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IMO-2021-notes_1
|
Let $n \ge 100$ be an integer.
Ivan writes the numbers $n, n+1, \dots, 2n$ each on different cards.
He then shuffles these $n+1$ cards, and divides them into two piles.
Prove that at least one of the piles contains two cards such that
the sum of their numbers is a perfect square.
|
We will find three cards $a < b < c$ such that
\begin{align*}
b+c &= (2k+1)^2 \\
c+a &= (2k)^2 \\
a+b &= (2k-1)^2
\end{align*}
for some integer $k$.
Solving for $a$, $b$, $c$ gives
\begin{align*}
a &= \frac{(2k)^2+(2k-1)^2-(2k+1)^2}{2} = 2k^2 - 4k \\
b &= \frac{(2k+1)^2+(2k-1)^2-(2k)^2}{2} = 2k^2 + 1 \\
c &= \frac{(2k+1)^2+(2k)^2-(2k-1)^2}{2} = 2k^2 + 4k
\end{align*}
We need to show that when $n \ge 100$, one can find a suitable $k$.
Let
\begin{align*}
I_k &\coloneq \left\{ n \in \ZZ \mid n \le a < b < c \le 2n \right\} \\
&= \{ n \in \ZZ \mid k^2+2k \le n \le 2k^2-4k \}
\end{align*}
be the interval such that when $n \in I_k$,
the problem dies for that choice of $k$.
It would be sufficient to show these intervals $I_k$
cover all the integers $\ge 100$.
Starting from $I_9 = \left\{ 99 \le n \le 126 \right\}$,
we have
\[ k \ge 9 \implies 2k^2 - 4k \ge (k+1)^2 + 2(k+1) \]
which means the right endpoint of $I_k$
exceeds the left endpoint of $I_{k+1}$.
Hence for $n \ge 99$ in fact the problem is true.
\begin{remark*}
The problem turns out to be false for $n = 98$, surprisingly.
The counterexample is for one pile to be
\[ \{98,100,102,\dots,126\}
\cup \{129,131,135,\dots,161 \}
\cup \{162, 164, \dots, 196\}. \]
\end{remark*}
|
IMO-2021-notes_2
|
Show that the inequality
\[\sum_{i=1}^n \sum_{j=1}^n \sqrt{|x_i-x_j|}
\le \sum_{i=1}^n \sum_{j=1}^n \sqrt{|x_i+x_j|} \]
holds for all real numbers $x_1$, $x_2$, \dots, $x_n$.
|
The proof is by induction on $n \ge 1$ with the base cases $n=1$ and
$n=2$ being easy to verify by hand.
In the general situation, consider replacing the tuple $(x_i)_i$
with $(x_i+t)_i$ for some parameter $t \in \RR$.
The inequality becomes
\[\sum_{i=1}^n \sum_{j=1}^n \sqrt{|x_i-x_j|}
\le \sum_{i=1}^n \sum_{j=1}^n \sqrt{|x_i+x_j+2t|}. \]
The left-hand side is independent of $t$.
\begin{claim*}
The right-hand side, viewed as a function $F(t)$ of $t$,
is minimized when $2t = -(x_i + x_j)$ for some $i$ and $j$.
\end{claim*}
\begin{proof}
Since $F(t)$ is the sum of piecewise concave functions,
it is hence itself piecewise concave.
Moreover $F$ increases without bound if $|t| \to \infty$.
On each of the finitely many intervals on which $F(t)$ is
concave, the function is minimized at its endpoints.
Hence the minimum value must occur at one of the endpoints.
\end{proof}
If $t = -x_i$ for some $i$, this is the same as shifting all the
variables so that $x_i = 0$.
In that case, we may apply induction on $n-1$ variables,
deleting the variable $x_i$.
If $t = -\frac{x_i+x_j}{2}$, then notice
\[ x_i + t = -(x_j + t) \]
so it's the same as shifting all the variables such that $x_i = -x_j$.
In that case, we may apply induction on $n-2$ variables,
after deleting $x_i$ and $x_j$.
|
IMO-2021-notes_3
|
Let $D$ be an interior point of the acute triangle $ABC$
with $AB > AC$ so that $\angle DAB = \angle CAD$.
The point $E$ on the segment $AC$ satisfies $\angle ADE =\angle BCD$,
the point $F$ on the segment $AB$ satisfies $\angle FDA =\angle DBC$,
and the point $X$ on the line $AC$ satisfies $CX = BX$.
Let $O_1$ and $O_2$ be the circumcenters of the triangles
$ADC$ and $EXD$, respectively.
Prove that the lines $BC$, $EF$, and $O_1O_2$ are concurrent.
|
\emph{This solution was contributed by Abdullahil Kafi}.
\begin{claim*}
Quadrilateral $BCEF$ is cyclic.
\end{claim*}
\begin{proof}
Let $D'$ be the isogonal conjugate of the point $D$. The
angle condition implies quadrilateral $CEDD'$ and $BFDD'$
are cyclic. By power of point we have \[ AE\cdot AC=AD\cdot AD'=AF\cdot AB \]
So $BCEF$ is cyclic.
\end{proof}
\begin{claim*}
Line $ZD$ is tangent to the circles $(BCD)$ and $(DEF)$
where $Z=EF\cap BC$.
\end{claim*}
\begin{proof}
Let $\angle CAD=\angle BAD=\alpha$, $\angle BCD=\beta$,
$\angle DBC=\gamma$, $\angle ACD=\phi$,
$\angle ABD=\epsilon$.
From $\triangle ABC$ we have
$2\alpha+\beta+\gamma+\phi+\epsilon=180^\circ$.
Let $\ell$ be a line tangent to $(BCD)$ and $K$ be a
point on it in the same side of $AD$ as $C$ and
$L=AD\cap BC$. From our labeling we have,
\begin{align*}
\angle AFE &= \beta + \phi \qquad \angle BFD =
\alpha + \gamma \qquad \angle DFE = \alpha + \phi
\qquad \angle CDL = \alpha + \phi
\end{align*}
Now $\angle CDJ = 180^\circ - \gamma - \beta - (\alpha + \phi) = \alpha + \epsilon$.
So $\angle DFE = \angle EDK = \alpha + \epsilon$, which
means $\ell$ is also tangent to $(DEF)$. Now by the
radical center theorem we have $\ell$ passes through
$Z$.
\end{proof}
Let $M$ be the Miquel point of the cyclic quadrilateral
$BCEF$. From the Miquel configuration we have $A$, $M$, $Z$
are collinear and $(AFEM)$, $(ZCEM)$ are cyclic.
\begin{claim*}
Points $B$, $X$, $M$, $E$ are cyclic.
\end{claim*}
\begin{proof}
Notice that $\angle EMB = 180^\circ - \angle AMB -\angle EMZ$
$=$ $180^\circ - 2\angle ACB = \angle EXB$.
\end{proof}
Let $N$ be the other intersection of circles $(ACD)$ and
$(DEX)$ and let $R$ be the intersection of $AC$ and $BM$.
\begin{center}
\begin{asy}
/*
Converted from GeoGebra by User:Azjps using Evan's magic cleaner
https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py
*/
import graph;
size(14cm);
pen ffxfqq = rgb(1.,0.49803,0.);
pen qqwuqq = rgb(0.,0.39215,0.);
pen ffdxqq = rgb(1.,0.84313,0.);
pen qqffff = rgb(0.,1.,1.);
draw((12.,-7.5)--(3.88816,0.36923), linewidth(0.4) + red);
draw((3.88816,0.36923)--(0.5,-7.5), linewidth(0.4) + red);
draw((2.73129,-2.31767)--(4.74594,-3.92841), linewidth(0.4));
draw((4.74594,-3.92841)--(5.47993,-1.17493), linewidth(0.4));
draw(circle((6.25,-6.90419), 5.78078), linewidth(0.4) + red);
draw((6.25,5.85474)--(12.,-7.5), linewidth(0.4));
draw((-9.73375,-7.5)--(5.47993,-1.17493), linewidth(0.4) + red);
draw((-9.73375,-7.5)--(12.,-7.5), linewidth(0.4) + red);
draw((-9.73375,-7.5)--(4.74594,-3.92841), linewidth(0.4));
draw(circle((0.03259,-2.63473), 4.88766), linewidth(0.4) + ffxfqq);
draw(circle((6.84056,0.75675), 5.13208), linewidth(0.4) + qqwuqq);
draw(circle((13.09242,3.87122), 11.42357), linewidth(0.4) + ffxfqq);
draw(circle((6.25,-5.31169), 6.15233), linewidth(0.4) + red);
draw(circle((3.88816,-1.22327), 1.59250), linewidth(0.4) + red);
draw((3.88816,0.36923)--(-9.73375,-7.5), linewidth(0.4));
draw(circle((9.81689,-0.52472), 7.30892), linewidth(0.4) + ffdxqq);
draw((2.50861,-0.42772)--(12.,-7.5), linewidth(0.4));
draw((1.83901,1.90685)--(4.74594,-3.92841), linewidth(0.4));
draw((6.25,5.85474)--(3.88816,0.36923), linewidth(0.4));
draw((0.5,-7.5)--(4.74594,-3.92841), linewidth(0.4));
draw((4.74594,-3.92841)--(12.,-7.5), linewidth(0.4));
draw(shift((-9.73375,-7.5))*xscale(14.91368)*yscale(14.91368)*arc((0,0),1,3.18577,50.41705), linewidth(0.4) + dotted + qqffff);
dot((12.,-7.5),linewidth(3.pt));
label("$B$", (12.47142,-8.47771), NE);
dot((0.5,-7.5),linewidth(3.pt));
label("$C$", (-0.29392,-8.47771), NE);
dot((3.88816,0.36923),linewidth(3.pt));
label("$A$", (3.42645,1.11336), NE);
dot((6.25,5.85474),linewidth(3.pt));
label("$X$", (5.78156,6.47208), NE);
dot((4.74594,-3.92841),linewidth(3.pt));
label("$D$", (4.65520,-5.03038), NE);
dot((5.47993,-1.17493),linewidth(3.pt));
label("$F$", (5.64503,-0.86628), NE);
dot((2.73129,-2.31767),linewidth(3.pt));
label("$E$", (1.58333,-2.43635), NE);
dot((-9.73375,-7.5),linewidth(3.pt));
label("$Z$", (-10.77243,-8.20465), NE);
dot((2.50861,-0.42772),linewidth(3.pt));
label("$M$", (1.20788,-0.21777), NE);
dot((1.83901,1.90685),linewidth(3.pt));
label("$N$", (0.79829,2.47864), NE);
dot((3.29328,-1.01240),linewidth(3.pt));
label("$R$", (3.63125,-1.10520), NE);
\end{asy}
\end{center}
\begin{claim*}
Points $B$, $D$, $M$, $N$ are cyclic.
\end{claim*}
\begin{proof}
By power of point we have
\[
\opname{Pow}(R, (ACD)) = RC \cdot RA = RM \cdot RB
= RE \cdot RX = \opname{Pow}(R, (DEX)).
\]
Hence $R$ lies on the radical axis of $(ACD)$ and
$(DEX)$, so $N$, $R$, $D$ are collinear. Also
\[ RN \cdot RD = RA \cdot RC = RM \cdot RB \] So $BDMN$
is cyclic.
\end{proof}
Notice that $(ACD)$, $(BDMN)$, $(DEX)$ are coaxial so their
centers are collinear. Now we just need to prove the
centers of $(ACD)$, $(BDMN)$ and $Z$ are collinear. To
prove this, take a circle $\omega$ with radius $ZD$
centered at $Z$. Notice that by power of point
\[ ZC \cdot ZB = ZD^2 = ZE \cdot ZF = ZM \cdot ZA \]
which means inversion circle $\omega$ swaps $(ACD)$ and $(BDMN)$.
So the centers of $(ACD)$ and $(BDMN)$ must
have to be collinear with the center of inversion circle, as desired.
|
IMO-2021-notes_4
|
Let $\Gamma$ be a circle with center $I$, and $ABCD$ a convex quadrilateral
such that each of the segments $AB$, $BC$, $CD$ and $DA$ is tangent to $\Gamma$.
Let $\Omega$ be the circumcircle of the triangle $AIC$.
The extension of $BA$ beyond $A$ meets $\Omega$ at $X$,
and the extension of $BC$ beyond $C$ meets $\Omega$ at $Z$.
The extensions of $AD$ and $CD$ beyond $D$ meet $\Omega$ at $Y$ and $T$, respectively.
Prove that
\[ AD + DT + TX + XA = CD + DY + YZ + ZC. \]
|
Let $PQRS$ be the contact points of $\Gamma$ an $\ol{AB}$, $\ol{BC}$,
$\ol{CD}$, $\ol{DA}$.
\begin{center}
\begin{asy}
/*
Converted from GeoGebra by User:Azjps using Evan's magic cleaner
https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py
*/
pair I = (0.,0.);
pair P = (-0.40442,0.91457);
pair Q = (-0.35670,-0.93421);
pair R = (0.99998,0.00468);
pair S = (0.70710,0.70710);
pair A = (0.22244,1.19177);
pair B = (-2.62593,-0.06777);
pair C = (1.00681,-1.45484);
pair D = (0.99806,0.41615);
pair X = (2.38106,2.14630);
pair Z = (1.47042,-1.63185);
pair Y = (2.86072,-1.44650);
pair T = (0.99083,1.96044);
pair E = (0.99283,1.53243);
pair F = (4.01929,-2.60507);
size(12.41979cm);
pen qqwuqq = rgb(0.,0.39215,0.);
pen fuqqzz = rgb(0.95686,0.,0.6);
pen zzttqq = rgb(0.6,0.2,0.);
pen cqcqcq = rgb(0.75294,0.75294,0.75294);
draw(P--Q--R--S--cycle, linewidth(0.6) + zzttqq);
draw(circle(I, 1.), linewidth(0.6) + qqwuqq);
draw(circle((1.92609,0.25714), 1.94318), linewidth(0.6) + fuqqzz);
draw(P--Q, linewidth(0.6) + zzttqq);
draw(Q--R, linewidth(0.6) + zzttqq);
draw(R--S, linewidth(0.6) + zzttqq);
draw(S--P, linewidth(0.6) + zzttqq);
draw(circle((0.50340,-0.72742), 0.88462), linewidth(0.6) + fuqqzz);
draw(circumcircle(I,P,S), dotted + fuqqzz);
draw(A--F, linewidth(0.6) + qqwuqq);
draw(B--X, linewidth(0.6) + qqwuqq);
draw(B--F, linewidth(0.6) + qqwuqq);
draw(C--T, linewidth(0.6) + qqwuqq);
dot("$I$", I, dir(160));
dot("$P$", P, dir((-17.392, 6.881)));
dot("$Q$", Q, dir((-17.176, -15.513)));
dot("$R$", R, dir((2.333, 5.318)));
dot("$S$", S, dir((2.248, 5.460)));
dot("$A$", A, dir((-11.357, 7.426)));
dot("$B$", B, dir((-5.839, 9.792)));
dot("$C$", C, dir((2.205, -22.196)));
dot("$D$", D, dir(45));
dot("$X$", X, dir(80));
dot("$Z$", Z, dir((-3.145, -21.676)));
dot("$Y$", Y, dir(90));
dot("$T$", T, dir((-10.053, 11.473)));
dot("$E$", E, dir((10.253, -19.990)));
dot("$F$", F, dir((2.446, 4.708)));
\end{asy}
\end{center}
\begin{claim*}
We have $\triangle IQZ \cong \triangle IRT$.
Similarly, $\triangle IPX \cong \triangle ISY$.
\end{claim*}
\begin{proof}
By considering $(CQIR)$ and $(CITZ)$,
there is a spiral similarity similarity
mapping $\triangle IQZ$ to $\triangle IRT$.
Since $IQ = IR$, it is in fact a congruence.
\end{proof}
This congruence essentially solves the problem.
First, it implies:
\begin{claim*}
$TX = YZ$.
\end{claim*}
\begin{proof}
Because we saw $IX = IY$ and $IT = IZ$.
\end{proof}
Then, we can compute
\begin{align*}
AD + DT + XA
&= AD + (RT - RD) + (XP-AP) \\
&= (AD-RD-AP) + RT + XP = RT + XP
\end{align*}
and
\begin{align*}
CD + DY + ZC &= CD + (SY-SD) + (ZQ-QC) \\
&= (CD-SD-QC) + SY + ZQ = SY + ZQ
\end{align*}
but $ZQ = RT$ and $XP = SY$, as needed.
|
IMO-2021-notes_5
|
Two squirrels, Bushy and Jumpy, have collected $2021$ walnuts for the winter.
Jumpy numbers the walnuts from $1$ through $2021$, and digs $2021$ little holes
in a circular pattern in the ground around their favourite tree.
The next morning Jumpy notices that Bushy had placed one walnut into each hole,
but had paid no attention to the numbering.
Unhappy, Jumpy decides to reorder the walnuts by performing a sequence of 2021 moves.
In the $k$th move, Jumpy swaps the positions of the two walnuts adjacent to walnut $k$.
Prove that there exists a value of $k$ such that, on the $k$th move,
Jumpy swaps some walnuts $a$ and $b$ such that $a<k<b$.
|
Assume for contradiction no such $k$ exists.
We will use a so-called ``threshold trick''.
This process takes exactly $2021$ steps.
Right after the $k$th move, we consider a situation where
we color walnut $k$ red as well, so at the $k$th step there are $k$ ones.
For brevity, a non-red walnut is called black.
An example is illustrated below with $2021$ replaced by $6$.
\begin{center}
\begin{asy}
size(14cm);
dotfactor *= 1.5;
picture base;
picture pA, pB, pC, pD, pE, pF, pG;
pen old = black;
pen used = mediumred;
pen active = red;
draw(pA, unitcircle);
label(pA, "Initial", origin);
dot(pA, "$1$", dir( 0), dir( 0), old);
dot(pA, "$4$", dir( 60), dir( 60), old);
dot(pA, "$2$", dir(120), dir(120), old);
dot(pA, "$5$", dir(180), dir(180), old);
dot(pA, "$3$", dir(240), dir(240), old);
dot(pA, "$6$", dir(300), dir(300), old);
draw(pB, unitcircle);
dot(pB, "$\boxed{1}$", dir( 0), dir( 0), active);
dot(pB, "$6$", dir( 60), dir( 60), old);
dot(pB, "$2$", dir(120), dir(120), old);
dot(pB, "$5$", dir(180), dir(180), old);
dot(pB, "$3$", dir(240), dir(240), old);
dot(pB, "$4$", dir(300), dir(300), old);
draw(pB, dir(60)--dir(300), blue+dotted, Arrows(TeXHead), Margins);
draw(pC, unitcircle);
dot(pC, "$1$", dir( 0), dir( 0), used);
dot(pC, "$5$", dir( 60), dir( 60), old);
dot(pC, "$\boxed{2}$", dir(120), dir(120), active);
dot(pC, "$6$", dir(180), dir(180), old);
dot(pC, "$3$", dir(240), dir(240), old);
dot(pC, "$4$", dir(300), dir(300), old);
draw(pC, dir(60)--dir(180), blue+dotted, Arrows(TeXHead), Margins);
draw(pD, unitcircle);
dot(pD, "$1$", dir( 0), dir( 0), used);
dot(pD, "$5$", dir( 60), dir( 60), old);
dot(pD, "$2$", dir(120), dir(120), used);
dot(pD, "$4$", dir(180), dir(180), old);
dot(pD, "$\boxed{3}$", dir(240), dir(240), active);
dot(pD, "$6$", dir(300), dir(300), old);
draw(pD, dir(180)--dir(300), blue+dotted, Arrows(TeXHead), Margins);
draw(pE, unitcircle);
dot(pE, "$1$", dir( 0), dir( 0), used);
dot(pE, "$5$", dir( 60), dir( 60), old);
dot(pE, "$3$", dir(120), dir(120), used);
dot(pE, "$\boxed{4}$", dir(180), dir(180), active);
dot(pE, "$2$", dir(240), dir(240), used);
dot(pE, "$6$", dir(300), dir(300), old);
draw(pE, dir(120)--dir(240), blue+dotted, Arrows(TeXHead), Margins);
draw(pF, unitcircle);
dot(pF, "$1$", dir( 0), dir( 0), used);
dot(pF, "$\boxed{5}$", dir( 60), dir( 60), active);
dot(pF, "$3$", dir(120), dir(120), used);
dot(pF, "$4$", dir(180), dir(180), used);
dot(pF, "$2$", dir(240), dir(240), used);
dot(pF, "$6$", dir(300), dir(300), old);
draw(pF, dir(120)--dir(0), blue+dotted, Arrows(TeXHead), Margins);
draw(pG, unitcircle);
dot(pG, "$1$", dir( 0), dir( 0), used);
dot(pG, "$5$", dir( 60), dir( 60), used);
dot(pG, "$3$", dir(120), dir(120), used);
dot(pG, "$4$", dir(180), dir(180), used);
dot(pG, "$2$", dir(240), dir(240), used);
dot(pG, "$\boxed{6}$", dir(300), dir(300), active);
draw(pG, dir(60)--dir(240), blue+dotted, Arrows(TeXHead), Margins);
add(shift(-3,3)*pA);
add(shift(0,3)*pB);
add(shift(3,3)*pC);
add(shift(6,3)*pD);
add(shift(0,0)*pE);
add(shift(3,0)*pF);
add(shift(6,0)*pG);
\end{asy}
\end{center}
\begin{claim*}
At each step, the walnut that becomes red is between two non-red or two red walnuts.
\end{claim*}
\begin{proof}
By definition.
\end{proof}
On the other hand, if there are $2021$ walnuts,
one obtains a parity obstruction to this simplified process:
\begin{claim*}
After the first step, there is always a consecutive block of black walnuts positive even length.
\end{claim*}
\begin{proof}
After the first step, there is a block of $2020$ black walnuts.
Thereafter, note that a length $2$ block of black walnuts can never be changed.
Meanwhile for even lengths at least $4$, if one places a red walnut inside it,
the even length block splits into an odd length block and an even length block.
\end{proof}
\begin{remark*}
The statement is true with $2021$ replaced by any odd number, and false for any even number.
The motivation comes from the following rephrasing of the problem:
\begin{quote}
Start with all $0$'s and at each step change a $0$ between
two matching numbers from a $0$ to a $1$.
\end{quote}
Although the coloring (or $0$/$1$) argument may appear to lose information
at first, I think it should be \emph{equivalent} to the original process;
the ``extra'' information comes down to the choice of
which walnut to color red at each step.
\end{remark*}
|
IMO-2021-notes_6
|
Let $m \ge 2$ be an integer,
$A$ a finite set of integers (not necessarily positive)
and $B_1$, $B_2$, \dots, $B_m$ subsets of $A$.
Suppose that, for every $k=1,2,\dots,m$,
the sum of the elements of $B_k$ is $m^k$.
Prove that $A$ contains at least $\frac{m}{2}$ elements.
\end{enumerate}
|
If $0 \le X < m^{m+1}$ is a multiple of $m$, then write it in base $m$ as
\[ X = \sum_{i=1}^m c_i m^i \qquad c_i \in \{0,1,2,\dots,m-1\} \]
Then swapping the summation to over $A$ through the $B_i$'s gives
\[ X = \sum_{i = 1}^n \left( \sum_{b \in B_i} b \right) c_i
= \sum_{a \in A} f_a(X) a
\quad\text{where}\quad
f_a(X) \coloneq \sum_{i : a \in B_i} c_i.
\]
Evidently, $0 \le f_a(X) \le n(m-1)$ for any $a$ and $X$.
So, setting $|A| = n$, the right-hand side of the display takes on at most
$\left( n(m-1) + 1 \right)^n$ distinct values.
This means
\[ m^m \le \left( n(m-1) \right)^n \]
which implies $n \ge m/2$.
\begin{remark*}
[Motivation comments from USJL]
In linear algebra terms,
we have some $n$-dimensional 0/1 vectors $\vec{v_1}$, \dots, $\vec{v_m}$
and an $n$-dimensional vector $\vec a$
such that $\vec{v_i} \cdot \vec a = m^i$ for $i=1, \dots, m$.
The intuition is that if $n$ is too small,
then there should be lots of linear dependences between $\vec{v_i}$.
In fact, \emph{Siegel's lemma} is a result that says,
if there are many more vectors than the dimension of the ambient space,
there exist linear dependences whose coefficients are not-too-big integers.
On the other hand, any linear dependence between $m$, $m^2$, \dots, $m^m$
is going to have coefficients that are pretty big;
at least one of them needs to exceed $m$.
Applying Siegel's lemma turns out to solve the problem
(and is roughly equivalent to the solution above).
\end{remark*}
\begin{remark*}
In \url{https://aops.com/community/p23185192},
\texttt{dgrozev} shows the stronger bound
$n \ge \left(\frac{2}{3}+\frac{c}{\log m} \right)m$ elements,
for some absolute constant $c > 0$.
\end{remark*}
|
IMO-2022-notes_1
|
The Bank of Oslo issues two types of coin: aluminum (denoted $A$) and bronze
(denoted $B$). Marianne has $n$ aluminum coins and $n$ bronze coins arranged in a
row in some arbitrary initial order.
A chain is any subsequence of consecutive coins of the same type.
Given a fixed positive integer $k \leq 2n$,
Gilberty repeatedly performs the following operation:
he identifies the longest chain containing the $k$\ts{th} coin from the left
and moves all coins in that chain to the left end of the row.
For example, if $n=4$ and $k=4$, the process starting
from the ordering $AABBBABA$ would be
$AABBBABA \to BBBAAABA \to AAABBBBA \to BBBBAAAA \to \dotsb$.
Find all pairs $(n,k)$ with $1 \leq k \leq 2n$
such that for every initial ordering,
at some moment during the process,
the leftmost $n$ coins will all be of the same type.
|
Answer: $n \le k \le \left\lceil \frac 32 n \right\rceil$.
Call a maximal chain a \emph{block}.
Then the line can be described as a sequence of blocks: it's one of:
\begin{align*}
\underbrace{A\dots A}_{e_1}
\underbrace{B\dots B}_{e_2}
\underbrace{A\dots A}_{e_3}
\dots
\underbrace{A\dots A}_{e_m} & \text{ for odd $m$} \\
\underbrace{A\dots A}_{e_1}
\underbrace{B\dots B}_{e_2}
\underbrace{A\dots A}_{e_3}
\dots
\underbrace{B\dots B}_{e_m} & \text{ for even $m$}
\end{align*}
or the same thing with the roles of $A$ and $B$ flipped.
The main claim is the following:
\begin{claim*}
The number $m$ of blocks will never increase after an operation.
Moreover, it stays the same if and only if
\begin{itemize}
\ii $k \le e_1$; or
\ii $m$ is even and $e_m \ge 2n+1-k$.
\end{itemize}
\end{claim*}
\begin{proof}
This is obvious, just run the operation and see!
\end{proof}
The problem asks for which values of $k$ we always reach $m=2$ eventually;
we already know that it's non-increasing.
We consider a few cases:
\begin{itemize}
\ii If $k < n$, then any configuration with $e_1 = n-1$ will never change.
\ii If $k > \left\lceil 3n/2 \right\rceil$,
then take $m=4$ and $e_1 = e_2 = \left\lfloor n/2 \right\rfloor$
and $e_3 = e_4 = \left\lceil n/2 \right\rceil$.
This configuration retains $m=4$ always:
the blocks simply rotate.
\ii Conversely, suppose $k \ge n$ has the property that $m > 2$ stays fixed.
If after the first three operations $m$ hasn't changed,
then we must have $m \ge 4$ even, and $e_m, e_{m-1}, e_{m-2} \ge 2n+1 - k$.
Now,
\[ n \ge e_m + e_{m-2} \ge 2(2n+1-k) \implies k \ge \frac 32 n + 1 \]
so this completes the proof.
\end{itemize}
|
IMO-2022-notes_2
|
Find all functions $f \colon \RR^+ \to \RR^+$ such that for each $x \in \RR^+$,
there is exactly one $y \in \RR^+$ satisfying \[ xf(y)+yf(x) \leq 2. \]
|
The answer is $f(x) \equiv 1/x$ which obviously works (here $y=x$).
For the converse, assume we have $f$ such that
each $x \in \RR^+$ has a \emph{friend} $y$ with $xf(y)+yf(x)\le2$.
By symmetry $y$ is also the friend of $x$.
\begin{claim*}
In fact every number is its own friend.
\end{claim*}
\begin{proof}
Assume for contradiction $a \neq b$ are friends.
Then we know that $af(a) + af(a) > 2 \implies f(a) > \frac 1a$.
Analogously, $f(b) > \frac 1b$.
However, we then get
\[ 2 \ge a f(b) + b f(a) > \frac ab + \frac ba \overset{\text{AMGM}}{\ge} 2 \]
which is impossible.
\end{proof}
The problem condition now simplifies to saying
\[ f(x) \le \frac1x \text{ for all $x$}, \qquad
xf(y) + yf(x) > 2 \text{ for all $x \neq y$}. \]
In particular, for any $x>0$ and $\eps > 0$ we have
\begin{align*}
2 &< xf(x+\eps) + (x+\eps)f(x) \le \frac{x}{x+\eps} + (x+\eps) f(x) \\
\implies f(x) &> \frac{x+2\eps}{(x+\eps)^2}
= \frac{1}{x + \frac{\eps^2}{x+2\eps}}.
\end{align*}
Since this holds for all $\eps > 0$ this forces $f(x) \ge \frac1x$ as well.
We're done.
\begin{remark*}
Alternatively, instead of using $x+\eps$,
it also works to consider $y = \frac{1}{f(x)}$.
For such a $y$, we would have
\[ xf\left( \frac{1}{f(x)} \right) + \frac{1}{f(x)} \cdot f(x)
= xf\left( \frac{1}{f(x)} \right) + 1
\leq x \cdot f(x) + 1 \leq 1 + 1 = 2 \]
which gives a similar contradiction.
\end{remark*}
|
IMO-2022-notes_3
|
Let $k$ be a positive integer and let $S$ be a finite set of odd prime numbers.
Prove that there is at most one way (up to rotation and reflection)
to place the elements of $S$ around the circle such that the product
of any two neighbors is of the form $x^2+x+k$ for some positive integer $x$.
|
We replace ``positive integer $x$'' with ``nonnegative integer $x$'',
and say numbers of the form $x^2+x+k$ are \emph{good}.
We could also replace ``nonnegative integer $x$'' with ``integer $x$''
owing to the obvious map $x \mapsto 1-x$.
\begin{claim*}
If $p$ is an odd prime, there are at most two odd primes $q$ and $r$
less than $p$ for which $pq = x^2+x+k$ and $pr = y^2+y+k$ are good.
Moreover, if the above occurs and $x,y \ge 0$,
then $x+y+1=p$ and $xy \equiv k \pmod p$.
\end{claim*}
\begin{proof}
The equation $T^2+T+k \equiv 0 \pmod{p}$ has at most two solutions
modulo $p$, i.e.\ at most two solutions in the interval $[0,p-1]$.
Because $0 \le x,y < p$ from $p > \max(q,r)$ and $k > 0$,
the first half follows.
For the second half,
Vieta also gives $x+y \equiv -1 \pmod p$ and $xy \equiv k \pmod p$,
and we know $0 < x+y < 2p$.
\end{proof}
\begin{claim*}
If two such primes do exist as above, then $qr$ is also good (!).
\end{claim*}
\begin{proof}
Let $pq = x^2+x+k$ and $pr = y^2+y+k$ for $x,y \ge 0$ as before.
Fix $\alpha \in \CC$ such that $\alpha^2 + \alpha + k = 0$;
then for any $n \in \ZZ$, we have
\[ n^2 + n + k = \opname{Norm}(n-\alpha). \]
Hence
\[
pq \cdot pr = \opname{Norm}\Big((x-\alpha)(y-\alpha)\Big)
= \opname{Norm}\Big( (xy-k) - (x+y+1)\alpha\Big)
\]
But $\opname{Norm}(p) = p^2$,
so combining with the second half of the previous claim gives
\[ qr = \opname{Norm}(\frac1p(xy-k)-\alpha) \] as needed.
\end{proof}
These two claims imply the claim directly by induction on $|S|$,
since one can now delete the largest element of $S$.
\begin{remark*}
To show that the condition is not vacuous,
the author gave a ring of $385$ primes for $k=41$;
see \url{https://aops.com/community/p26068963}.
\end{remark*}
|
IMO-2022-notes_4
|
Let $ABCDE$ be a convex pentagon such that $BC=DE$.
Assume that there is a point $T$ inside $ABCDE$
with $TB=TD$, $TC=TE$ and $\angle ABT = \angle TEA$.
Let line $AB$ intersect lines $CD$ and $CT$ at points $P$ and $Q$, respectively.
Assume that the points $P$, $B$, $A$, $Q$ occur on their line in that order.
Let line $AE$ intersect $CD$ and $DT$ at points $R$ and $S$, respectively.
Assume that the points $R$, $E$, $A$, $S$ occur on their line in that order.
Prove that the points $P$, $S$, $Q$, $R$ lie on a circle.
|
The conditions imply
\[ \triangle BTC \cong \triangle DTE,
\qquad\text{and}\qquad
\triangle BTY \overset{-}{\sim} \triangle ETX. \]
Define $K = \ol{CT} \cap \ol{AE}$, $L = \ol{DT} \cap \ol{AB}$,
$X = \ol{BT} \cap \ol{AE}$, $Y = \ol{ET} \cap \ol{BY}$.
\begin{center}
\begin{asy}
size(12cm);
pair Y = dir(116.9642725);
pair X = dir(76.9642725);
pair B = dir(175.9642725);
pair E = dir(24.9642725);
pair A = extension(X, E, B, Y);
pair T = extension(X, B, E, Y);
pair D = T+(B-T)*dir(83);
pair C = T+(E-T)*dir(-83);
pair Q = extension(C, T, A, B);
pair S = extension(D, T, A, E);
pair P = extension(C, D, A, B);
pair R = extension(C, D, A, E);
pair L = extension(D, T, A, B);
pair K = extension(C, T, A, E);
filldraw(B--T--C--cycle, opacity(0.1)+yellow, 0.1+yellow);
filldraw(D--T--E--cycle, opacity(0.1)+yellow, 0.1+yellow);
filldraw(B--T--Q--cycle, opacity(0.1)+lightred, red);
filldraw(E--T--S--cycle, opacity(0.1)+lightcyan, blue);
draw(T--X, blue);
draw(T--Y, red);
draw(circumcircle(L, K, Q), gray);
draw(circumcircle(P, Q, R), dotted);
draw(T--D, red+dashed);
draw(T--C, blue+dashed);
draw(K--L, gray);
draw(B--P--R--E, gray);
draw(B--C, dotted);
draw(D--E, dotted);
write(angle(D-C));
dot("$Y$", Y, dir(Y));
dot("$X$", X, dir(60));
dot("$B$", B, dir(B));
dot("$E$", E, dir(E));
dot("$A$", A, dir(A));
dot("$T$", T, 1.8*dir(280));
dot("$D$", D, dir(270));
dot("$C$", C, dir(270));
dot("$Q$", Q, dir(Q));
dot("$S$", S, dir(S));
dot("$P$", P, dir(P));
dot("$R$", R, dir(R));
dot("$L$", L, 1.4*dir(200));
dot("$K$", K, 1.4*dir(355));
/* TSQ Source:
!size(12cm);
Y = dir 116.9642725
X = dir 76.9642725 R60
B = dir 175.9642725
E = dir 24.9642725
A = extension X E B Y
T = extension X B E Y 1.8R280
D = T+(B-T)*dir(83) R270
C = T+(E-T)*dir(-83) R270
Q = extension C T A B
S = extension D T A E
P = extension C D A B
R = extension C D A E
L = extension D T A B 1.4R200
K = extension C T A E 1.4R355
B--T--C--cycle 0.1 yellow / 0.1 yellow
D--T--E--cycle 0.1 yellow / 0.1 yellow
B--T--Q--cycle 0.1 lightred / red
E--T--S--cycle 0.1 lightcyan / blue
T--X blue
T--Y red
circumcircle L K Q gray
circumcircle P Q R dotted
T--D red dashed
T--C blue dashed
K--L gray
B--P--R--E gray
B--C dotted
D--E dotted
!write(angle(D-C));
*/
\end{asy}
\end{center}
\begin{claim*}
[Main claim]
We have
\[ \triangle BTQ \overset{-}{\sim} \triangle ETS,
\qquad\text{and}\qquad
BY:YL:LQ = EX:XK:KS. \]
In other words, $TBYLQ \overset{-}{\sim} TEXKS$.
\end{claim*}
\begin{proof}
We know $\triangle BTY \overset{-}{\sim} \triangle ETX$.
Also, $\dang BTL = \dang BTD = \dang CTE = \dang KTE$
and $\dang BTQ = \dang BTC = \dang DTE = \dang STE$.
\end{proof}
It follows from the claim that:
\begin{itemize}
\ii $TL/TQ = TK/TS$, ergo $TL \cdot TS = TK \cdot TQ$,
so $KLSQ$ is cyclic; and
\ii $TC/TK = TE/TK = TB/TL = TD/TL$, so $\ol{KL} \parallel \ol{PCDR}$.
