problem_id stringlengths 16 19 | problem stringlengths 69 2.04k | solution stringlengths 60 23.9k |
|---|---|---|
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 &... |
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 ... |
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_... | \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\... |
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 $\Omega$... | 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,... |
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 a... | 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 exampl... |
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
... |
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 repea... | 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{ fo... |
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 ... |
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... |
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 ... | 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)... |
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(... |
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},
mea... | 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 len... |
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.... |
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 $L$.
... | \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}
... |
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
\[ P(a_n)=a_{n+1}a_{n+2}\dotsm a_{n+k} \]
... | 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 <... |
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.... |
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... | 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\... |
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 i... | 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... |
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 \alp... |
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 ... |
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 $a_1$, $a_3$, $a_5$, \dot... | 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... |
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 tang... | 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 $\t... |
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.
... | 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 en... |
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... | 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... |
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 positi... | 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 $... |
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 $MN$ ... | 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.
\beg... |
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 ... |
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 s... |
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 $x_n$ s... | 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 wher... |
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 e... | 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{st... |
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 ... |
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 necess... | 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} = 2... |
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 segme... | 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... |
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... |
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
i... | 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 \\
\... |
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... | 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 = extens... |
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 ... |
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 ... |
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 $\ell(P_1)$, $\ell(P_2)$, $\ell(P_3)$
fo... | 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... |
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 $W_{n-1}$.
Pr... | 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 $n$... |
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... |
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... |
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)... |
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... |
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_{\t... |
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... | 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);
pai... |
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... | 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... |
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(1... |
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... |
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 a... | 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... |
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 t... | 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;
... |
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... |
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 ... | 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), di... |
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 t... |
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 Deno... | 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 opp... |
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}
... |
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 $[1, b^{2M}]$ f... |
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 ... | 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 other... |
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 ... | 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{X... |
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... | 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 initia... |
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 $a = 3c$... |
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 $M$ de... | 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}
... |
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) ... |
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( ... |
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 rectan... | 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... |
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 tri... | 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);
... |
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 the... |
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... | 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} = \empty... |
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 l... |
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 $Q$ a... | 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 ... |
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
\... |
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} \\
&\... |
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 ... |
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 & ... |
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)... |
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{... |
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 $R_1$ travelin... | 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 swa... |
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, t... |
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 ... |
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 $R$ differ... | 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... |
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(... |
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 g... |
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 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 678... |
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 b... | 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 t... |
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}
Su... |
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 $E$-symmedia... |
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 ... |
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 ... | 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 ... |
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
w... | 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 ... |
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 ... | 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 oppos... |
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 \time... | \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 th... |
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... |
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 possib... | 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}.
... |
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... |
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( ... |
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... |
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}... | 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... |
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... |
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
\d... |
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