problem_id stringlengths 16 19 | problem stringlengths 69 2.04k | solution stringlengths 60 23.9k |
|---|---|---|
JMO-2021-notes_6 | Let $n \ge 4$ be an integer.
Find all positive real solutions to the following
system of $2n$ equations:
\begin{align*}
a_1 &= \frac{1}{a_{2n}} + \frac{1}{a_{2}}, & a_2 &= a_1 + a_3, \\[1ex]
a_3 &= \frac{1}{a_{2}} + \frac{1}{a_{4}}, & a_4 &= a_3 + a_5, \\[1ex]
a_5 &= \frac{1}{a_{4}} + \frac{1}{a_{6}}, & a_6 &= a_... | The answer is that the only solution is
$(1,2,1,2,\dots,1,2)$ which works.
We will prove $a_{2k}$ is a constant sequence,
at which point the result is obvious.
\paragraph{First approach (Andrew Gu).}
Apparently, with indices modulo $2n$, we should have
\[ a_{2k} = \frac{1}{a_{2k-2}}
+ \frac{2}{a_{2k}} + \frac{1}{a_... |
JMO-2022-notes_1 | For which positive integers $m$ does there exist an infinite sequence in $\ZZ/m\ZZ$
which is both an arithmetic progression and a geometric progression,
but is nonconstant? | Answer: $m$ must \emph{not} be squarefree.
The problem is essentially asking when there exists
a nonconstant arithmetic progression in $\ZZ/m\ZZ$
which is also a geometric progression.
Now,
\begin{itemize}
\ii If $m$ is squarefree, then consider three $(s-d, d, s+d)$ in arithmetic progression.
It's geometric if an... |
JMO-2022-notes_2 | Let $a$ and $b$ be positive integers.
Every cell of an $(a+b+1)\times (a+b+1)$ grid is colored either amber or bronze
such that there are at least $a^2+ab-b$ amber cells
and at least $b^2+ab-a$ bronze cells.
Prove that it is possible to choose $a$ amber cells and $b$ bronze cells
such that no two of the $a+b$ chosen ce... | \begin{claim*}
There exists a transversal $T_a$ with at least $a$ amber cells.
Analogously, there exists a transversal $T_b$ with at least $b$ bronze cells.
\end{claim*}
\begin{proof}
If one picks a random transversal, the expected value
of the number of amber cells is at least
\[ \frac{a^2+ab-b}{a+b+1} = (a-... |
JMO-2022-notes_3 | Let $b\geq 2$ and $w\geq 2$ be fixed integers, and $n=b+w$.
Given are $2b$ identical black rods and $2w$ identical white rods,
each of side length $1$.
We assemble a regular $2n$-gon using these rods
so that parallel sides are the same color.
Then, a convex $2b$-gon $B$ is formed by translating the black rods,
and a c... | We are going to prove that one may swap a black rod with an adjacent white rod
(as well as the rods parallel to them)
without affecting the difference in the areas of $B-W$.
Let $\vec u$ and $\vec v$ denote the originally black and white vectors
that were adjacent on the $2n$-gon and are now going to be swapped.
Let $\... |
JMO-2022-notes_4 | Let $ABCD$ be a rhombus, and let $K$ and $L$ be points
such that $K$ lies inside the rhombus, $L$ lies outside the rhombus,
and $KA = KB = LC = LD$.
Prove that there exist points $X$ and $Y$ on lines $AC$ and $BD$
such that $KXLY$ is also a rhombus. | To start, notice that $\triangle AKB \cong \triangle DLC$ by SSS.
Then by the condition $K$ lies inside the rhombus while $L$ lies outside it,
we find that the two congruent triangles are just translations of each other
(i.e.\ they have the same orientation).
\paragraph{First solution.}
Let $M$ be the midpoint of $\ol... |
JMO-2022-notes_5 | Find all pairs of primes $(p, q)$ for which $p-q$ and $pq-q$ are both perfect squares. | The answer is $(3,2)$ only.
This obviously works so we focus on showing it is the only one.
\paragraph{Approach using difference of squares (from author).}
Set
\begin{align*}
a^2 &= p-q \\
b^2 &= pq-q.
\end{align*}
Note that $0 < a < p$, and $0 < b < p$ (because $q \le p$).
Now subtracting gives
\[ \underbrace{(b-... |
JMO-2022-notes_6 | Let $a_0$, $b_0$, $c_0$ be complex numbers, and define
\begin{align*}
a_{n+1} &= a_n^2 + 2b_nc_n \\
b_{n+1} &= b_n^2 + 2c_na_n \\
c_{n+1} &= c_n^2 + 2a_nb_n
\end{align*}
for all nonnegative integers $n$.
Suppose that $\max{\{|a_n|, |b_n|, |c_n|\}} \leq 2022$ for all $n \ge 0$.
Prove that \[ |a_0|^2 + |b_0|^2 + |c... | For brevity, set $s_n \coloneq |a_n|^2 + |b_n|^2 + |c_n|^2$.