\end{itemize}
With these two bullets, we're done by Reim theorem.
|
IMO-2022-notes_5
|
Find all triples $(a,b,p)$ of positive integers with $p$ prime and
\[ a^p=b!+p. \]
|
The answer is $(2,2,2)$ and $(3,4,3)$ only, which work.
In what follows we assume $a \ge 2$.
\begin{claim*}
We have $b \le 2p-2$, and hence $a < p^2$.
\end{claim*}
\begin{proof}
For the first half, assume first for contradiction that $b \ge 2p$.
Then $b!+p \equiv p \pmod{p^2}$, so $\nu_p(b!+p)=1$,
but $\nu_p(a^p)=1$ never occurs.
We can also rule out $b = 2p-1$ since that would give
\[ (2p-1)!+p = p \left[ (p-1)! (p+1)(p+2)\dots(2p-1) + 1 \right] \]
By Wilson theorem the inner bracket is $(-1)^2+1 \equiv 2 \pmod p$
exactly, contradiction for $p > 2$.
And when $p=2$, $3!+2=8$ is not a perfect square.
The second half follows as $a^p \le (2p-2)!+p < p^{2p}$.
(Here we used the crude estimate
$(2p-2)! = \prod_{k=1}^{p-1}k \cdot (2p-1-k) < (p(p-1))^{p-1}$).
\end{proof}
\begin{claim*}
We have $a \ge p$, and hence $b \ge p$.
\end{claim*}
\begin{proof}
For the first half, assume for contradiction that $p > a$.
Then
\[ b! + p = a^p \ge a^{p-1} + p \ge a^a + p > a! + p \implies b > a. \]
Then taking modulo $a$ now gives $0 \equiv 0 + p \pmod{a}$,
which is obviously impossible.
The second half follows from $b! = a^p-p \ge p! - p > (p-1)!$.
\end{proof}
\begin{claim*}
We have $a=p$ exactly.
\end{claim*}
\begin{proof}
We know $p \ge b$ hence $p \mid b!+p$, so let $a = pk$ for $k < p$.
Then $k \mid b!$ yet $k \nmid a^p-p$, contradiction.
\end{proof}
Let's get the small $p$ out of the way:
\begin{itemize}
\ii For $p=2$, checking $2 \le b \le 3$ gives $(a,b)=(2,2)$ only.
\ii For $p=3$, checking $3 \le b \le 5$ gives $(a,b)=(3,4)$ only.
\end{itemize}
Once $p \ge 5$, if $b! = p^p - p = p(p^{p-1}-1)$
then applying Zsigmondy gets a prime factor $q \equiv 1 \pmod{p-1}$
which divides $p^{p-1}-1$.
Yet $q \le b \le 2p-2$ and $q \neq p$, contradiction.
|
IMO-2022-notes_6
|
Let $n$ be a positive integer.
A \emph{Nordic square} is an $n \times n$ board
containing all the integers from $1$ to $n^2$
so that each cell contains exactly one number.
An \emph{uphill path} is a sequence of one or more cells such that:
\begin{enumerate}
\ii the first cell in the sequence is a \emph{valley},
meaning the number written is less than all its orthogonal neighbors,
\ii each subsequent cell in the sequence is orthogonally
adjacent to the previous cell, and
\ii the numbers written in the cells in the sequence are in increasing order.
\end{enumerate}
Find, as a function of $n$, the smallest possible total number
of uphill paths in a Nordic square.
\end{enumerate}
|
Answer: $2n^2-2n+1$.
\paragraph{Bound.}
The lower bound is the ``obvious'' one:
\begin{itemize}
\ii For any pair of adjacent cells, say $a > b$,
one can extend it to a downhill path (the reverse of an uphill path)
by walking downwards until one reaches a valley.
This gives $2n(n-1)=2n^2-2n$ uphill paths of length $\ge 2$.
\ii There is always at least one uphill path of length $1$,
namely the single cell $\{1\}$ (or indeed any valley).
\end{itemize}
\paragraph{Construction.}
For the construction, the ideas it build a tree $T$ on the grid
such that no two cells not in $T$ are adjacent.
An example of such a grid is shown below for $n=15$ with $T$ in yellow
and cells not in $T$ in black; it generalizes to any $3 \mid n$,
and then to any $n$ by deleting the last $n \bmod 3$ rows
and either/both of the leftmost/rightmost column.
\begin{center}
\begin{asy}
size(11cm);
defaultpen(fontsize(8pt));
filldraw(box( (0,0),(15,-15) ), black, black+2);
int L = 0;
void go(int x, int y) {
++L;
filldraw(shift(x,-y-1)*unitsquare, paleyellow,black);
label("$"+(string)L+"$", (x+0.5,-y-0.5));
}
for (int m=0; m<=2; ++m) {
go(6*m, 0);
go(6*m+1, 0);
go(6*m+2, 0);
for (int j=0; j<7; ++j) {
go(6*m+1, 2*j+1);
go(6*m+1, 2*j+2);
go(6*m, 2*j+2);
go(6*m+2, 2*j+2);
}
if (m != 2) {
go(6*m+3, 0);
go(6*m+3, 1);
go(6*m+4, 1);
go(6*m+5, 1);
go(6*m+5, 0);
for (int j=1; j<7; ++j) {
go(6*m+4, 2*j);
go(6*m+4, 2*j+1);
go(6*m+3, 2*j+1);
go(6*m+5, 2*j+1);
}
go(6*m+4,14);
}
}
\end{asy}
\end{center}
Place $1$ anywhere in $T$ and then place all the small numbers at most $|T|$
adjacent to previously placed numbers (example above).
Then place the remaining numbers outside $T$ arbitrarily.
By construction, as $1$ is the only valley, any uphill path must start from $1$.
And by construction, it may only reach a given pair of terminal cells in one
way, i.e.\ the downhill paths we mentioned are the only one.
End proof.
|
IMO-2023-notes_1
|
Determine all composite integers $n>1$ that satisfy the following property:
if $d_1 < d_2 < \dots < d_k$ are all the positive divisors of $n$ with
then $d_i$ divides $d_{i+1} + d_{i+2}$ for every $1 \leq i \leq k - 2$.
|
The answer is prime powers.
\paragraph{Verification that these work.}
When $n = p^e$, we get $d_i = p^{i-1}$.
The $i$\ts{th} relationship reads \[ p^{i-1} \mid p^i + p^{i+1} \]
which is obviously true.
\paragraph{Proof that these are the only answers.}
Conversely, suppose $n$ has at least two distinct prime divisors.
Let $p < q$ denote the two smallest ones,
and let $p^e$ be the largest power of $p$ which both divides $n$
and is less than $q$, hence $e \ge 1$.
Then the smallest factors of $n$ are $1$, $p$, \dots, $p^e$, $q$.
So we are supposed to have
\[ \frac{n}{q} \mid \frac{n}{p^e} + \frac{n}{p^{e-1}}
= \frac{(p+1)n}{p^e} \]
which means that the ratio
\[ \frac{q(p+1)}{p^e} \]
needs to be an integer, which is obviously not possible.
|
IMO-2023-notes_2
|
Let $ABC$ be an acute-angled triangle with $AB < AC$.
Let $\Omega$ be the circumcircle of $ABC$.
Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$.
The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E \neq A$.
The line through $D$ parallel to $BC$ meets line $BE$ at $L$.
Denote the circumcircle of triangle $BDL$ by $\omega$.
Let $\omega$ meet $\Omega$ again at $P \neq B$.
Prove that the line tangent to $\omega$ at $P$ meets line $BS$
on the internal angle bisector of $\angle BAC$.
|
\begin{claim*}
We have $LPS$ collinear.
\end{claim*}
\begin{proof}
Because $\dang LPB = \dang LDB = \dang CBD = \dang CBS = \dang SCB = \dang SPB$.
\end{proof}
Let $F$ be the antipode of $A$, so $AMFS$ is a rectangle.
\begin{claim*}
We have $PDF$ collinear. (This lets us erase $L$.)
\end{claim*}
\begin{proof}
Because $\dang SPD = \dang LPD = \dang LBD = \dang SBE = \dang FCS = \dang FPS$.
\end{proof}
Let us define $X = \ol{AM} \cap \ol{BS}$ and complete chord $\ol{PXQ}$.
We aim to show that $\ol{PXQ}$ is tangent to $(PDLB)$.
\begin{center}
\begin{asy}
/*
Converted from GeoGebra by User:Azjps using Evan's magic cleaner
https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py
*/
pair C = (0.5,0.);
pair B = (-4.5,0.);
pair A = (-3.54989,4.84867);
pair M = (-2.,-1.19129);
pair S = (-2.,5.24638);
pair X = (-3.07373,2.99308);
pair E = (-3.54989,-0.79358);
pair D = (-3.54989,1.99384);
pair L = (-6.88710,1.99384);
pair P = (-4.96929,3.27021);
pair Q = (1.20076,2.36814);
pair F = (-0.45010,-0.79358);
size(9cm);
pen qqffff = rgb(0.,1.,1.);
pen yqqqqq = rgb(0.50196,0.,0.);
pen zzttqq = rgb(0.6,0.2,0.);
pen ffxfqq = rgb(1.,0.49803,0.);
pen qqwuqq = rgb(0.,0.39215,0.);
draw(A--B--C--cycle, linewidth(0.6) + zzttqq);
draw(A--M, linewidth(0.6) + qqffff);
draw(B--S, linewidth(0.6) + qqffff);
draw(circle((-2.,2.02754), 3.21884), linewidth(0.6) + yqqqqq);
draw(A--B, linewidth(0.6) + zzttqq);
draw(B--C, linewidth(0.6) + zzttqq);
draw(C--A, linewidth(0.6) + zzttqq);
draw(circle((-5.21849,1.56567), 1.72266), linewidth(0.6) + ffxfqq);
draw(A--E, linewidth(0.6) + yqqqqq);
draw(D--L, linewidth(0.6) + qqwuqq);
draw(P--Q, linewidth(0.6) + ffxfqq);
draw(L--E, linewidth(0.6) + ffxfqq);
draw(E--Q, linewidth(0.6) + qqwuqq);
draw(E--F, linewidth(0.6) + yqqqqq);
draw(S--M, linewidth(0.6) + yqqqqq);
draw(A--F, linewidth(0.6));
draw(P--F, dashed);
draw(S--L, dashed);
dot("$C$", C, dir((2.343, -22.443)));
dot("$B$", B, dir((-19.721, -25.483)));
dot("$A$", A, dir((-11.383, 10.955)));
dot("$M$", M, dir((-8.308, -21.862)));
dot("$S$", S, dir((3.090, 6.140)));
dot("$X$", X, dir((3.315, 5.773)));
dot("$E$", E, dir((-12.903, -21.357)));
dot("$D$", D, dir(20));
dot("$L$", L, dir((-24.946, -1.451)));
dot("$P$", P, dir((-12.308, 13.017)));
dot("$Q$", Q, dir(20));
dot("$F$", F, dir((1.604, -19.837)));
\end{asy}
\end{center}
\begin{claim*}
[Main projective claim]
We have $XP = XA$.
\end{claim*}
\begin{proof}
Introduce $Y = \ol{PDF} \cap \ol{AM}$.
Note that
\[ -1 = (SM;EF) \overset{A}{=} (S,X;D,\ol{AF} \cap \ol{ES}) \overset{F}{=} (\infty X;YA) \]
where $\infty = \ol{AM} \cap \ol{SF}$ is at infinity (because $AMSF$ is a rectangle).
Thus, $XY = XA$.
\begin{center}
\begin{asy}
/*
Converted from GeoGebra by User:Azjps using Evan's magic cleaner
https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py
*/
pair C = (0.5,0.);
pair B = (-4.5,0.);
pair A = (-3.54989,4.84867);
pair M = (-2.,-1.19129);
pair S = (-2.,5.24638);
pair X = (-3.07373,2.99308);
pair E = (-3.54989,-0.79358);
pair D = (-3.54989,1.99384);
pair P = (-4.96929,3.27021);
pair Q = (1.20076,2.36814);
pair K = (10.67519,1.99384);
pair F = (-0.45010,-0.79358);
pair Y = (-2.59758,1.13749);
import graph;
size(9cm);
pen qqffff = rgb(0.,1.,1.);
pen yqqqqq = rgb(0.50196,0.,0.);
pen zzttqq = rgb(0.6,0.2,0.);
pen ffxfqq = rgb(1.,0.49803,0.);
pen qqwuqq = rgb(0.,0.39215,0.);
pen ffqqff = rgb(1.,0.,1.);
draw(circle((-2.,2.02754), 3.21884), linewidth(0.6) + yqqqqq);
draw(A--E, linewidth(0.6) + yqqqqq);
draw(E--F, linewidth(0.6) + yqqqqq);
draw(S--M, linewidth(0.6) + yqqqqq);
draw(A--F, linewidth(0.6));
draw(A--M, blue);
draw(S--D, blue);
draw(D--F, red);
draw(D--P, gray+dotted);
draw(P--Q, gray+dotted);
/*
draw(A--M, linewidth(0.6) + qqffff);
draw(B--S, linewidth(0.6) + qqffff);
draw(P--F, linewidth(0.6) + ffqqff);
draw(P--Q, linewidth(0.6) + ffxfqq);
*/
dot("$A$", A, dir((-10.485, 9.780)));
dot("$M$", M, dir((-7.272, -19.720)));
dot("$S$", S, dir((3.090, 5.242)));
dot("$X$", X, dir(310));
dot("$E$", E, dir((-11.866, -19.423)));
dot("$D$", D, dir(20));
dot("$P$", P, dir((-16.384, -0.247)));
dot("$F$", F, dir((1.466, -18.041)));
dot("$Y$", Y, dir((2.746, 5.774)));
dot(extension(A,F,B,S));
\end{asy}
\end{center}
Since $\triangle APY$ is also right, we get $XP = XA$.
\end{proof}
\begin{proof}[Alternative proof of claim without harmonic bundles,
from Solution 9 of the marking scheme]
With $Y = \ol{PDF} \cap \ol{AM}$ defined as before, note that
$\ol{AE} \parallel \ol{SM}$ and $\ol{AM} \parallel \ol{SF}$ (as $AMFS$ is a rectangle)
gives respectively the similar triangles
\[ \triangle AXD \sim \triangle MXS, \qquad \triangle XDY \sim \triangle SDF. \]
From this we conclude
\[ \frac{AX}{XD} = \frac{AX+XM}{XD+SX} = \frac{AM}{SD} = \frac{SF}{SD} = \frac{XY}{XD}. \]
So $AX = XY$ and as before we conclude $XP = XA$.
\end{proof}
From $XP = XA$, we conclude that $\arc{PM}$ and $\arc{AQ}$ have the same measure.
Since $\arc{AS}$ and $\arc{EM}$ have the same measure,
it follows $\arc{PE}$ and $\arc{SQ}$ have the same measure.
The desired tangency then follows from
\[ \dang QPL = \dang QPS = \dang PQE = \dang PFE = \dang PDL. \]
\begin{remark*}
[Logical ordering]
This solution is split into two phases:
the ``synthetic phase'' where we do a bunch of angle chasing, and the
``projective phase'' where we use cross-ratios because I like projective.
For logical readability (so we write in only one logical direction),
the projective phase is squeezed in two halves of the synthetic phase,
but during an actual solve it's expected to complete
the whole synthetic phase first (i.e.\ to reduce the problem to show $XP=XA$).
\end{remark*}
\begin{remark*}
There are quite a multitude of approaches for this problem;
the marking scheme for this problem at the actual IMO had 13 different solutions.
\end{remark*}
|
IMO-2023-notes_3
|
For each integer $k\geq 2$, determine all infinite sequences of positive integers
$a_1$, $a_2$, \dots\ for which there exists a polynomial $P$ of the form
\[ P(x)=x^k+c_{k-1}x^{k-1}+\dots + c_1 x+c_0, \]
where $c_0$, $c_1$, \dots, $c_{k-1}$ are non-negative integers, such that
\[ P(a_n)=a_{n+1}a_{n+2}\dotsm a_{n+k} \]
for every integer $n\geq 1$.
|
The answer is $a_n$ being an arithmetic progression.
Indeed, if $a_n = d(n-1) + a_1$ for $d \ge 0$ and $n \ge 1$, then
\[ a_{n+1} a_{n+2} \dots a_{n+k} = (a_n+d)(a_n+2d)\dots(a_n+kd) \]
so we can just take $P(x) = (x+d)(x+2d) \dots (x+kd)$.
The converse direction takes a few parts.
\begin{claim*}
Either $a_1 < a_2 < \dotsb$ or the sequence is constant.
\end{claim*}
\begin{proof}
Note that
\begin{align*}
P(a_{n-1}) &= a_{n}a_{n+1}\dotsm a_{n+k-1} \\
P(a_n) &= a_{n+1}a_{n+2}\dotsm a_{n+k} \\
\implies a_{n+k} &= \frac{P(a_n)}{P(a_{n-1})} \cdot a_n.
\end{align*}
Now the polynomial $P$ is strictly increasing over $\NN$.
So assume for contradiction there's an index $n$ such that $a_n < a_{n-1}$.
Then in fact the above equation shows $a_{n+k} < a_n < a_{n-1}$.
Then there's an index $\ell \in [n+1,n+k]$ such that
$a_\ell < a_{\ell-1}$, and also $a_\ell < a_n$.
Continuing in this way, we can an infinite descending subsequence of $(a_n)$,
but that's impossible because we assumed integers.
Hence we have $a_1 \le a_2 \le \dotsb$.
Now similarly, if $a_n = a_{n-1}$ for any index $n$, then $a_{n+k} = a_n$,
ergo $a_{n-1} = a_n = a_{n+1} = \dots = a_{n+k}$.
So the sequence is eventually constant, and then by downwards induction,
it is fully constant.
\end{proof}
\begin{claim*}
There exists a constant $C$ (depending only $P$, $k$)
such that we have $a_{n+1} \leq a_n + C$.
\end{claim*}
\begin{proof}
Let $C$ be a constant such that $P(x) < x^k + Cx^{k-1}$ for all $x \in \NN$
(for example $C = c_0 + c_1 + \dots + c_{k-1} + 1$ works).
We have
\begin{align*}
a_{n+k} &= \frac{P(a_n)}{a_{n+1} a_{n+2} \dots a_{n+k-1}} \\
&< \frac{P(a_n)}{(a_n+1)(a_n+2)\dots(a_n+k-1)} \\
&< \frac{a_n^k + C \cdot a_n^{k-1}}{(a_n+1)(a_n+2)\dots(a_n+k-1)} \\
&< a_n + C + 1. \qedhere
\end{align*}
\end{proof}
Assume henceforth $a_n$ is nonconstant, and hence unbounded.
For each index $n$ and term $a_n$ in the sequence,
consider the associated differences
$d_1 = a_{n+1} - a_n$, $d_2 = a_{n+2} - a_{n+1}$, \dots, $d_k = a_{n+k}-a_{n+k-1}$,
which we denote by
\[ \Delta(n) \coloneq (d_1, \dots, d_k).\]
This $\Delta$ can only take up to $C^k$ different values.
So in particular, some tuple $(d_1, \dots, d_n)$
must appear infinitely often as $\Delta(n)$; for that tuple, we obtain
\[ P(a_N) = (a_N+d_1)(a_N+d_1+d_2) \dots (a_N+d_1+\dots+d_k) \]
for infinitely many $N$.
But because of that, we actually must have
\[ P(X) = (X+d_1)(X+d_1+d_2) \dots (X+d_1+\dots+d_k). \]
However, this \emph{also} means that \emph{exactly} one output to $\Delta$
occurs infinitely often (because that output is determined by $P$).
Consequently, it follows that $\Delta$ is eventually constant.
For this to happen, $a_n$ must eventually coincide with an arithmetic
progression of some common difference $d$,
and $P(X) = (X+d)(X+2d) \dots (X+kd)$.
Finally, this implies by downwards induction that $a_n$ is
an arithmetic progression on all inputs.
|
IMO-2023-notes_4
|
Let $x_1$, $x_2$, \dots, $x_{2023}$ be pairwise different positive real numbers such that
\[ a_n = \sqrt{(x_1+x_2+\dots+x_n)
\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_n}\right)} \]
is an integer for every $n=1,2,\dots,2023$. Prove that $a_{2023} \geq 3034$.
|
Note that $a_{n+1} > \sqrt{\sum_1^n x_i \sum_1^n \frac{1}{x_i}} = a_n$ for all $n$,
so that $a_{n+1} \geq a_n + 1$.
Observe $a_1 = 1$.
We are going to prove that \[ a_{2m+1} \geq 3m+1 \qquad \text{for all } m \geq 0 \]
by induction on $m$, with the base case being clear.
We now present two variations of the induction.
The first shorter solution compares $a_{n+2}$ directly to $a_n$,
showing it increases by at least $3$.
Then we give a longer approach that compares $a_{n+1}$ to $a_n$,
and shows it cannot increase by $1$ twice in a row.
\paragraph{Induct-by-two solution.}
Let $u = \sqrt{\frac{x_{n+1}}{x_{n+2}}} \neq 1$.
Note that by using Cauchy-Schwarz with three terms:
\begin{align*}
a_{n+2}^2 &= \Bigg[ (x_1+\dots+x_n)+x_{n+1}+x_{n+2} \Bigg]
\Bigg[ \left(\frac{1}{x_1}+\dots+\frac{1}{x_n}\right)
+\frac{1}{x_{n+2}} + \frac{1}{x_{n+1}} \Bigg] \\
&\geq \left( \sqrt{ (x_1+\dots+x_n)\left(\frac{1}{x_1}+\dots+\frac{1}{x_n}\right)}
+ \sqrt{\frac{x_{n+1}}{x_{n+2}}} + \sqrt{\frac{x_{n+2}}{x_{n+1}}} \right)^2 \\
&= \left( a_n + u + \frac 1u \right)^2. \\
\implies a_{n+2} &\ge a_n + u + \frac 1u > a_n + 2
\end{align*}
where the last equality $u + \frac 1u > 2$ is by AM-GM, strict as $u \neq 1$.
It follows that $a_{n+2} \geq a_n + 3$, completing the proof.
\paragraph{Induct-by-one solution.}
The main claim is:
\begin{claim*}
It's impossible to have
$a_n = c$, $a_{n+1} = c+1$, $a_{n+2} = c+2$ for any $c$ and $n$.
\end{claim*}
\begin{proof}
Let $p = x_{n+1}$ and $q = x_{n+2}$ for brevity.
Let $s = \sum_1^n x_i$ and $t = \sum_1^n \frac{1}{x_n}$, so $c^2 = a_n^2 = st$.
From $a_n = c$ and $a_{n+1} = c+1$ we have
\begin{align*}
(c+1)^2 &= a_{n+1}^2 = (p+s)\left( \frac 1p+t \right) \\
&= st + pt + \frac1ps + 1 = c^2 + pt + \frac1ps + 1 \\
&\overset{\text{AM-GM}}{\geq} c^2 + 2\sqrt{st} + 1 = c^2 + 2\sqrt{c^2} + 1 = (c+1)^2.
\end{align*}
Hence, equality must hold in the AM-GM we must have exactly
\[ p t = \frac 1p s = c. \]
If we repeat the argument again on $a_{n+1}=c+1$ and $a_{n+2}=c+2$, then
\[ q \left( \frac 1p + t \right) = \frac 1q \left( p + s \right) = c + 1. \]
However this forces $\frac pq = \frac qp = 1$ which is impossible.
\end{proof}
|
IMO-2023-notes_5
|
Let $n$ be a positive integer.
A \emph{Japanese triangle} consists of $1 + 2 + \dots + n$ circles arranged in an
equilateral triangular shape such that for each $1 \le i \le n$,
the $i$\ts{th} row contains exactly $i$ circles, exactly one of which is colored red.
A \emph{ninja path} in a Japanese triangle is a sequence of $n$ circles
obtained by starting in the top row, then repeatedly going from a circle to
one of the two circles immediately below it and finishing in the bottom row.
Here is an example of a Japanese triangle with $n = 6$,
along with a ninja path in that triangle containing two red circles.
\begin{center}
\begin{asy}
size(4cm);
pair X = dir(240); pair Y = dir(0);
path c = scale(0.5)*unitcircle;
int[] t = {0,0,2,2,3,0};
for (int i=0; i<=5; ++i) {
for (int j=0; j<=i; ++j) {
filldraw(shift(i*X+j*Y)*c, (t[i]==j) ? lightred : white);
draw(shift(i*X+j*Y)*c);
}
}
draw((0,0)--(X+Y)--(2*X+Y)--(3*X+2*Y)--(4*X+2*Y)--(5*X+2*Y),linewidth(1.5));
path q = (3,-3sqrt(3))--(-3,-3sqrt(3));
draw(q,Arrows(TeXHead, 1));
label("$n = 6$", q, S);
\end{asy}
\end{center}
In terms of $n$, find the greatest $k$ such that in each Japanese triangle
there is a ninja path containing at least $k$ red circles.
|
The answer is
\[ k = \left\lfloor \log_2(n) \right\rfloor + 1. \]
\paragraph{Construction.}
It suffices to find a Japanese triangle for $n = 2^e-1$
with the property that at most $e$ red circles in any ninja path.
The construction shown below for $e=4$ obviously generalizes,
and works because in each of the sets $\{1\}$, $\{2,3\}$, $\{4,5,6,7\}$,
\dots, $\{2^{e-1},\dots,2^e-1\}$, at most one red circle can be taken.
(These sets are colored in different shades of red for visual clarity).
\begin{center}
\begin{asy}
size(8cm);
pair X = dir(240); pair Y = dir(0);
path c = scale(0.5)*unitcircle;
fill(shift(0*X+0*Y)*c, heavyred);
fill(shift(1*X+0*Y)*c, red);
fill(shift(2*X+2*Y)*c, red);
fill(shift(3*X+0*Y)*c, magenta);
fill(shift(4*X+2*Y)*c, magenta);
fill(shift(5*X+4*Y)*c, magenta);
fill(shift(6*X+6*Y)*c, magenta);
fill(shift(7*X+0*Y)*c, lightred);
fill(shift(8*X+2*Y)*c, lightred);
fill(shift(9*X+4*Y)*c, lightred);
fill(shift(10*X+6*Y)*c, lightred);
fill(shift(11*X+8*Y)*c, lightred);
fill(shift(12*X+10*Y)*c, lightred);
fill(shift(13*X+12*Y)*c, lightred);
fill(shift(14*X+14*Y)*c, lightred);
for (int i=0; i<15; ++i) {
for (int j=0; j<=i; ++j) {
draw(shift(i*X+j*Y)*c);
}
}
\end{asy}
\end{center}
\paragraph{Bound.}
Conversely, we show that in any Japanese triangle,
one can find a ninja path containing at least
\[ k = \left\lfloor \log_2(n) \right\rfloor + 1. \]
The following short solution was posted at \url{https://aops.com/community/p28134004},
apparently first found by the team leader for Iran.
We construct a rooted binary tree $T_1$ on the set of all circles as follows.
For each row, other than the bottom row:
\begin{itemize}
\ii Connect the red circle to both circles under it;
\ii White circles to the left of the red circle in its row are connected to the left;
\ii White circles to the right of the red circle in its row are connected to the right.
\end{itemize}
The circles in the bottom row are all leaves of this tree.
For example, the $n=6$ construction in the beginning gives the tree
shown on the left half of the figure below:
\begin{center}
\begin{asy}
size(11cm);
pair X = dir(240); pair Y = dir(0);
path c = scale(0.3)*unitcircle;
int[] t = {0,0,2,2,3,0};
for (int i=0; i<5; ++i) {
for (int j=0; j<=i; ++j) {
if (j <= t[i]) {
draw( (i*X+j*Y)--((i+1)*X+j*Y), blue+1.4);
}
if (j >= t[i]) {
draw( (i*X+j*Y)--((i+1)*X+(j+1)*Y), blue+1.4);
}
}
}
for (int i=0; i<=5; ++i) {
for (int j=0; j<=i; ++j) {
filldraw(shift(i*X+j*Y)*c, (t[i]==j) ? lightred : white, black);
}
}
label("$T_1$", 6*X+3*Y, fontsize(14pt));
transform sh = shift(7,0);
draw(sh*((5*X+0*Y)--(0*X+0*Y)--(2*X+2*Y)--(3*X+2*Y)--(4*X+3*Y)), blue+1.4);
for (int i=0; i<=5; ++i) {
filldraw(sh*shift(i*X+t[i]*Y)*c, lightred, black);
}
label("$T_2$", sh*(6*X+3*Y), fontsize(14pt));
\end{asy}
\end{center}
Now focus on only the red circles, as shown in the right half of the figure.
We build a new rooted tree $T_2$ where each red circle is joined to the
red circle below it if there was a path of (zero or more)
white circles in $T_1$ between them.
Then each red circle has at most $2$ direct descendants in $T_2$.
Hence the depth of the new tree $T_2$ exceeds $\log_2(n)$, which produces the desired path.
\paragraph{Another recursive proof of bound, communicated by Helio Ng.}
We give another proof that $\lfloor \log_2 n\rfloor + 1$ is always achievable.
Define $f(i, j)$ to be the maximum number of red circles contained in
the portion of a ninja path from $(1, 1)$ to $(i, j)$,
including the endpoints $(1, 1)$ and $(i, j)$.
(If $(i,j)$ is not a valid circle in the triangle, define $f(i, j)=0$ for convenience.)
An example is shown below with the values of $f(i,j)$ drawn in the circles.
\begin{center}
\begin{asy}
unitsize(7mm);
pair X = dir(240); pair Y = dir(0);
int[] t = {0,0,2,0,2,4,6};
for (int i=0; i<=6; ++i) {
for (int j=0; j<=i; ++j) {
filldraw(shift(i*X+j*Y)*scale(0.5)*unitcircle, (t[i]==j) ? lightred : white, black);
}
}
label("$1$",0*X+0*Y);
label("$2$",1*X+0*Y);
label("$1$",1*X+1*Y);
label("$2$",2*X+0*Y);
label("$2$",2*X+1*Y);
label("$2$",2*X+2*Y);
label("$3$",3*X+0*Y);
label("$2$",3*X+1*Y);
label("$2$",3*X+2*Y);
label("$2$",3*X+3*Y);
label("$3$",4*X+0*Y);
label("$3$",4*X+1*Y);
label("$3$",4*X+2*Y);
label("$2$",4*X+3*Y);
label("$2$",4*X+4*Y);
label("$3$",5*X+0*Y);
label("$3$",5*X+1*Y);
label("$3$",5*X+2*Y);
label("$3$",5*X+3*Y);
label("$3$",5*X+4*Y);
label("$2$",5*X+5*Y);
label("$3$",6*X+0*Y);
label("$3$",6*X+1*Y);
label("$3$",6*X+2*Y);
label("$3$",6*X+3*Y);
label("$3$",6*X+4*Y);
label("$3$",6*X+5*Y);
label("$3$",6*X+6*Y);
\end{asy}
\end{center}
We have that
\[ f(i, j) = \max \left\{f(i-1,j-1), f(i,j-1) \right\} +
\begin{cases}
1 & \text{if $(i,j)$ is red} \\
0 & \text{otherwise}
\end{cases} \]
since every ninja path passing through $(i, j)$ also passes through
either $(i-1,j-1)$ or $(i,j-1)$. Now consider the quantity
$S_j = f(0, j) + \dots + f(j, j)$. We obtain the following recurrence:
\begin{claim*}
$S_{j+1} \geq S_j + \left\lceil \frac{S_j}{j} \right\rceil + 1$.
\end{claim*}
\begin{proof}
Consider a maximal element $f(m, j)$ of $ \{f(0, j), \dots, f(j, j) \}$.
We perform the following manipulations:
\begin{align*}
S_{j+1}
&= \sum_{i=0}^{j+1} \max \left\{f(i-1,j), f(i,j) \right\} + \sum_{i=0}^{j+1}
\begin{cases}
1 & \text{if $(i,j+1)$ is red} \\
0 & \text{otherwise}
\end{cases} \\
&= \sum_{i=0}^{m} \max \left\{f(i-1,j), f(i,j) \right\} +
\sum_{i=m+1}^{j} \max \left\{f(i-1,j), f(i,j) \right\} + 1 \\
&\geq \sum_{i=0}^{m} f(i,j) + \sum_{i=m+1}^{j} f(i-1,j) + 1 \\
&= S_j + f(m, j) + 1 \\
&\geq S_j + \left\lceil \frac{S_j}{j} \right\rceil + 1
\end{align*}
where the last inequality is due to Pigeonhole.
\end{proof}
This is actually enough to solve the problem.
Write $n = 2^c + r$, where $0 \leq r \leq 2^c - 1$.
\begin{claim*}
$S_n \geq cn + 2r + 1$. In particular, $\left\lceil \frac{S_n}{n} \right\rceil \geq c + 1$.
\end{claim*}
\begin{proof}
First note that $S_n \geq cn + 2r + 1$ implies
$\left\lceil \frac{S_n}{n} \right\rceil \geq c + 1$ because
\[ \left\lceil \frac{S_n}{n} \right\rceil \geq \left\lceil \frac{cn + 2r + 1}{n} \right\rceil
= c + \left\lceil \frac{2r + 1}{n} \right\rceil = c + 1.
\]
We proceed by induction on $n$.
The base case $n = 1$ is clearly true as $S_1 = 1$.
Assuming that the claim holds for some $n=j$, we have
\begin{align*}
S_{j+1} &\geq S_j + \left\lceil \frac{S_j}{j} \right\rceil + 1 \\
&\geq cj + 2r + 1 + c + 1 + 1 \\
&= c(j+1) + 2(r+1) + 1
\end{align*}
so the claim is proved for $n=j+1$ if $j+1$ is not a power of $2$.
If $j+1 = 2^{c+1}$, then by writing
$c(j+1) + 2(r+1) + 1 = c(j+1) + (j+1) + 1 = (c+2)(j+1) + 1$, the claim is also proved.
\end{proof}
Now $\left\lceil \frac{S_n}{n} \right\rceil \geq c + 1$ implies the existence of
some ninja path containing at least $c+1$ red circles, and we are done.
|
IMO-2023-notes_6
|
Let $ABC$ be an equilateral triangle.
Let $A_1$, $B_1$, $C_1$ be interior points of $ABC$
such that $BA_1=A_1C$, $CB_1=B_1A$, $AC_1=C_1B$, and
\[ \angle BA_1C + \angle CB_1A + \angle AC_1B = 480\dg. \]
Let $A_2 = \ol{BC_1} \cap \ol{CB_1}$, $B_2 = \ol{CA_1} \cap \ol{AC_1}$,
$C_2 = \ol{AB_1} \cap \ol{BA_1}$.
Prove that if triangle $A_1B_1C_1$ is scalene,
then the circumcircles of triangles $AA_1A_2$, $BB_1B_2$, and $CC_1C_2$
all pass through two common points.
\end{enumerate}
|
This is the second official solution from the marking scheme,
also communicated to me by Michael Ren.
Define $O$ as the center of $ABC$ and set the angles
\begin{align*}
\alpha &\coloneq \angle A_1CB = \angle CBA_1 \\
\beta &\coloneq \angle ACB_1 = \angle B_1AC \\
\gamma &\coloneq \angle C_1AB = \angle C_1BA
\end{align*}
so that
\[ \alpha + \beta + \gamma = 30\dg. \]
In particular, $\max(\alpha,\beta,\gamma) < 30\dg$,
so it follows that $A_1$ lies inside $\triangle OBC$, and similarly for the others.
This means for example that $C_1$ lies between $B$ and $A_2$, and so on.
