vertex $1$,
and then color the triangles with the $m$ colors in cyclic order.
For example, when $n = 9$ and $m = 3$, a coloring with red, green, blue would be:
\begin{center}
\begin{asy}
size(6cm);
pair A(int i) { return dir(40*i); }
pen[] fillcolors = {palered, palegreen, palecyan};
pen[] labelcolors = {brown, darkgreen, deepblue};
string[] labels = {"R", "Green", "Blue", "Red", "Green", "Blue", "R"};
for (int i=0; i<=6; ++i) {
filldraw(A(0)--A(i+1)--A(i+2)--cycle, fillcolors[i-3*(i#3)], black+1.5);
label(labels[i], incenter(A(0),A(i+1),A(i+2)),
labelcolors[i-3*(i#3)] + fontsize(14pt));
}
\end{asy}
\end{center}
To see this works one can just do the shoelace calculation:
fix a residue $r \bmod m$ corresponding to one of the colors.
Then
\begin{align*}
\sum_{\substack{0 \le j < n \\ j \equiv r \bmod m}} \operatorname{Area}(\omega^0, \omega^j, \omega^{j+1})
&= \sum_{\substack{0 \le j < n \\ j \equiv r \bmod m}} T(j, j+1) \\
&= \sum_{\substack{0 \le j < n \\ j \equiv r \bmod m}}
\frac{(\omega^j-1)(\omega^{j+1}-1)(\omega^{-(j+1)}-\omega^{-j})}{4i} \\
&= \frac{\omega-1}{4i} \sum_{\substack{0 \le j < n \\ j \equiv r \bmod m}}
(\omega^{-j}-1)(\omega^j-\omega^{-1}) \\
&= \frac{\omega-1}{4i} \left( \frac{n}{m} \left( 1 + \omega\inv \right)
+ \sum_{\substack{0 \le j < n \\ j \equiv r \bmod m}} (\omega^{-1-j}-\omega^j) \right).
\end{align*}
(We allow degenerate triangles where $j \in \{0,m-1\}$ with area zero
to simplify the notation above.)
However, if $m$ is a proper divisor of $m$, then
$\sum_j \omega^j = \omega^r(1+\omega^m+\omega^{2m}+\dots+\omega^{n-m}) = 0$.
For the same reason,
$\sum_j \omega^{-1-j} = 0$.
So the inner sum vanishes, and the total area of this color equals
\[ \sum_{\substack{0 \le j < n \\ j \equiv r \bmod m}}
\operatorname{Area}(\omega^0, \omega^j, \omega^{j+1})
= \frac nm \frac{(\omega-1)(\omega^{-1}+1)}{4i}
= \frac nm \cdot \frac{\omega-\omega^{-1}}{4i}
. \]
Because the right-hand side does not depend on the residue $r$,
this shows all colors have equal area.
\paragraph{Proof of necessity.}
It's obvious that $m < n$ (in fact $m \le n-2$).
So we focus on just showing $m \mid n$.
Repeating the same calculation as above, we find that if there was a valid
triangulation and coloring, the total area of each color would equal
\[ S \coloneq \frac nm \cdot \frac{\omega-\omega^{-1}}{4i}. \]
However:
\begin{claim*}
The number $4i \cdot S$ is not an algebraic integer when $m \nmid n$.
\end{claim*}
\begin{proof}
This is easiest to see if one knows the advanced result that
$K \coloneq \QQ(\omega)$ is a number field
whose ring of integers is known to be $\mathcal{O}_K = \ZZ[\omega]$,
in which case it follows right away.
\end{proof}
\begin{remark*}
We spell out the details in the proof a bit more explicitly here.
It's enough to show that
$\omega \cdot 4i \cdot S = \frac nm \omega^2 - \frac nm$
is not an algebraic integer for completeness.
Take $K = \mathbb Q(\omega)$ of degree $d \coloneq \varphi(n) \ge 2$;
as a $\mathbb Q$-module, it obeys
$K = \mathbb Q \oplus \omega \mathbb Q \oplus \dots \oplus \omega^{d-1} \mathbb Q$.
The theorem we are quoting is that, as $\ZZ$-modules, we have
$\mathcal{O}_K = \mathbb Z[\omega] =
\mathbb Z \oplus \omega \mathbb Z \oplus \dots \oplus \omega^{d-1} \mathbb Z$
i.e.\ $\mathcal{O}_K$ contains exactly those numbers in $K$
for which the canonical $\mathbb Q$-coefficients happen to be integers.
(This is quite famous, but for a print reference,
see Neukirch's \emph{Algebraic Number Theory}, Section I.10.)
And $\omega \cdot 4i \cdot S$ fails this criteria, since $\frac nm \notin \ZZ$.
\end{remark*}
However, for any $a$ and $b$, the number
\[ 4i \cdot T(a,b) = (\omega^a-1)(\omega^b-1)(\omega^{-b}-\omega^{-a}) \]
is an algebraic integer.
Since a finite sum of algebraic integers is also an algebraic integer,
the sum of expressions of the form $4i \cdot T(a,b)$ will never equal $4i \cdot S$.
\begin{remark*}
If one wants to avoid citing the fact that $\mathcal{O}_K = \ZZ[\omega]$,
then one can instead note that $T(a,b)$ is actually always divisible by
$(\omega-1)(\omega^{-1}+1) = \omega - \omega^{-1}$ over the algebraic integers
(at least one of $\{\omega^a-1, \omega^b-1, \omega^{-a} - \omega^{-b}\}$
is a multiple of $\omega+1$, by casework on $a,b \bmod 2$).
Then one using $\frac{4i}{(\omega-1)(\omega^{-1}+1)}$ as the scaling factor
instead of $4i$, one sees that we actually need $\frac nm$ to be an algebraic integer.
But a rational number is an algebraic integer if and only if it's a rational integer;
so $\frac nm$ is an algebraic integer exactly when $m$ divides $n$.
\end{remark*}
|
USAMO-2024-notes_4
|
Let $m$ and $n$ be positive integers.
A circular necklace contains $mn$ beads, each either red or blue.