Note that the $s_n$ are real numbers.
\begin{claim*}
[Key miraculous identity]
We have
\[ s_{n+1} - s_n^2 = 2 |a_n \ol{b_n} + b_n \ol{c_n} + c_n \ol{a_n}|^2. \]
\end{claim*}
\begin{proof}
We prove this by mechanical calculation. First,
\begin{align*... |
JMO-2023-notes_1 | Find all triples of positive integers $(x,y,z)$ satisfying
\[ 2(x+y+z+2xyz)^2 = (2xy+2yz+2zx+1)^2 + 2023. \] | Answer: $(3,3,2)$ and permutations.
The solution hinges upon the following claim:
\begin{claim*}
The identity
\[ 2(x+y+z+2xyz)^2 - (2xy+2yz+2zx+1)^2 = (2x^2-1)(2y^2-1)(2z^2-1) \]
is true.
\end{claim*}
\begin{proof}
This can be proved by manually expanding; we show where it ``came from''.
In algebraic numb... |
JMO-2023-notes_2 | In an acute triangle $ABC$, let $M$ be the midpoint of $\ol{BC}$.
Let $P$ be the foot of the perpendicular from $C$ to $AM$.
Suppose that the circumcircle of triangle $ABP$
intersects line $BC$ at two distinct points $B$ and $Q$.
Let $N$ be the midpoint of $\ol{AQ}$.
Prove that $NB = NC$. | We show several different approaches.
In all solutions, let $D$ denote the foot of the altitude from $A$.
\begin{center}
\begin{asy}
size(10cm);
pair A = dir(115);
pair B = dir(210);
pair C = dir(330);
pair D = foot(A, B, C);
pair M = midpoint(B--C);
pair P = foot(C, A, M);
pair Q = 2*M-D;
filldraw(A--B--C--cycle, o... |
JMO-2023-notes_3 | Consider an $n$-by-$n$ board of unit squares for some odd positive integer $n$.
We say that a collection $C$ of identical dominoes is a
maximal grid-aligned configuration on the board if $C$ consists of $(n^2-1)/2$
dominoes where each domino covers exactly two neighboring squares
and the dominoes don't overlap: $C$ the... | The answer is that
\[ k(C) \le \left( \frac{n+1}{2} \right)^2. \]
\begin{remark*}
[Comparison with USAMO version]
In the USAMO version of the problem,
students instead are asked to find all possible values of $k(C)$.
The answer is
$k(C) \in \left\{ 1, 2, \dots, \left( \frac{n-1}{2} \right)^2 \right\}
\cup \... |
JMO-2023-notes_5 | Positive integers $a$ and $N$ are fixed,
and $N$ positive integers are written on a blackboard.
Alice and Bob play the following game.
On Alice's turn, she must replace some integer $n$ on the board with $n+a$,
and on Bob's turn he must replace some even integer $n$ on the board with $n/2$.
Alice goes first and they al... | For $N=1$, there is nothing to prove since each player has only one option each turn.
We address $N \ge 2$ only henceforth.
Let $S$ denote the numbers on the board.
\begin{claim*}
When $N \ge 2$, if $\nu_2(x) < \nu_2(a)$ for all $x \in S$,
the game must terminate no matter what either player does.
\end{claim*}
\be... |
JMO-2023-notes_6 | Isosceles triangle $ABC$, with $AB=AC$, is inscribed in circle $\omega$.
Let $D$ be an arbitrary point inside $BC$ such that $BD \neq DC$.
Ray $AD$ intersects $\omega$ again at $E$ (other than $A$).
Point $F$ (other than $E$) is chosen on $\omega$ such that $\angle DFE = 90\dg$.
Line $FE$ intersects rays $AB$ and $AC$ ... | We present three solutions.
\paragraph{Angle chasing solution.}
Note that $(BDA)$ and $(CDA)$ are congruent,
since $BA=CA$ and $\angle BDA + \angle CDA = 180\dg$.
So these two circles are reflections around line $ED$.
Moreover, $(DEF)$ is obviously also symmetric around line $ED$.
\begin{center}
\begin{asy}
size(10cm)... |
JMO-2024-notes_1 | Let $ABCD$ be a cyclic quadrilateral with $AB=7$ and $CD=8$.
Points $P$ and $Q$ are selected on line segment $AB$ so that $AP=BQ=3$.
Points $R$ and $S$ are selected on line segment $CD$ so that $CR=DS=2$.
Prove that $PQRS$ is a cyclic quadrilateral. | Here are three possible approaches.
\paragraph{The one-liner.}
The four points $P$, $Q$, $R$, $S$ have equal power $-12$ with respect to $(ABCD)$.
So in fact they're on a circle concentric with $(ABCD)$.
\paragraph{The external power solution.}
We distinguish between two cases.
\subparagraph{Case where $AB$ and $CD$... |
JMO-2024-notes_2 | Let $m$ and $n$ be positive integers.
Let $S$ be the set of lattice points $(x,y)$ with $1 \le x \le 2m$ and $1 \le y \le 2n$.
A configuration of $mn$ axis-parallel rectangles is called \emph{happy}
if each point of $S$ is the vertex of exactly one rectangle.