Therefore the polygon $A_2C_1B_2A_1C_2B_1$ is convex.
\begin{center}
\begin{asy}
size(14cm);
pair A = dir(90);
pair B = dir(210);
pair C = dir(330);
real s = abs(B-C)/2;
pair A_1 = midpoint(B--C) + dir(A) * s*Tan(11);
pair B_1 = midpoint(C--A) + dir(B) * s*Tan(13);
pair C_1 = midpoint(A--B) + dir(C) * s*Tan(6);
pair A_1 = A_1;
pair B_1 = B_1;
pair C_1 = C_1;
pair O = origin;
filldraw(A--B--C--cycle, opacity(0.1)+lightcyan, blue);
draw(A--A_1, blue);
draw(B--B_1, blue);
draw(C--C_1, blue);
pair A_2 = extension(B, C_1, B_1, C);
pair B_2 = extension(C, A_1, C_1, A);
pair C_2 = extension(A, B_1, B, A_1);
draw(B--A_2--C, lightred);
draw(C--B_2--A, lightred);
draw(A--C_2--B, lightred);
markangle("$\alpha$", n=1, radius=32.0, C, B, A_1, deepgreen);
markangle("$\alpha$", n=1, radius=32.0, A_1, C, B, deepgreen);
markangle("$\beta$", n=2, radius=56.0, A, C, B_1, deepgreen);
markangle("$\beta$", n=2, radius=56.0, B_1, A, C, deepgreen);
markangle("$\gamma$", n=3, radius=64.0, B, A, C_1, deepgreen);
markangle("$\gamma$", n=3, radius=64.0, C_1, B, A, deepgreen);
draw(circumcircle(B_1, C_1, B_2), gray);
draw(circumcircle(C_1, A_1, C_2), gray);
draw(circumcircle(A_1, B_1, A_2), gray);
pair Z = A_1;
pair X = -Z+2*foot(circumcenter(A_1, A_2, B_1), A, O);
pair Y = -Z+2*foot(circumcenter(A_1, A_2, C_1), A, O);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$A_1$", A_1, dir(A_1));
dot("$B_1$", B_1, dir(B_1));
dot("$C_1$", C_1, dir(C_1));
dot("$O$", O, 1.4*dir(240));
dot("$A_2$", A_2, dir(A_2));
dot("$B_2$", B_2, dir(B_2));
dot("$C_2$", C_2, dir(320));
dot("$X$", X, 1.7*dir(295));
dot("$Y$", Y, dir(55));
/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
!size(14cm);
A = dir 90
B = dir 210
C = dir 330
!real s = abs(B-C)/2;
!pair A_1 = midpoint(B--C) + dir(A) * s*Tan(11);
!pair B_1 = midpoint(C--A) + dir(B) * s*Tan(13);
!pair C_1 = midpoint(A--B) + dir(C) * s*Tan(6);
A_1 = A_1
B_1 = B_1
C_1 = C_1
O 1.4R240 = origin
A--B--C--cycle / 0.1 lightcyan / blue
A--A_1 / blue
B--B_1 / blue
C--C_1 / blue
A_2 = extension B C_1 B_1 C
B_2 = extension C A_1 C_1 A
C_2 320 = extension A B_1 B A_1
B--A_2--C / lightred
C--B_2--A / lightred
A--C_2--B / lightred
!markangle("$\alpha$", n=1, radius=32.0, C, B, A_1, deepgreen);
!markangle("$\alpha$", n=1, radius=32.0, A_1, C, B, deepgreen);
!markangle("$\beta$", n=2, radius=56.0, A, C, B_1, deepgreen);
!markangle("$\beta$", n=2, radius=56.0, B_1, A, C, deepgreen);
!markangle("$\gamma$", n=3, radius=64.0, B, A, C_1, deepgreen);
!markangle("$\gamma$", n=3, radius=64.0, C_1, B, A, deepgreen);
circumcircle B_1 C_1 B_2 / gray
circumcircle C_1 A_1 C_2 / gray
circumcircle A_1 B_1 A_2 / gray
Z := A_1
X 1.7R295 = -Z+2*foot (circumcenter A_1 A_2 B_1) A O
Y 55 = -Z+2*foot (circumcenter A_1 A_2 C_1) A O
*/
\end{asy}
\end{center}
We start by providing the ``interpretation'' for the $480\dg$ angle in the statement:
\begin{claim*}
Point $A_1$ is the circumcenter of $\triangle A_2BC$, and similarly for the others.
\end{claim*}
\begin{proof}
We have $\angle BA_1C = 180\dg - 2\alpha$, and
\begin{align*}
\angle BA_2C &= 180\dg - \angle CBC_1 - \angle B_1CB \\
&= 180\dg - \left( 60\dg - \gamma \right) - \left( 60\dg - \beta \right) \\
&= 60\dg + \beta + \gamma = 90\dg - \alpha = \half \angle BA_1 C.
\end{align*}
Since $A_1$ lies inside $\triangle BA_2C$, it follows $A_1$ is exactly the circumcenter.
\end{proof}
\begin{claim*}
Quadrilateral $B_2C_1B_1C$ can be inscribed in a circle, say $\gamma_a$.
Circles $\gamma_b$ and $\gamma_c$ can be defined similarly.
Finally, these three circles are pairwise distinct.
\end{claim*}
\begin{proof}
Using directed angles now, we have
\[ \dang B_2B_1C_2 = 180\dg-\dang AB_1B_2
= 180\dg - 2 \dang ACB = 180\dg-2(60\dg-\alpha) = 60\dg+2\alpha. \]
By the same token, $\dang B_2C_1C_2 = 60\dg+2\alpha$.
This establishes the existence of $\gamma_a$.
The proof for $\gamma_b$ and $\gamma_c$ is the same.
Finally, to show the three circles are distinct, it would be enough to
verify that the convex hexagon $A_2C_1B_2A_1C_2B_1$ is not cyclic.
Assume for contradiction it was cyclic. Then
\[ 360\dg = \angle C_2A_1B_2 + \angle B_2C_1A_2 + \angle A_2B_1C_2
= \angle BA_1C + \angle CB_1A + \angle AC_1B = 480\dg \]
which is absurd.
This contradiction eliminates the degenerate case, so the three circles are distinct.
\end{proof}
For the remainder of the solution, let $\opname{Pow}(P, \omega)$
denote the power of a point $P$ with respect to a circle $\omega$.
Let line $AA_1$ meet $\gamma_b$ and $\gamma_c$ again at $X$ and $Y$,
and set $k_a \coloneq \frac{AX}{AY}$.
Consider the locus of all points $P$ such that
\[ \mathcal C_a \coloneq \Big\{ \text{points } P \text{ in the plane satisfying }
\opname{Pow}(P, \gamma_b) = k_a \opname{Pow}(P, \gamma_c) \Big\}. \]
We recall the \emph{coaxiality lemma}\footnote{We quickly outline a proof of this lemma:
in the Cartesian coordinate system, the expression $\opname{Pow}((x,y), \omega)$
is an expression of the form $x^2 + y^2 + {\bullet} x + {\bullet} y + {\bullet}$
for some constants ${\bullet}$ whose value does not matter.
Substituting this into the equation
$\frac{k_a \opname{Pow}(P, \gamma_c) - \opname{Pow}(P, \gamma_b)}{k_a-1} = 0$
gives the equation of a circle provided $k_a \neq 1$,
and when $k_a = 1$, one instead recovers the radical axis.},
which states that (given $\gamma_b$ and $\gamma_c$ are not concentric)
the locus $\mathcal C_a$ must be either a circle (if $k_a \neq 1$) or a line (if $k_a=1$).
On the other hand, $A_1$, $A_2$, and $A$ all obviously lie on $\mathcal C_a$.
(For $A_1$ and $A_2$, the powers are both zero, and for the point $A$,
we have $\opname{Pow}(P, \gamma_b) = AX \cdot AA_1$
and $\opname{Pow}(P, \gamma_c) = AY \cdot AA_1$.)
So $\mathcal C_a$ must be exactly the circumcircle of $\triangle AA_1A_2$
from the problem statement.
We turn to evaluating $k_a$ more carefully.
First, note that
\[ \angle A_1XB_1 = \angle A_1B_2B_1 = \angle CB_2B_1
= 90\dg - \angle B_2AC = 90\dg - (60\dg-\gamma) = 30\dg + \gamma. \]
Now using the law of sines, we derive
\begin{align*}
\frac{AX}{AB_1} &= \frac{\sin \angle AB_1X}{\sin \angle AXB_1}
= \frac{\sin(\angle A_1XB_1 - \angle XAB_1)}{\sin \angle A_1XB_1} \\
&= \frac{\sin\left( (30\dg+\gamma)-(30\dg-\beta) \right)}{\sin (30\dg+\gamma)}
= \frac{\sin(\beta+\gamma)}{\sin (30\dg+\gamma)}.
\end{align*}
Similarly, $AY = AC_1 \cdot \frac{\sin(\beta+\gamma)}{\sin(30\dg+\beta)}$, so
\[ k_a = \frac{AX}{AY}
= \frac{AB_1}{AC_1} \cdot \frac{\sin(30\dg+\beta)}{\sin(30\dg+\gamma)}. \]
Now define analogous constants $k_b$ and $k_c$
and circles $\mathcal C_b$ and $\mathcal C_c$.
Owing to the symmetry of the expressions, we have the key relation
\[ k_a k_b k_c = 1. \]
In summary, the three circles in the problem statement may be described as
\begin{align*}
\mathcal C_a &= (AA_1A_2)
= \left\{ \text{points $P$ such that } \opname{Pow}(P, \gamma_b) = k_a \opname{Pow}(P, \gamma_c) \right\} \\
\mathcal C_b &= (BB_1B_2)
= \left\{ \text{points $P$ such that } \opname{Pow}(P, \gamma_c) = k_b \opname{Pow}(P, \gamma_a) \right\} \\
\mathcal C_c &= (CC_1C_2)
= \left\{ \text{points $P$ such that } \opname{Pow}(P, \gamma_a) = k_c \opname{Pow}(P, \gamma_b) \right\}.
\end{align*}
Since $k_a$, $k_b$, $k_c$ have product $1$,
it follows that any point on at least two of the circles
must lie on the third circle as well.
The convexity of hexagon $A_2C_1B_2A_1C_2B_1$ mentioned earlier ensures these
any two of these circles do intersect at two different points, completing the solution.
|
IMO-2024-notes_1
|
Find all real numbers $\alpha$ so that, for every positive integer $n$, the integer
\[ \left\lfloor \alpha \right\rfloor + \left\lfloor 2 \alpha \right\rfloor
+ \left\lfloor 3 \alpha \right\rfloor + \dots + \left\lfloor n \alpha \right\rfloor \]
is divisible by $n$.
|
The answer is that $\alpha$ must be an even integer.
Let $S(n, \alpha)$ denote the sum in question.
\paragraph{Analysis for $\alpha$ an integer.}
If $\alpha$ is an integer, then the sum equals
\[ S(n, \alpha) = (1+2+\dots+n) \alpha = \frac{n(n+1)}{2} \cdot \alpha \]
which is obviously a multiple of $n$ if $2 \mid \alpha$;
meanwhile, if $\alpha$ is an odd integer then $n = 2$ gives a counterexample.
\paragraph{Main case.}
Suppose $\alpha$ is not an integer; we show the desired condition can never be true.
Note that replacing $\alpha$ with $\alpha \pm 2$ changes by
\[ S(n \pm 2, \alpha) - S(n, \alpha) = 2(1+2+\dots+n) = n(n+1) \equiv 0 \pmod n \]
for every $n$.
Thus, by shifting appropriately we may assume $-1 < \alpha < 1$ and $\alpha \notin \ZZ$.
\begin{itemize}
\ii If $0 < \alpha < 1$,
then let $m \ge 2$ be the smallest integer such that $m \alpha \ge 1$.
Then
\[ S(m, \alpha) = \underbrace{0 + \dots + 0}_{m-1\text{ terms}} + 1 = 1 \]
is not a multiple of $m$.
\ii If $-1 < \alpha < 0$,
then let $m \ge 2$ be the smallest integer such that $m \alpha \le -1$.
Then
\[ S(m, \alpha) = \underbrace{(-1) + \dots + (-1)}_{m-1\text{ terms}} + 0 = -(m-1) \]
is not a multiple of $m$.
\end{itemize}
|
IMO-2024-notes_2
|
For which pairs of positive integers $(a,b)$ is the sequence
\[ \gcd(a^n+b, b^n+a) \qquad n = 1, 2, \dotsc \]
eventually constant?
|
The answer is $(a,b)=(1,1)$ only, which obviously works since the sequence is always $2$.
Conversely, assume the sequence
\[ x_n \coloneq \gcd(a^n+b, b^n+a) \] is eventually constant.
The main crux of the other direction is to consider
\[ M \coloneq ab+1. \]
\begin{remark*}
[Motivation]
The reason to consider the number is the same technique used in IMO 2005/4,
namely the idea to consider ``$n = -1$''.
The point is that the two rational numbers
\[ \frac 1a + b = \frac{ab+1}{b}, \qquad \frac 1b + a = \frac{ab+1}{a} \]
have a large common factor: we could write ``$x_{-1} = ab + 1$'', loosely speaking.
Now, the sequence is really only defined for $n \ge 1$,
so one should instead take $n \equiv -1 \pmod{\varphi(M)}$
--- and this is exactly what we do.
\end{remark*}
Obviously $\gcd(a,M) = \gcd(b,M) = 1$.
Let $n$ be a sufficiently large multiple of $\varphi(M)$
so that \[ x_{n-1} = x_n = x_{n+1} = \dotsb. \]
We consider the first three terms;
the first one is the ``key'' one that gets the bulk of the work,
and the rest is bookkeeping and extraction.
\begin{itemize}
\ii Consider $x_{n-1}$.
Note that
\[ a (a^{n-1} + b) = a^n + ab \equiv 1 + (-1) \equiv 0 \pmod M \]
and similarly $b (b^n + a) \equiv 0 \pmod M$.
Hence $M \mid x_{n-1}$.
\ii Consider $x_n$, which is now known to be divisible by $M$. Note that
\begin{align*}
0 &\equiv a^n + b \equiv 1 + b \pmod M \\
0 &\equiv b^n + a \equiv 1 + a \pmod M.
\end{align*}
So $a \equiv b \equiv -1 \pmod M$.
\ii Consider $x_{n+1}$, which is now known to be divisible by $M$. Note that
\[ 0 \equiv a^{n+1} + b \equiv b^{n+1} + a \equiv a + b \pmod M. \]
We knew $a \equiv b \equiv -1 \pmod M$,
hence this means $0 \equiv 2 \pmod M$, so $M = 2$.
\end{itemize}
From $M = 2$ we then conclude $a = b = 1$, as desired.
\begin{remark*}
[No alternate solutions known]
At the time nobody seems to know any solution not depending critically
on $M = ab+1$ (or prime numbers dividing $M$, etc.).
They vary in execution once some term of the form $x_{k\varphi(n)-1}$ is taken,
but avoiding the key idea altogether does not currently seem possible.
A good example to consider for ruling out candidate ideas is $(a,b) = (18,9)$.
\end{remark*}
|
IMO-2024-notes_3
|
Let $a_1$, $a_2$, $a_3$, \dots\ be an infinite sequence of positive integers,
and let $N$ be a positive integer.
Suppose that, for each $n > N$,
the number $a_n$ is equal to the number of times $a_{n-1}$ appears
in the list $(a_1, a_2, \dots, a_{n-1})$.
Prove that at least one of the sequences $a_1$, $a_3$, $a_5$, \dots\
and $a_2$, $a_4$, $a_6$, \dots\ is eventually periodic.
|
We present the solution from ``gigamilkmen'tgeg''
in \url{https://aops.com/community/p31224483},
with some adaptation from the first shortlist official solution as well.
Set $M \coloneq \max(a_1, \dots, a_N)$.
\paragraph{Setup.}
We will visualize the entire process as follows.
We draw a stack of towers labeled $1$, $2$, \dots, each initially empty.
For $i=1,2,\dots$, we imagine the term $a_i$ as adding a block $B_i$ to tower $a_i$.
Then there are $N$ initial blocks placed, colored \emph{red}.
The rest of the blocks are colored \emph{yellow}:
if the last block $B_i$ was added to a tower that then reaches height $a_{i+1}$,
the next block $B_{i+1}$ is added to tower $a_{i+1}$.
We'll say $B_i$ \emph{contributes} to the tower containing $B_{i+1}$.
In other words, the yellow blocks $B_i$ for $i > N$
are given coordinates $B_i = (a_i, a_{i+1})$ for $i>N$.
Note in particular that in towers $M+1$, $M+2$, \dots, the blocks are all yellow.
\begin{center}
\begin{asy}
unitsize(0.85cm);
int[] a = {1,2,2,2,2,2,3,4,1,2,6,1,3,2,7,1,4,2,8,1,5,1,6,2,9,1,7,2,10};
int N = 8;
filldraw(shift(0,0)*unitsquare, palered, black+1);
filldraw(shift(1,0)*unitsquare, palered, black+1);
filldraw(shift(2,0)*unitsquare, palered, black+1);
filldraw(shift(3,0)*unitsquare, palered, black+1);
filldraw(shift(1,1)*unitsquare, palered, black+1);
filldraw(shift(1,2)*unitsquare, palered, black+1);
filldraw(shift(1,3)*unitsquare, palered, black+1);
filldraw(shift(1,4)*unitsquare, palered, black+1);
for (int i=N; i<a.length-1; ++i) {
filldraw(shift(a[i]-1, a[i+1]-1)*unitsquare, paleyellow, black+1);
}
draw((0,0)--(4,0)--(4,1)--(2,1)--(2,5)--(1,5)--(1,1)--(0,1)--cycle, brown+2);
label("$\boxed{N}$", (a[N-1]-0.5, a[N]-0.5), fontsize(14pt));
void draw_arrow(int k, pen p) {
draw((a[k-1]-0.5,a[k]-0.5)--(a[k]-0.5,a[k+1]-0.5),
p, EndArrow(TeXHead), Margin(2,2));
}
draw((4,10)--(4,-0.7), deepgreen);
label("$M$", (4,-0.7), dir(-90), deepgreen);
for (int j=1; j<=9; ++j) {
label("$"+(string)j+"$", (j-0.5,0), dir(-90), blue+fontsize(10pt));
}
// Figure 1: setup
for (int i=8; i<=14; ++i) { draw_arrow(i, black+1.0); }
for (int i=N; i<a.length-1; ++i) {
label("$"+(string)(i+1)+"$", (a[i]-0.5, a[i+1]-0.5), fontsize(12pt));
}
\end{asy}
\end{center}
We let $h_\ell$ denote the height of the $\ell$\ts{th} tower at a given time $n$.
(This is an abuse of notation and we should write $h_\ell(n)$ at time $n$,
but $n$ will always be clear from context.)
\paragraph{Up to alternating up and down.}
We start with two independent easy observations:
the set of numbers that occur infinitely often is downwards closed,
and consecutive terms cannot both be huge.
\begin{center}
\begin{asy}
unitsize(0.85cm);
int[] a = {1,2,2,2,2,2,3,4,1,2,6,1,3,2,7,1,4,2,8,1,5,1,6,2,9,1,7,2,10,1,8,2,11,1,9,2,12};
int N = 8;
// Red X region
fill(box((4,4),(11,12)), lightgray);
draw((4,4)--(11,12), red+1.4);
draw((11,4)--(4,12), red+1.4);
filldraw(shift(0,0)*unitsquare, palered, black+1);
filldraw(shift(1,0)*unitsquare, palered, black+1);
filldraw(shift(2,0)*unitsquare, palered, black+1);
filldraw(shift(3,0)*unitsquare, palered, black+1);
filldraw(shift(1,1)*unitsquare, palered, black+1);
filldraw(shift(1,2)*unitsquare, palered, black+1);
filldraw(shift(1,3)*unitsquare, palered, black+1);
filldraw(shift(1,4)*unitsquare, palered, black+1);
for (int i=N; i<a.length-1; ++i) {
filldraw(shift(a[i]-1, a[i+1]-1)*unitsquare, paleyellow, black+1);
}
draw((0,0)--(4,0)--(4,1)--(2,1)--(2,5)--(1,5)--(1,1)--(0,1)--cycle, brown+2);
label("$\boxed{N}$", (a[N-1]-0.5, a[N]-0.5), fontsize(14pt));
void draw_arrow(int k, pen p) {
draw((a[k-1]-0.5,a[k]-0.5)--(a[k]-0.5,a[k+1]-0.5),
p, EndArrow(TeXHead), Margin(2,2));
}
draw((2,0)--(2,-0.7), deepgreen);
draw((4,12)--(4,-0.7), deepgreen);
label("$L$", (2,-0.7), dir(-90), deepgreen);
label("$M$", (4,-0.7), dir(-90), deepgreen);
draw((4,4)--(11,4), deepgreen);
for (int j=1; j<=11; ++j) {
label("$"+(string)j+"$", (j-0.5,0), dir(-90), blue+fontsize(10pt));
}
// Figure 2: claim with C
filldraw(shift(a[18]-1, a[19]-1)*unitsquare, yellow, black+2);
filldraw(shift(a[19]-1, a[20]-1)*unitsquare, yellow, black+2);
filldraw(shift(a[30]-1, a[31]-1)*unitsquare, yellow, black+2);
filldraw(shift(a[31]-1, a[32]-1)*unitsquare, yellow, black+2);
draw_arrow(19, blue+1.1);
draw_arrow(31, blue+1.1);
for (int i=N; i<a.length-1; ++i) {
label("$"+(string)(i+1)+"$", (a[i]-0.5, a[i+1]-0.5), fontsize(12pt));
}
\end{asy}
\end{center}
\begin{claim*}
If the $(k+1)$\ts{st} tower grows arbitrarily high, so does tower $k$.
In fact, there exists a constant $C$ such that $h_{k} \ge h_{k+1} - C$ at all times.
\end{claim*}
\begin{proof}
Suppose $B_n$ is a yellow block in tower $k+1$.
Then with at most finitely many exceptions, $B_{n-1}$ is a yellow block at height $k+1$,
and the block $B_r$ right below $B_{n-1}$ is also yellow;
then $B_{r+1}$ is in tower $k$.
Hence, with at most finitely many exceptions, the map
\[ B_n \mapsto B_{n-1} \mapsto B_r \mapsto B_{r+1} \]
provides an injective map taking each yellow block in tower $k+1$
to a yellow block in tower $k$.
(The figure above shows $B_{32} \to B_{31} \to B_{19} \to B_{20}$ as an example.)
\end{proof}
\begin{claim*}
If $a_n > M$ then $a_{n+1} \le M$.
\end{claim*}
\begin{proof}
Assume for contradiction there's a first moment where $a_n > M$ and $a_{n+1} > M$,
meaning the block $B_n$ was added to an all-yellow tower past $M$
that has height exceeding $M$.
(This is the X'ed out region in the figure above.)
In $B_n$'s tower, every (yellow) block (including $B_n$)
was contributed by a block placed in different towers at height $a_n > M$.
So before $B_n$, there were already $a_{n+1} > M$ towers of height more than $M$.
This contradicts minimality of $n$.
\end{proof}
It follows that the set of indices with $a_n \le M$
has arithmetic density at least half, so certainly
at least some of the numbers must occur infinitely often.
Of the numbers in $\{1,2,\dots,M\}$,
define $L$ such that towers $1$ through $L$ grow unbounded
but towers $L+1$ through $M$ do not.
Then we can pick a larger threshold $N' > N$ such that
\begin{itemize}
\ii Towers $1$ through $L$ have height greater than $(M,N)$;
\ii Towers $L+1$ through $M$ will receive no further blocks;
\ii $a_{N'} \le L$.
\end{itemize}
After this threshold, the following statement is true:
\begin{claim*}
[Alternating small and big]
The terms $a_{N'}$, $a_{N' + 2}$, $a_{N' + 4}$, \dots\ are all at most $L$ while
the terms $a_{N' + 1}$, $a_{N' + 3}$, $a_{N' + 5}$, \dots\ are all greater than $M$.
\end{claim*}
\paragraph{Automaton for $n \equiv N' \pmod 2$.}
From now on we always assume $n > N'$.
When $n \equiv N' \pmod 2$, i.e., when $a_n$ is small, we define the state
\[ S(n) = (h_1, h_2, \dots, h_L; a_n). \]
For example, in the figure below, we illustrate how
\[ S(34) = (9,11;a_{34}=1) \longrightarrow S(36) = (9,12;a_{36}=2) \]
\begin{center}
\begin{asy}
unitsize(0.85cm);
int[] a = {1,2,2,2,2,2,3,4,1,2,6,1,3,2,7,1,4,2,8,1,5,1,6,2,9,1,7,2,10,1,8,2,11,1,9,2,12};
int N = 8;
filldraw(shift(0,0)*unitsquare, palered, black+1);
filldraw(shift(1,0)*unitsquare, palered, black+1);
filldraw(shift(2,0)*unitsquare, palered, black+1);
filldraw(shift(3,0)*unitsquare, palered, black+1);
filldraw(shift(1,1)*unitsquare, palered, black+1);
filldraw(shift(1,2)*unitsquare, palered, black+1);
filldraw(shift(1,3)*unitsquare, palered, black+1);
filldraw(shift(1,4)*unitsquare, palered, black+1);
for (int i=N; i<a.length-1; ++i) {
filldraw(shift(a[i]-1, a[i+1]-1)*unitsquare, paleyellow, black+1);
}
draw((0,0)--(4,0)--(4,1)--(2,1)--(2,5)--(1,5)--(1,1)--(0,1)--cycle, brown+2);
label("$\boxed{N}$", (a[N-1]-0.5, a[N]-0.5), fontsize(14pt));
void draw_arrow(int k, pen p) {
draw((a[k-1]-0.5,a[k]-0.5)--(a[k]-0.5,a[k+1]-0.5),
p, EndArrow(TeXHead), Margin(2,2));
}
draw((2,0)--(2,-0.7), deepgreen);
draw((4,12)--(4,-0.7), deepgreen);
label("$L$", (2,-0.7), dir(-90), deepgreen);
label("$M$", (4,-0.7), dir(-90), deepgreen);
draw((4,4)--(11,4), deepgreen);
for (int j=1; j<=11; ++j) {
label("$"+(string)j+"$", (j-0.5,0), dir(-90), blue+fontsize(10pt));
}
// Figure 3: automaton
filldraw(shift(a[33]-1, a[34]-1)*unitsquare, yellow, black+2);
filldraw(shift(a[34]-1, a[35]-1)*unitsquare, yellow, black+2);
filldraw(shift(a[35]-1, a[36]-1)*unitsquare, yellow, black+2);
draw_arrow(34, deepgreen+1.1);
draw_arrow(35, deepgreen+1.1);
for (int i=N; i<a.length-1; ++i) {
label("$"+(string)(i+1)+"$", (a[i]-0.5, a[i+1]-0.5), fontsize(12pt));
}
\end{asy}
\end{center}
The final element $a_n$ simply reminds us which tower was most recently incremented.
At this point we can give a complete description of how to move from $S(n)$ to $S(n+2)$:
\begin{itemize}
\ii The intermediate block $B_{n+1}$ is placed in the tower
corresponding to the height $a_{n+1}$ of $B_n$;
\ii That tower will have height $a_{n+2}$ equal to the number of towers
with height at least $a_{n+1}$; that is, it equals the cardinality of the set
\[ \{ i \colon h_i \ge h_{a_n} \} \]
\ii We increment $h_{a_{n+2}}$ by $1$ and update $a_n$.
\end{itemize}
For example, the illustrated $S(34) \to S(36)$
corresponds to the block $B_{34}$ at height $h_1$ in tower $1$
giving the block $B_{35}$ at height $2$ in tower $h_1$,
then block $B_{36}$ at height $h_2 + 1$ being placed in tower $2$.
\paragraph{Pigeonhole periodicity argument.}
Because only the \emph{relative} heights matter in the automata above,
if we instead define
\[ T(n) = (h_1-h_2, h_2-h_3, \dots, h_{L-1}-h_L; a_n). \]
then $T(n+2)$ can be determined from just $T(n)$.
So it would be sufficient to show $T(n)$ only takes on finitely many values
to show that $T(n)$ (and hence $a_n$) is eventually periodic.
Since we have the bound $h_{k+1} \le h_k + C$,
we are done upon proving the following lower bound:
\begin{claim*}
For every $1 \le \ell < L$ and $n > N'$,
we have $h_\ell \le h_{\ell+1} + C \cdot (L-1)$.
\end{claim*}
\begin{proof}
Assume for contradiction that there is some moment $n > N'$ such that
\[ h_\ell > h_{\ell+1} + C \cdot (L-1) \]
and WLOG assume that $h_\ell$ was just updated at the moment $n$.
Together with $h_{k+1} \le h_k + C$ for all $k$ and triangle inequality, we conclude
\[ \min(h_1, \dots, h_\ell) > q \coloneq \max(h_{\ell+1}, \dots, h_L). \]
We find that the blocks now in fact alternate between being placed
among the first $\ell$ towers and in towers with indices greater than $q$ thereafter.
Hence the heights $h_{\ell+1}$, \dots, $h_L$ never grow after this moment.
This contradicts the definition of $L$.
\end{proof}
\begin{remark*}
In fact, it can be shown that the period is actually exactly $L$,
meaning the periodic part will be exactly a permutation of $(1,2,\dots,L)$.
For any $L$, it turns out there is indeed a permutation achieving that periodic part.
\end{remark*}
|
IMO-2024-notes_4
|
Let triangle $ABC$ with incenter $I$ satisfying $AB < AC < BC$.
Let $X$ be a point on line $BC$, different from $C$,
such that the line through $X$ and parallel to $AC$ is tangent to the incircle.
Similarly, let $Y$ be a point on line $BC$, different from $B$,
such that the line through $Y$ and parallel to $AB$ is tangent to the incircle.
Line $AI$ intersects the circumcircle of triangle $ABC$ again at $P$.
Let $K$ and $L$ be the midpoints of $AC$ and $AB$, respectively.
Prove that $\angle KIL + \angle YPX = 180^{\circ}$.
|
Let $T$ be the reflection of $A$ over $I$, the most important point to add
since it gets rid of $K$ and $L$ as follows.
\begin{claim*}
We have $\angle KIL = \angle BTC$,
and lines $TX$ and $TY$ are tangent to the incircle.
\end{claim*}
\begin{proof}
The first part is true since $\triangle BTC$ is the image of $\triangle KIL$
under a homothety of ratio $2$.
The second part is true because lines $AB$, $AC$, $TX$, $TY$
determine a rhombus with center $I$.
\end{proof}
We thus delete $K$ and $L$ from the picture altogether; they aren't needed anymore.
\begin{center}
\begin{asy}
size(11cm);
pair A = dir(105);
pair B = dir(200);
pair C = dir(340);
pair P = dir(270);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
pair I = incenter(A, B, C);
filldraw(incircle(A, B, C), opacity(0.1)+lightcyan, blue);
draw(A--B--C--cycle, blue);
pair E = foot(I, C, A);
pair F = foot(I, A, B);
pair T = 2*I-A;
pair U = 2*I-E;
pair V = 2*I-F;
draw(U--T--V, deepgreen);
draw(B--T--C, red);
pair K = midpoint(A--B);
pair L = midpoint(A--C);
draw(K--I--L, red+dashed);
pair X = extension(U, T, B, C);
pair Y = extension(V, T, B, C);
draw(A--P, gray);
draw(circumcircle(B, X, P), gray+dashed);
draw(circumcircle(C, Y, P), gray+dashed);
dot("$A$", A, dir(A));
dot("$B$", B, dir(160));
dot("$C$", C, dir(20));
dot("$P$", P, dir(45));
dot("$I$", I, dir(250));
dot("$T$", T, dir(225));
dot("$K$", K, dir(K));
dot("$L$", L, dir(L));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(310));
/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
!size(11cm);
A = dir 105
B 160 = dir 200
C 20 = dir 340
P 45 = dir 270
unitcircle / 0.1 lightcyan / blue
I 250 = incenter A B C
incircle A B C / 0.1 lightcyan / blue
A--B--C--cycle / blue
E := foot I C A
F := foot I A B
T 225 = 2*I-A
U := 2*I-E
V := 2*I-F
U--T--V / deepgreen
B--T--C / red
K = midpoint A--B
L = midpoint A--C
K--I--L / red dashed
X = extension U T B C
Y 310 = extension V T B C
A--P / gray
circumcircle B X P / gray dashed
circumcircle C Y P / gray dashed
*/
\end{asy}
\end{center}
\begin{claim*}
We have $BXPT$ and $CYPT$ are cyclic.
\end{claim*}
\begin{proof}
$\dang TYC = \dang TYB = \dang ABC = \dang APC = \dang TPC$ and similarly.
(Some people call this Reim's theorem.)
\end{proof}
To finish, observe that
\[ \dang CTB = \dang CTP + \dang PTB = \dang CYP + \dang PXB
= \dang XYP + \dang XYP = \dang XPY \]
as desired.
(The length conditions $AC > AB > BC$ ensure that $B$, $X$, $Y$, $C$ are collinear
in that order, and that $T$ lies on the opposite side of $\ol{BC}$ as $A$.
Hence the directed equality $\dang CTB = \dang XPY$
translates to the undirected $\angle BTC + \angle XPY = 180\dg$.)
|
IMO-2024-notes_5
|
Turbo the snail is in the top row of a grid with $2024$ rows and $2023$ columns
and wants to get to the bottom row.
However, there are $2022$ hidden monsters, one in every row except the first and last,
with no two monsters in the same column.
Turbo makes a series of attempts to go from the first row to the last row.
On each attempt, he chooses to start on any cell in the first row,
then repeatedly moves to an orthogonal neighbor.
(He is allowed to return to a previously visited cell.)
If Turbo reaches a cell with a monster,
his attempt ends and he is transported back to the first row to start a new attempt.
The monsters do not move between attempts, and Turbo remembers whether or not each cell
he has visited contains a monster.
If he reaches any cell in the last row, his attempt ends and Turbo wins.
Find the smallest integer $n$ such that Turbo has a strategy which guarantees
being able to reach the bottom row in at most $n$ attempts,
regardless of how the monsters are placed.
|
Surprisingly the answer is $n = 3$ for \emph{any} grid size $s \times (s-1)$ when $s \ge 4$.
We prove this in that generality.
\paragraph{Proof that at least three attempts are needed.}
When Turbo first moves into the second row, Turbo could encounter a monster $M_1$ right away.
Then on the next attempt, Turbo must enter the third row in different column as $M_1$,
and again could encounter a monster $M_2$ right after doing so.
This means no strategy can guarantee fewer than three attempts.
\paragraph{Strategy with three attempts.}
On the first attempt, we have Turbo walk through the entire second row
until he finds the monster $M_1$ in it.
Then we get two possible cases.
\subparagraph{Case where $M_1$ is not on the edge.}
In the first case, if that monster $M_1$ is not on the edge of the row,
then Turbo can trace two paths below it as shown below.
At least one of these paths works, hence three attempts is sufficient.
\begin{center}
\begin{asy}
usepackage("amssymb");
unitsize(0.7cm);
pen gr = gray+linetype("4 2");
int n = 6;
for (int i=0; i<=n-1; ++i) {
draw((0,i)--(n,i), gr);
}
for (int i=0; i<=n; ++i) {
draw((i,0)--(i,n-1), gr);
}
draw(box((0,-1), (n,0)), black);
draw(box((0,n-1), (n,n)), black);
draw(box((0,-1), (n,n)), black);
label((n/2,n-0.5), "Starting row");
label((n/2,-0.5), "Goal row");
label((2.5,4.5), "$M_1$", red);
dotfactor *= 2;
dot((1.5,4.5), deepgreen);
dot((3.5,4.5), deepgreen);
draw((1.5,4.5)--(1.5,3.5)--(2.3,3.5)--(2.3,0.2), deepgreen+1.2, EndArrow(TeXHead));
draw((3.5,4.5)--(3.5,3.5)--(2.7,3.5)--(2.7,0.2), deepgreen+1.2, EndArrow(TeXHead));
\end{asy}
\end{center}
\subparagraph{Case where $M_1$ is on the edge.}
WLOG, $M_1$ is in the leftmost cell.
Then Turbo follows the green staircase pattern shown in the left figure below.
If the staircase is free of monsters, then Turbo wins on the second attempt.
Otherwise, if a monster $M_2$ is encountered on the staircase,
Turbo has found a safe path to the left of $M_2$;
then Turbo can use this to reach the column $M_1$ is in, and escape from there.
This is shown in purple in the center and right figure
(there are two slightly different cases depending on whether $M_2$
was encountered going east or south).
\begin{center}
\begin{asy}
usepackage("amssymb");
unitsize(0.65cm);
pen gr = gray+linetype("4 2");
int n = 6;
dotfactor *= 2;
picture pic1, pic2, pic3;
picture[] pics = {pic1, pic2, pic3};
for (int j=0; j<3; ++j) {
for (int i=0; i<=n-1; ++i) {
draw(pics[j], (0,i)--(n,i), gr);
}
for (int i=0; i<=n; ++i) {
draw(pics[j], (i,0)--(i,n-1), gr);
}
draw(pics[j], box((0,-1), (n,0)), black);
draw(pics[j], box((0,n-1), (n,n)), black);
draw(pics[j], box((0,-1), (n,n)), black);
label(pics[j], "Starting row", (n/2,n-0.5));
label(pics[j], "Goal row", (n/2,-0.5));
}
label(pic1, "$M_1$", (0.5,4.5), red);
dot(pic1, (1.5,4.5), deepgreen);
draw(pic1, (1.5,4.5)--(2.5,4.5)--(2.5,3.5)--(3.5,3.5)--(3.5,2.5)
--(4.5,2.5)--(4.5,1.5)--(5.5,1.5)--(5.5,0.2), deepgreen+1.2, EndArrow(TeXHead));
label(pic2, "$M_1$", (0.5,4.5), red);
label(pic2, "$M_2$", (4.5,2.5), red);
dot(pic2, (1.5,4.5), deepgreen);
draw(pic2, (3.5,2.5)--(4.5,2.5)--(4.5,1.5)--(5.5,1.5)--(5.5,0.2), deepgreen+dashed);
draw(pic2, (1.5,4.5)--(2.5,4.5)--(2.5,3.5)--(3.5,3.5)--(3.5,2.5)--(0.5,2.5)--(0.5,0.2),
purple+1.5, EndArrow(TeXHead));
label(pic3, "$M_1$", (0.5,4.5), red);
label(pic3, "$M_2$", (4.5,1.5), red);
dot(pic3, (1.5,4.5), deepgreen);
draw(pic3, (3.5,2.5)--(4.5,2.5)--(4.5,1.5)--(5.5,1.5)--(5.5,0.2), deepgreen+dashed);
draw(pic3, (1.5,4.5)--(2.5,4.5)--(2.5,3.5)--(3.5,3.5)--(3.5,1.5)--(0.5,1.5)--(0.5,0.2),
purple+1.5, EndArrow(TeXHead));
add(pic1);
add(shift(7,0)*pic2);
add(shift(14,0)*pic3);
\end{asy}
\end{center}
Thus the problem is solved in three attempts, as promised.