It turned out that no matter how the necklace was cut into $m$ blocks of $n$
consecutive beads, each block had a distinct number of red beads.
Determine, with proof, all possible values of the ordered pair $(m, n)$.
|
The answer is $m \leq n+1$ only.
\paragraph{Proof the task requires $m \le n+1$.}
Each of the $m$ blocks has a red bead count between $0$ and $n$,
each of which appears at most once, so $m \le n+1$ is needed.
\paragraph{Construction when $m=n+1$.}
For concreteness, here is the construction for $n=4$, which obviously generalizes.
The beads are listed in reading order as an array with $n+1$ rows and $n$ columns.
Four of the blue beads have been labeled $B_1$, \dots, $B_n$
to make them easier to track as they move.
\[
T_0 =
\begin{bmatrix}
{\color{red}R} & {\color{red}R} & {\color{red}R} & {\color{red}R} \\
{\color{red}R} & {\color{red}R} & {\color{red}R} & {\color{blue}B_1} \\
{\color{red}R} & {\color{red}R} & {\color{blue}B} & {\color{blue}B_2} \\
{\color{red}R} & {\color{blue}B} & {\color{blue}B} & {\color{blue}B_3} \\
{\color{blue}B} & {\color{blue}B} & {\color{blue}B} & {\color{blue}B_4}
\end{bmatrix}.
\]
To prove this construction works,
it suffices to consider the $n$ cuts $T_0$, $T_1$, $T_2$, \dots, $T_{n-1}$
made where $T_i$ differs from $T_{i-1}$ by having the cuts one bead later
also have the property each row has a distinct red count:
\[
T_1 =
\begin{bmatrix}
{\color{red}R} & {\color{red}R} & {\color{red}R} & {\color{red}R} \\
{\color{red}R} & {\color{red}R} & {\color{blue}B_1} & {\color{red}R} \\
{\color{red}R} & {\color{blue}B} & {\color{blue}B_2} & {\color{red}R} \\
{\color{blue}B} & {\color{blue}B} & {\color{blue}B_3} & {\color{blue}B} \\
{\color{blue}B} & {\color{blue}B} & {\color{blue}B_4} & {\color{red}R}
\\
\end{bmatrix}
\quad
T_2 =
\begin{bmatrix}
{\color{red}R} & {\color{red}R} & {\color{red}R} & {\color{red}R} \\
{\color{red}R} & {\color{blue}B_1} & {\color{red}R} & {\color{red}R} \\
{\color{blue}B} & {\color{blue}B_2} & {\color{red}R} & {\color{blue}B} \\
{\color{blue}B} & {\color{blue}B_3} & {\color{blue}B} & {\color{blue}B} \\
{\color{blue}B} & {\color{blue}B_4} & {\color{red}R} & {\color{red}R}
\end{bmatrix}
\quad
T_3 =
\begin{bmatrix}
{\color{red}R} & {\color{red}R} & {\color{red}R} & {\color{red}R} \\
{\color{blue}B_1} & {\color{red}R} & {\color{red}R} & {\color{blue}B} \\
{\color{blue}B_2} & {\color{red}R} & {\color{blue}B} & {\color{blue}B} \\
{\color{blue}B_3} & {\color{blue}B} & {\color{blue}B} & {\color{blue}B} \\
{\color{blue}B_4} & {\color{red}R} & {\color{red}R} & {\color{red}R}
\end{bmatrix}.
\]
We can construct a table showing for each $1 \le k \le n+1$
the number of red beads which are in the $(k+1)$st row of $T_i$ from the bottom:
\[
\begin{array}{c|cccc}
k & T_0 & T_1 & T_2 & T_3 \\ \hline
k=4 & 4 & 4 & 4 & 4 \\
k=3 & 3 & 3 & 3 & 2 \\
k=2 & 2 & 2 & 1 & 1 \\
k=1 & 1 & 0 & 0 & 0 \\
k=0 & 0 & 1 & 2 & 3
\end{array}.
\]
This suggests following explicit formula for the entry of the $(i,k)$th cell
which can then be checked straightforwardly:
\[
\#(\text{red cells in $k$th row of $T_i$})
=
\begin{cases}
k & k > i \\
k-1 & i \ge k > 0 \\
i & k = 0.
\end{cases}
\]
And one can see for each $i$, the counts are all distinct
(they are $(i, 0, 1, \dots, k-1, k+1, \dots, k)$ from bottom to top).
This completes the construction.
\paragraph{Construction when $m < n+1$.}
Fix $m$.
Take the construction for $(m,m-1)$ and add $n+1-m$ cyan beads to the start of each row;
for example, if $n = 7$ and $m = 5$ then the new construction is
\[
T =
\begin{bmatrix}
{\color{cyan}C} & {\color{cyan}C}& {\color{cyan}C} &
{\color{red}R} & {\color{red}R} & {\color{red}R} & {\color{red}R} \\
{\color{cyan}C} & {\color{cyan}C}& {\color{cyan}C} &
{\color{red}R} & {\color{red}R} & {\color{red}R} & {\color{blue}B_1} \\
{\color{cyan}C} & {\color{cyan}C}& {\color{cyan}C} &
{\color{red}R} & {\color{red}R} & {\color{blue}B} & {\color{blue}B_2} \\
{\color{cyan}C} & {\color{cyan}C}& {\color{cyan}C} &
{\color{red}R} & {\color{blue}B} & {\color{blue}B} & {\color{blue}B_3} \\
{\color{cyan}C} & {\color{cyan}C}& {\color{cyan}C} &
{\color{blue}B} & {\color{blue}B} & {\color{blue}B} & {\color{blue}B_4}
\end{bmatrix}.
\]
This construction still works for the same reason (the cyan beads
do nothing for the first $n+1-m$ shifts, then one reduces to the previous case).
If we treat cyan as a shade of blue, this finishes the problem.
|
USAMO-2024-notes_5
|
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 $BEM$.
|
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.
|
USAMO-2024-notes_6
|
Let $n > 2$ be an integer and let $\ell \in \{1, 2,\dots, n\}$.
A collection $A_1$, \dots, $A_k$ of (not necessarily distinct) subsets of
$\{1, 2,\dots, n\}$ is called $\ell$-large if $|A_i| \ge \ell$ for all $1 \le i \le k$.
Find, in terms of $n$ and $\ell$, the largest real number $c$ such that the inequality
\[ \sum_{i=1}^k\sum_{j=1}^k x_ix_j \frac{|A_i\cap A_j|^2}{|A_i|\cdot|A_j|}
\ge c\left(\sum_{i=1}^k x_i\right)^2 \]
holds for all positive integer $k$,
all nonnegative real numbers $x_1$, $x_2$, \dots, $x_k$,
and all $\ell$-large collections $A_1$, $A_2$, \dots, $A_k$
of subsets of $\{1,2,\dots,n\}$.
\end{enumerate}
|
The answer turns out to be
\[ c = \frac{n+\ell^2-2\ell}{n(n-1)}. \]
The following solution using indicator functions was
suggested by the language reasoning system developed by Thang Luong
and the team at Google DeepMind (\url{https://dpmd.ai/imo-silver}).
\paragraph{Rewriting the sum.}
Abbreviate $[n] \coloneq \{1, \dots, n\}$,
and WLOG assume $\sum x_i = 1$.
To prove the bound, we first rewrite the left-hand side using indicator functions
and switch the order of summation as
\begin{align*}
\text{LHS}
&= \sum_{i=1}^k\sum_{j=1}^k \frac{x_i}{|A_i|} \frac{x_j}{|A_j|} |A_i\cap A_j|^2 \\
&= \sum_{i=1}^k\sum_{j=1}^k \frac{x_i}{|A_i|} \frac{x_j}{|A_j|}
\left( \sum_{p \in [n]} \sum_{q \in [n]} \mathbf{1}_{p \in A_i} \mathbf{1}_{q \in A_j} \right) \\
&= \sum_{p \in [n]} \sum_{q \in [n]} \sum_{i=1}^k\sum_{j=1}^k
\frac{x_i \mathbf{1}_{p \in A_i}}{|A_i|} \cdot \frac{x_j \mathbf{1}_{q \in A_j}}{|A_j|}.
\end{align*}
Hence, if we define the $n^2$ variables
\[ v_{p,q} \coloneq \sum_{i : p,q \in A_i} \frac{x_i}{|A_i|} \]
for each $(p,q) \in [n]^2$
then the previous expression simplifies to just
\[ \text{LHS} = \sum_{p \in [n]} \sum_{q \in [n]} v_{p,q}^2. \]
\paragraph{Proof of bound.}
We split the sum into $p=q$ versus $p \neq q$ and bound by QM-AM
separately in the obvious way.
For the terms $p=q$,
\begin{align*}
\sum_{p \in [n]} v_{p,p}
&= \sum_{i=1}^k |A_i| \cdot \frac{x_i}{|A_i|} = \sum_{i=1}^k x_i = 1 \\
\implies \sum_{p \in [n]} v_{p,p}^2
&\ge \frac{\left( \sum_{p \in [n]} v_{p,p} \right)^2}{n} = \frac 1n.
\end{align*}
When $p \neq q$ we instead have
\begin{align*}
\sum_{p \neq q \in [n]} v_{p,q}
&= \sum_{i=1}^k 2 \binom{|A_i|}{2} \cdot \frac{x_i}{|A_i|}
= \sum_{i=1}^k (|A_i|-1) x_i \ge \sum_{i=1}^k (\ell-1) x_i = \ell-1 \\
\implies \sum_{p \neq q \in [n]} v_{p,q}^2
&\ge \frac{\left( \sum_{p \neq q \in [n]} v_{p,q} \right)^2}{n(n-1)}
\ge \frac{(\ell-1)^2}{n(n-1)}.
\end{align*}
Combining the two bounds gives
\[ \text{LHS} \ge \frac 1n + \frac{(\ell-1)^2}{n(n-1)} = \frac{n+\ell^2-2\ell}{n(n-1)} \]
which matches the answer claimed above.
\paragraph{Construction.}
Equality occurs if $v_{p,p}$ is the same for all $p$
and $v_{p,q}$ is the same for all $p \neq q$,
and $|A_i| = \ell$ for all $i$.
This occurs if one takes, say, $k = \binom{n}{\ell}$ and $A_i$ to be
all the subsets of size $\ell$ each exactly once.
\begin{remark*}
Originally, the write-up I used considered vectors in $\RR^{n^2}$
where entries will be indexed by ordered pairs $(p,q) \in \{1, \dots, n\}^2$.
It was equivalent to the solution above, but looking back,
I think this makes the problem look harder than it actually is.
For comparison, here is how the notation with vectors works, using
$\left< \bullet,\bullet \right>$ for dot product, and $\lVert \bullet \rVert$ for the vector norm.
For $i=1,\dots,n$ define $\mathbf{v}_i$ by
\[ \mathbf{v}_i[p,q] \coloneq \begin{cases}
\frac{1}{|A_i|} & p \in A_i \text{ and } q \in A_i \\
0 & \text{otherwise};
\end{cases}
\qquad \mathbf{v} \coloneq \sum_i x_i \mathbf{v}_i.
\]
Then
\begin{align*}
\sum_i \sum_j x_ix_j \frac{|A_i\cap A_j|^2}{|A_i||A_j|}
&= \sum_i \sum_j x_ix_j \left< \mathbf{v}_i, \mathbf{v}_j\right> \\
&= \left< \mathbf{v}, \mathbf{v} \right> = \left\lVert \mathbf{v} \right\rVert^2.
\end{align*}
Define two more vectors $\mathbf{e}$ and $\mathbf{1}$;
the vector $\mathbf{e}$ has $1$ in the $(p,q)$\ts{th} component if $p=q$, and $0$
otherwise, while $\mathbf{1}$ has all-ones.