\paragraph{Extended remark: all working strategies look similar to this.}
As far as we know, all working strategies are variations of the above.
In fact, we will try to give a description of the space of possible strategies,
although this needs a bit of notation.
\begin{definition*}
For simplicity, we only use $s$ even in the figures below.
We define the \emph{happy triangle} as the following cells:
\begin{itemize}
\item All $s-1$ cells in the first row (which has no monsters).
\item The center $s-3$ cells in the second row.
\item The center $s-5$ cells in the third row.
\item \dots
\item The center cell in the $\frac s2$\textsuperscript{th} row.
\end{itemize}
\end{definition*}
For $s=12$, the happy triangle is the region shaded in the thick border below.
\begin{center}
\begin{asy}
usepackage("amssymb");
unitsize(0.7cm);
pen gr = gray+linetype("4 2");
void setup(int n) {
for (int i=0; i<=n-1; ++i) {
draw((0,i)--(n,i), gr);
}
for (int i=0; i<=n; ++i) {
draw((i,0)--(i,n-1), gr);
}
draw(box((0,-1), (n,0)), black);
draw(box((0,n-1), (n,n)), black);
draw(box((0,-1), (n,n)), black);
label((n/2,n-0.5), "Starting row");
label((n/2,-0.5), "Goal row");
path p = (0,n);
for (int i=0; i<n/2; ++i) {
p = p--(i,n-1-i)--(i+1,n-1-i);
}
for (int i=(n+1)#2; i<n; ++i) {
p = p--(i,i)--(i+1,i);
}
p = p--(n,n)--cycle;
filldraw(p, opacity(0.15)+yellow, blue+1.8);
}
setup(11);
\end{asy}
\end{center}
\begin{definition*}
Given a cell, define a \emph{shoulder} to be the cell directly northwest or northeast of it.
Hence there are two shoulders of cells outside the first and last column,
and one shoulder otherwise.
\end{definition*}
Then solutions roughly must distinguish between these two cases:
\begin{itemize}
\item \textbf{Inside happy triangle:}
If the first monster $\color{red}M_1$ is found in the \emph{happy triangle},
and there is a safe path found by Turbo to the two shoulders
(marked $\color{green!60!black}\bigstar$ in the figure),
then one can finish in two more moves by considering the two paths from $\color{green!60!black}\bigstar$
that cut under the monster $\color{red}M_1$; one of them must work.
This slightly generalizes the easier case in the solution above
(which focuses only on the case where $\color{red}M_1$ is in the first row).
\begin{center}
\begin{asy}
usepackage("amssymb");
unitsize(0.7cm);
pen gr = gray+linetype("4 2");
void setup(int n) {
for (int i=0; i<=n-1; ++i) {
draw((0,i)--(n,i), gr);
}
for (int i=0; i<=n; ++i) {
draw((i,0)--(i,n-1), gr);
}
draw(box((0,-1), (n,0)), black);
draw(box((0,n-1), (n,n)), black);
draw(box((0,-1), (n,n)), black);
label((n/2,n-0.5), "Starting row");
label((n/2,-0.5), "Goal row");
path p = (0,n);
for (int i=0; i<n/2; ++i) {
p = p--(i,n-1-i)--(i+1,n-1-i);
}
for (int i=(n+1)#2; i<n; ++i) {
p = p--(i,i)--(i+1,i);
}
p = p--(n,n)--cycle;
filldraw(p, opacity(0.15)+yellow, blue+1.8);
}
setup(7);
label((2.5,4.5), "$M_1$", red);
label((1.5,5.5), "$\bigstar$", deepgreen);
label((3.5,5.5), "$\bigstar$", deepgreen);
draw((1.5,5.5)--(1.5,3.5)--(2.3,3.5)--(2.3,0.2), deepgreen+1.2, EndArrow(TeXHead));
draw((3.5,5.5)--(3.5,3.5)--(2.7,3.5)--(2.7,0.2), deepgreen+1.2, EndArrow(TeXHead));
\end{asy}
\end{center}
\item \textbf{Outside happy triangle:}
Now suppose the first monster $\color{red}M_1$ is outside the \emph{happy triangle}.
Of the two shoulders, take the one closer to the center
(if in the center column, either one works; if only one shoulder, use it).
If there is a safe path to that shoulder,
then one can take a staircase pattern towards the center, as shown in the figure.
In that case, the choice of shoulder and position guarantees the staircase
reaches the bottom row, so that if no monster is along this path, the algorithm ends.
Otherwise, if one encounters a second monster along the staircase,
then one can use the third trial to cut under the monster $\color{red}M_1$.
\begin{center}
\begin{asy}
usepackage("amssymb");
unitsize(0.7cm);
pen gr = gray+linetype("4 2");
void setup(int n) {
for (int i=0; i<=n-1; ++i) {
draw((0,i)--(n,i), gr);
}
for (int i=0; i<=n; ++i) {
draw((i,0)--(i,n-1), gr);
}
draw(box((0,-1), (n,0)), black);
draw(box((0,n-1), (n,n)), black);
draw(box((0,-1), (n,n)), black);
label((n/2,n-0.5), "Starting row");
label((n/2,-0.5), "Goal row");
path p = (0,n);
for (int i=0; i<n/2; ++i) {
p = p--(i,n-1-i)--(i+1,n-1-i);
}
for (int i=(n+1)#2; i<n; ++i) {
p = p--(i,i)--(i+1,i);
}
p = p--(n,n)--cycle;
filldraw(p, opacity(0.15)+yellow, blue+1.8);
}
setup(9);
label((1.5,3.5), "$M_1$", red);
label((2.5,4.5), "$\bigstar$", deepgreen);
draw((2.5,4.5)--(2.5,3.5)--(3.5,3.5)--(3.5,2.5)--(4.5,2.5)--(4.5,1.5)--(5.5,1.5)--(5.5,0.2),
deepgreen+1.2, EndArrow(TeXHead));
\end{asy}
\qquad
\begin{asy}
usepackage("amssymb");
unitsize(0.7cm);
pen gr = gray+linetype("4 2");
void setup(int n) {
for (int i=0; i<=n-1; ++i) {
draw((0,i)--(n,i), gr);
}
for (int i=0; i<=n; ++i) {
draw((i,0)--(i,n-1), gr);
}
draw(box((0,-1), (n,0)), black);
draw(box((0,n-1), (n,n)), black);
draw(box((0,-1), (n,n)), black);
label((n/2,n-0.5), "Starting row");
label((n/2,-0.5), "Goal row");
path p = (0,n);
for (int i=0; i<n/2; ++i) {
p = p--(i,n-1-i)--(i+1,n-1-i);
}
for (int i=(n+1)#2; i<n; ++i) {
p = p--(i,i)--(i+1,i);
}
p = p--(n,n)--cycle;
filldraw(p, opacity(0.15)+yellow, blue+1.8);
}
setup(9);
label((1.5,3.5), "$M_1$", red);
label((2.5,4.5), "$\bigstar$", deepgreen);
draw((2.5,4.5)--(2.5,3.5)--(3.5,3.5)--(3.5,2.5)--(4.5,2.5)--(4.5,1.5)--(5.5,1.5)--(5.5,0.2),
deepgreen+dashed, EndArrow(TeXHead));
label((5.5,1.5), "$M_2$", red);
draw((2.5,4.5)--(2.5,3.5)--(3.5,3.5)--(3.5,2.5)--(4.5,2.5)--(4.5,1.5)--(1.5,1.5)--(1.5,0.2),
purple+1.5, EndArrow(TeXHead));
\end{asy}
\end{center}
\end{itemize}
We now prove the following proposition:
in any valid strategy for Turbo,
in the case where Turbo first encounters a monster upon leaving the happy triangle,
the second path \emph{must} outline the same staircase shape.
The monsters pre-commit to choosing their pattern to be
\emph{either} a NW-SE diagonal or NE-SW diagonal, with a single one-column gap;
see figure below for an example.
Note that this forces any valid path for Turbo to pass through the particular gap.
\begin{center}
\begin{asy}
usepackage("amssymb");
unitsize(0.7cm);
pen gr = gray+linetype("4 2");
void setup(int n) {
for (int i=0; i<=n-1; ++i) {
draw((0,i)--(n,i), gr);
}
for (int i=0; i<=n; ++i) {
draw((i,0)--(i,n-1), gr);
}
draw(box((0,-1), (n,0)), black);
draw(box((0,n-1), (n,n)), black);
draw(box((0,-1), (n,n)), black);
label((n/2,n-0.5), "Starting row");
label((n/2,-0.5), "Goal row");
path p = (0,n);
for (int i=0; i<n/2; ++i) {
p = p--(i,n-1-i)--(i+1,n-1-i);
}
for (int i=(n+1)#2; i<n; ++i) {
p = p--(i,i)--(i+1,i);
}
p = p--(n,n)--cycle;
filldraw(p, opacity(0.15)+yellow, blue+1.8);
}
setup(11);
for (int i=0; i<7; ++i) {
label((i+0.5,9.5-i), "$M$", red);
}
for (int i=7; i<10; ++i) {
label((i+1.5,9.5-i), "$M$", red);
}
\end{asy}
\end{center}
We may assume without loss of generality that Turbo first encounters a monster $M_1$
when Turbo first leaves the happy triangle, and that this forces an NW-SE configuration.
\begin{center}
\begin{asy}
usepackage("amssymb");
unitsize(0.7cm);
pen gr = gray+linetype("4 2");
void setup(int n) {
for (int i=0; i<=n-1; ++i) {
draw((0,i)--(n,i), gr);
}
for (int i=0; i<=n; ++i) {
draw((i,0)--(i,n-1), gr);
}
draw(box((0,-1), (n,0)), black);
draw(box((0,n-1), (n,n)), black);
draw(box((0,-1), (n,n)), black);
label((n/2,n-0.5), "Starting row");
label((n/2,-0.5), "Goal row");
path p = (0,n);
for (int i=0; i<n/2; ++i) {
p = p--(i,n-1-i)--(i+1,n-1-i);
}
for (int i=(n+1)#2; i<n; ++i) {
p = p--(i,i)--(i+1,i);
}
p = p--(n,n)--cycle;
filldraw(p, opacity(0.15)+yellow, blue+1.8);
}
setup(13);
label((0.5,11.5), "($M$)", red+fontsize(9pt));
label((1.5,10.5), "($M$)", red+fontsize(9pt));
label((2.5,9.5), "($M$)", red+fontsize(9pt));
label((3.5,8.5), "$M_1$", red);
label((4.5,7.5), "X", brown);
label((11.5,1.5), "$M_2$", red);
for (int i=1; i<=6; ++i) {
label((11.5-i,1.5+i), "$\clubsuit$", deepgreen);
}
\end{asy}
\end{center}
Then the following is true:
\begin{proposition*}
The strategy of Turbo on the second path \emph{must}
visit every cell in ``slightly raised diagonal'' marked with
$\color{green!60!black}\clubsuit$ in the figure above
in order from top to bottom, until it encounters a second Monster $M_2$
(or reaches the bottom row and wins anyway).
It's both okay and irrelevant if Turbo visits other cells above this diagonal,
but the marked cells must be visited from top to bottom in that order.
\end{proposition*}
\begin{proof}
If Turbo tries to sidestep by visiting the cell southeast of $M_1$
(marked {\color{brown}X} in the Figure),
then Turbo clearly cannot finish after this (for $s$ large enough).
Meanwhile, suppose Turbo tries to ``skip'' one of the $\color{green!60!black}\clubsuit$,
say in column $C$, then the gap could equally well be in the column to the left of $C$.
This proves the proposition.
\end{proof}
\begin{remark*}
[Memories of safe cells are important, not just monster cells]
Here is one additional observation that one can deduce from this.
We say a set $\mathcal S$ of revealed monsters is called \emph{obviously winnable} if,
based on only the positions of the monsters
(and not the moves or algorithm that were used to obtain them),
one can identify a guaranteed winning path for Turbo using only $\mathcal S$.
For example, two monsters in adjacent columns which are not diagonally
adjacent is obviously winnable.
Then no strategy can guarantee obtaining an obviously winnable set in $2$ moves
(or even $k$ moves for any constant $k$, if $s$ is large enough in terms of $k$).
So any valid strategy must \emph{also} use the \emph{memory} of identified safe cells
that do not follow just from the revealed monster positions.
\end{remark*}
|
IMO-2024-notes_6
|
A function $f \colon \QQ \to \QQ$ is called \emph{aquaesulian}
if the following property holds: for every $x,y \in \mathbb{Q}$,
\[ f(x+f(y)) = f(x) + y \quad \text{or} \quad f(f(x)+y) = x + f(y). \]
Show that there exists an integer $c$ such that for any aquaesulian function $f$
there are at most $c$ different rational numbers of the
form $f(r) + f(-r)$ for some rational number $r$,
and find the smallest possible value of $c$.
\end{enumerate}
|
We will prove that
\[ \left\{ f(x) + f(-x) \mid x \in \QQ \right\} \]
contains at most $2$ elements
and give an example where there are indeed $2$ elements.
We fix the notation $x \to y$ to mean that $f(x+f(y)) = f(x)+y$.
So the problem statement means that either $x \to y$ or $y \to x$ for all $x$, $y$.
In particular, we always have $x \to x$, and hence
\[ f(x+f(x)) = x+f(x) \]
for every $x$.
\paragraph{Construction.}
The function
\[ f(x) = \left\lfloor 2x \right\rfloor - x \]
can be seen to satisfy the problem conditions.
Moreover, $f(0)+f(0) = 0$ but $f(1/3)+f(-1/3) = -1$.
\begin{remark*}
Here is how I (Evan) found the construction.
Let $h(x) \coloneq x+f(x)$, and let $S \coloneq h(\QQ) = \{h(x) \mid x \in \QQ\}$.
Hence $f$ is the identity on all of $S$.
If we rewrite the problem condition in terms of $h$ instead of $f$,
it asserts that at least one of the equations
\begin{align*}
h(x+h(y)-y) &= h(x)+h(y) \\
h(y+h(x)-x) &= h(x)+h(y)
\end{align*}
is true.
In particular, $S$ is closed under addition.
Now, the two trivial solutions for $h$ are $h(x) = 2x$ and $h(x) = 0$.
To get a nontrivial construction, we must also have $S \neq \{0\}$ and $S \neq \QQ$.
So a natural guess is to take $S = \ZZ$.
And indeed $h(x) = \left\lfloor 2x \right\rfloor$ works fine.
\end{remark*}
\begin{remark*}
This construction is far from unique. For example,
$f(x) = 2\left\lfloor x \right\rfloor - x = \left\lfloor x \right\rfloor - \{x\}$
seems to have been more popular to find.
\end{remark*}
\paragraph{Proof (communicated by Abel George Mathew).}
We start by proving:
\begin{claim*}
$f$ is injective.
\end{claim*}
\begin{proof}
Suppose $f(a) = f(b)$. WLOG $a \to b$. Then
\[ f(a)+a = f(a+f(a)) = f(a+f(b)) = f(a)+b \implies a=b. \qedhere. \]
\end{proof}
\begin{claim*}
Suppose $s \to r$.
Then either $f(r) + f(-r) = 0$ or $f(f(s)) = s+f(r)+f(-r)$.
\end{claim*}
\begin{proof}
Take the given statement with $x = s+f(r)$ and $y = -r$; then
\begin{align*}
x + f(y) &= s + f(r) + f(-r) \\
y + f(x) &= f(s+f(r)) - r = f(s).
\end{align*}
Because $f$ is injective, if $x \to y$ then $f(r) + f(-r) = 0$.
Meanwhile, if $y \to x$ then indeed $f(f(s)) = s + f(r) + f(-r)$.
\end{proof}
Finally, suppose $a$ and $b$ are different numbers for which
$f(a)+f(-a)$ and $f(b)+f(-b)$ are both nonzero.
Again, WLOG $a \to b$.
Then
\[ f(a) + f(-a) \overset{a \to a}{=} f(f(a))-a \overset{a \to b}{=} f(b) + f(-b). \]
This shows at most two values can occur.
\begin{remark*}
The above solution works equally well for $f \colon \RR \to \RR$.
But the choice of $\QQ$ permits some additional alternate solutions.
\end{remark*}
\begin{remark*}
After showing $f$ injective,
a common lemma proved is that $-f(-f(x)) = x$, i.e.\ $f$ is an involution.
This provides some alternative paths for solutions.
\end{remark*}
|
IMO-2025-notes_1
|
A line in the plane is called \emph{sunny}
if it is not parallel to any of the $x$–axis, the $y$–axis, or the line $x+y=0$.
Let $n \ge 3$ be a given integer.
Determine all nonnegative integers $k$ such that there exist $n$ distinct lines
in the plane satisfying both of the following:
\begin{itemize}
\ii for all positive integers $a$ and $b$ with $a+b\le n+1$,
the point $(a,b)$ lies on at least one of the lines; and
\ii exactly $k$ of the $n$ lines are sunny.
\end{itemize}
|
The answer is $0$, $1$, or $3$ sunny lines.
In what follows, we draw the grid as equilateral instead of a right triangle;
this has no effect on the problem statement but is more symmetric.
We say a \emph{long line} is one of the three lines at the edge of the grid,
i.e.\ one of the (non-sunny) lines passing through $n$ points.
The main claim is the following.
\begin{claim*}
If $n \ge 4$, any set of $n$ lines must have at least one long line.
\end{claim*}
\begin{proof}
Consider the $3(n-1)$ points on the outer edge of the grid.
If there was no long line, each of the $n$ lines passes through at most two such points.
So we obtain $2n \ge 3(n-1)$, which forces $n \le 3$.
\end{proof}
Hence, by induction we may repeatedly delete a long line without changing the number
of sunny lines until $n = 3$ (and vice-versa: given a construction for smaller $n$
we can increase $n$ by one and add a long line).
We now classify \emph{all} the ways to cover
the $1+2+3=6$ points in an $n=3$ grid with $3$ lines.
\begin{center}
\begin{asy}
dotfactor *= 2;
unitsize(2cm);
pair A = dir(90);
pair B = dir(210);
pair C = dir(330);
pair X = midpoint(B--C);
pair Y = midpoint(C--A);
pair Z = midpoint(A--B);
picture tri;
dot(tri, A);
dot(tri, B);
dot(tri, C);
dot(tri, X);
dot(tri, Y);
dot(tri, Z);
picture left_case;
picture right_case;
add(left_case, tri);
add(right_case, tri);
draw(left_case, (1.2*A-0.2*B)--(1.2*B-0.2*A), red, Arrows);
draw(left_case, circle((C+X+Y)/3, 0.75), blue);
label(left_case, "Long line present", 1.2*dir(-90));
draw(right_case, (1.3*A-0.3*X)--(1.3*X-0.3*A), deepgreen, Arrows);
draw(right_case, (1.3*B-0.3*Y)--(1.3*Y-0.3*B), deepgreen, Arrows);
draw(right_case, (1.3*C-0.3*Z)--(1.3*Z-0.3*C), deepgreen, Arrows);
label(right_case, "No long line", 1.2*dir(-90));
add(left_case);
add(shift(3,0)*right_case);
\end{asy}
\end{center}
\begin{itemize}
\ii If there is a long line (say, the red one in the figure),
the remaining $1+2=3$ points (circled in blue) are covered with two lines.
One of the lines passes through $2$ points and must not be sunny;
the other line may or may not be sunny.
Hence in this case the possible counts of sunny lines are $0$ or $1$.
\ii If there is no long line, each of the three lines passes through at most $2$ points.
But there are $6$ total lines, so in fact each line must pass through
\emph{exactly} two points.
The only way to do this is depicted in the figure in the right.
In this case there are $3$ sunny lines.
\end{itemize}
This proves that $0$, $1$, $3$ are the only possible answers.
\begin{remark*}
The concept of a sunny line is not that important to the problem.
The proof above essentially classifies all the ways to cover
the $1+2+\dots+n$ points with exactly $n$ lines.
Namely, one should repeatedly take a long line and decrease $n$ until $n=3$,
and then pick one of the finitely many cases for $n=3$.
The count of sunny lines just happens to be whatever is possible for $n=3$,
since long lines are not sunny.
\end{remark*}
|
IMO-2025-notes_2
|
Let $\Omega$ and $\Gamma$ be circles with centres $M$ and $N$, respectively,
such that the radius of $\Omega$ is less than the radius of $\Gamma$.
Suppose $\Omega$ and $\Gamma$ intersect at two distinct points $A$ and $B$.
Line $MN$ intersects $\Omega$ at $C$ and $\Gamma$ at $D$,
so that $C$, $M$, $N$, $D$ lie on $MN$ in that order.
Let $P$ be the circumcenter of triangle $ACD$.
Line $AP$ meets $\Omega$ again at $E \neq A$ and meets $\Gamma$ again at $F \neq A$.
Let $H$ be the orthocenter of triangle $PMN$.
Prove that the line through $H$ parallel to $AP$ is tangent
to the circumcircle of triangle $BEF$.
|
Throughout the solution, we define
\begin{align*}
\alpha &\coloneq \dang DCA = \dang BCD \implies \dang PAD = \dang CAB = 90\dg - \alpha \\
\beta &\coloneq \dang ADC = \dang CDB \implies \dang CAP = \dang BAD = 90\dg - \beta.
\end{align*}
Ignore the points $H$, $M$, $N$ for now and focus on the remaining ones.
\begin{claim*}
We have $\ol{CE} \parallel \ol{AD}$ and $\ol{DF} \parallel \ol{AC}$.
\end{claim*}
\begin{proof}
$\dang AEC = \dang ABC = \dang CAB = 90\dg - \alpha$.
\end{proof}
Hence, if we let $A' \coloneq \ol{CE} \cap \ol{DF}$, we have a parallelogram $ACA'D$.
Note in particular that $\ol{BA'} \parallel \ol{CD}$.
\begin{center}
\begin{asy}
size(15cm);
pair A = dir(117);
pair C = dir(150);
pair D = dir(30);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
pair B = -A+2*foot(A, C, D);
pair M = circumcenter(A, B, C);
pair N = circumcenter(A, B, D);
pair P = origin;
draw(circumcircle(A, C, B), gray);
draw(circumcircle(A, D, B), gray);
draw(C--D, blue);
draw(A--B, blue);
pair H = orthocenter(P, M, N);
pair A_prime = C+D-A;
pair T = M+N-A;
pair E = extension(A, P, M, T);
pair F = extension(A, P, N, T);
draw(A--F, lightred);
draw(C--A_prime, lightred);
draw(F--D, lightred);
draw(M--H--N, lightred);
draw(circumcircle(A_prime, E, F), gray);
draw(A_prime--B, blue);
draw(circumcircle(B, E, F), purple);
draw(M--T, dashed);
draw(N--F, dashed);
draw(C--A--D, lightred);
markangle("$\alpha$", n=1, radius=18.0, D,C,A, deepgreen);
markangle("$\beta$", n=2, radius=30.0, A,D,C, deepgreen);
markangle("$\alpha$", n=1, radius=18.0, C,D,F, deepgreen);
markangle("$\beta$", n=2, radius=30.0, E,C,D, deepgreen);
markangle("$90^{\circ}-\alpha$", n=3, radius=14.0, C,A,B, deepcyan);
markangle("$90^{\circ}-\alpha$", n=3, radius=14.0, E,A,D, deepcyan);
dot("$A$", A, dir(A));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$B$", B, dir(225));
dot("$M$", M, dir(125));
dot("$N$", N, dir(N));
dot("$P$", P, dir(250));
dot("$H$", H, dir(270));
dot("$A'$", A_prime, dir(315));
dot("$T$", T, dir(230));
dot("$E$", E, dir(260));
dot("$F$", F, dir(315));
/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
!size(15cm);
A = dir 117
C = dir 150
D = dir 30
unitcircle / 0.1 lightcyan / blue
B 225 = -A+2*foot A C D
M 125 = circumcenter A B C
N = circumcenter A B D
P 250 = origin
circumcircle A C B / gray
circumcircle A D B / gray
C--D / blue
A--B / blue
H 270 = orthocenter P M N
A' 315 = C+D-A
T 230 = M+N-A
E 260 = extension A P M T
F 315 = extension A P N T
A--F / lightred
C--A' / lightred
F--D / lightred
M--H--N / lightred
circumcircle A' E F / gray
A'--B / blue
circumcircle B E F / purple
M--T / dashed
N--F / dashed
C--A--D / lightred
!markangle("$\alpha$", n=1, radius=18.0, D,C,A, deepgreen);
!markangle("$\beta$", n=2, radius=30.0, A,D,C, deepgreen);
!markangle("$\alpha$", n=1, radius=18.0, C,D,F, deepgreen);
!markangle("$\beta$", n=2, radius=30.0, E,C,D, deepgreen);
!markangle("$90^{\circ}-\alpha$", n=3, radius=14.0, C,A,B, deepcyan);
!markangle("$90^{\circ}-\alpha$", n=3, radius=14.0, E,A,D, deepcyan);
*/
\end{asy}
\end{center}
Next, let $T$ denote the circumcenter of $\triangle A'EF$.
(This will be the tangency point later in the problem.)
\begin{claim*}
Point $T$ also lies on $\ol{BA'}$ and is also the arc midpoint of $\widehat{EF}$ on $(BEF)$.
\end{claim*}
\begin{proof}
We compute the angles of $\triangle A'EF$:
\begin{align*}
\dang FEA' &= \dang AEC = \dang ABC = \dang CAB = 90\dg - \alpha \\
\dang A'FE &= \dang DFA = \dang DBA = \dang BAD = 90\dg - \beta \\
\dang EA'F &= \alpha + \beta.
\end{align*}
Then, since $T$ is the circumcenter, it follows that:
\[ \dang EA'T = \dang 90\dg - \dang A'FE = \beta = \dang A'CD = \dang CA'B. \]
This shows that $T$ lies on $\ol{BA'}$.
Also, we have $\dang ETF = 2 \dang EA'F = 2(\alpha+\beta)$ and
\begin{align*}
\dang EBF &= \dang EBA + \dang ABF = \dang ECA + \dang ADF \\
&= \dang A'CA + \dang ADA' = (\alpha+\beta) + (\alpha+\beta) = 2(\alpha+\beta)
\end{align*}
which proves that $T$ also lies on $\ol{AB'}$.
\end{proof}
We then bring $M$ and $N$ into the picture as follows:
\begin{claim*}
Point $T$ lies on both lines $ME$ and $NF$.
\end{claim*}
\begin{proof}
To show that $F$, $T$, $N$ are collinear,
note that $\triangle FEA' \sim \triangle FAD$ via a homothety at $F$.
This homothety maps $T$ to $N$.
\end{proof}
We now deal with point $H$ using two claims.
\begin{claim*}
We have $\ol{MH} \parallel \ol{AD}$ and $\ol{NH} \parallel \ol{AC}$.
\end{claim*}
\begin{proof}
Note that $\ol{MH} \perp \ol{PN}$,
but $\ol{PN}$ is the perpendicular bisector of $\ol{AD}$,
so in fact $\ol{MH} \parallel \ol{AD}$.
Similarly, $\ol{NH} \parallel \ol{AC}$.
\end{proof}
\begin{claim*}
Lines $\ol{MH}$ and $\ol{NH}$ bisect $\angle NMT$ and $\angle MNT$.
In fact, point $H$ is the incenter of $\triangle TMN$,
and $\dang NTH = \dang HTM = 90\dg-(\alpha+\beta)$.
\end{claim*}
\begin{proof}
Hence, $\dang HMN = \dang A'CD = \dang ADC = \beta$.
But $\dang TMN = \dang CME = 2 \dang CAE = -2(90\dg-2\beta) = 2\beta$.
That proves $\ol{MH}$ bisects $\angle NMT$; the other one is similar.
To show that $H$ is an incenter (rather than an excenter)
and get the last angle equality, we need to temporarily undirect our angles.
Assume WLOG that $\triangle ACD$ is directed counterclockwise.
The problem condition that $C$ and $D$ are the farther intersections of line $MN$
mean that $\angle NHM = \angle CAD > 90\dg$.
We are also promised $C$, $M$, $N$, $D$ are collinear in that order.
Hence the reflections of line $MN$ over lines $MH$ and $NH$, which meet at $T$,
should meet at a point for which $T$ lies on the same side as $H$.
In other words, $\triangle MTN$ is oriented counterclockwise and contains $H$.
Working with undirected $\alpha = \angle DCA$ and $\beta = \angle ADC$ with $\alpha + \beta < 90\dg$,
\[ \angle NTH = \angle HTM = \half \angle NTM = \half(180\dg-2(\alpha+\beta))
= 90\dg - (\alpha + \beta). \]
This matches the claim and finishes the result.
\end{proof}
Now
\[ \dang NFA = 90\dg - \dang ADF = 90\dg-(\alpha+\beta) = \dang NTH \]
so $\ol{HT} \parallel \ol{AP}$.
And since $TE = TF$, we have the tangency requested too now, as desired.
\begin{remark*}
There are many other ways to describe the point $T$.
For example, $AMTN$ is a parallelogram and $MBTN$ is an isosceles trapezoid.
In coordination, we joked that it was impossible to write a false conjecture.
\end{remark*}
|
IMO-2025-notes_3
|
A function $f \colon \NN \to \NN$ is said to be \emph{bonza} if
\[ f(a)\quad\text{divides}\quad b^a-f(b)^{f(a)} \]
for all positive integers $a$ and $b$.
Determine the smallest real constant $c$ such that $f(n) \leq cn$
for all bonza functions $f$ and all positive integers $n$.
|
The answer is $c=4$.
Let $P(a,b)$ denote the given statement $f(a) \mid b^a - f(b)^{f(a)}$.
\begin{claim*}
We have $f(n) \mid n^n$ for all $n$.
\end{claim*}
\begin{proof}
Take $P(n,n)$.
\end{proof}
\begin{claim*}
Unless $f = \id$, we have $f(p) = 1$ for all odd primes $p$.
\end{claim*}
\begin{proof}
Consider any prime $q$ with $f(q) > 1$.
Then $f(q)$ is a power of $q$, and for each $n$ we get
\[ P(q,n) \implies q \mid f(q) \mid n^q - f(n)^{f(q)}. \]
Fermat's little theorem now gives
$n^q \equiv n \pmod q$ and $f(n)^{f(q)} \equiv f(n) \pmod q$
(since $f(q)$ is a power of $q$), and therefore $q \mid n - f(n)$.
Hence, unless $f$ is the identity function,
only finitely many $q$ could have $f(q) > 1$.
Now let $p$ be any odd prime,
and let $q$ be a large prime such that $q \not\equiv 1 \pmod p$
(possible for all $p > 2$, say by Dirichlet).
Then
\[ P(p,q) \implies f(p) \mid q^p - 1^p. \]
The RHS is $q^p - 1 \equiv q - 1 \not\equiv 0 \pmod p$, so $f(p) = 1$.
\end{proof}
\begin{claim*}
We have $f(n) \mid 2^\infty$ for all $n$.
\end{claim*}
\begin{proof}
If $p \mid f(n)$ is odd then $P(n,p)$ gives $p \mid f(n) \mid p^n-1^n$,
contradiction.
\end{proof}
(In particular, we now know $f(n) = 1$ for all odd $n$, though we don't use this.)
\begin{claim*}
We have $f(n) \le 2^{\nu_2(n)+2}$ for all $n$.
\end{claim*}
\begin{proof}
Consider $P(n,5) \implies f(n) \mid 5^n - 1^n$.
It's well-known that $\nu_2(5^n-1) = \nu_2(n)+2$ for all $n$.
\end{proof}
This immediately shows $f(n) \le 4n$ for all $n$, hence $c=4$ in the problem statement works.
For the construction, the simplest one seems to be
\[
f(n) = \begin{cases}
1 & n \text{ odd} \\
16 & n = 4 \\
2 & n \text{ even}, n \neq 4
\end{cases}
\]
which is easily checked to work and has $f(4) = 16$.
\begin{remark*}
With a little more case analysis we can classify all functions $f$.
The two trivial solutions are $f(n) = n$ and $f(n) = 1$;
the others are described by writing $f(n) = 2^{e(n)}$ for any function $e$ satisfying
\begin{itemize}
\ii $e(n) = 0$ for odd $n$;
\ii $1 \le e(2) \le 2$;
\ii $1 \le e(n) \le \nu_2(n)+2$ for even $n > 2$.
\end{itemize}
This basically means that there are almost no additional constraints
beyond what is suggested by the latter two claims.
\end{remark*}
|
IMO-2025-notes_4
|
An infinite sequence $a_1$, $a_2$, \dots\ consists of positive integers
has each of which has at least three proper divisors.
Suppose that for each $n\geq 1$,
$a_{n+1}$ is the sum of the three largest proper divisors of $a_n$.
Determine all possible values of $a_1$.
|
The answer is $a_1 = 12^e \cdot 6 \cdot \ell$
for any $e, \ell \ge 0$ with $\gcd(\ell, 10) = 1$.
Let $\mathbf{S}$ denote the set of positive integers with at least three divisors.
For $x \in \mathbf{S}$, let $\psi(x)$ denote the sum of the three largest ones,
so that $\psi(a_n) = a_{n+1}$.
\paragraph{Proof that all such $a_1$ work.}
Let $x = 12^e \cdot 6 \cdot \ell \in \mathbf{S}$ with $\gcd(\ell, 10) = 1$.
As $\frac12+\frac13+\frac14 = \frac{13}{12}$, we get
\[ \psi(x) = \begin{cases}
x & e = 0 \\
\frac{13}{12} x & e > 0
\end{cases} \]
so by induction on the value of $e$ we see that $\psi(x) \in \mathbf{S}$
(the base $e=0$ coming from $\psi$ fixing $x$).
\paragraph{Proof that all $a_1$ are of this form.}
In what follows $x$ is always an element of $\mathbf{S}$,
not necessarily an element of the sequence.
\begin{claim*}
Let $x \in \mathbf{S}$. If $2 \mid \psi(x)$ then $2 \mid x$.
\end{claim*}
\begin{proof}
If $x$ is odd then every divisor of $x$ is odd,
so $\psi(x)$ is the sum of three odd numbers.
\end{proof}
\begin{claim*}
Let $x \in \mathbf{S}$. If $6 \mid \psi(x)$ then $6 \mid x$.
\end{claim*}
\begin{proof}
We consider only $x$ even because of the previous claim.
We prove the contrapositive that $3 \nmid x \implies 6 \nmid \psi(x)$ (for even $x$).
\begin{itemize}
\ii If $4 \mid x$, then letting $d$ be the third largest proper divisor of $x$,
\[ \psi(x) = \frac x2 + \frac x4 + d = \frac 34 x + d \equiv d \not\equiv 0 \pmod 3. \]
\ii Otherwise, let $p \mid x$ be the smallest prime dividing $x$, with $p > 3$.
If the third-smallest nontrivial divisor of $x$ is $2p$, then
\[ \psi(x) = \frac x2 + \frac xp + \frac{x}{2p} = \frac{3}{2p} x + \frac x2
\equiv \frac x2 \not\equiv 0 \pmod 3. \]
If the third-smallest nontrivial divisor of $x$ is instead an odd prime $q$, then
\[ \psi(x) = \frac x2 + \frac xp + \frac{x}{q} \equiv 1+0+0 \equiv 1 \pmod 2. \qedhere \]
\end{itemize}
\end{proof}
To tie these two claims into the problem, we assert:
\begin{claim*}
Every $a_i$ must be divisible by $6$.
\end{claim*}
\begin{proof}
The idea is to combine the previous two claims
(which have no dependence on the sequence) with a size argument.
\begin{itemize}
\ii For odd $x \in \mathbf{S}$ note that $\psi(x) < \left( \frac13+\frac15+\frac17 \right) x < x$
and $\psi(x)$ is still odd.
So if any $a_i$ is odd the sequence is strictly decreasing and that's impossible.
Hence, we may assume $a_1$, $a_2$, \dots\ are all even.
\ii If $x \in \mathbf{S}$ is even but $3 \nmid x$ then $\psi(x) < \left( \half+\frac14+\frac15 \right) x < x$
and $\psi(x)$ is still not a multiple of $3$.
So if any $a_i$ is not divisible by $3$ the sequence is again strictly decreasing.
\qedhere
\end{itemize}
\end{proof}
On the other hand, if $x$ is a multiple of $6$, we have the following formula for $\psi(x)$:
\[ \psi(x) = \begin{cases}
\frac{13}{12}x & 4 \mid x \\
\frac{31}{30}x & 4 \nmid x \text{ but } 5 \mid x \\
x & 4 \nmid x \text{ and } 5 \nmid x.
\end{cases} \]
Looking back on our sequence of $a_i$ (which are all multiples of $6$),
the center case cannot happen with our $a_i$, because $\frac{31}{30}x$
is odd when $x \equiv 2 \pmod 4$.
Hence in actuality
\[ a_{n+1} = \frac{13}{12} a_n \quad\text{or}\quad a_{n+1} = a_n \]
for every $n$.