In that case, note that
\begin{align*}
\left< \mathbf{e}, \mathbf{v} \right> &= \sum_i x_i \left< \mathbf{e}, \mathbf{v}_i \right> = \sum_i x_i \\
\left< \mathbf{1}, \mathbf{v} \right> &= \sum_i x_i \left< \mathbf{1}, \mathbf{v}_i \right> = \sum_i x_i |A_i|.
\end{align*}
That means for any positive real constants $\alpha$ and $\beta$,
by Cauchy-Schwarz for vectors, we should have
\begin{align*}
\left\lVert \alpha \mathbf{e} + \beta \mathbf{1} \right\rVert \left\lVert \mathbf{v} \right\rVert
&\ge \left< \alpha \mathbf{e} + \beta \mathbf{1}, \mathbf{v} \right>
= \alpha \left< \mathbf{e}, \mathbf{v}\right> + \beta \left< \mathbf{1}, \mathbf{v} \right> \\
&= \alpha \cdot \sum x_i + \beta \cdot \sum x_i |A_i|
\ge (\alpha + \ell \beta) \sum x_i.
\end{align*}
Set $\mathbf{w} \coloneq \alpha \mathbf{e} + \beta \mathbf{1}$ for brevity.
Then
\[ \mathbf{w}[p,q] =
\begin{cases}
\alpha + \beta & \text{if } p = q \\
\beta & \text{if }p \neq q
\end{cases}
\implies \left\lVert \mathbf{w} \right\rVert
= \sqrt{n \cdot (\alpha + \beta)^2 + (n^2-n) \cdot \beta^2}. \]
Therefore, we get an lower bound
\[ \frac{\left\lVert \mathbf{v} \right\rVert}{\sum x_i}
\ge \frac{\alpha + \ell \beta}{\sqrt{n \cdot (\alpha + \beta)^2 + (n^2-n) \cdot \beta^2}} \]
Letting $\alpha = n - \ell$ and $\beta = \ell-1$ gives a proof that the constant
\[ c = \frac{((n-\ell)+\ell(\ell-1))^2}{n \cdot (n-1)^2 + (n^2-n) \cdot (\ell-1)^2}
= \frac{(n+\ell^2-2\ell)^2}{n(n-1)\left( n+\ell^2-2\ell \right)}
= \frac{n+\ell^2-2\ell}{n(n-1)} \]
makes the original inequality always true.
(The choice of $\alpha:\beta$ is suggested by the example below.)
Meanwhile, for the construction (where $k = \binom{n}{\ell}$, etc.)
it would be sufficient to show that $\mathbf{w}$
and $\mathbf{v}$ are scalar multiples, so that the above Cauchy-Schwarz is equality.
However, we can compute
\[
\mathbf{w}[p,q] =
\begin{cases}
n-1 & \text{if } p = q \\
\ell-1 & \text{if }p \neq q
\end{cases},
\qquad
\mathbf{v}[p,q] =
\begin{cases}
\binom{n-1}{\ell-1} \cdot \frac{1}{\ell} & \text{if } p = q \\
\binom{n-2}{\ell-2} \cdot \frac{1}{\ell} & \text{if } p \neq q
\end{cases}
\]
which are indeed scalar multiples.
\end{remark*}
|
USAMO-2025-notes_1
|
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*}
|
USAMO-2025-notes_2
|
Let $n > k \ge 1$ be integers.
Let $P(x) \in \RR[x]$ be a polynomial of degree $n$
with no repeated roots and $P(0) \neq 0$.
Suppose that for any real numbers $a_0$, \dots, $a_k$
such that the polynomial $a_k x^k + \dots + a_1 x + a_0$ divides $P(x)$,
the product $a_0 a_1 \dots a_k$ is zero.
Prove that $P(x)$ has a nonreal root.
|
By considering any $k+1$ of the roots of $P$, we may as well assume WLOG that $n = k+1$.
Suppose that $P(x) = (x+r_1) \dots (x+r_n) \in \RR[x]$ has $P(0) \neq 0$.
Then the problem hypothesis is that each of the $n$ polynomials
(of degree $n-1$) given by
\begin{align*}
P_1(x) &= (x+r_2)(x+r_3)(x+r_4) \dots (x+r_n) \\
P_2(x) &= (x+r_1)(x+r_3)(x+r_4) \dots (x+r_n) \\
P_3(x) &= (x+r_1)(x+r_2)(x+r_4) \dots (x+r_n) \\
&\vdotswithin= \\
P_n(x) &= (x+r_1)(x+r_2)(x+r_3) \dots (x+r_{n-1})
\end{align*}
has at least one coefficient equal to zero.
(Explicitly, $P_i(x) = \frac{P(x)}{x+r_i}$.)
We'll prove that at least one $r_i$ is not real.
Obviously the leading and constant coefficients of each $P_i$ are nonzero,
and there are $n-2$ other coefficients to choose between.
So by pigeonhole principle, we may assume, say,
that $P_1$ and $P_2$ share the position of a zero coefficient, say
the $x^k$ one, for some $1 \le k < n-1$.
\begin{claim*}
If $P_1$ and $P_2$ both have $x^k$ coefficient equal to zero, then the polynomial
\[ Q(x) = (x+r_3)(x+r_4) \dots (x+r_n) \]
has two consecutive zero coefficients, namely $b_k = b_{k-1} = 0$.
\end{claim*}
\begin{proof}
Invoking Vieta formulas, suppose that
\[ Q(x) = x^{n-2} + b_{n-3} x^{n-3} + \dots + b_0. \]
(And let $b_{n-2} = 1$.)
Then the fact that the $x^k$ coefficient of $P_1$ and $P_2$ are both zero means
\[ r_1 b_k + b_{k-1} = r_2 b_k + b_{k-1} = 0 \]
and hence that $b_k = b_{k-1} = 0$ (since the $r_i$ are nonzero).
\end{proof}
To solve the problem, we use:
\begin{lemma*}
If $F(x) \in \RR[x]$ is a polynomial with two consecutive zero coefficients,
it cannot have all distinct real roots.
\end{lemma*}
Here are two possible proofs of the lemma I know (there are more).
\begin{proof}
[First proof using Rolle's theorem]
Say $x^t$ and $x^{t+1}$ coefficients of $F$ are both zero.