Let $T$ be the smallest index such that $a_T = a_{T+1} = a_{T+2} = \dotsb$
(it must exist because we cannot multiply by $\frac{13}{12}$ forever).
Then we can exactly describe the sequence by
\[ a_n = a_1 \cdot \left( \frac{13}{12} \right)^{\min(n,T)-1}. \]
Hence $a_1 = \left( \frac{12}{13} \right)^{T-1} a_T$,
and since $a_T$ is a multiple of $6$ not divisible by $4$ or $5$,
it follows $a_1$ has the required form.
|
IMO-2025-notes_5
|
Alice and Bazza are playing the \emph{inekoalaty game},
a two‑player game whose rules depend on a positive real number $\lambda$ which is known to both players.
On the $n$th turn of the game (starting with $n=1$) the following happens:
\begin{itemize}
\ii If $n$ is odd, Alice chooses a nonnegative real number $x_n$ such that
\[ x_1 + x_2 + \cdots + x_n \le \lambda n. \]
\ii If $n$ is even, Bazza chooses a nonnegative real number $x_n$ such that
\[ x_1^2 + x_2^2 + \cdots + x_n^2 \le n. \]
\end{itemize}
If a player cannot choose a suitable $x_n$, the game ends and the other player wins.
If the game goes on forever, neither player wins.
All chosen numbers are known to both players.
Determine all values of $\lambda$ for which Alice has a winning strategy
and all those for which Bazza has a winning strategy.
|
The answer is that Alice has a winning strategy for $\lambda > 1/\sqrt2$,
and Bazza has a winning strategy for $\lambda < 1/\sqrt2$.
(Neither player can guarantee winning for $\lambda = 1/\sqrt2$.)
We divide the proof into two parts.
\paragraph{Alice's strategy when $\lambda \ge 1/\sqrt2$.}
Consider the strategy where Alice always plays $x_{2i+1} = 0$ for $i=0, \dots, k-1$.
In this situation, when $n = 2k+1$ we have
\begin{align*}
\sum_1^{2k} x_i &= 0 + x_2 + 0 + x_4 + \dots + 0 + x_{2k} \\
&\le k \cdot \sqrt{\frac{x_2^2 + \dots + x_{2k}^2}{k}}
= \sqrt{2} \cdot k < \lambda \cdot (2k+1)
\end{align*}
and so the choices for $x_{2k+1}$ are
\[ x_{2k+1} \in [0, \lambda \cdot (2k+1) - \sqrt{2} k] \]
which is nonempty.
Hence Alice can't ever lose with this strategy.
But suppose further $\lambda > \frac{1}{\sqrt 2}$; we show Alice can win.
Choose $k$ large enough that
\[ \sqrt{2} \cdot k < \lambda \cdot (2k+1) - \sqrt{2k+2}. \]
Then on the $(2k+1)$st turn, Alice can (after playing $0$ on all earlier turns)
play a number greater than $\sqrt{2k+2}$ and cause Bazza to lose.
\paragraph{Bazza strategy when $\lambda \le 1/\sqrt2$.}
Consider the strategy where Bazza always plays $x_{2i+2} = \sqrt{2 - x_{2i+1}^2}$
for all $i = 0, \dots, k-1$ (i.e.\ that is, the largest possible value Bazza can play).
To analyze Bazza's choices on each of his turns, we first need to estimate $x_{2k+1}$.
We do this by writing
\begin{align*}
\lambda \cdot (2k+1) &\ge x_1 + x_2 + \dots + x_{2k+1} \\
&= \left(x_1 + \sqrt{2-x_1^2}\right) + \left(x_3 + \sqrt{2-x_3^2}\right) \\
&\qquad+ \dots + \left(x_{2k-1} + \sqrt{2-x_{2k}^2}\right) + x_{2k+1} \\
&\ge {\underbrace{\sqrt{2} + \dots + \sqrt{2}}_{k \text{ times}}} + x_{2k+1}
= \sqrt 2 \cdot k + x_{2k+1}
\end{align*}
where we have used the fact that $t + \sqrt{2-t^2} \ge 2$ for all $t \ge 0$.
This means that
\[ x_{2k+1} \le \lambda \cdot (2k+1) - \sqrt 2 k < \sqrt 2. \]
And $x_1^2 + x_2^2 + \dots + x_{2k}^2 + x_{2k+1}^2 = (2 + \dots + 2) + x_{2k+1}^2$,
Bazza can indeed choose $x_{2k+2} = \sqrt{2-x_{2k+1}^2}$ and always has a move.
But suppose further $\lambda < 1/\sqrt2$.
Then the above calculation \emph{also} shows that Alice couldn't
have made a valid choice for large enough $k$,
since $\lambda \cdot (2k+1) - \sqrt 2 k < 0$ for large $k$.
\begin{remark*}
In the strategies above,
we saw that Alice prefers to always play $0$
and Bazza prefers to always play as large as possible.
One could consider what happens in the opposite case:
\begin{itemize}
\ii If Alice tries to always play the largest number possible,
her strategy still wins for $\lambda > 1$.
\ii If Bazza tries to always play $0$,
Alice can win no matter the value for $\lambda > 0$.
\end{itemize}
\end{remark*}
|
IMO-2025-notes_6
|
Consider a $2025 \times 2025$ grid of unit squares.
Matilda wishes to place on the grid some rectangular tiles, possibly of different sizes,
such that each side of every tile lies on a grid line
and every unit square is covered by at most one tile.
Determine the minimum number of tiles Matilda needs to place
so that each row and each column of the grid has exactly one unit square
that is not covered by any tile.
\end{enumerate}
|
The answer is $2112 = 2025 + 2 \cdot 45 - 3$.
In general, the answer turns out to be $\left\lceil n + 2 \sqrt n - 3 \right\rceil$,
but when $n$ is not a perfect square the solution is more complicated.
\begin{remark*}
The 2017 Romanian Masters in Math asked the same problem where the tiles
are replaced by \emph{sticks}, i.e.\ $1 \times k$ tiles.
The answer to that problem is completely different and as far as I know
there is no connection at all to the present IMO problem.
\end{remark*}
\paragraph{Construction.}
We show a general construction when $n = k^2$ illustrated below for $k=5$;
it generalizes readily.
There are a total of $(k-1)^2$ tiles which are $k \times k$ squares and another
$4(k-1)$ tiles on the boundary, giving a total of
\[ (k-1)^2 + 4(k-1) = k^2 + 2k - 3 \]
tiles as promised.
\begin{center}
\begin{asy}
unitsize(0.4cm);
int n = 5;
int N = n*n;
path outer = box((0,0),(N,N));
pair P(int i, int j) {
return (n*i+j,n*j-i+n-1);
}
for (int i=0; i<N; ++i) {
draw((0,i)--(N,i), dotted);
draw((i,0)--(i,N), dotted);
}
for (int i=-n; i<2*n; ++i) {
for (int j=-n; j<2*n; ++j) {
fill(shift(P(i,j))*unitsquare);
filldraw(shift(P(i,j)+(1,0))*scale(n)*unitsquare, opacity(0.2)+yellow, red+1.2);
}
}
clip(outer);
draw(outer, blue+1.8);
\end{asy}
\end{center}
\begin{remark*}
Ironically, the construction obtaining the answer
in the floor pattern at Sunshine Coast airport,
the closest airport to the site of the exam.
See \href{https://i0.wp.com/havewheelchairwilltravel.net/wp-content/uploads/2019/10/IMG_6815.jpg}{this image}
or \href{https://stea.com.au/wp-content/uploads/2020/12/sunshine-coast-airport-design-stea-architectural-projects-20.jpg}{this image}.
\end{remark*}
\paragraph{Bound.}
There are several approaches (all hard), but
the shortest proof seems to be the following one that exploits
the Erd\"{o}s-Szekeres theorem; it is Solution 6 in the shortlist.
The theorem being quoted is:
\begin{theorem*}
[Erd\"{o}s-Szekeres]
Let $n \ge 1$ be an integer.
Given a permutation of $(1, \dots, n)$,
if a \emph{longest increasing subsequence (LIS)} has length $a$ and
a \emph{longest decreasing subsequence (LDS)} has length $b$, then $ab \ge n$.
\end{theorem*}
This is a stronger version of the theorem compared to another version
which instead just asserts that $\max(a,b) \ge \sqrt n$.
To apply this to the present problem, take the $n$ uncovered squares
which we henceforth call ``black'' as a permutation.
Then consider both an LIS of length $a$ and an LDS of length $b$.
We do the following artistic illustration:
\begin{itemize}
\ii Draw the LIS as a broken line,
then connect it to the southwest and northwest corner of the board.
\ii Similarly, draw the LDS as a broken line,
then connect it to the northwest and southeast corner of the board.
\ii These two steps partition the board into four quadrants,
which we call north, east, south, west.
\ii For each black cell in the north quadrant,
write an $\mathbf{N}$ in the cell above it
(for the cell in the first row, this will be off the board).
Do the same for $\mathbf{E}$ (east), $\mathbf{S}$ (south), $\mathbf{W}$ (west).
\ii Some black cells are in multiple quadrants (i.e.\ part of the LIS/LDS).
Write all letters in that case.
\end{itemize}
The figure below shows two examples of the process,
each for a board with $n=9$, for two choices of LIS and LDS.
The cells in the LIS and LDS have been marked with green circles,
and the boundaries of the quadrants are drawn in green lines.
In the left example, the LIS and LDS have a black square in common
(that cell has all four directions labeled).
In the right example, the LIS and do not have common squares.
\begin{center}
\begin{asy}
unitsize(0.65cm);
int n = 3;
int N = n*n;
path outer = box((0,0),(N,N));
pair P(int i, int j) {
return (n*i+j,n*j-i+n-1);
}
pair Q(int i, int j) {
return P(i,j)+(0.5,0.5);
}
for (int i=0; i<N; ++i) {
draw((0,i)--(N,i), dotted);
draw((i,0)--(i,N), dotted);
}
for (int i=-n; i<2*n; ++i) {
for (int j=-n; j<2*n; ++j) {
fill(shift(P(i,j))*unitsquare);
filldraw(shift(P(i,j)+(1,0))*scale(n)*unitsquare, opacity(0.2)+yellow, red+1.2);
}
}
clip(outer);
draw(outer, blue+1.8);
draw((0,0)--Q(0,0)--Q(2,1)--Q(2,2)--(N,N), deepgreen+1.5);
draw((0,N)--Q(0,2)--Q(1,2)--Q(2,1)--(N,0), deepgreen+1.5);
real r = 0.4;
filldraw(circle(Q(0,0), r), lightgreen, deepgreen+1.5);
filldraw(circle(Q(2,1), r), lightgreen, deepgreen+1.5);
filldraw(circle(Q(2,2), r), lightgreen, deepgreen+1.5);
filldraw(circle(Q(0,2), r), lightgreen, deepgreen+1.5);
filldraw(circle(Q(1,2), r), lightgreen, deepgreen+1.5);
label("$\mathbf{N}$", (2.5,9.5));
label("$\mathbf{N}$", (5.5,8.5));
label("$\mathbf{N}$", (8.5,7.5));
label("$\mathbf{N}$", (7.5,4.5));
label("$\mathbf{W}$", (1.5,8.5));
label("$\mathbf{W}$", (0.5,5.5));
label("$\mathbf{W}$", (-0.5,2.5));
label("$\mathbf{W}$", (4.5,7.5));
label("$\mathbf{W}$", (3.5,4.5));
label("$\mathbf{W}$", (6.5,3.5));
label("$\mathbf{E}$", (9.5,6.5));
label("$\mathbf{E}$", (8.5,3.5));
label("$\mathbf{S}$", (7.5,2.5));
label("$\mathbf{S}$", (0.5,1.5));
label("$\mathbf{S}$", (3.5,0.5));
label("$\mathbf{S}$", (6.5,-0.5));
\end{asy}
\qquad
\begin{asy}
unitsize(0.65cm);
int n = 3;
int N = n*n;
path outer = box((0,0),(N,N));
pair P(int i, int j) {
return (n*i+j,n*j-i+n-1);
}
pair Q(int i, int j) {
return P(i,j)+(0.5,0.5);
}
for (int i=0; i<N; ++i) {
draw((0,i)--(N,i), dotted);
draw((i,0)--(i,N), dotted);
}
for (int i=-n; i<2*n; ++i) {
for (int j=-n; j<2*n; ++j) {
fill(shift(P(i,j))*unitsquare);
filldraw(shift(P(i,j)+(1,0))*scale(n)*unitsquare, opacity(0.2)+yellow, red+1.2);
}
}
clip(outer);
draw(outer, blue+1.8);
draw((0,0)--Q(0,0)--Q(0,1)--Q(2,2)--(N,N), deepgreen+1.5);
draw((0,N)--Q(0,2)--Q(1,1)--Q(2,1)--(N,0), deepgreen+1.5);
real r = 0.4;
filldraw(circle(Q(0,0), r), lightgreen, deepgreen+1.5);
filldraw(circle(Q(0,1), r), lightgreen, deepgreen+1.5);
filldraw(circle(Q(2,2), r), lightgreen, deepgreen+1.5);
filldraw(circle(Q(0,2), r), lightgreen, deepgreen+1.5);
filldraw(circle(Q(1,1), r), lightgreen, deepgreen+1.5);
filldraw(circle(Q(2,1), r), lightgreen, deepgreen+1.5);
label("$\mathbf{N}$", (2.5,9.5));
label("$\mathbf{N}$", (5.5,8.5));
label("$\mathbf{N}$", (8.5,7.5));
label("$\mathbf{W}$", (1.5,8.5));
label("$\mathbf{W}$", (0.5,5.5));
label("$\mathbf{W}$", (-0.5,2.5));
label("$\mathbf{E}$", (9.5,6.5));
label("$\mathbf{E}$", (8.5,3.5));
label("$\mathbf{E}$", (5.5,4.5));
label("$\mathbf{S}$", (7.5,2.5));
label("$\mathbf{S}$", (1.5,4.5));
label("$\mathbf{S}$", (0.5,1.5));
label("$\mathbf{S}$", (3.5,0.5));
label("$\mathbf{S}$", (4.5,3.5));
label("$\mathbf{S}$", (6.5,-0.5));
\end{asy}
\end{center}
We observe that:
\begin{claim*}
In this algorithm, the total number of letters written is exactly
\[
C \coloneq \begin{cases}
n + a + b + 1 & \text{if the LIS and LDS intersect} \\
n + a + b & \text{otherwise}.
\end{cases}
\]
\end{claim*}
\begin{proof}
This is obvious. Each black square contributes at least one letter.
Each black square on exactly one of the LIS and LDS contributes one extra letter.
And a black square on both contributes $4$ letters instead of $1+1+1$.
\end{proof}
Note by AM-GM we have $a + b \ge 2 \sqrt{ab} \ge 2 \sqrt n$,
so we have a bound of
\[ C \ge n + 2 \sqrt n + \eps \quad\text{where}\quad
\eps \coloneq \begin{cases}
1 & \text{if the LIS and LDS intersect} \\
0 & \text{otherwise}.
\end{cases} \]
To relate $C$ to the number of tiles,
the critical claim is the following, which is by construction:
\begin{claim*}
None of Matilda's tiles can have more than one letter written in any cell.
\end{claim*}
\begin{proof}
This follows from the construction.
\end{proof}
We split into two cases based on $\eps$.
\begin{itemize}
\ii When $\eps = 1$, at most four letters go off the grid
(one for each direction), the number of tiles is at least $C - 4 \ge n + 2 \sqrt n - 3$.
\ii Suppose $\eps = 0$.
Then $C - 4 \ge n + 2 \sqrt n - 4$.
However, we make the additional observation here that the tile
where the LIS and LDS meet has no letters on it either;
hence there are at least $1 + (C-4) \ge n + 2 \sqrt n - 3$ tiles.
\end{itemize}
\begin{remark*}
\href{https://aops.com/community/p35341804}{USJL} mentions that when $n = ab$,
it is in fact always possible to guarantee $\eps = 1$.
Moreover, when $a \neq b$, the AM-GM inequality is strict.
This gives a way to avoid the additional observation needed for $\eps = 0$ above.
\end{remark*}
|
JMO-2010-notes_1
|
Let $P(n)$ be the number of permutations $(a_1, \dots, a_n)$
of the numbers $(1, 2, \dots, n)$ for which $ka_k$ is a perfect square
for all $1 \leq k \leq n$.
Find with proof the smallest $n$ such that $P (n)$ is a multiple of $2010$.
|
The answer is $n= 4489$.
We begin by giving a complete description of $P(n)$:
\begin{claim*}
We have
\[ P(n) = \prod_{c \text{ squarefree}} \left\lfloor \sqrt{\frac nc} \right\rfloor! \]
\end{claim*}
\begin{proof}
Every positive integer can be uniquely expressed in the form $c \cdot m^2$
where $c$ is a squarefree integer and $m$ is a perfect square.
So we may, for each squarefree positive integer $c$, define the set
\[ S_c = \left\{ c \cdot 1^2, c \cdot 2^2, c \cdot 3^2, \dots \right\}
\cap \{1, 2, \dots, n\} \]
and each integer from $1$ through $n$ will be in exactly one $S_c$.
Note also that
\[ \left\lvert S_c \right\rvert = \left\lfloor
\sqrt{\frac nc} \right\rfloor. \]
Then, the permutations in the problem are exactly
those which send elements of $S_c$ to elements of $S_c$.
In other words,
\[ P(n) = \prod_{c \text{ squarefree}} |S_c|!
= \prod_{c \text{ squarefree}} \left\lfloor \sqrt{\frac nc} \right\rfloor!
\qedhere \]
\end{proof}
We want the smallest $n$ such that $2010$ divides $P(n)$.
\begin{itemize}
\ii Note that $P(67^2)$ contains $67!$ as a term,
which is divisible by $2010$, so $67^2$ is a candidate.
\ii On the other hand, if $n < 67^2$,
then no term in the product for $P(n)$ is divisible by the prime $67$.
\end{itemize}
So $n = 67^2 = 4489$ is indeed the minimum.
|
JMO-2010-notes_2
|
Let $n > 1$ be an integer. Find, with proof, all sequences
$x_1$, $x_2$, \dots, $x_{n-1}$ of positive integers
with the following three properties:
\begin{enumerate}[(a)]
\ii $x_1 < x_2 < \dotsb < x_{n-1}$;
\ii $x_i + x_{n-i} = 2n$ for all $i = 1, 2, \dots , n - 1$;
\ii given any two indices $i$ and $j$ (not necessarily distinct)
for which $x_i + x_j < 2n$,
there is an index $k$ such that $x_i + x_j = x_k$.
\end{enumerate}
|
The answer is $x_k = 2k$ only, which obviously work,
so we prove they are the only ones.
Let $x_1 < x_2 < \dots < x_n$ be
any sequence satisfying the conditions.
Consider:
\[ x_1 + x_1 < x_1 + x_2 < x_1 + x_3 < \dots
< x_1 + x_{n-2}. \]
All these are results of condition (c),
since $x_1 + x_{n-2} < x_1 + x_{n-1} = 2n$.
So each of these must be a member of the sequence.
However, there are $n-2$ of these terms,
and there are exactly $n-2$ terms greater
than $x_1$ in our sequence.
Therefore, we get the one-to-one correspondence below:
\begin{align*}
x_2 &= x_1 + x_1 \\
x_3 &= x_1 + x_2 \\
& \vdotswithin= \\
x_{n-1} &= x_1 + x_{n-2}
\end{align*}
It follows that $x_2 = 2x_1$,
so that $x_3 = 3x_1$ and so on.
Therefore, $x_m = m x_1$.
We now solve for $x_1$ in condition (b)
to find that $x_1=2$ is the only solution,
and the desired conclusion follows.
|
JMO-2010-notes_3
|
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.
|
JMO-2010-notes_4
|
A triangle is called a \emph{parabolic} triangle
if its vertices lie on a parabola $y = x^2$.
Prove that for every nonnegative integer $n$,
there is an odd number $m$ and a parabolic triangle
with vertices at three distinct points
with integer coordinates with area $(2^nm)^2$.
|
For $n=0$, take instead $(a,b) = (1,0)$.
For $n > 0$, consider a triangle
with vertices at $(a,a^2)$, $(-a,a^2)$ and $(b, b^2)$.
Then the area of this triangle was equal to
\[ \half (2a) \left( b^2-a^2 \right) = a(b^2-a^2). \]
To make this equal $2^{2n} m^2$,
simply pick $a = 2^{2n}$,
and then pick $b$ such that $b^2-m^2 = 2^{4n}$,
for example $m = 2^{4n-2}-1$ and $b=2^{4n-2}+1$.
|
JMO-2010-notes_5
|
Two permutations $a_1,a_2,\dots,a_{2010}$
and $b_1,b_2,\dots,b_{2010}$ of the numbers $1,2,\dots,2010$
are said to intersect if $a_k=b_k$ for some value of $k$ in the range $1\le k\le 2010$.
Show that there exist $1006$ permutations of the numbers $1,2,\dots,2010$
such that any other such permutation is guaranteed to
intersect at least one of these $1006$ permutations.
|
A valid choice is the following $1006$ permutations:
\[
\begin{array}{cccc ccc | ccccc}
1 & 2 & 3 & \dotsb & 1004 & 1005 & 1006 & 1007 & 1008 & \dotsb & 2009 & 2010 \\
2 & 3 & 4 & \dotsb & 1005 & 1006 & 1 & 1007 & 1008 & \dotsb & 2009 & 2010 \\
3 & 4 & 5 & \dotsb & 1006 & 1 & 2 & 1007 & 1008 & \dotsb & 2009 & 2010 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots &
\vdots& \vdots & \vdots & \vdots & \vdots \\
1004 & 1005 & 1006 & \dotsb & 1001 & 1002 & 1003 & 1007 & 1008 & \dotsb & 2009 & 2010 \\
1005 & 1006 & 1 & \dotsb & 1002 & 1003 & 1004 & 1007 & 1008 & \dotsb & 2009 & 2010 \\
1006 & 1 & 2 & \dotsb & 1003 & 1004 & 1005 & 1007 & 1008 & \dotsb & 2009 & 2010 \\
\end{array}
\]
This works. Indeed, any permutation should have one of $\{1, 2, \dots, 1006\}$
somewhere in the first $1006$ positions,
so one will get an intersection.
\begin{remark*}
In fact, the last $1004$ entries do not matter with this
construction, and we chose to leave them as $1007, 1008, \dots, 2010$
only for concreteness.
\end{remark*}
\begin{remark*}
Using Hall's marriage lemma one may prove that
the result becomes false with $1006$ replaced by $1005$.
\end{remark*}
|
JMO-2010-notes_6
|
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.
\end{enumerate}
|
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.
|
JMO-2011-notes_1
|
Find all positive integers $n$ such that $2^n+12^n+2011^n$ is a perfect square.
|
The answer $n=1$ works, because $2^1+12^1+2011^1=45^2$.
We prove it's the only one.
\begin{itemize}
\ii If $n \ge 2$ is even, then modulo $3$ we have
$2^n+12^n+2011^n \equiv 1+0+1 \equiv 2 \pmod 3$
so it is not a square.
\ii If $n \ge 3$ is odd, then modulo $4$ we have
$2^n+12^n+2011^n \equiv 0+0+3 \equiv 3 \pmod 4$
so it is not a square.
\end{itemize}
This completes the proof.
|
JMO-2011-notes_2
|
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.
|
JMO-2011-notes_3
|
For a point $P = (a,a^2)$ in the coordinate plane,
let $\ell(P)$ denote the line passing through $P$ with slope $2a$.
Consider the set of triangles with vertices of the form
$P_1 = (a_1, a_1^2)$, $P_2 = (a_2, a_2^2)$, $P_3 = (a_3, a_3^2)$,
such that the intersection of the lines $\ell(P_1)$, $\ell(P_2)$, $\ell(P_3)$
form an equilateral triangle $\Delta$.
Find the locus of the center of $\Delta$ as $P_1P_2P_3$ ranges over all such triangles.
|
The answer is the line $y = -1/4$.
I did not find this problem inspiring,
so I will not write out most of the boring calculations
since most solutions are just going to be
``use Cartesian coordinates and grind all the way through''.
The ``nice'' form of the main claim is as follows
(which is certainly overkill for the present task,
but is too good to resist including):
\begin{claim*}
[Naoki Sato]
In general, the orthocenter of $\Delta$ lies on
the directrix $y = -1/4$ of the parabola
(even if the triangle $\Delta$ is not equilateral).
\end{claim*}
\begin{proof}
By writing out the equation $y = 2a_i x - a_i^2$ for $\ell(P_i)$,
we find the vertices of the triangle are located at
\[
\left( \frac{a_1+a_2}{2}, a_1a_2 \right); \quad
\left( \frac{a_2+a_3}{2}, a_2a_3 \right); \quad
\left( \frac{a_3+a_1}{2}, a_3a_1 \right).
\]
The coordinates of the orthocenter can be checked explicitly to be
\[ H = \left( \frac{a_1 + a_2 + a_3 + 4 a_1 a_2 a_3}{2},
- \frac 14 \right). \]
An advanced synthetic proof of this fact is given at
\url{https://aops.com/community/p2255814}.
\end{proof}
This claim already shows that every point lies on $y = -1/4$.
We now turn to showing that, even when restricted to equilateral triangles,
we can achieve every point on $y = -1/4$.
In what follows $a = a_1$, $b = a_2$, $c = a_3$ for legibility.
\begin{claim*}
Lines $\ell(a)$, $\ell(b)$, $\ell(c)$
form an equilateral triangle if and only if
\begin{align*}
a+b+c &= -12abc \\
ab+bc+ca &= -\frac34.
\end{align*}
Moreover, the $x$-coordinate of the equilateral triangle is $\frac13(a+b+c)$.
\end{claim*}
\begin{proof}
The triangle is equilateral if and only if the centroid and orthocenter coincide, i.e.
\[ \left( \frac{a+b+c}{3}, \frac{ab+bc+ca}{3} \right) = G = H =
\left( \frac{a+b+c+4abc}{2}, -\frac14 \right). \]
Setting the $x$ and $y$ coordinates equal,
we derive the claimed equations.
\end{proof}
Let $\lambda$ be any real number.
We are tasked to show that
\[ P(X) = X^3 - 3\lambda \cdot X^2 - \frac34 X + \frac{\lambda}{4} \]
has three real roots (with multiplicity);
then taking those roots as $(a,b,c)$
yields a valid equilateral-triangle triple
whose $x$-coordinate is exactly $\lambda$, be the previous claim.
To prove that, pick the values
\begin{align*}
P(-\sqrt{3}/2) &= -2\lambda \\
P(0) &= \tfrac14 \lambda \\
P(\sqrt{3}/2) &= -2\lambda.
\end{align*}
The intermediate value theorem
(at least for $\lambda \neq 0$) implies that $P$
should have at least two real roots now,
and since $P$ has degree $3$, it has all real roots.
That's all.
|
JMO-2011-notes_4
|
A \emph{word} is defined as any finite string of letters.
A word is a \emph{palindrome} if it reads the same backwards and forwards.
Let a sequence of words $W_0$, $W_1$, $W_2$, \dots\ be defined as follows:
$W_0 = a$, $W_1 = b$, and for $n \ge 2$,
$W_n$ is the word formed by writing $W_{n-2}$ followed by $W_{n-1}$.
Prove that for any $n \ge 1$, the word formed by writing
$W_1$, $W_2$, $W_3$, \dots, $W_n$ in succession is a palindrome.
|
To aid in following the solution, here are the first several words:
\begin{align*}
W_0 &= a \\
W_1 &= b \\
W_2 &= ab \\
W_3 &= bab \\
W_4 &= abbab \\
W_5 &= bababbab \\
W_6 &= abbabbababbab \\
W_7 &= bababbababbabbababbab
\end{align*}
We prove that $W_1 W_2 \dots W_n$ is a palindrome by induction on $n$.
The base cases $n=1,2,3,4$ can be verified by hand.
For the inductive step, we let $\ol{X}$ denote the word $X$ written
backwards. Then
\begin{align*}
W_1 W_2 \dots W_{n-3} W_{n-2} W_{n-1} W_n
&\overset{\text{IH}}{=} (\ol{W_{n-1}} \ol{W_{n-2}} \ol{W_{n-3}}
\dots \ol{W_2} \ol{W_1}) W_n \\
&= (\ol{W_{n-1}} \ol{W_{n-2}} \ol{W_{n-3}}
\dots \ol{W_2} \ol{W_1}) W_{n-2} W_{n-1} \\
&= \ol{W_{n-1}} \ol{W_{n-2}} (\ol{W_{n-3}}
\dots \ol{W_2} \ol{W_1}) W_{n-2} W_{n-1}
\end{align*}
with the first equality being by the induction hypothesis.
By induction hypothesis again the inner parenthesized term
is also a palindrome, and so this completes the proof.
|
JMO-2011-notes_5
|
Points $A$, $B$, $C$, $D$, $E$ lie on a circle $\omega$ and point $P$ lies outside the circle.
The given points are such that
(i) lines $PB$ and $PD$ are tangent to $\omega$,
(ii) $P$, $A$, $C$ are collinear,
and (iii) $\ol{DE} \parallel \ol{AC}$.
Prove that $\ol{BE}$ bisects $\ol{AC}$.
|
We present two solutions.
\paragraph{First solution using harmonic bundles.}
Let $M = \ol{BE} \cap \ol{AC}$
and let $\infty$ be the point at infinity
along $\ol{DE} \parallel \ol{AC}$.
\begin{center}
\begin{asy}
size(8cm);
pair B = dir(100);
pair D = dir(210);
pair E = dir(330);
pair P = 2*B*D/(B+D);
pair A = OP(P--(P+8*(E-D)), unitcircle);
pair M = extension(B, E, A, P);
pair C = 2*M-A;
filldraw(unitcircle, opacity(0.1)+lightblue, lightblue);
draw(P--A, lightblue);
draw(B--P--D, lightblue);
draw(D--E, heavygreen);
draw(B--E, heavygreen);
draw(A--E--C, heavygreen);
draw(B--A--D--C--cycle, heavycyan);
dot("$B$", B, dir(B));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$P$", P, dir(P));
dot("$A$", A, dir(A));
dot("$M$", M, dir(45));
dot("$C$", C, dir(100));
/* TSQ Source:
B = dir 100
D = dir 210
E = dir 330
P = 2*B*D/(B+D)
A = IP P--(P+8*(E-D)) unitcircle
M = extension B E A P R45
C = 2*M-A R100
unitcircle 0.1 lightblue / lightblue
P--A lightblue
B--P--D lightblue
D--E heavygreen
B--E heavygreen
A--E--C heavygreen
B--A--D--C--cycle heavycyan
*/
\end{asy}
\end{center}
Note that $ABCD$ is harmonic, so
\[ -1 = (AC;BD) \overset{E}{=} (AC;M\infty) \]
implying $M$ is the midpoint of $\ol{AC}$.
\paragraph{Second solution using complex numbers (Cynthia Du).}
Suppose we let $b$, $d$, $e$ be free on unit circle,
so $p = \frac{2bd}{b+d}$.
Then $d/c = a/e$, and $a + c = p + ac \ol p$.
Consequently,
\begin{align*}
ac &= de \\
\half(a + c) &= \frac{bd}{b+d} + de \cdot \frac{1}{b+d}
= \frac{d(b+e)}{b+d}. \\
\frac{a+c}{2ac} &= \frac{(b+e)}{e(b+d)}.
\end{align*}
From here it's easy to see
\[ \frac{a+c}{2} + \frac{a+c}{2ac} \cdot be = b + e \]
which is what we wanted to prove.
|
JMO-2011-notes_6
|
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.
\end{enumerate}
|
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*}
|
JMO-2012-notes_1
|
Given a triangle $ABC$, let $P$ and $Q$ be points
on segments $\ol{AB}$ and $\ol{AC}$, respectively, such that $AP=AQ$.
Let $S$ and $R$ be distinct points on segment $\ol{BC}$
such that $S$ lies between $B$ and $R$,
$\angle BPS=\angle PRS$, and $\angle CQR=\angle QSR$.
Prove that $P$, $Q$, $R$, $S$ are concyclic.
|
Assume for contradiction that $(PRS)$ and $(QRS)$ are distinct.
Then $\ol{RS}$ is the radical axis of these two circles.
\begin{center}
\begin{asy}
size(12cm);
pair P = dir(155);
pair Q = dir(55);
pair S = dir(245);
pair R = dir(295);
pair A = 2*P*Q/(P+Q);
pair B = extension(A, P, R, S);
pair C = extension(A, Q, R, S);
draw(unitcircle, dashed);
markangle(B,P,S,radius=18,n=1,deepgreen,StickIntervalMarker(1,deepgreen));
markangle(P,R,S,radius=18,n=1,deepgreen,StickIntervalMarker(1,deepgreen));
markangle(R,Q,C,radius=12,n=2,purple);
markangle(R,S,Q,radius=12,n=2,purple);
filldraw(A--B--C--cycle, opacity(0.1)+gray, brown);
draw(A--P, red+1.4);
draw(A--Q, red+1.4);
draw(S--P--R--Q--cycle, black);
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
dot("$S$", S, dir(S));
dot("$R$", R, dir(R));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
!size(12cm);
P = dir 155
Q = dir 55
S = dir 245
R = dir 295
A = 2*P*Q/(P+Q)
B = extension A P R S
C = extension A Q R S
unitcircle / dashed
!markangle(B,P,S,radius=18,n=1,deepgreen,StickIntervalMarker(1,deepgreen));
!markangle(P,R,S,radius=18,n=1,deepgreen,StickIntervalMarker(1,deepgreen));
!markangle(R,Q,C,radius=12,n=2,purple);
!markangle(R,S,Q,radius=12,n=2,purple);
A--B--C--cycle / 0.1 gray / brown
A--P / red 1.4
A--Q / red 1.4
S--P--R--Q--cycle / black
*/
\end{asy}
\end{center}
However, the given angle conditions give two tangencies:
\begin{itemize}
\item the given $\angle BPS = \angle PRS$ means $\ol{AP}$ is tangent to $(PRS)$;
\item the given $\angle CQR = \angle QSR$ means $\ol{AQ}$ is tangent to $(QRS)$.
\end{itemize}
Since $AP = AQ$, it follows the point $A$ has equal power $AP^2 = AQ^2$ to both circles.
Hence $A$ should lie on the radical axis $\ol{RS}$,
which is impossible since $A$ does not lie on line $BC$.
This contradiction finishes the problem.
|
JMO-2012-notes_2
|
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 $a_k^2 = F_k = F_{k-1} + F_{k-2} \ge F_j + F_i = a_j^2 + a_i^2$.
|
JMO-2012-notes_3
|
For $a,b,c > 0$ prove that
\[ \frac{a^3+3b^3}{5a+b}
+ \frac{b^3+3c^3}{5b+c}
+ \frac{c^3+3a^3}{5c+a}
\ge \frac23(a^2+b^2+c^2). \]
|
Here are two possible approaches.
\paragraph{Cauchy-Schwarz approach.}
Apply Titu lemma to get
\[ \sum_{\text{cyc}} \frac{a^3}{5a+b}
= \sum_{\text{cyc}} \frac{a^4}{5a^2+ab}
\ge \frac{(a^2+b^2+c^2)^2}{\sum_{\text{cyc}} (5a^2+ab)}
\ge \frac{a^2+b^2+c^2}{6} \]
where the last step follows
from the identity $\sum_{\text{cyc}} (5a^2+ab) \le 6(a^2+b^2+c^2)$.
Similarly,
\[ \sum_{\text{cyc}} \frac{b^3}{5a+b}
= \sum_{\text{cyc}} \frac{b^4}{5ab+b^2}
\ge \frac{(a^2+b^2+c^2)^2}{\sum_{\text{cyc}} (5ab+b^2)}
\ge \frac{a^2+b^2+c^2}{6} \]
using the fact that $\sum_{\text{cyc}} 5ab+b^2 \le 6(a^2+b^2+c^2)$.
Therefore, adding the first display to
three times the second display implies the result.
\paragraph{Second linearization approach.}
The main magical claim is:
\begin{claim*}
We have
\[ \frac{a^3+3b^3}{5a+b} \geq \frac{25}{36}b^2 - \frac{1}{36}a^2. \]
\end{claim*}
\begin{proof}
Let $x=a/b > 0$. The desired inequality is equivalent to
\[ \frac{x^3+3}{5x+1} \ge \frac{25-x^2}{36}. \]
However,
\begin{align*}
36(x^3+3) - (5x+1)(25-x^2)
&= 41x^3 + x^2 - 125x + 83 \\
&= (x-1)^2(41x+83) \ge 0. \qedhere
\end{align*}
\end{proof}
Sum the claim cyclically to finish.
\begin{remark*}
[Derivation of the main claim]
The overall strategy is to hope for a constant $k$ such that
\[ \frac{a^3+3b^3}{5a+b} \geq ka^2 + \left(\frac{2}{3}-k\right)b^2. \]
is true.
Letting $x=a/b$ as above and expanding,
we need a value $k$ such that the cubic polynomial
\[ P(x) \coloneq (x^3+3)-(5x+1)\left( kx^2 + \left( \frac 23 - k \right) \right)
= (1-5k)x^3 - kx^2 + \left( 5k-\frac{10}{3} \right)x + \left( k + \frac{7}{3} \right) \]
is nonnegative everywhere.