Assume for contradiction all the roots of $F$ are real and distinct.
Then by Rolle's theorem, every higher-order derivative of $F$
should have this property too.
However, the $t$th order derivative of $F$ has a double root of $0$, contradiction.
\end{proof}
\begin{proof}
[Second proof using Descartes rule of signs]
The number of (nonzero) roots of $F$ is bounded above by
the number of sign changes of $F(x)$ (for the positive roots)
and the number of sign changes of $F(-x)$ (for the negative roots).
Now consider each pair of consecutive nonzero coefficients in $F$,
say ${\star} x^i$ and ${\star} x^j$ for $i > j$.
\begin{itemize}
\ii If $i-j=1$, then this sign change will only count for one of $F(x)$ or $F(-x)$
\ii If $i-j \ge 2$, then the sign change could count towards both $F(x)$ or $F(-x)$
(i.e.\ counted twice), but also there is at least one zero coefficient between them.
\end{itemize}
Hence if $b$ is the number of nonzero coefficients of $F$,
and $z$ is the number of \emph{consecutive runs} of zero coefficients of $F$,
then the number of real roots is bounded above by
\[ 1 \cdot (b-1-z) + 2 \cdot z = b - 1 + z \le \deg F. \]
However, if $F$ has \emph{two} consecutive zero coefficients, then the inequality is strict.
\end{proof}
\begin{remark*}
The final claim has appeared before apparently
in the HUST Team Selection Test for the Vietnamese Math Society's
undergraduate olympiad; see \url{https://aops.com/community/p33893374} for citation.
\end{remark*}
|
USAMO-2025-notes_3
|
Alice the architect and Bob the builder play a game.
First, Alice chooses two points $P$ and $Q$ in the plane
and a subset $\mathcal{S}$ of the plane, which are announced to Bob.
Next, Bob marks infinitely many points in the plane,
designating each a city.
He may not place two cities within distance at most one unit of each other,
and no three cities he places may be collinear.
Finally, roads are constructed between the cities as follows:
each pair $A$, $B$ of cities is connected with a road along the line segment $AB$
if and only if the following condition holds:
\begin{quote}
For every city $C$ distinct from $A$ and $B$,
there exists $R\in \mathcal{S}$ such that $\triangle PQR$ is directly similar
to either $\triangle ABC$ or $\triangle BAC$.
\end{quote}
Alice wins the game if (i) the resulting roads allow for travel
between any pair of cities via a finite sequence of roads and (ii) no two roads cross.
Otherwise, Bob wins. Determine, with proof, which player has a winning strategy.
|
The answer is that Alice wins.
Let's define a \emph{Bob-set} $V$ to be a set of points in the plane with no three collinear
and with all distances at least $1$.
The point of the problem is to prove the following fact.
\begin{claim*}
Given a Bob-set $V \subseteq \RR^2$, consider the \emph{Bob-graph}
with vertex set $V$ defined as follows:
draw edge $ab$ if and only if the disk with diameter $\ol{ab}$
contains no other points of $V$ on or inside it.
Then the Bob-graph is (i) connected, and (ii) planar.
\end{claim*}
Proving this claim shows that Alice wins since Alice can specify $\mathcal{S}$
to be the set of points outside the disk of diameter $PQ$.
\begin{proof}
[Proof that every Bob-graph is connected]
Assume for contradiction the graph is disconnected.
Let $p$ and $q$ be two points in different connected components.
Since $pq$ is not an edge, there exists a third point $r$
inside the disk with diameter $\ol{pq}$.
Hence, $r$ is in a different connected component
from at least one of $p$ or $q$ --- let's say point $p$.
Then we repeat the same argument on the disk with diameter $\ol{pr}$
to find a new point $s$, non-adjacent to either $p$ or $r$.
See the figure below, where the X'ed out dashed edges indicate
points which are not only non-adjacent but in different connected components.
\begin{center}
\begin{asy}
size(6cm);
pair p = (-1,0);
pair q = (1,0);
pair r = (0.4,0.7);
filldraw(unitcircle, opacity(0.1)+blue, blue);
filldraw(circle((p+r)/2, abs(p-r)/2), opacity(0.1)+red, red);
pair s = (-0.3,0.6);
draw(p--q, dashed+blue);
draw(p--r, dashed+red);
draw(s--r, dashed);
dot("$p$", p, dir(p));
dot("$q$", q, dir(q));
dot("$r$", r, dir(r));
dot("$s$", s, dir(s));
picture X;
real eps = 0.05;
draw(X, (-eps,-eps)--(eps,eps), black+1.4);
draw(X, (eps,-eps)--(-eps,eps), black+1.4);
add(X);
add(shift(midpoint(p--r))*X);
add(shift(midpoint(s--r))*X);
label("$\delta_1$", (0,0), 2*dir(45), deepgreen);
label("$\delta_2$", midpoint(p--r), 2*dir(-10), deepgreen);
label("$\delta_3$", midpoint(s--r), 2*dir(100), deepgreen);
\end{asy}
\end{center}
In this way we generate an infinite sequence of distances
$\delta_1$, $\delta_2$, $\delta_3$, \dots\ among the non-edges in the picture above.
By the ``Pythagorean theorem'' (or really the inequality for it), we have
\[ \delta_i^2 \le \delta_{i-1}^2 - 1 \]
and this eventually generates a contradiction for large $i$,
since we get $0 \le \delta_i^2 \le \delta_1^2 - (i-1)$.
\end{proof}
\begin{proof}
[Proof that every Bob-graph is planar]
Assume for contradiction edges $ac$ and $bd$ meet, meaning $abcd$ is a convex quadrilateral.
WLOG assume $\angle bad \ge 90\dg$ (each quadrilateral has an angle at least $90\dg$).
Then the disk with diameter $\ol{bd}$ contains $a$, contradiction.
\end{proof}
\begin{remark*}
In real life, the Bob-graph is actually called the \href{https://w.wiki/DVnd}{Gabriel graph}.
Note that we never require the Bob-set to be infinite;
the solution works unchanged for finite Bob-sets.
However, there are approaches that work for finite Bob-sets that don't work for
infinite sets, such as the \href{https://www.graphclasses.org/classes/gc_1224.html}{relative neighbor graph},
in which one joins $a$ and $b$ iff there is no $c$ such that
$d(a,b) \le \max \{d(a,c), d(b,c)\}$.
In other words, edges are blocked by triangles where $ab$ is the longest edge
(rather than by triangles where $ab$ is the longest edge of a right or obtuse triangle
as in the Gabriel graph).
The relative neighbor graph has fewer edges than the Gabriel graph, so it is planar too.
When the Bob-set is finite, the relative distance graph is still connected.
The same argument above works where the distances now satisfy
\[ \delta_1 > \delta_2 > \dots \]
instead, and since there are finitely many distances one arrives at a contradiction.
However for infinite Bob-sets the descending condition is insufficient,
and connectedness actually fails altogether.
A counterexample (communicated to me by Carl Schildkraut) is to start by taking
$A_n \approx (2n,0)$ and $B_n \approx (2n+1, \sqrt3)$ for all $n \ge 1$,
then perturb all the points slightly so that
\begin{align*}
B_1A_1 &> A_1A_2 > A_2B_1 > B_1B_2 > B_2A_2 \\
&> A_2A_3 > A_3B_2 > B_2B_3 > B_3A_3 \\
&> \dotsb.
\end{align*}
A cartoon of the graph is shown below.
\begin{center}
\begin{asy}
dotfactor *= 1.5;
pair A1 = (0,0);
pair A2 = (2,0);
pair A3 = (4,0);
pair A4 = (6,0);
pair B1 = (1,3**0.5);
pair B2 = (3,3**0.5);
pair B3 = (5,3**0.5);
pair B4 = (7,3**0.5);
dot("$A_1$", A1, dir(-90), blue);
dot("$A_2$", A2, dir(-90), blue);
dot("$A_3$", A3, dir(-90), blue);
dot("$B_1$", B1, dir(90), blue);
dot("$B_2$", B2, dir(90), blue);
dot("$B_3$", B3, dir(90), blue);
label("$\dots$", A4, blue);
label("$\dots$", B4, blue);
draw("$2.08$", A1--A2, dir(-90), blue, Margins);
draw("$2.04$", A2--A3, dir(-90), blue, Margins);
draw(A3--A4, blue, Margins);
draw("$2.06$", B1--B2, dir(90), blue, Margins);
draw("$2.02$", B2--B3, dir(90), blue, Margins);
draw(B3--B4, blue, Margins);
draw(rotate(60)*"$2.09$", B1--A1, dotted);
draw(rotate(-60)*"$2.07$", B1--A2, dotted);
draw(rotate(60)*"$2.05$", B2--A2, dotted);
draw(rotate(-60)*"$2.03$", B2--A3, dotted);
draw(rotate(60)*"$2.01$", B3--A3, dotted);
\end{asy}
\end{center}
In that case, $\{A_n\}$ and $\{B_n\}$ will be disconnected from each other:
none of the edges $A_nB_n$ or $B_nA_{n+1}$ are formed.
In this case the relative neighbor graph consists of the edges
$A_1A_2A_3A_4 \dotsm$ and $B_1 B_2 B_3 B_4 \dotsm$.
That's why for the present problem, the inequality
\[ \delta_i^2 \le \delta_{i-1}^2 - 1 \]
plays such an important role, because it causes the (squared) distances
to decrease appreciably enough to give the final contradiction.
\end{remark*}
|
USAMO-2025-notes_4
|
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 $CX = CY$.
|
USAMO-2025-notes_5
|
Find all positive integers $k$ such that: for every positive integer $n$, the sum
\[ \binom n0^k + \binom n1^k + \dots + \binom nn^k \]
is divisible by $n+1$.
|
The answer is all even $k$.
Let's abbreviate $S(n) \coloneq \binom n0^k + \dots + \binom nn^k$ for the sum in the problem.
\paragraph{Proof that even $k$ is necessary.}
Choose $n=2$. We need $3 \mid S(2) = 2+2^k$,
which requires $k$ to be even.
\begin{remark*}
It's actually not much more difficult to just use $n = p-1$ for prime $p$,
since $\binom{p-1}{i} \equiv (-1)^i \pmod p$.
Hence $S(p-1) \equiv 1 + (-1)^k + 1 + (-1)^k + \dots + 1 \pmod p$,
and this also requires $k$ to be even.
This special case is instructive in figuring out the proof to follow.
\end{remark*}
\paragraph{Proof that $k$ is sufficient.}
From now on we treat $k$ as fixed, and we let $p^e$ be a prime fully dividing $n+1$.
The basic idea is to reduce from $n+1$ to $(n+1)/p$ by an induction.
\begin{remark*}
Here is a concrete illustration that makes it clear what's going on.
Let $p = 5$.
When $n = p-1 = 4$, we have
\[ S(4) = 1^k + 4^k + 6^k + 4^k + 1^k \equiv 1 + 1 + 1 + 1 + 1 \equiv 0 \pmod 5. \]
When $n = p^2-1 = 24$, the $25$ terms of $S(24)$ in order are, modulo $25$,
\begin{align*}
S(24) &\equiv 1^k + 1^k + 1^k + 1^k + 1^k\\
&+ 4^k + 4^k + 4^k + 4^k + 4^k \\
&+ 6^k + 6^k + 6^k + 6^k + 6^k \\
&+ 4^k + 4^k + 4^k + 4^k + 4^k \\
&+ 1^k + 1^k + 1^k + 1^k + 1^k \\
&= 5(1^k + 4^k + 6^k + 4^k + 1^k).
\end{align*}
The point is that $S(24)$ has five copies of $S(4)$, modulo $25$.