Since $P(1) = 0$ necessarily, in order for $P(1-\eps)$ and $P(1+\eps)$
to both be nonnegative (for small $\eps$),
the polynomial $P$ must have a double root at $1$, meaning
the first derivative $P'(1) = 0$ needs to vanish.
In other words, we need
\[ 3(1-5k) - 2k + \left( 5k - \frac{10}{3} \right) = 0. \]
Solving gives $k = -1/36$.
One then factors out the repeated root $(x-1)^2$ from the resulting $P$.
\end{remark*}
|
JMO-2012-notes_4
|
Let $\alpha$ be an irrational number with $0 < \alpha < 1$,
and draw a circle in the plane whose circumference has length $1$.
Given any integer $n\ge 3$,
define a sequence of points $P_1$, $P_2$, \dots, $P_n$ as follows.
First select any point $P_1$ on the circle, and for $2\le k\le n$ define $P_k$
as the point on the circle for which the length of arc $P_{k-1}P_k$ is $\alpha$,
when travelling counterclockwise around the circle from $P_{k-1}$ to $P_k$.
Suppose that $P_a$ and $P_b$ are the nearest adjacent points on either side of $P_n$.
Prove that $a+b\le n$.
|
No points coincide since $\alpha$ is irrational.
Assume for contradiction that $n < a+b < 2n$.
Then it follows that \[ \ol{P_n P_{a+b-n}} \parallel \ol{P_a P_b} \]
as shown below.
\begin{center}
\begin{asy}
size(9cm);
transform t = shift(3.4,0);
pair A = dir(170);
pair B = -conj(A);
pair C = dir(130);
pair D = -conj(C);
draw(unitcircle);
draw(A--B, red);
draw(C--D, blue);
dot("$P_a$", A, A);
dot("$P_b$", B, B);
dot("$P_n$", C, C);
dot("$P_{a+b-n}$", D, D);
draw(t*unitcircle);
draw(t*(A--B), red);
draw(t*(C--D), blue);
dot("$P_a$", t*A, A);
dot("$P_b$", t*B, B);
dot("$P_{a+b-n}$", t*C, C);
dot("$P_n$", t*D, D);
\end{asy}
\end{center}
This is an obvious contradiction
since then $P_{a+b-n}$ is contained in the arc
$\arc{P_a P_b}$ of the circle through $P_n$.
|
JMO-2012-notes_5
|
For distinct positive integers $a, b < 2012$,
define $f(a, b)$ to be the number of integers $k$ with $1 \le k < 2012$
such that the remainder when $ak$ divided by $2012$ is
greater than that of $bk$ divided by $2012$.
Let $S$ be the minimum value of $f(a, b)$, where $a$ and $b$ range
over all pairs of distinct positive integers less than $2012$.
Determine $S$.
|
The answer is $S = 502$ (not $503$!).
\begin{claim*}
If $\gcd(k, 2012) = 1$, then necessarily either $k$ or $2012-k$ will counts towards $S$.
\end{claim*}
\begin{proof}
First note that both $ak$, $bk$ are nonzero modulo $2012$.
Note also that $ak \not\equiv bk \pmod{2012}$.
So if $r_a$ is the remainder of $ak \pmod{2012}$,
then $2012-r_a$ is the remainder of $a(2012-k) \pmod{2012}$
Similarly we can consider $r_b$ and $2012-r_b$.
As mentioned already, we have $r_a \neq r_b$.
So either $r_a > r_b$ or $2012-r_a > 2012-r_b$.
\end{proof}
This implies $S \ge \half \varphi(2012) = 502$.
But this can actually be achieved by taking $a = 4$ and $b = 1010$, since
\begin{itemize}
\ii If $k$ is even, then $ak \equiv bk \pmod{2012}$
so no even $k$ counts towards $S$; and
\ii If $k \equiv 0 \pmod{503}$, then $ak \equiv 0 \pmod{2012}$
so no such $k$ counts towards $S$.
\end{itemize}
This gives the final answer $S \ge 502$.
\begin{remark*}
A similar proof works with $2012$ replaced by any $n$
and will give an answer of $\half\varphi(n)$.
For composite $n$, one uses the Chinese remainder theorem to pick distinct
$a$ and $b$ not divisible by $n$ such that $\operatorname{lcm}(a-b, a) = n$.
\end{remark*}
|
JMO-2012-notes_6
|
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.
\end{enumerate}
|
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*}
|
JMO-2013-notes_1
|
Are there integers $a$ and $b$
such that $a^5b+3$ and $ab^5+3$
are both perfect cubes of integers?
|
No, there do not exist such $a$ and $b$.
We prove this in two cases.
\begin{itemize}
\ii Assume $3 \mid ab$.
WLOG we have $3 \mid a$,
but then $a^5b+3 \equiv 3 \pmod 9$, contradiction.
\ii Assume $3 \nmid ab$.
Then $a^5b+3$ is a cube not divisible by $3$,
so it is $\pm 1 \bmod 9$,
and we conclude
\[ a^5b \in \{5,7\} \pmod 9. \]
Analogously \[ ab^5 \in \{5,7\} \pmod 9. \]
We claim however these two equations cannot hold simultaneously.
Indeed $(ab)^6 \equiv 1 \pmod 9$ by Euler's theorem,
despite $5 \cdot 5 \equiv 7$, $5 \cdot 7 \equiv 8$,
$7 \cdot 7 \equiv 4$ mod $9$.
\end{itemize}
|
JMO-2013-notes_2
|
Each cell of an $m\times n$ board is filled with some nonnegative integer.
Two numbers in the filling are said to be
\emph{adjacent} if their cells share a common side.
The filling is called a \emph{garden} if it
satisfies the following two conditions:
\begin{enumerate}
\item[(i)] The difference between
any two adjacent numbers is either $0$ or $1$.
\item[(ii)] If a number is less than or equal to
all of its adjacent numbers, then it is equal to $0$.
\end{enumerate}
Determine the number of distinct gardens in terms of $m$ and $n$.
|
The numerical answer is $2^{mn}-1$.
But we claim much more, by giving an explicit description of all gardens:
\begin{quote}
Let $S$ be any nonempty subset of the $mn$ cells.
Suppose we fill each cell $\theta$
with the minimum (taxicab) distance
from $\theta$ to some cell in $S$
(in particular, we write $0$ if $\theta \in S$).
Then
\begin{itemize}
\ii This gives a garden, and
\ii All gardens are of this form.
\end{itemize}
\end{quote}
Since there are $2^{mn}-1$ such nonempty subsets $S$,
this would finish the problem.
An example of a garden with $|S| = 3$ is shown below.
\[
\begin{bmatrix}
2 & 1 & 2 & 1 & \mathbf{\color{red}0} & 1 \\
1 & \mathbf{\color{red}0} & 1 & 2 & 1 & 2 \\
1 & 1 & 2 & 3 & 2 & 3 \\
\mathbf{\color{red}0} & 1 & 2 & 3 & 3 & 4 \\
\end{bmatrix}
\]
It is actually fairly easy to see that this
procedure always gives a garden;
so we focus our attention on showing that every
garden is of this form.
Given a garden, note first that it has at least
one cell with a zero in it --- by considering
the minimum number across the entire garden.
Now let $S$ be the (thus nonempty) set of
cells with a zero written in them.
We contend that this works, i.e.\ the following sentence holds:
\begin{claim*}
If a cell $\theta$ is labeled $d$,
then the minimum distance from that cell
to a cell in $S$ is $d$.
\end{claim*}
\begin{proof}
The proof is by induction on $d$, with $d = 0$ being by definition.
Now, consider any cell $\theta$ labeled $d \ge 1$.
Every neighbor of $\theta$ has label at least $d-1$,
so any path will necessarily take $d-1$ steps after leaving $\theta$.
Conversely, there is \emph{some} $d-1$ adjacent to $\theta$ by (ii).
Stepping on this cell and using the minimal path
(by induction hypothesis) gives us a path to
a cell in $S$ with length \emph{exactly} $d$.
So the shortest path does indeed have distance $d$, as desired.
\end{proof}
|
JMO-2013-notes_3
|
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.
|
JMO-2013-notes_4
|
Let $f(n)$ be the number of ways to write $n$ as a sum of powers of $2$,
where we keep track of the order of the summation.
For example, $f(4)=6$ because $4$ can be written
as $4$, $2+2$, $2+1+1$, $1+2+1$, $1+1+2$, and $1+1+1+1$.
Find the smallest $n$ greater than $2013$ for which $f(n)$ is odd.
|
The answer is $2047$.
For convenience, we agree that $f(0) = 1$.
Then by considering cases on the first number in the representation,
we derive the recurrence
\[ f(n) = \sum_{k=0}^{\left\lfloor \log_2 n \right\rfloor} f(n-2^k).
\qquad (\heartsuit) \]
We wish to understand the parity of $f$. The first few values are
\begin{align*}
f(0) &= 1 \\
f(1) &= 1 \\
f(2) &= 2 \\
f(3) &= 3 \\
f(4) &= 6 \\
f(5) &= 10 \\
f(6) &= 18 \\
f(7) &= 31.
\end{align*}
Inspired by the data we make the key claim that
\begin{claim*}
$f(n)$ is odd if and only if $n+1$ is a power of $2$.
\end{claim*}
\begin{proof}
We call a number \emph{repetitive} if it is zero or its binary representation
consists entirely of $1$'s.
So we want to prove that $f(n)$ is odd if and only if $n$ is repetitive.
This only takes a few cases:
\begin{itemize}
\ii If $n = 2^k$, then $(\heartsuit)$
has exactly two repetitive terms on the right-hand side, namely $0$ and $2^k-1$.
\ii If $n = 2^k + 2^\ell - 1$, then $(\heartsuit)$
has exactly two repetitive terms on the right-hand side,
namely $2^{\ell+1}-1$ and $2^{\ell}-1$.
\ii If $n = 2^k-1$, then $(\heartsuit)$
has exactly one repetitive terms on the right-hand side, namely $2^{k-1}-1$.
\ii For other $n$,
there are no repetitive terms at all on the right-hand side of $(\heartsuit)$.
\end{itemize}
Thus the induction checks out.
\end{proof}
So the final answer to the problem is $2047$.
|
JMO-2013-notes_5
|
Quadrilateral $XABY$ is inscribed in the semicircle $\omega$ with
diameter $\ol{XY}$.
Segments $AY$ and $BX$ meet at $P$.
Point $Z$ is the foot of the perpendicular from $P$ to line $\ol{XY}$.
Point $C$ lies on $\omega$ such that line $XC$ is perpendicular to line $AZ$.
Let $Q$ be the intersection of segments $AY$ and $XC$.
Prove that
\[\dfrac{BY}{XP}+\dfrac{CY}{XQ}=\dfrac{AY}{AX}.\]
|
Let $\beta = \angle YXP$ and $\alpha = \angle PYX$ and set $XY = 1$.
We do not direct angles in the following solution.
\begin{center}
\begin{asy}
size(10cm);
pair X = Drawing("X", dir(180), dir(225));
pair Y = Drawing("Y", dir(0), dir(315));
pair A = Drawing("A", dir(130), dir(135));
pair B = Drawing("B", dir(94), dir(94));
pair P = Drawing("P", IP(X--B,A--Y), dir(90));
pair Z = Drawing("Z", foot(P,X,Y), dir(-90));
pair C = 2 * foot(origin, A+Y-Z, Y) - Y;
Drawing("C", C, C);
pair Q = Drawing("Q", IP(A--Y,X--C), dir(80));
draw(X--A--B--Y--cycle);
draw(A--Y--C--X--B);
draw(P--Z--A);
draw(arc(origin,1,0,180));
markangle("$\beta$", Y, X, P);
markangle("$\alpha$", A, Y, X);
\end{asy}
\end{center}
Observe that
\[ \angle AZX = \angle APX = \alpha + \beta \]
since $APZX$ is cyclic.
In particular, $\angle CXY = 90^\circ- (\alpha + \beta)$.
It is immediate that
\[ BY = \sin \beta, \quad CY = \cos \left( \alpha + \beta \right),
\quad AY = \cos \alpha, \quad AX = \sin \alpha. \]
The Law of Sines on $\triangle XPY$ gives $XP = XY \frac{\sin\alpha}{\sin(\alpha+\beta)}$,
and on $\triangle XQY$ gives $XQ = XY \frac{\sin\alpha}{\sin(90+\beta)} = \frac{\sin\alpha}{\cos\beta}$.
So, the given is equivalent to
\[ \frac{\sin\beta}{\frac{\sin\alpha}{\sin(\alpha+\beta)}}
+ \frac{\cos(\alpha+\beta)}{\frac{\sin\alpha}{\cos\beta}}
= \frac{\cos\alpha}{\sin\alpha} \]
which is equivalent to $\cos\alpha = \cos\beta\cos(\alpha+\beta) + \sin\beta\sin(\alpha+\beta)$.
This is obvious, because the right-hand side is just $\cos\left( (\alpha+\beta)-\beta \right)$.
|
JMO-2013-notes_6
|
Find all real numbers $x,y,z \ge 1$ satisfying
\[ \min \left( \sqrt{x+xyz}, \sqrt{y+xyz}, \sqrt{z+xyz} \right)
= \sqrt{x-1} + \sqrt{y-1} + \sqrt{z-1}. \]
\end{enumerate}
|
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 $t > 0$.
|
JMO-2014-notes_1
|
Let $a$, $b$, $c$ be real numbers greater than or equal to $1$.
Prove that
\[ \min \left( \frac{10a^2-5a+1}{b^2-5b+10},
\frac{10b^2-5b+1}{c^2-5c+10},
\frac{10c^2-5c+1}{a^2-5a+10} \right) \le abc. \]
|
Notice that
\[ \frac{10a^2-5a+1}{a^2-5a+10} \le a^3 \]
since it rearranges to $(a-1)^5 \ge 0$.
Cyclically multiply to get
\[
\prod_{\text{cyc}} \left( \frac{10a^2-5a+1}{b^2-5b+10} \right)
\le (abc)^3
\]
and the minimum is at most the geometric mean.
|
JMO-2014-notes_2
|
Let $\triangle ABC $ be a non-equilateral,
acute triangle with $\angle A = 60\dg$,
and let $O$ and $H$ denote the circumcenter and orthocenter
of $\triangle{ABC}$, respectively.
\begin{enumerate}[(a)]
\ii Prove that line $OH$ intersects both segments $AB$ and $AC$
at two points $P$ and $Q$, respectively.
\ii Denote by $s$ and $t$ the respective areas of triangle $APQ$
and quadrilateral $BPQC$.
Determine the range of possible values for $s/t$.
\end{enumerate}
|
We begin with some synthetic work.
Let $I$ denote the incenter, and recall (``fact 5'')
that the arc midpoint $M$ is the center of $(BIC)$,
which we denote by $\gamma$.
Now we have that
\[ \angle BOC = \angle BIC = \angle BHC = 120\dg. \]
Since all three centers lie inside $ABC$ (as it was acute),
and hence on the opposite side of $\ol{BC}$ as $M$,
it follows that $O$, $I$, $H$ lie on minor arc $BC$ of $\gamma$.
We note this implies (a) already,
as line $OH$ meets line $BC$ outside of segment $BC$.
\begin{center}
\begin{asy}
pair A = dir(50);
pair B = dir(210);
pair C = dir(330);
pair M = dir(270);
pair I = incenter(A, B, C);
pair O = circumcenter(A, B, C);
pair H = orthocenter(A, B, C);
pair P = extension(O, H, A, B);
pair Q = extension(O, H, A, C);
filldraw(A--P--Q--cycle, opacity(0.2)+lightgreen, deepgreen);
filldraw(B--P--Q--C--cycle, opacity(0.2)+lightcyan, deepgreen);
filldraw(circumcircle(A, B, C), opacity(0.1)+yellow, red);
filldraw(CP(M, I), opacity(0.1)+lightblue, dotted+blue);
draw(A--M, red);
filldraw(A--O--M--H--cycle, opacity(0.1)+orange, lightred);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$M$", M, dir(M));
dot("$I$", I, dir(20));
dot("$O$", O, dir(90));
dot("$H$", H, dir(50));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
/* TSQ Source:
A = dir 50
B = dir 210
C = dir 330
M = dir 270
I = incenter A B C R20
O = circumcenter A B C R90
H = orthocenter A B C R50
P = extension O H A B
Q = extension O H A C
A--P--Q--cycle 0.2 lightgreen / deepgreen
B--P--Q--C--cycle 0.2 lightcyan / deepgreen
circumcircle A B C 0.1 yellow / red
CP M I 0.1 lightblue / dotted blue
A--M red
A--O--M--H--cycle 0.1 orange / lightred
*/
\end{asy}
\end{center}
\begin{claim*}
Triangle $APQ$ is equilateral with side length $\frac{b+c}{3}$.
\end{claim*}
\begin{proof}
Let $R$ be the circumradius.
We have $R = OM = OA = MH$, and even $AH = 2R \cos A = R$,
so $AOMH$ is a rhombus.
Thus $\ol{OH} \perp \ol{AM}$ and in this way
we derive that $\triangle APQ$ is isosceles, hence equilateral.
Finally, since $\angle PBH = 30\dg$, and $\angle BPH = 120\dg$,
it follows that $\triangle BPH$ is isosceles and $BP = PH$.
Similarly, $CQ = QH$.
So $b+c = AP + BP + AQ + QC = AP + AQ + PQ$ as needed.
\end{proof}
Finally, we turn to the boring task of
extracting the numerical answer.
We have
\[ \frac{s}{s+t} = \frac{[APQ]}{[ABC]}
= \frac{\frac{\sqrt3}4 \left( \frac{b+c}{3} \right)^2}%
{\frac{\sqrt3}4 bc}
= \frac{b^2+2bc+c^2}{9bc}
= \frac19 \left( 2 + \frac bc + \frac cb \right). \]
So the problem is reduced to analyzing the behavior of $b/c$.
For this, we imagine fixing $\Gamma$ the circumcircle of $ABC$,
as well as the points $B$ and $C$.
Then as we vary $A$ along the ``topmost'' arc of measure $120\dg$,
we find $b/c$ is monotonic with values $1/2$ and $2$ at endpoints,
and by continuity all values $b/c \in (1/2,2)$ can be achieved.
So \[ \half < \frac bc < 2
\implies 4/9 < \frac{s}{s+t} < 1/2
\implies 4/5 < \frac st < 1 \]
as needed.
|
JMO-2014-notes_3
|
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 $f(x) \equiv x^2$.
|
JMO-2014-notes_4
|
Let $b \ge 2$ be a fixed integer,
and let $s_b(n)$ denote the sum of the base-$b$ digits of $n$.
Show that there are infinitely many positive
integers that cannot be represented in the from $n + s_b(n)$
where $n$ is a positive integer.
|
For brevity let $f(n) = n + s_b(n)$.
Select any integer $M$.
Observe that $f(x) \ge b^{2M}$ for any $x \ge b^{2M}$,
but also $f(b^{2M}-k) \ge b^{2M}$ for $k = 1, 2, \dots, M$,
since the base-$b$ expansion of $b^{2M}-k$ will start out with
at least $M$ digits $b-1$.
Thus $f$ omits at least $M$ values in $[1, b^{2M}]$ for any $M$.
|
JMO-2014-notes_5
|
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 $A$.
|
JMO-2014-notes_6
|
Let $ABC$ be a triangle with incenter $I$,
incircle $\gamma$ and circumcircle $\Gamma$.
Let $M$, $N$, $P$ be the midpoints of $\ol{BC}$, $\ol{CA}$, $\ol{AB}$
and let $E$, $F$ be the tangency points of $\gamma$
with $\ol{CA}$ and $\ol{AB}$, respectively.
Let $U$, $V$ be the intersections of line $EF$
with line $MN$ and line $MP$, respectively,
and let $X$ be the midpoint of arc $BAC$ of $\Gamma$.
\begin{enumerate}[(a)]
\ii Prove that $I$ lies on ray $CV$.
\ii Prove that line $XI$ bisects $\ol{UV}$.
\end{enumerate}
\end{enumerate}
|
The fact that $I = \ol{BU} \cap \ol{CV}$ and is Lemma 1.45 from EGMO.
As for (b), we note:
\begin{claim*}
Line $IX$ is a symmedian of $\triangle IBC$.
\end{claim*}
\begin{proof}
Recall that $(BIC)$ has circumcenter
coinciding with the antipode of $X$ (by ``Fact 5'').
So this follows from the fact
that $\ol{XB}$ and $\ol{XC}$ are tangent.
\end{proof}
Since $BVUC$ is cyclic with diagonals intersecting at $I$,
and $IX$ is symmedian of $\triangle IBC$,
it is median of $\triangle IUV$, as needed.
\begin{remark*}
[Alternate solution to (b) by Gunmay Handa]
It's well known that $X$ is the midpoint of $\ol{I_b I_c}$
(by considering the nine-point circle of the excentral triangle).
However, $\ol{UV} \parallel \ol{I_b I_c}$
and $I = \ol{I_b U} \cap \ol{I_c V}$, implying the result.
\end{remark*}
|
JMO-2015-notes_1
|
Given a sequence of real numbers,
a move consists of choosing two terms
and replacing each with their arithmetic mean.
Show that there exists a sequence of $2015$ distinct real numbers
such that after one initial move
is applied to the sequence --- no matter what move --- there
is always a way to continue with a finite sequence of moves
so as to obtain in the end a constant sequence.
|
One valid example of a sequence is $0$, $1$, \dots, $2014$.
We will show how to achieve the all-$1007$ sequence
based on the first move.
Say two numbers are \emph{opposites} if their average is $1007$.
We consider $1007$ as its own opposite.
We consider two cases:
\begin{itemize}
\ii First, suppose the first initial move did
\emph{not} involve the number $1007$.
Suppose the two numbers changed were $a$ and $b$,
replaced by $c = \half(a+b)$ twice.
\begin{itemize}
\ii If $a$ and $b$ are opposites, we
simply operate on all the other pairs of opposites.
\ii Otherwise let $a'$ and $b'$ be the opposites of $a$ and $b$,
so all four of $a$, $b$, $a'$, $b'$ are distinct.
Then operate on $a'$ and $b'$ to get $c' = 2014-c$.
We work with only these four numbers ande replace them as
follows:
\[
\begin{array}{cccc}
\half(a+b) & \half(a+b) & a' & b' \\
\half(a+b) & \half(a+b) & \half(a'+b') & \half(a'+b') \\
1007 & \half(a+b) & 1007 & \half(a'+b') \\
1007 & 1007 & 1007 & 1007
\end{array}
\]
Finally, we operate on the remaining $1005$ pairs of opposites.
\end{itemize}
\ii Now suppose the first initial move involved the number $1007$
and some $a$.
Let $k$ be any number other than $a$ or its opposite,
and let $a'$, $k'$ be the opposites of $a$ and $k$.
We work with only these five numbers:
and replace them in the following way:
\[
\begin{array}{ccccc}
\half(a+1007) & \half(a+1007) & a' & k & k' \\
\half(a+1007) & \half(a+1007) & a' & 1007 & 1007 \\
\half(a+1007) & \half(a+1007) & \half(a'+1007) &
\half(a'+1007) & 1007 \\
1007 & \half(a+1007) & 1007 &
\half(a'+1007) & 1007 \\
1007 & 1007 & 1007 & 1007 & 1007 \\
\end{array}
\]
Finally, we operate on the remaining $1005$ pairs of opposites.
\end{itemize}
\begin{remark*}
In fact, the same proof basically works for any sequence
with average $m$ such that $m$ is in the sequence,
and every term has an opposite.
However for ``most'' sequences one expects
the result to not be possible.
As a simple example,
the goal is impossible for $(0, 1, \dots, 2013, 2015)$
since the average of the terms is $1007 + \frac{1}{2015}$,
but in the process the only denominators
ever generated are powers of $2$.
This narrows the search somewhat.
\end{remark*}
|
JMO-2015-notes_2
|
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).
|
JMO-2015-notes_3
|
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.
|
JMO-2015-notes_4
|
Find all functions $f \colon \QQ \to \QQ$ such that
\[ f(x)+f(t)=f(y)+f(z) \]
for all rational numbers $x<y<z<t$ that form an arithmetic progression.
|
Answer: any linear function $f$.
These work.
Here is one approach: for any $a$ and $d > 0$
\begin{align*}
f(a) + f(a+3d) &= f(a+d) + f(a+2d) \\
f(a-d) + f(a+2d) &= f(a) + f(a+d) \\
\intertext{which imply}
f(a-d) + f(a+3d) &= 2f(a+d).
\end{align*}
Thus we conclude that for arbitrary $x$ and $y$ we have
\[ f(x) + f(y) = 2f\left( \frac{x+y}{2} \right) \]
thus $f$ satisfies Jensen functional equation over $\QQ$, so linear.
The solution can be made to avoid appealing to Jensen's functional equation;
here is a presentation of such a solution based on the official ones.
Let $d > 0$ be a positive integer, and let $n$ be an integer.
Consider the two equations
\begin{align*}
f\left( \frac{2n-1}{2d} \right) + f\left( \frac{2n+2}{2d} \right)
&= f\left( \frac{2n}{2d} \right) + f\left( \frac{2n+1}{2d} \right) \\
f\left( \frac{2n-2}{2d} \right) + f\left( \frac{2n+1}{2d} \right)
&= f\left( \frac{2n-1}{2d} \right) + f\left( \frac{2n}{2d} \right)
\end{align*}
Summing them and simplifying implies that
\[ f\left( \frac{n-1}{d} \right) + f\left( \frac{n+1}{d} \right)
= 2 f \left( \frac nd \right) \]
or equivalently
\[ f\left( \frac nd \right) - f\left( \frac{n-1}{d} \right)
= f\left( \frac{n+1}{d} \right) - f\left( \frac nd \right). \]
This implies that on the set of rational numbers
with denominator dividing $d$, the function $f$ is linear.
In particular, we should have
$f\left( \frac nd \right) = f(0) + \frac nd f(1)$
since $\frac nd$, $0$, $1$ have denominators dividing $d$.
This is the same as saying $f(q) = f(0) + q (f(1)-f(0))$
for any $q \in \QQ$, which is what we wanted to prove.
|
JMO-2015-notes_5
|
Let $ABCD$ be a cyclic quadrilateral.
Prove that there exists a point $X$ on segment $\ol{BD}$
such that $\angle BAC=\angle XAD$ and $\angle BCA=\angle XCD$
if and only if there exists a point $Y$ on segment $\ol{AC}$
such that $\angle CBD=\angle YBA$ and $\angle CDB=\angle YDA$.
|
Both conditions are equivalent to $ABCD$ being harmonic.
Here is a complex solution. Extend $U$ and $V$ and shown. Thus $u = bd/a$ and $v = bd/c$.
\begin{center}
\begin{asy}
size(4.5cm);
pair A = Drawing("A", dir(110), dir(110));
pair B = Drawing("B", dir(160), dir(160));
pair D = Drawing("D", dir( 20), dir( 20));
pair C = Drawing("C", dir(240), dir(240));
draw(unitcircle);
draw(A--B--C--D--cycle);
draw(B--D);
pair U = -conj(A);
pair V = -conj(C);
Drawing("U", U, U);
Drawing("V", V, V);
draw(C--U);
draw(A--V);
Drawing("X", extension(C,U,A,V), dir(0));
draw(A--U, dashed);
draw(C--V, dashed);
draw(A--C);
\end{asy}
\end{center}
Note $\ol{AV} \cap \ol{CU}$ lies on the
perpendicular bisector of $\ol{BD}$ unconditionally.
Then $X$ exists as described if and only if
the midpoint of $\ol{BD}$ lies on $\ol{AV}$.
In complex numbers this is $a + v = m + av \ol m$, or
\[
a + \frac{bd}{c} = \frac{b+d}{2} + \frac{abd}{c} \cdot \frac{b+d}{2bd}
\iff 2(ac+bd) = (b+d)(a+c)
\]
which is symmetric.
|
JMO-2015-notes_6
|
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?
\end{enumerate}
|
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.
|
JMO-2016-notes_1
|
The isosceles triangle $\triangle ABC$, with $AB=AC$,
is inscribed in the circle $\omega$.
Let $P$ be a variable point on the arc $BC$
that does not contain $A$,
and let $I_B$ and $I_C$ denote the incenters of
triangles $\triangle ABP$ and $\triangle ACP$, respectively.
Prove that as $P$ varies, the circumcircle
of triangle $\triangle PI_{B}I_{C}$ passes through a fixed point.
|
Let $M$ be the midpoint of arc $BC$ not containing $A$.
We claim $M$ is the desired fixed point.
\begin{center}
\begin{asy}
pair A = dir(90);
pair M_B = dir(152);
pair B = M_B*M_B/A;
pair M_C = A*A/M_B;
pair C = A*A/B;
pair P = dir(250);
draw(unitcircle, blue);
filldraw(A--B--P--cycle, opacity(0.1)+lightblue, blue);
filldraw(A--C--P--cycle, opacity(0.1)+lightblue, blue);
draw(B--C, blue);
pair M = -A;
pair I_B = incenter(P, A, B);
pair I_C = incenter(P, A, C);
draw(I_B--P--I_C, blue);
draw(A--P, blue);
filldraw(incircle(A, B, P), opacity(0.05)+heavycyan, heavycyan+dotted);
filldraw(incircle(A, C, P), opacity(0.05)+heavycyan, heavycyan+dotted);
filldraw(M_B--I_B--M--cycle, opacity(0.2)+yellow, orange+1);
filldraw(M_C--I_C--M--cycle, opacity(0.2)+yellow, orange+1);
draw(circumcircle(P, I_B, I_C), red+dashed);
dot("$A$", A, dir(A));
dot("$M_B$", M_B, dir(M_B));
dot("$B$", B, dir(B));
dot("$M_C$", M_C, dir(M_C));
dot("$C$", C, dir(C));
dot("$P$", P, dir(P));
dot("$M$", M, dir(M));
dot("$I_B$", I_B, dir(160));
dot("$I_C$", I_C, dir(170));
/* TSQ Source:
A = dir 90
M_B = dir 152
B = M_B*M_B/A
M_C = A*A/M_B
C = A*A/B
P = dir 250
unitcircle blue
A--B--P--cycle 0.1 lightblue / blue
A--C--P--cycle 0.1 lightblue / blue
B--C blue
M = -A
I_B = incenter P A B R160
I_C = incenter P A C R170
I_B--P--I_C blue
A--P blue
incircle A B P 0.05 heavycyan / heavycyan dotted
incircle A C P 0.05 heavycyan / heavycyan dotted
M_B--I_B--M--cycle 0.2 yellow / orange+1
M_C--I_C--M--cycle 0.2 yellow / orange+1
circumcircle P I_B I_C red dashed
*/
\end{asy}
\end{center}
Let $M_B$, $M_C$ be the second intersections of $PI_B$ and $PI_C$ with circumcircle.
\begin{claim*}
We have $\triangle I_{B}M_{B}M \cong \triangle I_{C}M_{C}M$.
\end{claim*}
\begin{proof}
Note that
\begin{align*}
M_{B}I_{B} &= M_{B}B = M_{C}C = M_{C}I_{C} \\
MM_B &= MM_C \\
\angle I_{B}M_{B}M &= \frac{1}{2} \widehat{PM} = \angle I_{C}M_{C}M.
\end{align*}
This implies the desired congruence.
\end{proof}
Since $\angle MPA = 90^\circ$ and ray $PA$ bisects $\angle I_{B}PI_{C}$,
the conclusion $MI_B = MI_C$ finishes the problem.
\begin{remark}
Complex in the obvious way DOES NOT WORK,
because the usual claim (``the fixed point is arc midpoint'') is FALSE
if the hypothesis that $P$ lies in the interior of the arc is dropped.
See figure below.
\begin{center}
\begin{asy}
pair A = dir(90);
pair M_B = dir(152);
pair B = M_B*M_B/A;
pair M_C = A*A/M_B;
pair C = A*A/B;
pair P = dir(140);
draw(unitcircle, blue);
filldraw(A--B--P--cycle, opacity(0.1)+lightblue, blue);
filldraw(A--C--P--cycle, opacity(0.1)+lightblue, blue);
draw(B--C, blue);
pair M = -A;
pair I_B = incenter(P, A, B);
pair I_C = incenter(P, A, C);
draw(A--P, blue);
filldraw(incircle(A, B, P), opacity(0.05)+heavycyan, heavycyan+dotted);
filldraw(incircle(A, C, P), opacity(0.05)+heavycyan, heavycyan+dotted);
draw(circumcircle(P, I_B, I_C), red+dashed);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$P$", P, dir(P));
dot("$M$", M, dir(M));
dot("$I_B$", I_B, dir(100));
dot("$I_C$", I_C, dir(340));
/* TSQ Source:
A = dir 90
M_B := dir 152
B = M_B*M_B/A
M_C := A*A/M_B
C = A*A/B
P = dir 140
unitcircle blue
A--B--P--cycle 0.1 lightblue / blue
A--C--P--cycle 0.1 lightblue / blue
B--C blue
M = -A
I_B = incenter P A B R100
I_C = incenter P A C R340
A--P blue
incircle A B P 0.05 heavycyan / heavycyan dotted
incircle A C P 0.05 heavycyan / heavycyan dotted
circumcircle P I_B I_C red dashed
*/
\end{asy}
\end{center}
Fun story, I pointed this out to Zuming during grading;
I was the only one that realized the subtlety.
\end{remark}
|
JMO-2016-notes_2
|
Prove that there exists a positive integer $n < 10^6$
such that $5^n$ has six consecutive zeros in its decimal representation.
|
We will prove that $\boxed{n = 20 + 2^{19} = 524308}$ fits the bill.
First, we claim that
\[ 5^n \equiv 5^{20} \pmod{5^{20}} \qquad\text{and}\qquad
5^n \equiv 5^{20} \pmod{2^{20}}. \]
Indeed, the first equality holds since both sides are $0 \pmod{5^{20}}$,
and the second by $\varphi(2^{20}) = 2^{19}$ and Euler's theorem.
Hence \[ 5^n \equiv 5^{20} \pmod{10^{20}}. \]
In other words, the last $20$ digits of $5^n$
will match the decimal representation of $5^{20}$, with leading zeros.
However, we have
\[ 5^{20} = \frac{1}{2^{20}} \cdot 10^{20}
< \frac{1}{1000^2} \cdot 10^{20} = 10^{-6} \cdot 10^{20} \]
and hence those first six of those $20$ digits will all be zero.
This completes the proof!
(To be concrete, it turns out that $5^{20} = 95367431640625$
and so the last $20$ digits of $5^n$ will be $00000095367431640625$.)
\begin{remark*}
Many of the first posts in the JMO 2016 discussion thread
(see \url{https://aops.com/community/c5h1230514})
claimed that the problem was ``super easy''.
In fact, the problem was solved by only about 10\% of contestants.
\end{remark*}
\paragraph{Authorship comments.}
This problem was inspired by the observation $5^8 \equiv 5^4 \pmod{10^4}$,
i.e.\ that $5^8$ ended with $0625$.
I noticed this one day back in November, when I was lying on my bed after
a long afternoon and was mindlessly computing powers of $5$ in my head
because I was too tired to do much else.
When I reached $5^8$ I noticed for the first time that the ending
$0625$ was actually induced by $5^4$.
(Given how much MathCounts I did, I really should have known this earlier!)
Thinking about this for a few more seconds,
I realized one could obtain arbitrarily long strings of $0$'s
by using a similar trick modulo larger powers of $10$.
This surprised me, because I would have thought that if this was true,
then I would have learned about it back in my contest days.
However, I could not find any references,
and I thought the result was quite nice,
so I submitted it as a proposal for the JMO,
where I thought it might be appreciated.
The joke about six consecutive zeros is due to Zuming Feng.
|
JMO-2016-notes_3
|
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*}
|
JMO-2016-notes_4
|
Find, with proof, the least integer $N$ such that
if any $2016$ elements are removed
from the set $\{1, 2, \dots, N\}$,
one can still find $2016$ distinct numbers
among the remaining elements with sum $N$.
|
The answer is
\[ N = 2017 + 2018 + \dots + 4032
= 1008 \cdot 6049 = 6097392. \]
\paragraph{Proof that $N \ge 6097392$ is necessary.}
To see that $N$ must be at least this large,
consider the situation
when $1$, $2$, \dots, $2016$ are removed.
Among the remaining elements,
any sum of $2016$ elements is certainly
at least $2017 + 2018 + \dots + 4032$.
\paragraph{Proof that $N = 6097392$ does in fact work.}
Consider the $3024$ pairs of numbers
$(1, 6048)$, $(2, 6047)$, \dots, $(3024, 3025)$.
Regardless of which $2016$ elements of $\{1, 2, \dots, N\}$ are deleted,
at least $3024 - 2016 = 1008$ of these pairs have both elements remaining.
Since each pair has sum $6049$, we can take these pairs to be the desired numbers.
|
JMO-2016-notes_5
|
Let $\triangle ABC$ be an acute triangle, with $O$ as its circumcenter.
Point $H$ is the foot of the perpendicular from $A$ to line $BC$,
and points $P$ and $Q$ are the feet of the perpendiculars
from $H$ to the lines $AB$ and $AC$, respectively.
Given that \[ AH^2 = 2AO^2, \]
prove that the points $O$, $P$, and $Q$ are collinear.
|
We present two approaches.
\paragraph{First approach (synthetic).}
First, since $AP \cdot AB = AH^2 = AQ \cdot AC$, it follows that $PQCB$ is cyclic.
Consequently, we have $AO \perp PQ$.
\begin{center}
\begin{asy}
size(8cm);
pair A = dir(110);
pair H = A+1.414*dir(-90);
pair U = H+6*dir(0);
pair V = H-6*dir(0);
pair B = OP(U--V, unitcircle);
pair C = IP(U--V, unitcircle);
pair P = foot(H, A, B);
pair Q = foot(H, A, C);
pair D = -A;
draw(A--B--C--cycle, blue);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
draw(A--H);
draw(P--H--Q);
pair K = circumcenter(A, B, C);
draw(P--Q, blue);
draw(A--D--C, red);
filldraw(circumcircle(P, Q, B), opacity(0.1)+yellow, dashed+orange);
filldraw(circumcircle(Q, C, D), opacity(0.1)+yellow, dashed+orange);
dot("$A$", A, dir(A));
dot("$H$", H, dir(H));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(70));
dot("$D$", D, dir(D));
dot("$K$", K, dir(60));
/* Source generated by TSQ */
\end{asy}
\end{center}
Let $K$ be the foot of $A$ onto $PQ$,
and let $D$ be the point diametrically opposite $A$.
Thus $A$, $K$, $O$, $D$ are collinear.
Since quadrilateral $KQCD$ is cyclic ($\angle QKD = \angle QCD = 90^\circ$),
we have
\[ AK \cdot AD = AQ \cdot AC = AH^2
\implies AK = \frac{AH^2}{AD} = \frac{AH^2}{2AO} = AO \]
so $K = O$.
\paragraph{Second approach (coordinates), with Joshua Hsieh.}
We impose coordinates with $H$ at the origin
and $A = (0,a)$, $B = (-b,0)$, $C = (c,0)$, for $a,b,c > 0$.
\begin{claim*}
The circumcenter has coordinates $(\frac{c-b}{2}, \frac a2 - \frac{bc}{2a})$.
\end{claim*}
\begin{proof}
This is a known lemma but but we reproduce its proof for completeness.
It uses the following steps:
\begin{itemize}
\ii By power of a point, the second intersection
of line $AH$ with the circumcircle is $(0, -\frac{bc}{a})$.
\ii Since the orthocenter is the reflection of this point across line $BC$,
the orthocenter is given exactly by $(0, \frac{bc}{a})$.
\ii The centroid is is $\frac{\vec A + \vec B + \vec C}{3} = (\frac{c-b}{3}, \frac a3)$.
\ii Since $\vec H - \vec O = 3( \vec G - \vec O)$
according to the Euler line, we have $\vec O = \frac 32 \vec G - \frac 12 \vec H$.
This gives the desired formula. \qedhere
\end{itemize}
\end{proof}
Note that $HQ = \frac{HA \cdot HC}{AC} = \frac{ac}{\sqrt{a^2+c^2}}$.
If we let $T$ be the foot from $Q$ to $BC$,
then $\triangle HQT \overset{\sim}{+} \triangle AHC$
and so the $x$-coordinate of $Q$ is given by
$HQ \cdot \frac{AH}{AC} = \frac{a^2c}{a^2+c^2}$.
Repeating the analogous calculation for $Q$ and $P$ gives
\begin{align*}
Q &= \left( \frac{a^2 c}{a^2+c^2}, \frac{ac^2}{a^2+c^2} \right) \\
P &= \left( -\frac{a^2 b}{a^2+b^2}, \frac{ab^2}{a^2+b^2} \right).
\end{align*}
Then, $O$, $P$, $Q$ are collinear if and only if the
following shoelace determinant vanishes
(with denominators cleared out):
\begin{align*}
0
&= \det
\begin{bmatrix}
-a^2b & ab^2 & a^2+b^2 \\
a^2c & ac^2 & a^2+c^2 \\
a(c-b) & a^2-bc & 2a
\end{bmatrix}
= a \det
\begin{bmatrix}
-ab & ab^2 & a^2+b^2 \\
ac & ac^2 & a^2+c^2 \\
c-b & a^2-bc & 2a
\end{bmatrix} \\
&= a \det
\begin{bmatrix}
-a(b+c) & a(b^2-c^2) & b^2-c^2 \\
ac & ac^2 & a^2+c^2 \\
c-b & a^2-bc & 2a
\end{bmatrix}
= a(b+c) \det
\begin{bmatrix}
-a & a(b-c) & b-c \\
ac & ac^2 & a^2+c^2 \\
c-b & a^2-bc & 2a
\end{bmatrix} \\
&= a(b+c) \cdot \big[ -a(a^2c^2-a^4+bc(a^2+c^2))
+ ac(b-c) \left( -a^2-bc \right)
- (b-c)^2 \cdot a^3 \big] \\
&= a^2(b+c)(a^4-a^2b^2-b^2c^2-c^2a^2).
\end{align*}
On the other hand,
\begin{align*}
AH^2 &= a^2 \\
2AO^2 &= 2 \left[ \left( \frac{c-b}{2} \right)^2
+ \left( - \frac{a}{2} - \frac{bc}{2a} \right)^2 \right]
= \frac{a^2+b^2+c^2 + \frac{b^2c^2}{a^2}}{2} \\
\implies AH^2 - 2AO^2 &= \half \left( a^2-b^2-c^2-\frac{b^2c^2}{a^2} \right).
\end{align*}
So the conditions are equivalent.
|
JMO-2016-notes_6
|
Find all functions $f \colon \RR \to \RR$ such that
for all real numbers $x$ and $y$,
\[ (f(x)+xy) \cdot f(x-3y) + (f(y)+xy) \cdot f(3x-y) = (f(x+y))^2. \]
\end{enumerate}
|
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.
|
JMO-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*}
|
JMO-2017-notes_2
|
Show that the Diophantine equation
\[ \left( 3x^3+xy^2 \right)\left( x^2y+3y^3 \right) = (x-y)^7 \]
has infinitely many solutions in positive integers,
and characterize all the solutions.
|
Let $x=da$, $y=db$, where $\gcd(a,b) = 1$ and $a > b$.
The equation is equivalent to
\[ (a-b)^7 \mid ab\left( a^2+3b^2 \right)\left( 3a^2+b^2 \right)
\qquad (\star) \]
with the ratio of the two becoming $d$.
\begin{claim*}
The equation $(\star)$ holds if and only if $a-b=1$.
\end{claim*}
\begin{proof}
Obviously if $a-b=1$ then $(\star)$ is true.
Conversely, suppose $(\star)$ holds.
\begin{itemize}
\ii If $a$ and $b$ are both odd, then $a^2+3b^2 \equiv 4 \pmod 8$.
Similarly $3a^2+b^2 \equiv 4 \pmod 8$.
Hence $2^4$ exactly divides right-hand side, contradiction.
\ii Now suppose $a-b$ is odd.
We have $\gcd(a-b,a) = \gcd(a-b,b) = 1$ by Euclid, but also
\[ \gcd(a-b, a^2+3b^2) = \gcd(a-b, 4b^2) = 1 \]
and similarly $\gcd(a-b, 3a^2+b^2) = 1$.
Thus $a-b$ is coprime to each of $a$, $b$,
$a^2+3b^2$, $3a^2+b^2$ and this forces $a-b = 1$.
\end{itemize}
\end{proof}
This therefore describes all solutions: namely,
for any $b \ge 1$, if we set $a = b+1$ and $d = ab(a^2+3b^2)(3a^2+b)$
then $(x,y) = (da,db)$ works and any solution is of this form.
\begin{remark*}
One can give different cosmetic representations of the same solution set.
For example, we could write $b = \half (n-1)$
and $a = \half (n+1)$ with $n > 1$ any \emph{odd} integer.
Then $d = ab(a^2+3b^2)(3a^2+b^2) =
\frac{(n-1)(n+1)(n^2+n+1)(n^2-n+1)}{4} = \frac{n^6-1}{4}$,
and hence the solution is
\[ (x,y) = (da, db) = \left( \frac{(n+1)(n^6-1)}{8},
\frac{(n-1)(n^6-1)}{8} \right) \]
which is a little simpler to write.
The smallest solutions are $(364,182)$, $(11718, 7812)$, \dots.
\end{remark*}
|
JMO-2017-notes_3
|
Let $ABC$ be an equilateral triangle and $P$ a point on its circumcircle.
Set $D = \ol{PA} \cap \ol{BC}$, $E = \ol{PB} \cap \ol{CA}$, $F = \ol{PC} \cap \ol{AB}$.
Prove that the area of triangle $DEF$ is twice the area of triangle $ABC$.
|
\paragraph{First solution (barycentric).}
We invoke barycentric coordinates on $ABC$.
Let $P = (u:v:w)$, with $uv+vw+wu = 0$ (circumcircle equation with $a=b=c$).
Then $D = (0:v:w)$, $E = (u:0:w)$, $F = (u:v:0)$.
Hence
\begin{align*}
\frac{[DEF]}{[ABC]}
&= \frac{1}{(u+v)(v+w)(w+u)}
\det
\begin{bmatrix}
0 & v & w \\
u & 0 & w \\
u & v & 0
\end{bmatrix}
\\
&= \frac{2uvw}{(u+v)(v+w)(w+u)} \\
&= \frac{2uvw}{(u+v+w)(uv+vw+wu)-uvw} \\
&= \frac{2uvw}{-uvw} = -2
\end{align*}
as desired (areas signed).
\paragraph{Second solution (``nice'' lengths).}
WLOG $ABPC$ is convex.
Let $x = AB = BC = CA$.
By Ptolemy's theorem and strong Ptolemy,
\begin{align*}
PA &= PB + PC \\
PA^2 &= PB \cdot PC + AB \cdot AC = PB \cdot PC + x^2 \\
\implies x^2 &+ PB^2 + PB \cdot PC + PC^2.
\end{align*}
Also, $PD \cdot PA = PB \cdot PC$ and similarly
since $\ol{PA}$ bisects $\angle BPC$ (causing $\triangle BPD \sim \triangle APC$).
Now $P$ is the Fermat point of $\triangle DEF$,
since $\angle DPF = \angle FPE = \angle EPD = 120\dg$. Thus
\begin{align*}
[DEF] &= \frac{\sqrt3}{4} \sum_{\text{cyc}} PE \cdot PF \\
&= \frac{\sqrt3}{4} \sum_{\text{cyc}}
\left( \frac{PA \cdot PC}{PB} \right)
\left( \frac{PA \cdot PB}{PC} \right) \\
&= \frac{\sqrt3}{4} \sum_{\text{cyc}} PA^2 \\
&= \frac{\sqrt3}{4} \left( (PB+PC)^2 + PB^2 + PC^2 \right) \\
&= \frac{\sqrt3}{4} \cdot 2x^2 = 2[ABC].
\end{align*}
|
JMO-2017-notes_4
|
Are there any triples $(a,b,c)$ of positive integers such that
$(a-2)(b-2)(c-2)+12$ is a prime number
that properly divides the positive number
$a^2+b^2+c^2+abc-2017$?
|
No such $(a,b,c)$.
Assume not.
Let $x=a-2$, $y=b-2$, $z=c-2$, hence $x,y,z \ge -1$.
\begin{align*}
a^2+b^2+c^2+abc-2017
&= (x+2)^2 + (y+2)^2 + (z+2)^2 \\ &
+ (x+2)(y+2)(z+2) - 2017 \\
&= (x+y+z+4)^2 + (xyz+12) - 45^2.
\end{align*}
Thus the divisibility relation becomes
\[ p = xyz+12 \mid \left( x+y+z+4 \right)^2 - 45^2 > 0 \]
so either
\begin{align*}
p &= xyz+12 \mid x+y+z-41 \\
p &= xyz+12 \mid x+y+z+49
\end{align*}
Assume $x \ge y \ge z$, hence $x \ge 14$ (since $x+y+z \ge 41$).
We now eliminate several edge cases
to get $x,y,z \neq -1$ and a little more:
\begin{claim*}
We have $x \ge 17$, $y \ge 5$, $z \ge 1$, and $\gcd(xyz, 6) = 1$.
\end{claim*}
\begin{proof}
First, we check that neither $y$ nor $z$ is negative.
\begin{itemize}
\ii If $x > 0$ and $y=z=-1$, then we want $p=x+12$
to divide either $x-43$ or $x+47$.
We would have $0 \equiv x-43 \equiv -55 \pmod p$
or $0\equiv x+47 \equiv 35 \pmod p$,
but $p > 11$ contradiction.
\ii If $x, y > 0$, and $z = -1$, then $p = 12-xy > 0$.
However, this is clearly incompatible with $x \ge 14$.
\end{itemize}
Finally, obviously $xyz \neq 0$ (else $p=12$).
So $p = xyz + 12 \ge 14 \cdot 1^2 +12 = 26$ or $p \ge 29$.
Thus $\gcd(6,p) = 1$ hence $\gcd(6,xyz)=1$.
We finally check that $y=1$ is impossible, which forces $y \ge 5$.
If $y=1$ and hence $z=1$ then $p=x+12$ should divide either
$x+51$ or $x-39$.
These give $39 \equiv 0 \pmod p$ or $25 \equiv 0 \pmod p$,
but we are supposed to have $p \ge 29$.
\end{proof}
In that situation $x+y+z-41$ and $x+y+z+49$ are both even,
so whichever one is divisible by $p$
is actually divisible by $2p$.
Now we deduce that:
\[ x+y+z+49 \ge 2p = 2xyz + 24 \implies 25 \ge 2xyz-x-y-z. \]
But $x \ge 17$ and $y \ge 5$ thus
\begin{align*}
2xyz-x-y-z &= z(2xy-1) - x - y \\
&\ge 2xy - 1 - x - y \\
&> (x-1)(y-1) > 60
\end{align*}
which is a contradiction.
Having exhausted all the cases we conclude no solutions exist.
\begin{remark*}
The condition that $x+y+z-41 > 0$ (which comes from ``properly divides'')
cannot be dropped. Examples of solutions in which $x+y+z-41 = 0$
include $(x,y,z) = (31,5,5)$ and $(x,y,z) = (29,11,1)$.
\end{remark*}
|
JMO-2017-notes_5
|
Let $O$ and $H$ be the circumcenter
and the orthocenter of an acute triangle $ABC$.
Points $M$ and $D$ lie on side $BC$
such that $BM = CM$ and $\angle BAD = \angle CAD$.
Ray $MO$ intersects the circumcircle of triangle $BHC$ in point $N$.
Prove that $\angle ADO = \angle HAN$.
|
Let $P$ and $Q$ be the arc midpoints of $\widehat{BC}$,
so that $ADMQ$ is cyclic (as $\dang QAD = \dang QMD = 90\dg$) .
Since it's known that $(BHC)$ and $(ABC)$ are reflections across line $BC$,
it follows $N$ is the reflection of the arc midpoint $P$ across $M$.
\begin{claim*}
Quadrilateral $ADNO$ is cyclic.
\end{claim*}
\begin{proof}
Since $PN \cdot PO = \half PN \cdot 2PO = PM \cdot PQ = PD \cdot PA$.
\end{proof}
\begin{center}
\begin{asy}
pair A = dir(130);
pair B = dir(220);
pair C = dir(320);
filldraw(unitcircle, opacity(0.2)+lightcyan, lightblue);
pair P = dir(-90);
pair Q = dir(90);
pair D = extension(A, P, B, C);
pair O = origin;
pair M = extension(B, C, O, P);
pair N = 2*M-P;
draw(A--B--C--cycle, lightblue);
draw(A--P--Q, lightblue);
draw(A--N--D--O--A, lightblue);
filldraw(A--D--N--O--cycle, opacity(0.1)+yellow, 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("$D$", D, dir(225));
dot("$O$", O, dir(315));
dot("$M$", M, dir(315));
dot("$N$", N, dir(315));
/* TSQ Source:
A = dir 130
B = dir 220
C = dir 320
unitcircle 0.1 lightcyan / lightblue
P = dir -90
Q = dir 90
D = extension A P B C R225
O = origin R315
M = extension B C O P R315
N = 2*M-P R315
A--B--C--cycle lightblue
A--P--Q lightblue
A--N--D--O--A lightblue
A--D--N--O--cycle 0.1 yellow / red
*/
\end{asy}
\end{center}
To finish, note that $\dang HAN = \dang ONA = \dang ODA$.
\begin{remark*}
The orthocenter $H$ is superficial
and can be deleted basically immediately.
One can reverse-engineer the fact that $ADNO$ is cyclic
from the truth of the problem statement.
\end{remark*}
\begin{remark*}
One can also show $ADNO$ concyclic by just
computing $\dang DAO = \dang PAO$
and $\dang DNO = \dang DPN = \dang APQ$
in terms of the angles of the triangle, or even more directly just because
\[ \dang DNO = \dang DNP = \dang NPD = \dang OPD = \dang ONA = \dang HAN. \]
\end{remark*}
|
JMO-2017-notes_6
|
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.
\end{enumerate}
|
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.
|
JMO-2018-notes_1
|
For each positive integer $n$, find the number of $n$-digit positive integers
for which no two consecutive digits are equal, and the last digit is a prime.
|
Almost trivial.
Let $a_n$ be the desired answer.
We have \[ a_n + a_{n-1} = 4 \cdot 9^{n-1} \]
for all $n$, by padding the $(n-1)$ digit numbers with a leading zero.
Since $a_0 = 0$, $a_1 = 4$, solving the recursion gives
\[ a_n = \frac 25 \left( 9^n - (-1)^n \right). \]
The end.
\begin{remark*}
For concreteness, the first few terms are $0$, $4$, $32$, $292$, \dots.
\end{remark*}
|
JMO-2018-notes_2
|
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.
|
JMO-2018-notes_3
|
Let $ABCD$ be a quadrilateral inscribed
in circle $\omega$ with $\ol{AC} \perp \ol{BD}$.
Let $E$ and $F$ be the reflections of $D$ over
$\ol{BA}$ and $\ol{BC}$, respectively, and let $P$ be
the intersection of $\ol{BD}$ and $\ol{EF}$.
Suppose that the circumcircles of $EPD$ and $FPD$
meet $\omega$ at $Q$ and $R$ different from $D$.
Show that $EQ = FR$.
|
Most of this problem is about realizing where
the points $P$, $Q$, $R$ are.
\paragraph{First solution (Evan Chen).}
Let $X$, $Y$, be the feet from $D$ to $\ol{BA}$, $\ol{BC}$,
and let $Z = \ol{BD} \cap \ol{AC}$.
By Simson theorem, the points $X$, $Y$, $Z$ are collinear.
Consequently, the point $P$ is the reflection of $D$ over $Z$,
and so we conclude $P$ is the orthocenter of $\triangle ABC$.
\begin{center}
\begin{asy}
size(10cm);
pair B = dir(100);
pair A = dir(210);
pair C = dir(330);
pair D = -A*C/B;
pair X = foot(D, B, A);
pair Y = foot(D, B, C);
pair Z = extension(A, C, B, D);
pair E = 2*X-D;
pair F = 2*Y-D;
pair P = A+B+C;
pair Q = -A*B/C;
pair R = -B*C/A;
filldraw(unitcircle, opacity(0.1)+lightred, red);
draw(A--B--C--D--cycle, red);
draw(A--C, red);
draw(B--P, red);
draw(A--X, red);
draw(X--Y, orange);
draw(E--F, orange);
draw(F--D--E, orange);
filldraw(circumcircle(P, F, D), opacity(0.1)+yellow, orange);
filldraw(circumcircle(P, E, D), opacity(0.1)+yellow, orange);
draw(A--R, red+dotted);
draw(C--Q, red+dotted);
draw(R--F, lightcyan);
draw(P--D, lightcyan);
draw(Q--E, lightcyan);
dot("$B$", B, dir(B));
dot("$A$", A, dir(180));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(80));
dot("$Z$", Z, dir(315));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$P$", P, dir(135));
dot("$Q$", Q, dir(110));
dot("$R$", R, dir(80));
/* TSQ Source:
!size(10cm);
B = dir 100
A = dir 210 R180
C = dir 330
D = -A*C/B
X = foot D B A
Y = foot D B C R80
Z = extension A C B D R315
E = 2*X-D
F = 2*Y-D
P = A+B+C R135
Q = -A*B/C R110
R = -B*C/A R80
unitcircle 0.1 lightred / red
A--B--C--D--cycle red
A--C red
B--P red
A--X red
X--Y orange
E--F orange
F--D--E orange
circumcircle P F D 0.1 yellow / orange
circumcircle P E D 0.1 yellow / orange
A--R red dotted
C--Q red dotted
R--F lightcyan
P--D lightcyan
Q--E lightcyan
*/
\end{asy}
\end{center}
Suppose now we extend ray $CP$ to meet $\omega$ again at $Q'$.
Then $\ol{BA}$ is the perpendicular bisector
of both $\ol{PQ'}$ and $\ol{DE}$;
consequently, $PQ'ED$ is an isosceles trapezoid.
In particular, it is cyclic, and so $Q' = Q$.
In the same way $R$ is the second intersection
of ray $\ol{AP}$ with $\omega$.
Now, because of the two isosceles trapezoids we have found, we conclude
\[ EQ = PD = FR \]
as desired.
\begin{remark*}
Alternatively, after identifying $P$,
one can note $\ol{BQE}$ and $\ol{BRF}$ are collinear.
Since $BE = BD = BF$,
upon noticing $BQ = BP = BR$ we are also done.
\end{remark*}
\paragraph{Second solution (Danielle Wang).}
Here is a solution which does not identify the point $P$ at all.
We know that $BE = BD = BF$, by construction.
\begin{center}
\begin{asy}
size(12cm);
pair B = dir(100);
pair A = dir(210);
pair C = dir(330);
pair D = -A*C/B;
pair X = foot(D, B, A);
pair Y = foot(D, B, C);
pair Z = extension(A, C, B, D);
pair E = 2*X-D;
pair F = 2*Y-D;
pair P = A+B+C;
pair Qp = -A*B/C;
pair Rp = -B*C/A;
filldraw(unitcircle, opacity(0.1)+lightred, red);
draw(A--B--C--D--cycle, red);
draw(A--C, red);
draw(B--D, heavycyan);
draw(A--X, red);
draw(E--F, orange);
draw(F--D--E, orange);
filldraw(circumcircle(P, F, D), opacity(0.1)+yellow, dotted);
filldraw(circumcircle(P, E, D), opacity(0.1)+yellow, dotted);
draw(E--B--F, heavycyan);
draw(Qp--D--Rp, orange);
draw(arc(B,abs(B-D),190,350), heavycyan);
dot("$B$", B, dir(B));
dot("$A$", A, dir(180));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot(X);
dot(Y);
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$P$", P, dir(135));
dot("$Q'$", Qp, dir(110));
dot("$R'$", Rp, dir(80));
/* TSQ Source:
!size(12cm);
B = dir 100
A = dir 210 R180
C = dir 330
D = -A*C/B
X .= foot D B A
Y .= foot D B C R80
Z := extension A C B D R315
E = 2*X-D
F = 2*Y-D
P = A+B+C R135
Q' = -A*B/C R110
R' = -B*C/A R80
unitcircle 0.1 lightred / red
A--B--C--D--cycle red
A--C red
B--D heavycyan
A--X red
E--F orange
F--D--E orange
circumcircle P F D 0.1 yellow / dotted
circumcircle P E D 0.1 yellow / dotted
E--B--F heavycyan
Qp--D--Rp orange
!draw(arc(B,abs(B-D),190,350), heavycyan);
*/
\end{asy}
\end{center}
\begin{claim*}
The points $B$, $Q$, $E$ are collinear.
Similarly the points $B$, $R$, $F$ are collinear.
\end{claim*}
\begin{proof}
Work directed modulo $180\dg$.
Let $Q'$ be the intersection of $\ol{BE}$ with $(ABCD)$.
Let $\alpha = \dang DEB = \dang BDE$
and $\beta = \dang BFD = \dang FDB$.
Observe that $BE = BD = BF$, so $B$ is the circumcenter
of $\triangle DEF$. Thus, $\dang DEP = \dang DEF = 90\dg - \beta$.
Then
\begin{align*}
\dang DPE &= \dang DEP + \dang PDE = (90\dg-\beta) + \alpha \\
&= \alpha - \beta + 90\dg \\
\dang DQ'B &= \dang DCB = \dang DCA + \dang ACB \\
&= \dang DBA - (90\dg - \dang DBC) = -(90\dg-\alpha) - (90\dg - (90\dg - \beta)) \\
&= \alpha - \beta + 90\dg.
\end{align*}
Thus $Q'$ lies on the desired circle, so $Q' = Q$.
\end{proof}
Now, by power of a point we have $BQ \cdot BE = BP \cdot BD = BR \cdot BF$,
so $BQ = BP = BR$. Hence $EQ = PD = FR$.
|
JMO-2018-notes_4
|
Find all real numbers $x$ for which
there exists a triangle $ABC$ with circumradius $2$,
such that $\angle ABC \ge 90\dg$, and
\[ x^4 + ax^3 + bx^2 + cx + 1 = 0 \]
where $a = BC$, $b = CA$, $c = AB$.
|
The answer is $x = -\half (\sqrt6 \pm \sqrt 2)$.
We prove this the only possible answer.
Evidently $x < 0$.
Now, note that
\[ a^2+c^2 \le b^2 \le 4b \]
since $b \le 4$ (the diameter of its circumcircle).
Then,
\begin{align*}
0 &= x^4 + ax^3 + bx^2 + cx + 1 \\
&= x^2 \left[ \left( x + \half a \right)^2
+ \left( \frac1x + \half c \right)^2
+ \left( b - \frac{a^2+c^2}{4} \right) \right] \\
&\ge 0+0+0 = 0.
\end{align*}
In order for equality to hold,
we must have $x = -\half a$, $1/x = -\half c$,
and $a^2+c^2 = b^2 = 4b$.
This gives us $b = 4$, $ac = 4$, $a^2+c^2=16$.
Solving for $a,c > 0$ implies
\[ \{a,c\} = \left\{ \sqrt6 \pm \sqrt 2 \right\}. \]
This gives the $x$ values claimed above;
by taking $a$, $b$, $c$ as deduced here,
we find they work too.
\begin{remark*}
Note that by perturbing $\triangle ABC$ slightly,
we see \emph{a priori} that the set of possible $x$
should consist of unions of intervals (possibly trivial).
So it makes sense to try inequalities no matter what.
\end{remark*}
|
JMO-2018-notes_5
|
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*}
|
JMO-2018-notes_6
|
Karl starts with $n$ cards labeled $1$, $2$, \dots, $n$
lined up in random order on his desk.
He calls a pair $(a,b)$ of cards \emph{swapped} if $a > b$
and the card labeled $a$ is to the left of the card labeled $b$.
Karl picks up the card labeled $1$ and inserts it back into the sequence in
the opposite position: if the card labeled $1$ had $i$ cards to its left,
then it now has $i$ cards to its right.
He then picks up the card labeled $2$ and reinserts it in the same manner,
and so on, until he has picked up and put
back each of the cards $1$, \dots, $n$ exactly once in that order.
For example, if $n = 4$, then one example of a process is
\[ 3142 \longrightarrow 3412 \longrightarrow 2341 \longrightarrow 2431 \longrightarrow 2341 \]
which has three swapped pairs both before and after.
Show that, no matter what lineup of cards Karl started with,
his final lineup has the same number of swapped pairs as the starting lineup.
\end{enumerate}
|
The official solution is really tricky.
Call the process $P$.
We define a new process $P'$ where, when re-inserting card $i$,
we additionally change its label from $i$ to $n+i$.
An example of $P'$ also starting with $3142$ is:
\[ 3142 \longrightarrow 3452 \longrightarrow 6345
\longrightarrow 6475 \longrightarrow 6785. \]
Note that now, each step of $P'$ preserves the number of inversions.
Moreover, the final configuration of $P'$ is the same as
the final configuration of $P$ with all cards incremented by $n$,
and of course thus has the same number of inversions. The end.
|
JMO-2019-notes_1
|
There are $a+b$ bowls arranged in a row,
numbered $1$ through $a+b$,
where $a$ and $b$ are given positive integers.
Initially, each of the first $a$ bowls contains an apple,
and each of the last $b$ bowls contains a pear.
A legal move consists of moving an apple from bowl $i$ to bowl $i+1$
and a pear from bowl $j$ to bowl $j-1$,
provided that the difference $i-j$ is even.
We permit multiple fruits in the same bowl at the same time.
The goal is to end up with the first $b$ bowls each containing a pear
and the last $a$ bowls each containing an apple.
Show that this is possible if and only if the product $ab$ is even.
|
First we show that if $ab$ is even then the goal is possible.
We prove the result by induction on $a+b$.
\begin{itemize}
\ii If $\min(a,b) = 0$ there is nothing to check.
\ii If $\min(a,b) = 1$, say $a=1$, then $b$ is even,
and we can swap the (only) leftmost apple
with the rightmost pear by working only with those fruits.
\ii Now assume $\min(a,b) \ge 2$ and $a+b$ is odd.
Then we can swap the leftmost apple with rightmost pear by working only with those fruits,
reducing to the situation of $(a-1, b-1)$ which is possible by induction
(at least one of them is even).
\ii Finally assume $\min(a,b) \ge 2$ and $a+b$ is even (i.e.\ $a$ and $b$ are both even).
Then we can swap the apple in position $1$ with the pear in position $a+b-1$,
and the apple in position $2$ with the pear in position $a+b$.
This reduces to the situation of $(a-2, b-2)$ which is also possible by induction.
\end{itemize}
Now we show that the result is impossible if $ab$ is odd.
Define
\begin{align*}
X &= \text{number apples in odd-numbered bowls} \\
Y &= \text{number pears in odd-numbered bowls}.
\end{align*}
Note that $X-Y$ does not change under this operation.
However, if $a$ and $b$ are odd,
then we initially have $X = \half(a+1)$ and $Y = \half(b-1)$,
while the target position has $X = \half(a-1)$ and $Y = \half(b+1)$.
So when $ab$ is odd this is not possible.
\begin{remark*}
Another proof that $ab$ must be even
is as follows.
First, note that apples only move right and pears only move left,
a successful operation must take exactly $ab$ moves.
So it is enough to prove that the \emph{number of moves}
made must be even.
However, the number of fruits in odd-numbered bowls
either increases by $+2$ or $-2$ in each move
(according to whether $i$ and $j$ are both even or both odd),
and since it ends up being the same at the end,
the number of moves must be even.
Alternatively, as pointed out in the official solutions,
one can consider the sums of squares of positions of fruits.
The quantity changes by
\[ \left[ (i+1)^2 + (j-1)^2 \right] - (i^2+j^2)
= 2(i-j) + 2 \equiv 2 \pmod 4 \]
at each step,
and eventually the sums of squares returns to zero, as needed.
\end{remark*}
|
JMO-2019-notes_2
|
For which pairs of integers $(a, b)$ do there exist functions
$f\colon \ZZ \to \ZZ$ and $g \colon \ZZ \to \ZZ$ obeying
\[ f(g(x)) = x + a \quad \text{and} \quad g(f(x)) = x + b \]
for all integers $x$?
|
The answer is if $a=b$ or $a=-b$.
In the former case, one can take $f(x) \equiv x+a$ and $g(x) \equiv x$.
In the latter case, one can take $f(x) \equiv -x+a$ and $g(x) = -x$.
Now we prove these are the only possibilities.
First:
\begin{claim*}
The functions $f$ and $g$ are bijections.
\end{claim*}
\begin{proof}
Surjectivity is obvious.
To see injective, note that if $f(u) = f(v)$
then $g(f(u)) = g(f(v)) \implies u+b = v+b \implies u=v$,
and similarly for $g$.
\end{proof}
Note also that for any $x$, we have
\begin{align*}
f(x+b) &= f(g(f(x))) = f(x) + a \\
g(x+a) &= g(f(g(x))) = g(x) + b.
\end{align*}
If either $a$ is zero or $b$ is zero,
we immediately get the other is zero, and hence done.
So assume $ab \neq 0$.
If $|b| > |a|$, then two of
\[ \{ f(0), f(1), \dots, f(b-1) \} \pmod{|a|} \]
coincide, which together with repeatedly applying the first equation
above will then give a contradiction to injectivity of $f$.
Similarly, if $|a| > |b|$ swapping the roles of $f$ and $g$
(and $a$ and $b$) will give a contradiction to injectivity of $g$.
This completes the proof.
\begin{remark*}
Here is a way to visualize the argument,
so one can see pictorially what is going on.
We draw two parallel number lines indexed by $\ZZ$.
Starting from $0$, we draw red arrow from $0$ to $f(0)$,
and then a blue arrow from $f(0)$ to $g(f(0)) = b$,
and then a red arrow from $b$ to $g(b) = f(0)+a$, and so on.
These arrows can be extended both directions,
leading to an infinite ``squaretooth'' wave.
The following is a picture of an example with $a,b > 0$.
\begin{center}
\begin{asy}
size(11cm);
usepackage("amsmath");
usepackage("amssymb");
draw( (-3,2)--(13,2), Arrows(TeXHead) );
draw( (-3,-2)--(13,-2), Arrows(TeXHead) );
label("$\mathbb Z$", (13, 2), dir(0));
label("$\mathbb Z$", (13, -2), dir(0));
pair bm1 = (-5,2);
pair b0 = (0,2);
pair b1 = (5,2);
pair b2 = (10,2);
pair b3 = (15,2);
pair am1 = (-1,-2);
pair a0 = (2.5,-2);
pair a1 = (6,-2);
pair a2 = (9.5,-2);
pair a3 = (13,-2);
dot("$0$", b0, dir(90));
dot("$b$", b1, dir(90));
dot("$2b$", b2, dir(90));
dot("$f(0)-a$", am1, dir(-90));
dot("$f(0)$", a0, dir(-90));
dot("$f(0)+a$", a1, dir(-90));
dot("$f(0)+2a$", a2, dir(-90));
draw(midpoint(bm1--am1)--am1, red, EndArrow, Margins);
label("$f$", midpoint(bm1--am1)--am1, dir(50), red);
draw(b0--a0, red, EndArrow, Margins);
label("$f$", b0--a0, dir(50), red);
draw(b1--a1, red, EndArrow, Margins);
label("$f$", b1--a1, dir(50), red);
draw(b2--a2, red, EndArrow, Margins);
label("$f$", b2--a2, dir(50), red);
draw(am1--b0, blue, EndArrow, Margins);
label("$g$", midpoint(am1--b0)--b0, dir(-50), blue);
draw(a0--b1, blue, EndArrow, Margins);
label("$g$", a0--b1, dir(-50), blue);
draw(a1--b2, blue, EndArrow, Margins);
label("$g$", a1--b2, dir(-50), blue);
draw(a2--midpoint(a2--b3), blue, EndArrow, Margins);
label("$g$", a2--midpoint(a2--b3), dir(-50), blue);
\end{asy}
\end{center}
The problem is essentially trying to decompose
our two copies of $\ZZ$ into multiple squaretooth waves.
For this to be possible, we expect the ``width'' of the waves
on the top and bottom must be the same --- i.e., that $|a| = |b|$.
\end{remark*}
\begin{remark*}
This also suggests how to classify all functions $f$ and $g$
satisfying the condition.
If $a = b = 0$ then any pair of functions $f$ and $g$
which are inverses to each other is okay.
There are thus uncountably many pairs of functions $(f,g)$ here.
If $a = b > 0$, then one sets $f(0)$, $f(1)$, \dots, $f(a-1)$
as any values which are distinct modulo $b$,
at which point $f$ and $g$ are uniquely determined.
An example for $a = b = 3$ is
\[ f(x) = \begin{cases}
x + 42 & x \equiv 0 \pmod 3 \\
x + 13 & x \equiv 1 \pmod 3 \\
x - 37 & x \equiv 2 \pmod 3,
\end{cases}
\qquad
g(x) = \begin{cases}
x - 39 & x \equiv 0 \pmod 3 \\
x + 40 & x \equiv 1 \pmod 3 \\
x - 10 & x \equiv 2 \pmod 3.
\end{cases} \]
The analysis for $a = b < 0$ and $a = -b$ are similar,
but we don't include the details here.
\end{remark*}
|
JMO-2019-notes_3
|
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.
|
JMO-2019-notes_4
|
Let $ABC$ be a triangle with $\angle B > 90\dg$
and let $E$ and $F$ be the feet of the altitudes from $B$ and $C$.
Can line $EF$ be tangent to the $A$-excircle?
|
We show it is not possible, by contradiction
(assuming $EF$ is indeed tangent).
Thus $BECF$ is a convex cyclic quadrilateral
inscribed in a circle with diameter $\ol{BC}$.
Note also that the $A$-excircle lies on the opposite side
from $A$ as line $EF$, since $A$, $E$, $C$ are collinear
in that order.
\paragraph{First solution by similarity.}
Note that $\triangle AEF$ is similar to $\triangle ABC$
(and oppositely oriented).