\end{remark*}
To make the pattern in the remark explicit, we prove the following lemma
on each \emph{individual} binomial coefficient.
\begin{lemma}
Suppose $p^e$ is a prime power which fully divides $n+1$. Then
\[ \binom{n}{i} \equiv \pm \binom{\frac{n+1}{p}-1}{\left\lfloor i/p \right\rfloor} \pmod{p^e}. \]
\end{lemma}
\begin{proof}
[Proof of lemma]
It's easiest to understand the proof by
looking at the cases $\left\lfloor i/p \right\rfloor \in \{0,1,2\}$ first.
\begin{itemize}
\ii For $0 \le i < p$, since $n \equiv -1 \mod p^e$, we have
\[ \binom{n}{i} = \frac{n(n-1) \dots (n-i+1)}{1 \cdot 2 \cdot \dots \cdot i}
\equiv \frac{(-1)(-2) \dots (-i)}{1 \cdot 2 \cdot \dots \cdot i} \equiv \pm 1 \pmod{p^e}. \]
\ii For $p \le i < 2p$ we have
\begin{align*}
\binom{n}{i}
&\equiv \pm 1 \cdot \frac{n-p+1}{p} \cdot \frac{(n-p)(n-p-1) \dots (n-i+1)}{(p+1)(p+2) \dots i} \\
&\equiv \pm 1 \cdot \frac{\frac{n-p+1}{p}}{1} \cdot \pm 1 \\
&\equiv \pm \binom{\frac{n+1}{p}-1}{1} \pmod{p^e}.
\end{align*}
\ii For $2p \le i < 3p$ the analogous reasoning gives
\begin{align*}
\binom ni
&\equiv \pm 1 \cdot \frac{n-p+1}{p} \cdot \pm 1 \cdot \frac{n-2p+1}{2p} \cdot \pm 1 \\
&\equiv \pm \frac{\left(\frac{n+1}{p}-1\right)\left( \frac{n+1}{p}-2 \right) }{1 \cdot 2} \\
&\equiv \pm \binom{\frac{n+1}{p}-1}{2} \pmod{p^e}.
\end{align*}
\end{itemize}
And so on. The point is that in general, if we write
\[ \binom ni = \prod_{1 \le j \le i} \frac{n-(j-1)}{j} \]
then the fractions for $p \nmid j$ are all $\pm 1 \pmod{p^e}$.
So only considers those $j$ with $p \mid j$; in that case
one obtains the claimed $\binom{\frac{n+1}{p}-1}{\left\lfloor i/p \right\rfloor}$ exactly
(even without having to take modulo $p^e$).
\end{proof}
From the lemma, it follows if $p^e$ is a prime power which fully divides $n+1$, then
\[ S(n) \equiv p \cdot S\left( \frac{n+1}{p}-1 \right) \pmod{p^e} \]
by grouping the $n+1$ terms (for $0 \le i \le n$)
into consecutive ranges of length $p$ (by the value of $\left\lfloor i/p \right\rfloor$).
\begin{remark*}
Actually, with the exact same proof (with better $\pm$ bookkeeping) one may show that
\[ n+1 \mid \sum_{i=0}^n \left( (-1)^i \binom ni \right)^k \]
holds for \emph{all} nonnegative integers $k$, not just $k$ even.
So in some sense this result is more natural than the one in the problem statement.
\end{remark*}
|
USAMO-2025-notes_6
|
Let $m$ and $n$ be positive integers with $m\geq n$.
There are $m$ cupcakes of different flavors arranged around a circle
and $n$ people who like cupcakes.
Each person assigns a nonnegative real number score to each cupcake,
depending on how much they like the cupcake.
Suppose that for each person $P$, it is possible to partition the circle
of $m$ cupcakes into $n$ groups of consecutive cupcakes
so that the sum of $P$'s scores of the cupcakes in each group is at least $1$.
Prove that it is possible to distribute the $m$ cupcakes to the $n$ people
so that each person $P$ receives cupcakes of total score at least $1$ with respect to $P$.
\end{enumerate}
|
Arbitrarily pick any one person --- call her Pip --- and her $n$ arcs.
The initial idea is to try to apply Hall's marriage lemma to match
the $n$ people with Pip's arcs (such that each such person is happy with their matched arc).
To that end, construct the obvious bipartite graph $\mathfrak{G}$
between the people and the arcs for Pip.
We now consider the following algorithm, which takes several steps.
\begin{itemize}
\ii If a perfect matching of $\mathfrak{G}$ exists, we're done!
\ii We're probably not that lucky.
Per Hall's condition, this means there is a \emph{bad set} $\mathcal{B}_1$ of people,
who are compatible with fewer than $|\mathcal{B}_1|$ of the arcs.
Then delete $\mathcal{B}_1$ and the neighbors of $\mathcal{B}_1$,
then try to find a matching on the remaining graph.
\ii If a matching exists now, terminate the algorithm.
Otherwise, that means there's another bad set $\mathcal{B}_2$
for the remaining graph.
We again delete $\mathcal{B}_2$ and the fewer than $\mathcal{B}_2$ neighbors.
\ii Repeat until some perfect matching $\mathcal{M}$ is possible in the remaining graph,
i.e.\ there are no more bad sets (and then terminate once that occurs).
Since Pip is a universal vertex, it's impossible to delete Pip,
so the algorithm does indeed terminate with nonempty $\mathcal{M}$.
\end{itemize}
A cartoon of this picture is shown below.