However, since they have the same $A$-exradius,
it follows they are congruent.
\begin{center}
\begin{asy}
size(6cm);
pair B = dir(213);
pair C = dir(47);
pair E = dir(115);
pair F = dir(280);
pair V = extension(F, E, B, C);
pair W = incenter(F, V, C);
pair J = 3.6*W-2.6*V;
pair T = foot(J, B, C);
pair A = extension(B, F, E, C);
filldraw(CP(J, T), opacity(0.1)+deepgreen, deepgreen);
pair U = OP(CP(J, T), CP(midpoint(B--J), J));
pair V = IP(CP(J, T), CP(midpoint(E--J), J));
fill(A--B--C--cycle, rgb(0.9,1,1));
draw(B--E, lightblue);
draw(C--F, lightblue);
draw(U--B--A--V, blue);
draw(B--C, blue);
draw(F--E, red);
draw(unitcircle, lightblue+dotted);
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$A$", A, dir(A));
/* TSQ Source:
!size(6cm);
B = dir 213
C = dir 47
E = dir 115
F = dir 280
V := extension F E B C
W := incenter F V C
J := 3.6*W-2.6*V
T := foot J B C
A = extension B F E C
CP J T 0.1 deepgreen / deepgreen
U := OP CP J T CP midpoint B--J J
V := IP CP J T CP midpoint E--J J
!fill(A--B--C--cycle, rgb(0.9,1,1));
B--E lightblue
C--F lightblue
U--B--A--V blue
B--C blue
F--E red
unitcircle lightblue dotted
*/
\end{asy}
\end{center}
Consequently we get $EF = BC$.
But this implies $BFCE$ is a rectangle, contradiction.
\paragraph{Second length solution by tangent lengths.}
By $t(\bullet)$ we mean the length of the tangent from $P$
to the $A$-excircle.
It is a classical fact for example that $t(A) = s$.
The main idea is to use the fact that
\[ a \cos A = EF = t(E) + t(F). \]
Here $EF = a \cos A$ follows from the extended law of sines
applied to the circle with diameter $\ol{BC}$,
since there we have $EF = BC \sin \angle ECF = a \sin \angle ACF = a \cos A$.
We may now compute
\begin{align*}
t(E) &= t(A) - AE = s - c \cos A \\
t(F) &= t(A) - AF = s - b \cos A.
\end{align*}
Therefore,
\begin{align*}
a \cos A = 2s - (b+c) \cos A \implies (a+b+c) \cos A &= 2s \\
\implies \cos A &= 1.
\end{align*}
This is an obvious contradiction.
\begin{remark*}
On the other hand, there really is an \emph{equality case}
with $A$ being some point at infinity (meaning $\cos A = 1$).
So, this problem is ``sharper'' than one might expect;
the answer is not ``obviously no''.
\end{remark*}
\paragraph{Third solution by Pitot and trigonometry.}
In fact, the $t(\bullet)$ notation from the previous
solution gives us a classical theorem
once we note the $A$-excircle is tangent to all four lines
$EF$, $BC$, $BF$ and $CE$:
\begin{claim*}
[Pitot theorem]
We have $BF + EF = BC + CE$.
\end{claim*}
\begin{proof}
Here is a proof for completeness.
By $t(B)$ we mean the length of the tangent from $B$ to
the $A$-excircle, and define $t(C)$, $t(E)$, $t(F)$ similarly.
Then
\begin{align*}
BF &= t(B) - t(F) & EF &= t(E) + t(F) \\
BC &= t(B) + t(C) & CE &= t(E) - t(C)
\end{align*}
and summing gives the result.
\end{proof}
\begin{center}
\begin{asy}
pair B = dir(180);
pair C = -B;
pair E = dir(40);
pair F = -E;
pair X = B+2*dir(F-B);
pair Y = C+2*dir(C-E);
pair J = extension(C, X, B, Y);
filldraw(CP(J, foot(J, B, C)), opacity(0.1)+deepgreen, deepgreen);
pair K1 = B+dir(E-C)*1.7;
pair K2 = C+dir(E-C)*1.7;
pair U = foot(J, B, F);
pair V = foot(J, C, E);
label("$A$", midpoint(K1--K2), dir(C-E));
filldraw(K1--B--C--K2--cycle, opacity(0.1)+lightcyan, invisible);
draw(K1--B--C--K2, blue);
draw(F--E, red);
draw(B--U, blue);
draw(C--V, blue);
draw(B--E, blue);
draw(C--F, blue);
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$J$", J, dir(-90));
/* TSQ Source:
B = dir 180
C = -B
E = dir 40
F = -E
X := B+2*dir(F-B)
Y := C+2*dir(C-E)
J = extension C X B Y R-90
CP J foot J B C 0.1 deepgreen / deepgreen
K1 := B+dir(E-C)*1.7
K2 := C+dir(E-C)*1.7
U := foot J B F
V := foot J C E
!label("$A$", midpoint(K1--K2), dir(C-E));
K1--B--C--K2--cycle 0.1 lightcyan / invisible
K1--B--C--K2 blue
F--E red
B--U blue
C--V blue
B--E blue
C--F blue
*/
\end{asy}
\end{center}
We now calculate all the lengths using trigonometry:
\begin{align*}
BC &= a \\
BF &= a \cos(180\dg-B) = a \cos (A+C) \\
CE &= a \cos C \\
EF &= BC \sin \angle ECF = a \sin \angle ACF = a \cos A.
\end{align*}
Thus, we apparently have
\[ \cos(A+C) + \cos A = 1 + \cos C \]
but this is impossible since $\cos(A+C) < \cos C$
(since $A+C = 180-B < 90\dg$) and $\cos A < 1$.
\paragraph{Fourth solution by Pitot and Ptolemy (Evan Chen).}
We give a trig-free way to finish from Pitot's theorem
\[ BF+EF = BC+CE. \]
Assume that $x = BF$, $y = CE$, and $BC = 1$;
then the above relation becomes
\[ 1 + y - x = BC + CE - BF = EF = EF \cdot 1 = xy + \sqrt{(1-x^2)(1-y^2)} \]
with the last step by Ptolemy's theorem.
This rearranges to give
\[ (1+y)(1-x) = \sqrt{(1-x^2)(1-y^2)}
\implies \frac{1+y}{1-y} = \frac{1+x}{1-x}
\implies x = y \]
but that means $BECF$ is a rectangle:
contradicting the fact that lines $BE$ and $CF$ meet at a point $A$.
\paragraph{Fifth solution, by angle chasing only!}
Let $J$ denote the $A$-excenter.
Then $J$ should be the intersection
of the internal bisectors of $\angle FEC$ and $\angle FBC$,
so it is the midpoint of arc $\widehat{FC}$
on the circle with diameter $\ol{BC}$.
\begin{center}
\begin{asy}
pair B = dir(180);
pair C = -B;
pair E = dir(40);
pair F = -E;
pair X = B+2*dir(F-B);
pair Y = C+2*dir(C-E);
pair J = extension(C, X, B, Y);
filldraw(CP(J, foot(J, B, C)), opacity(0.1)+deepgreen, deepgreen);
pair K1 = B+dir(E-C)*1.7;
pair K2 = C+dir(E-C)*1.7;
pair U = foot(J, B, F);
pair V = foot(J, C, E);
label("$A$", midpoint(K1--K2), dir(C-E));
filldraw(K1--B--C--K2--cycle, opacity(0.1)+lightcyan, invisible);
draw(K1--B--C--K2, blue);
draw(F--E, red);
draw(B--U, blue);
draw(C--V, blue);
draw(B--E, blue);
draw(C--F, blue);
filldraw(circumcircle(B, C, J), opacity(0.1)+yellow, orange+dashed);
draw(B--J--E, orange);
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$J$", J, dir(-90));
/* TSQ Source:
B = dir 180
C = -B
E = dir 40
F = -E
X := B+2*dir(F-B)
Y := C+2*dir(C-E)
J = extension C X B Y R-90
CP J foot J B C 0.1 deepgreen / deepgreen
K1 := B+dir(E-C)*1.7
K2 := C+dir(E-C)*1.7
U := foot J B F
V := foot J C E
!label("$A$", midpoint(K1--K2), dir(C-E));
K1--B--C--K2--cycle 0.1 lightcyan / invisible
K1--B--C--K2 blue
F--E red
B--U blue
C--V blue
B--E blue
C--F blue
circumcircle B C J 0.1 yellow / orange dashed
B--J--E orange
*/
\end{asy}
\end{center}
But now we get $\angle BJC = 90\dg$ from $J$ lying on this circle.
Yet $\angle BJC = 90\dg - \half \angle A$ in general,
so $\angle A = 0\dg$ which is impossible.
\paragraph{Sixth solution (Zuming Feng).}
This is similar to the preceding solution,
but phrased using contradiction and inequalities.
We let $X$ and $Y$ denote the tangency points of the $A$-excircle
on lines $AB$ and $AC$.
Moreover, let $J$ denote the $A$-excenter.
\begin{center}
\begin{asy}
size(8cm);
pair B = dir(213);
pair C = dir(47);
pair E = dir(115);
pair F = dir(280);
pair V = extension(F, E, B, C);
pair W = incenter(F, V, C);
pair J = 3.6*W-2.6*V;
pair T = foot(J, B, C);
pair A = extension(B, F, E, C);
filldraw(CP(J, T), opacity(0.1)+deepgreen, deepgreen);
pair X = OP(CP(J, T), CP(midpoint(B--J), J));
pair Y = IP(CP(J, T), CP(midpoint(E--J), J));
fill(A--B--C--cycle, rgb(0.9,1,1));
draw(B--E, lightblue);
draw(C--F, lightblue);
draw(X--B--A--Y, blue);
draw(B--C, blue);
draw(F--E, red);
draw(unitcircle, lightblue+dotted);
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$J$", J, dir(J));
dot("$A$", A, dir(A));
dot("$X$", X, dir(X));
dot("$Y$", Y, dir(Y));
/* TSQ Source:
!size(6cm);
B = dir 213
C = dir 47
E = dir 115
F = dir 280
V := extension F E B C
W := incenter F V C
J = 3.6*W-2.6*V
T := foot J B C
A = extension B F E C
CP J T 0.1 deepgreen / deepgreen
X = OP CP J T CP midpoint B--J J
Y = IP CP J T CP midpoint E--J J
!fill(A--B--C--cycle, rgb(0.9,1,1));
B--E lightblue
C--F lightblue
X--B--A--Y blue
B--C blue
F--E red
unitcircle lightblue dotted
*/
\end{asy}
\end{center}
Note that $AB > AE$ and $AX = AY$, therefore $BX < EY$.
By considering the right triangles $XBJ$ and $YEJ$
(which both have $JX = JY$),
we conclude $\tan \angle XBJ > \tan \angle YEJ$, thus
\[ \angle XBJ > \angle YEJ. \]
However, if line $EF$ was actually tangent to the $A$-excircle,
we would have
\[ 2\angle XBJ = \angle XBC = \angle FBC = \angle FEC
= \angle FEY = 2 \angle JEY \]
which is a contradiction.
\paragraph{Seventh solution, by complex numbers, for comedic effect (Evan Chen).}
Let us denote the tangency points of the $A$-excircle
with sides $BC$, $CA$, $AB$ as $x$, $y$, $z$.
Assume moreover that line $EF$ is tangent to the $A$-excircle
at a point $P$.
Also, for brevity let $s = xy+yz+zx$.
Then, we have
\begin{align*}
E = \frac{2py}{p+y} &= \half(b+y+y-y^2 \ol b)
= \frac{zx}{z+x} + y - \frac{y^2}{z+x} \\
\implies \frac{2}{\frac1p+\frac1y} &= \frac{xy+xz+zx-y^2}{z+x}
\implies \frac{\frac1p+\frac1y}{2} = \frac{x+z}{s-y^2}. \\
\intertext{Similarly by considering the point $F$,}
\frac{\frac1p+\frac1z}{2} &= \frac{x+y}{s-z^2}. \\
\intertext{Thus we can eliminate $P$ and obtain}
\implies \frac{\frac1y-\frac1z}{2} &= \frac{x+z}{s-y^2} - \frac{x+y}{s-z^2}
= \frac{-s(y-z) + x(y^2-z^2) + (y^3-z^3)}{(s-y^2)(s-z^2)} \\
\iff \frac{1}{2yz} &= \frac{s - x(y+z) - (y^2+yz+z^2)}{(s-y^2)(s-z^2)}
= \frac{-(y^2+z^2)}{(s-y^2)(s-z^2)} \\
\iff 0 &= (s-y^2)(s-z^2) + 2yz(y^2+z^2) \\
&= \left[ x(y+z) + y(z-y) \right]
\left[ x(y+z) + z(y-z) \right] + 2yz(y^2+z^2) \\
&= x^2(y+z)^2 - (y-z)^2 \cdot x(y+z) + yz(2y^2+2z^2 - (y-z)^2) \\
&= x^2(y+z)^2 - (y-z)^2 \cdot x(y+z) + yz(y+z)^2 \\
&= xyz(y+z) \left[ \frac xy + \frac xz
- \frac yz - \frac zy + 2
+ \frac yx + \frac zx \right].
\end{align*}
However, $\triangle XYZ$ is obtuse with $\angle X > 90\dg$,
we have $y+z \neq 0$.
Note that
\begin{align*}
\tfrac xy + \tfrac yx &= 2\opname{Re} \tfrac xy = 2 \cos(2 \angle XZY) \\
\tfrac xz + \tfrac zx &= 2\opname{Re} \tfrac xz = 2 \cos(2 \angle XYZ) \\
\tfrac yz + \tfrac zy &= 2\opname{Re} \tfrac yz < 2
\end{align*}
and since $\cos(2\angle XZY) + \cos(2\angle XYZ) > 0$
(say by sum-to-product), we are done.
|
JMO-2019-notes_5
|
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}
|
JMO-2020-notes_1
|
Let $n \ge 2$ be an integer.
Carl has $n$ books arranged on a bookshelf.
Each book has a height and a width.
No two books have the same height,
and no two books have the same width.
Initially, the books are arranged in
increasing order of height from left to right.
In a \emph{move}, Carl picks any two adjacent books
where the left book is wider and shorter than the right book,
and swaps their locations.
Carl does this repeatedly until no further moves are possible.
Prove that regardless of how Carl makes his moves,
he must stop after a finite number of moves, and when he does stop,
the books are sorted in increasing order of width from left to right.
|
We say that a pair of books $(A,B)$ is \emph{height-inverted}
if $A$ is to the left of $B$ and taller than $A$.
Similarly define \emph{width-inverted} pairs.
Note that every operation decreases the number of width-inverted pairs.
This proves the procedure terminates,
since the number of width-inverted pairs starts at $\binom n2$
and cannot increase indefinitely.
Now consider a situation where no more moves are possible.
Assume for contradiction two consecutive books $(A,B)$ are still width-inverted.
Since the operation isn't possible anymore, they are also height-inverted.
In particular, the operation could never have swapped $A$ and $B$.
But this contradicts the assumption there were no height-inverted pairs initially.
|
JMO-2020-notes_2
|
Let $\omega$ be the incircle of a fixed equilateral triangle $ABC$.
Let $\ell$ be a variable line that is tangent to $\omega$ and
meets the interior of segments $BC$ and $CA$ at points $P$ and $Q$, respectively.
A point $R$ is chosen such that $PR=PA$ and $QR=QB$.
Find all possible locations of the point $R$, over all choices of $\ell$.
|
Let $r$ be the inradius.
Let $T$ be the tangency point of $\ol{PQ}$
on arc $\widehat{DE}$ of the incircle, which we consider varying.
We define $R_1$ and $R_2$ to be the two intersections
of the circle centered at $P$ with radius $PA$,
and the circle centered at $Q$ with radius $QB$.
We choose $R_1$ to lie on the opposite side of $C$ as line $PQ$.
\begin{center}
\begin{asy}
size(11cm);
pair A = dir(150);
pair B = dir(30);
pair C = dir(270);
pair D = midpoint(B--C);
pair E = midpoint(A--C);
pair I = incenter(A, B, C);
draw(incircle(A, B, C), blue);
pair R_1 = dir(70);
pair P = extension(midpoint(A--R_1), I, B, C);
pair Q = extension(midpoint(B--R_1), I, A, C);
pair T = extension(R_1, I, P, Q);
filldraw(A--B--C--cycle, opacity(0.1)+yellow, blue);
draw(P--Q, blue);
pair Ap = 4*D;
pair Bp = 4*E;
pair R_2 = 2*T-R_1;
draw(R_2--P--R_1, gray);
draw(P--A, gray);
draw(R_2--R_1, heavycyan);
draw(A--Ap, deepgreen);
draw(B--Bp, deepgreen);
draw(arc(I,B,A), red+1);
draw(arc(I,Bp,Ap), red+1);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(10));
dot("$E$", E, dir(170));
dot("$I$", I, dir(100));
dot("$R_1$", R_1, dir(R_1));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
dot("$T$", T, dir(290));
dot("$A'$", Ap, dir(Ap));
dot("$B'$", Bp, dir(Bp));
dot("$R_2$", R_2, dir(R_2));
/* TSQ Source:
!size(11cm);
A = dir 150
B = dir 30
C = dir 270
D = midpoint B--C R10
E = midpoint A--C R170
I = incenter A B C R100
incircle A B C blue
R_1 = dir 70
P = extension midpoint A--R_1 I B C
Q = extension midpoint B--R_1 I A C
T = extension R_1 I P Q R290
A--B--C--cycle 0.1 yellow / blue
P--Q blue
A' = 4*D
B' = 4*E
R_2 = 2*T-R_1
R_2--P--R_1 gray
P--A gray
R_2--R_1 heavycyan
A--Ap deepgreen
B--Bp deepgreen
!draw(arc(I,B,A), red+1);
!draw(arc(I,Bp,Ap), red+1);
*/
\end{asy}
\end{center}
\begin{claim*}
The point $R_1$ is the unique point on ray $TI$ with $R_1I = 2r$.
\end{claim*}
\begin{proof}
Define $S$ to be the point on ray $TI$ with $SI = 2r$.
Note that there is a homothety at $I$ which maps $\triangle DTE$
to $\triangle ASB$, for some point $S$.
Note that since $TASD$ is an isosceles trapezoid,
it follows $PA = PS$.
Similarly, $QB = QS$.
So it follows that $S = R_1$.
\end{proof}
Since $T$ can be any point on the open arc $\widehat{DE}$,
it follows that the locus of $R_1$
is exactly the open $120\dg$ arc of $\widehat{AB}$
of the circle centered at $I$ with radius $2r$
(i.e.\ the circumcircle of $ABC$).
It remains to characterize $R_2$.
Since $TI = r$, $IR_1 = 2r$, it follows $TR_2 = 3r$ and $IR_2 = 4r$.
Define $A'$ on ray $DI$ such that $A'I = 4r$,
and $B'$ on ray $IE$ such that $B'I = 4r$.
Then it follows, again by homothety,
that the locus of $R_2$ is the $120\dg$ arc $\widehat{A'B'}$
of the circle centered at $I$ with radius $4r$.
In conclusion, the locus of $R$ is the two open $120\dg$ arcs we identified.
|
JMO-2020-notes_3
|
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*}
|
JMO-2020-notes_4
|
Let $ABCD$ be a convex quadrilateral inscribed in a circle and satisfying
\[ DA < AB = BC < CD. \]
Points $E$ and $F$ are chosen on sides $CD$ and $AB$
such that $\ol{BE} \perp \ol{AC}$ and $\ol{EF} \parallel \ol{BC}$.
Prove that $FB=FD$.
|
We present three approaches.
We note that in the second two approaches,
the result remains valid even if $AB \neq BC$,
as long $E$ is replaced by the point on $\ol{AC}$
satisfying $EA = EC$.
So the result is actually somewhat more general.
\paragraph{First solution by inscribed angle theorem.}
Since $\ol{EF} \parallel \ol{BC}$ we may
set $\theta = \angle FEB = \angle CBE = \angle EBF$.
This already implies $FE = FB$,
so we will in fact prove that $F$ is the circumcenter of $\triangle BED$.
\begin{center}
\begin{asy}
pair A = dir(175);
pair B = dir(270);
pair C = B*B/A;
pair D = dir(135);
pair E = extension(B, origin, C, D);
pair F = circumcenter(B, D, E);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
draw(D--F--B, lightred);
draw(F--E, lightred);
draw(B--D--C, blue);
draw(F--A--C, blue);
draw(E--B--C, blue);
draw(A--D, blue);
draw(CP(F, E), lightred);
markangle(1,7.0,E,B,F,deepgreen);
markangle(1,9.0,C,B,E,deepgreen);
markangle(1,9.0,F,E,B,deepgreen);
markangle(2,9.0,B,A,C,deepcyan);
markangle(2,9.0,B,D,C,deepcyan);
markangle(2,9.0,A,C,B,deepcyan);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$E$", E, dir(90));
dot("$F$", F, dir(250));
/* TSQ Source:
A = dir 175
B = dir 270
C = B*B/A
D = dir 135
E = extension B origin C D R90
F = circumcenter B D E R250
unitcircle 0.1 lightcyan / blue
D--F--B lightred
F--E lightred
B--D--C blue
F--A--C blue
E--B--C blue
A--D blue
CP F E lightred
!markangle(1,7.0,E,B,F,deepgreen);
!markangle(1,9.0,C,B,E,deepgreen);
!markangle(1,9.0,F,E,B,deepgreen);
!markangle(2,9.0,B,A,C,deepcyan);
!markangle(2,9.0,B,D,C,deepcyan);
!markangle(2,9.0,A,C,B,deepcyan);
*/
\end{asy}
\end{center}
Note that $\angle BDC = \angle BAC = 90\dg - \theta$.
However, $\angle BFE = 180\dg - 2 \theta$.
So by the inscribed angle theorem, $D$ lies on the circle
centered at $F$ with radius $FE = FB$, as desired.
\begin{remark*}
Another approach to the given problem is to show that $B$
is the $D$-excenter of $\triangle DAE$,
and $F$ is the arc midpoint of $\widehat{DAE}$
of the circumcircle of $\triangle DAE$.
In my opinion, this approach is much clumsier.
\end{remark*}
\paragraph{Second general solution by angle chasing.}
By Reim's theorem, $AFED$ is cyclic.
\begin{center}
\begin{asy}
pair A = dir(175);
pair B = dir(255);
pair C = dir(5);
pair D = dir(135);
pair E = extension(B, origin, C, D);
pair F = extension(A, B, E, E+B-C);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
draw(D--F, lightblue);
draw(A--E, lightblue);
draw(F--B--D--C, blue);
draw(F--A--C, blue);
draw(E--B--C, blue);
draw(A--D, blue);
draw(E--F, lightred);
draw(circumcircle(E, A, D), lightred);
markangle(2,13.0,C,A,E,deepgreen);
markangle(2,13.0,E,C,A,deepgreen);
markangle(2,13.0,D,B,F,deepgreen);
markangle(2,13.0,F,D,B,deepgreen);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$E$", E, dir(70));
dot("$F$", F, dir(F));
/* TSQ Source:
A = dir 175
B = dir 255
C = dir 5
D = dir 135
E = extension B origin C D R70
F = extension A B E E+B-C
unitcircle 0.1 lightcyan / blue
D--F lightblue
A--E lightblue
F--B--D--C blue
F--A--C blue
E--B--C blue
A--D blue
E--F lightred
circumcircle E A D lightred
!markangle(2,13.0,C,A,E,deepgreen);
!markangle(2,13.0,E,C,A,deepgreen);
!markangle(2,13.0,D,B,F,deepgreen);
!markangle(2,13.0,F,D,B,deepgreen);
*/
\end{asy}
\end{center}
Hence
\begin{align*}
\dang FDB &= \dang FDC - \dang BDC = \dang FAE - \dang FAC \\
&= \dang CAE = \dang ECA = \dang DCA = \dang DBA = \dang DBF
\end{align*}
as desired.
\paragraph{Third general solution by Pascal.}
Extend rays $AE$ and $DF$ to meet the circumcircle again at $G$ and $H$.
By Pascal's theorem on $HDCBAG$,
it follows that $E$, $F$, and $GH \cap BC$ are collinear,
which means that $\ol{EF} \parallel \ol{GH} \parallel \ol{BC}$.
\begin{center}
\begin{asy}
pair A = dir(175);
pair B = dir(255);
pair C = dir(5);
pair D = dir(135);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);
pair H = D*B/A;
pair G = A*C/D;
pair E = extension(A, G, D, C);
pair F = extension(A, B, D, H);
draw(A--B--C--D--cycle, blue);
draw(A--G, deepgreen);
draw(D--H, deepgreen);
draw(E--F, red);
draw(G--H, red);
draw(arc(origin,C,G), orange+1.5);
draw(arc(origin,D,A), orange+1.5);
draw(arc(origin,H,B), orange+1.5);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$H$", H, dir(H));
dot("$G$", G, dir(G));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
/* TSQ Source:
A = dir 175
B = dir 255
C = dir 5
D = dir 135
unitcircle 0.1 lightcyan / blue
H = D*B/A
G = A*C/D
E = extension A G D C
F = extension A B D H
A--B--C--D--cycle blue
A--G deepgreen
D--H deepgreen
E--F red
G--H red
!draw(arc(origin,C,G), orange+1.5);
!draw(arc(origin,D,A), orange+1.5);
!draw(arc(origin,H,B), orange+1.5);
*/
\end{asy}
\end{center}
Since $EA=EC$, it follows $DAGC$ in isosceles trapezoid.
But also $GHBC$ is an isosceles trapezoid.
Thus $\mathrm{m}\widehat{DA} = \mathrm{m}\widehat{GC}
= \mathrm{m}\widehat{BH}$,
so $DAHB$ is an isosceles trapezoid.
Thus $FD = FB$.
\begin{remark*}
Addicts of projective geometry can use Pascal
on $DBCAHG$ to finish rather than noting the equal arcs.
\end{remark*}
|
JMO-2020-notes_5
|
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 $P_n$.
|
JMO-2020-notes_6
|
Let $n \ge 2$ be an integer.
Let $P(x_1, x_2, \dots, x_n)$ be a nonconstant
$n$-variable polynomial with real coefficients.
Assuming that $P$ vanishes whenever two of its arguments are equal,
prove that $P$ has at least $n!$ terms.
\end{enumerate}
|
We present two solutions.
\paragraph{First solution using induction (by Ankan).}
Begin with the following observation:
\begin{claim*}
Let $1 \le i < j \le n$.
There is no term of $P$ which omits both $x_i$ and $x_j$.
\end{claim*}
\begin{proof}
Note that $P$ ought to become identically zero
if we set $x_i = x_j = 0$,
since it is zero for any choice of the remaining $n-2$ variables,
and the base field $\RR$ is infinite.
\end{proof}
\begin{remark*}
[Technical warning for experts]
The fact we used is not true if $\RR$ is replaced by a field
with finitely many elements, such as $\FF_p$,
even with one variable.
For example the one-variable polynomial $X^p-X$ vanishes
on every element of $\FF_p$,
by Fermat's little theorem.
\end{remark*}
We proceed by induction on $n \ge 2$ with the base case $n=2$ being clear.
Assume WLOG $P$ is not divisible by any of $x_1$, \dots, $x_n$,
since otherwise we may simply divide out this factor.
Now for the inductive step, note that
\begin{itemize}
\ii The polynomial $P(0, x_2, x_3, \dots, x_n)$
obviously satisfies the inductive hypothesis
and is not identically zero since $x_1 \nmid P$,
so it has at least $(n-1)!$ terms.
\ii Similarly, $P(x_1, 0, x_3, \dots, x_n)$
also has at least $(n-1)!$ terms.
\ii Similarly, $P(x_1, x_2, 0, \dots, x_n)$
also has at least $(n-1)!$ terms.
\ii \dots and so on.
\end{itemize}
By the claim, all the terms obtained in this way
came from different terms of the original polynomial $P$.
Therefore, $P$ itself has at least $n \cdot (n-1)! = n!$ terms.
\begin{remark*}
Equality is achieved by the
Vandermonde polynomial
$P = \prod_{1 \le i < j \le n} (x_i-x_j)$.
\end{remark*}
\paragraph{Second solution using Vandermonde polynomial (by Yang Liu).}
Since $x_i - x_j$ divides $P$ for any $i \neq j$, it follows that
$P$ should be divisible by the Vandermonde polynomial
\[ V = \prod_{i<j} (x_j-x_i)
= \sum_{\sigma} \opname{sgn}(\sigma) x_1^{\sigma(0)}
x_2^{\sigma(1)} \dots x_n^{\sigma(n-1)} \]
where the sum runs over all permutations $\sigma$ on $\{0, \dots, n-1\}$.
Consequently, we may write
\[ P = \sum_{\sigma} \opname{sgn}(\sigma)
x_1^{\sigma(0)} x_2^{\sigma(1)} \dots x_n^{\sigma(n-1)} Q \]
The main idea is that each of the $n!$ terms of the above
sum has a monomial not appearing in any of the other terms.
As an example, consider $x_1^{n-1} x_2^{n-2} \dots x_{n-1}^1 x_n^0$.
Among all monomial in $Q$, consider the monomial
$x_1^{e_1} x_2^{e_2} \dots x_n^{e_n}$
with the largest $e_1$, then largest $e_2$, \dots.
(In other words, take the lexicographically largest
$(e_1, \dots, e_n)$.)
This term
\[ x_1^{e_1 + (n-1)} x_2^{e_2 + (n-2)} \dots x_n^{e_n} \]
can't appear anywhere else because it is strictly lexicographically larger
than any other term appearing in any other expansion.
Repeating this argument with every $\sigma$ gives the conclusion.
|
JMO-2021-notes_1
|
Find all functions $f \colon \NN \to \NN$
which satisfy $f(a^2+b^2)=f(a)f(b)$ and
$f(a^2)=f(a)^2$ for all positive integers $a$ and $b$.
|
The answer is $f \equiv 1$ only, which works.
We prove it's the only one.
The bulk of the problem is:
\begin{claim*}
If $f(a)=f(b)=1$ and $a>b$, then $f(a^2-b^2)=f(2ab)=1$.
\end{claim*}
\begin{proof}
Write
\begin{align*}
1 = f(a)f(b) &= f(a^2+b^2) = \sqrt{f\left( (a^2+b^2)^2 \right)} \\
&= \sqrt{f\left( (a^2-b^2)^2 + (2ab)^2 \right)} \\
&= \sqrt{f(a^2-b^2) f(2ab)}. \qedhere
\end{align*}
\end{proof}
By setting $a=b=1$ in the given statement
we get $f(1) = f(2) = 1$.
Now a simple induction on $n$ shows $f(n) = 1$:
\begin{itemize}
\ii If $n = 2k$ take $(u,v) = (k,1)$ hence $2uv = n$.
\ii If $n = 2k+1$ take $(u,v) = (k+1,k)$ hence $u^2-v^2=n$.
\end{itemize}
|
JMO-2021-notes_2
|
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*}
|
JMO-2021-notes_3
|
An equilateral triangle $\Delta$ of side length $L > 0$ is given.
Suppose that $n$ equilateral triangles with side length $1$
and with non-overlapping interiors are drawn inside $\Delta$,
such that each unit equilateral triangle has sides parallel to $\Delta$,
but with opposite orientation.
Prove that \[ n \le \frac{2}{3} L^2. \]
|
We present the approach of Andrew Gu.
For each triangle, we draw a green regular hexagon
of side length $1/2$ as shown below.
\begin{center}
\begin{asy}
size(2cm);
filldraw(dir(0)--dir(180)--(dir(240)+dir(300))--cycle,
opacity(0.2)+lightred, black);
filldraw(dir(0)--dir(60)--dir(120)--dir(180)--dir(240)--dir(300)--cycle,
opacity(0.1)+green, deepgreen+1.5);
\end{asy}
\end{center}
\begin{claim*}
All the hexagons are disjoint and lie inside $\Delta$.
\end{claim*}
\begin{proof}
Annoying casework.
\end{proof}
Since each hexagon has area $\frac{3\sqrt{3}}{8}$ and
lies inside $\Delta$, we conclude
\[ \frac{3\sqrt3}{8} \cdot n \le \frac{\sqrt3}{4} L^2
\implies n \le \frac 23 L^2. \]
\begin{remark*}
The constant $\frac23$ is sharp and cannot be improved.
The following tessellation shows how to achieve the $\frac23$ density.
In the figure on the left,
one of the green hexagons is drawn in for illustration.
The version on the right has all the hexagons.
\begin{center}
\begin{asy}
size(11cm);
picture pic;
filldraw(pic, dir(0)--dir(180)--(dir(240)+dir(300))--cycle,
opacity(0.2)+lightred, black);
pair v = dir(180)+dir(240);
pair w = dir(0)+dir(-60);
for (int i=0; i<=7; ++i) {
for (int j=0; j<=7; ++j) {
if (abs(i-j) <= 4) {
add(shift(i*v+j*w)*pic);
}
}
}
filldraw(pic, (dir(0)--dir(60)--dir(120)--dir(180)--dir(240)--dir(300)--cycle),
opacity(0.1)+lightgreen, deepgreen+1.5);
for (int i=0; i<=7; ++i) {
for (int j=0; j<=7; ++j) {
if (abs(i-j) <= 4) {
add(shift(15,0)*shift(i*v+j*w)*pic);
}
}
}
filldraw(shift(0,-4*3**0.5)*(dir(0)--dir(60)--dir(120)--dir(180)--dir(240)--dir(300)--cycle),
opacity(0.1)+lightgreen, deepgreen+1.5);
\end{asy}
\end{center}
\end{remark*}
|
JMO-2021-notes_4
|
Carina has three pins, labeled $A$, $B$, and $C$, respectively,
located at the origin of the coordinate plane.
In a \emph{move}, Carina may move a pin to
an adjacent lattice point at distance $1$ away.
What is the least number of moves that Carina can make
in order for triangle $ABC$ to have area $2021$?
|
The answer is $128$.
Define the \textbf{bounding box} of triangle $ABC$ to be the smallest axis-parallel rectangle which
contains all three of the vertices $A$, $B$, $C$.
\begin{center}
\begin{asy}
pair A = (0,0);
pair X = (7,0);
pair Z = (0,3);
pair Y = X+Z-A;
pair B = (7,1);
pair C = (2,3);
filldraw(A--B--C--cycle, opacity(0.1)+lightcyan, blue);
filldraw(A--X--Y--Z--cycle, opacity(0.1)+lightred, red);
dot("$A$", A, dir(225), blue);
dot("$B$", B, dir(0), blue);
dot("$C$", C, dir(90), blue);
dot("$X$", X, dir(315), red);
dot("$Y$", Y, dir(45), red);
dot("$Z$", Z, dir(135), red);
\end{asy}
\end{center}
\begin{lemma*}
The area of a triangle $ABC$ is at most half the area of the bounding box.
\end{lemma*}
\begin{proof}
This can be proven by explicit calculation in coordinates.
Nonetheless, we outline a geometric approach.
By considering the smallest/largest $x$ coordinate and the smallest/largest $y$ coordinate,
one can check that some vertex of the triangle
must coincide with a corner of the bounding box
(there are four ``extreme'' coordinates across the $3\cdot2=6$ coordinates of our three points).
So, suppose the bounding box is $AXYZ$.
Imagine fixing $C$ and varying $B$ along the perimeter of the entire rectangle.
The area is a linear function of $B$, so the maximal area should be achieved when $B$
coincides with one of the vertices $\{A,X,Y,Z\}$.
But obviously the area of $\triangle ABC$ is
\begin{itemize}
\ii exactly $0$ if $B=A$,
\ii at most half the bounding box if $B\in\{X,Z\}$ by one-half-base-height,
\ii at most half the bounding box if $B=Y$, since $\triangle ABC$ is contained inside
either $\triangle AYZ$ or $\triangle AXZ$. \qedhere
\end{itemize}
\end{proof}
We now proceed to the main part of the proof.
\begin{claim*}
If $n$ moves are made, the bounding box has area at most $(n/2)^2$.
(In other words, a bounding box of area $A$ requires at least $\left\lceil 2\sqrt{A} \right\rceil$ moves.)
\end{claim*}
\begin{proof}
The sum of the width and height of the bounding box increases by at most $1$ each move,
hence the width and height have sum at most $n$.
So, by AM-GM, their product is at most $(n/2)^2$.
\end{proof}
This immediately implies $n \ge 128$,
since the bounding box needs to have area at least $4042 > 63.5^2$.
On the other hand, if we start all the pins at
the point $(3,18)$ then we can reach the following three points in $128$ moves:
\begin{align*}
A &= (0,0) \\
B &= (64,18) \\
C &= (3,64)
\end{align*}
and indeed triangle $ABC$ has area exactly $2021$.
\begin{remark*}
In fact, it can be shown that to obtain an area of $n/2$,
the bounding-box bound of $\left\lceil 2\sqrt n \right\rceil$ moves is best possible,
i.e.\ there will in fact exist a triangle with area $n/2$.
However, since this was supposed to be a JMO4 problem,
the committee made a choice to choose $n = 4042$
so that contestants only needed to give a single concrete triangle
rather than a general construction for all positive integers $n$.
\end{remark*}
|
JMO-2021-notes_5
|
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.
|
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