\begin{center}
\begin{asy}
size(8cm);
dotfactor *= 1.3;
real w = 3;
real eps = 0.4;
label("People", (-w,10));
label("Arcs of Pip", (w,10));
filldraw(box((-w-eps, 9+eps), (-w+eps,7-eps)), opacity(0.1)+lightred, red+1.2);
filldraw(box((-w-eps, 6+eps), (-w+eps,5-eps)), opacity(0.1)+lightred, orange+1.2);
filldraw(box((-w-eps, 4+eps), (-w+eps,2-eps)), opacity(0.1)+lightred, brown+1.2);
filldraw(box((w-eps, 9+eps), (w+eps,8-eps)), opacity(0.1)+lightred, red+1.2);
filldraw(box((w-eps, 7+eps), (w+eps,7-eps)), opacity(0.1)+lightred, orange+1.2);
filldraw(box((w-eps, 6+eps), (w+eps,5-eps)), opacity(0.1)+lightred, brown+1.2);
draw((-w+eps, 9+eps)--(w-eps, 9+eps), red+dashed);
draw((-w+eps, 7-eps)--(w-eps, 8-eps), red+dashed);
draw((-w+eps, 6+eps)--(w-eps, 9+eps), orange+dashed);
draw((-w+eps, 5-eps)--(w-eps, 7-eps), orange+dashed);
draw((-w+eps, 4+eps)--(w-eps, 9+eps), brown+dashed);
draw((-w+eps, 2-eps)--(w-eps, 5-eps), brown+dashed);
label((-w-eps, 8), "Bad set $\mathcal{B}_1$", dir(180), black);
label((-w-eps, 5.5), "Bad set $\mathcal{B}_2$", dir(180), black);
label((-w-eps, 3), "Bad set $\mathcal{B}_3$", dir(180), black);
draw((-w,1)--(w,1), deepgreen+1.3);
draw((-w,0)--(w,0), deepgreen+1.3);
label((0, 0.5), "Final perfect matching $\mathcal{M}$", deepgreen);
for (int i=0; i<10; ++i) {
dot((w,i), blue);
dot((-w,i), blue);
}
label("Pip", (-w,0), dir(180), blue);
\end{asy}
\end{center}
We commit to assigning each of person in $\mathcal{M}$ their matched arc
(in particular if there are no bad sets at all, the problem is already solved).
Now we finish the problem by induction on $n$
(for the remaining people) by simply deleting the arcs used up by $\mathcal{M}$.
To see why this deletion-induction works,
consider any particular person Quinn not in $\mathcal{M}$.
By definition, Quinn is not happy with any of the arcs in $\mathcal{M}$.
So when an arc $\mathcal{A}$ of $\mathcal{M}$ is deleted,
it had value less than $1$ for Quinn
so in particular it couldn't contain entirely any of Quinn's arcs.
Hence at most one endpoint among Quinn's arcs was in the deleted arc $\mathcal{A}$.
When this happens, this causes two arcs of Quinn to merge, and the merged value is
\[ (\ge 1) + (\ge 1) - (\le 1) \qquad \ge \qquad 1 \]
meaning the induction is OK.
See below for a cartoon of the deletion,
where Pip's arcs are drawn in blue while Quinn's arcs and scores are drawn in red
(in this example $n=3$).
\begin{center}
\begin{asy}
size(13cm);
usepackage("amsmath");
pair O = (0,0);
picture before;
picture after;
real r = 1;
real s = 0.9;
real t = 0.65;
draw(before, arc(O, s, -20, 80), red+1.3);
draw(before, arc(O, s, 100, 200), red+1.3);
draw(before, arc(O, s, 220, 320), red+1.3);
draw(before, "Pip arc to delete", arc(O, r, 40, 140), blue+1.3);
draw(before, rotate(-60)*"Pip arc", arc(O, r, 160, 260), blue+1.3);
draw(before, rotate( 60)*"Pip arc", arc(O, r, 280, 380), blue+1.3);
label(before, "$\boxed{0.2}$", t*dir(120), red);
label(before, "$\boxed{0.3}$", t*dir( 60), red);
label(before, "$\boxed{0.8}$", t*dir(180), red);
label(before, "$\boxed{0.43}$", t*dir(240), red);
label(before, "$\boxed{0.57}$", t*dir(300), red);
label(before, "$\boxed{0.7}$", t*dir( 0), red);
label(before, minipage("\centering Quinn's \\ values", 3cm), O, red);
draw(after, arc(O, s, -20, 200), red+1.3);
draw(after, arc(O, s, 220, 320), red+1.3);
draw(after, rotate(-60)*"Pip arc", arc(O, r, 160, 260), blue+1.3);
draw(after, rotate( 60)*"Pip arc", arc(O, r, 280, 380), blue+1.3);
label(after, "$\boxed{0.8}$", t*dir(180), red);
label(after, "$\boxed{0.43}$", t*dir(240), red);
label(after, "$\boxed{0.57}$", t*dir(300), red);
label(after, "$\boxed{0.7}$", t*dir( 0), red);
label(after, minipage("\centering Quinn's \\ values", 3cm), O, red);
add(before);
add(shift(3.2,0)*after);
\end{asy}
\end{center}
\begin{remark*}
This deletion argument can be thought of in some special cases even
before the realization of Hall, in the case where $\mathcal{M}$ has only one person (Pip).
This amounts to saying that if one of Pip's arcs isn't liked by anybody,
then that arc can be deleted and the induction carries through.
\end{remark*}
\begin{remark*}
Conversely, it should be reasonable to expect Hall's theorem to be helpful
even before finding the deletion argument.
While working on this problem, one of the first things I said was:
\begin{quote}
``We should let Hall do the heavy lifting for us:
find a way to make $n$ groups that satisfy Hall's condition,
rather than an assignment of $n$ groups to $n$ people.''
\end{quote}
As a general heuristic, for any type of ``compatible matching'' problem,
Hall's condition is usually the go-to tool.
(It is much easier to verify Hall's condition than actually find the matching yourself.)
Actually in most competition problems, if one realizes one is in a Hall setting,
one is usually close to finishing the problem.
This is a relatively rare example in which one needs an additional idea
to go alongside Hall's theorem.
\end{remark*}
|
sols-TST-IMO-2014_1
|
Let $ABC$ be an acute triangle, and let $X$ be a variable interior point
on the minor arc $BC$ of its circumcircle.
Let $P$ and $Q$ be the feet of the perpendiculars from $X$ to lines $CA$ and $CB$, respectively.
Let $R$ be the intersection of line $PQ$ and the perpendicular from $B$ to $AC$.
Let $\ell$ be the line through $P$ parallel to $XR$.
Prove that as $X$ varies along minor arc $BC$,
the line $\ell$ always passes through a fixed point.
|
The fixed point is the orthocenter,
since $\ell$ is a Simson line.
See Lemma 4.4 of \emph{Euclidean Geometry in Math Olympiads}.
|
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