Prove that the number of happy configurations is odd. | There are several possible approaches to the problem;
most of them involve pairing some of the happy configurations in various ways,
leaving only a few configurations which remain fixed.
We present the original proposer's solution and Evan's more complicated one.
\paragraph{Original proposer's solution.}
To this end, ... |
JMO-2024-notes_3 | A sequence $a_1$, $a_2$, \dots\ of positive integers is defined recursively by
$a_1 = 2$ and \[ a_{n+1} = a_n^{n+1} - 1 \qquad \text{ for } n \ge 1. \]
Prove that for every odd prime $p$ and integer $k$,
some term of the sequence is divisible by $p^k$. | We start with the following.
\begin{claim*}
Assume $n$ is a positive integer divisible by $p-1$.
Then either $a_{n-1} \equiv 0 \pmod p$ or $a_n \equiv 0 \pmod p$.
\end{claim*}
\begin{proof}
Suppose that $a_{n-1} \not\equiv 0 \pmod p$.
Then by Fermat's little theorem,
\[ a_n = a_{n-1}^n - 1 \equiv x^{p-1} - 1 ... |
JMO-2024-notes_4 | Let $n \geq 3$ be an integer.
Rowan and Colin play a game on an $n \times n$ grid of squares,
where each square is colored either red or blue.
Rowan is allowed to permute the rows of the grid and
Colin is allowed to permute the columns.
A grid coloring is orderly if:
\begin{itemize}
\ii no matter how Rowan permutes t... | The answer is $2n!+2$.
In fact, we can describe all the orderly colorings as follows:
\begin{itemize}
\ii The all-blue coloring.
\ii The all-red coloring.
\ii Each of the $n!$ colorings where every row/column has exactly one red cell.
\ii Each of the $n!$ colorings where every row/column has exactly one blue ce... |
JMO-2024-notes_5 | Solve over $\RR$ the functional equation
\[ f(x^2-y)+2yf(x) = f(f(x))+f(y). \] | The answer is $f(x) \equiv x^2$, $f(x) \equiv 0$, $f(x) \equiv -x^2$,
which obviously work.
Let $P(x,y)$ be the usual assertion.
\begin{claim*}
We have $f(0) = 0$ and $f$ even.
\end{claim*}
\begin{proof}
Combine $P(1,1/2)$ with $P(1,0)$ to get $f(0) = 0$.
Use $P(0,y)$ to deduce $f$ is even.
\end{proof}
\begin{c... |
JMO-2024-notes_6 | Point $D$ is selected inside acute triangle $ABC$
so that $\angle DAC = \angle ACB$ and $\angle BDC = 90^\circ + \angle BAC$.
Point $E$ is chosen on ray $BD$ so that $AE = EC$.
Let $M$ be the midpoint of $BC$.
Show that line $AB$ is tangent to the circumcircle of triangle $BEM$.
\end{enumerate} | This problem has several approaches and we showcase a collection of them.
\paragraph{The author's original solution.}
Complete isosceles trapezoid $ABQC$ (so $D \in \ol{AQ}$).
Reflect $B$ across $E$ to point $F$.
\begin{center}
\begin{asy}
size(7cm);
pair A = dir(185);
pair C = dir(355);
pair B = dir(115);
pair Q = A*... |
JMO-2025-notes_1 | Prove that if $f \colon \ZZ \to \ZZ$ is any function,
then there are infinitely many integers $c$ such that the function
$g(x) = f(x)+cx$ is not a bijection. | Assume for contradiction that there exists a finite ``bad'' set $S$
such that $f(x)+cx$ is bijective for all $c \notin S$.
The first observation is basically that given just $f(0)$ and $f(1)$,
or any two consecutive $f$-values, we can already find bad values of $c$.
\begin{claim*}
The numbers $f(0)-f(1)$, $f(1)-f(2)... |
JMO-2025-notes_2 | Fix positive integers $k$ and $d$.
Prove that for all sufficiently large odd positive integers $n$,
the digits of the base-$2n$ representation of $n^k$ are all greater than $d$. | The problem actually doesn't have much to do with digits:
the idea is to pick any length $\ell \le k$,
and look at the rightmost $\ell$ digits of $n^k$;
that is, the remainder upon division by $(2n)^\ell$.
We compute it exactly:
\begin{claim*}
Let $n \ge 1$ be an odd integer, and $k \ge \ell \ge 1$ integers.
Then
... |
JMO-2025-notes_3 | Let $m$ and $n$ be positive integers,
and let $\mathcal R$ be a $2m\times{2n}$ grid of unit squares.
A \emph{domino} is a $1\times2$ or $2\times{1}$ rectangle.
An \emph{up-right} path is a path from the lower-left corner of $\mathcal R$
to the upper-right corner of $\mathcal R$ formed by exactly $2m+2n$ edges of the gr... | Apply the usual black/white checkerboard coloring.
We define a \emph{staircase} to be the below above an up-right path
(equivalently, a Young diagram rotated $180\dg$).
The proof is composed of two steps.
One is rewriting the ``domino-tileable'' condition to one that is actually usable;
the other is the counting part.... |
JMO-2025-notes_4 | Let $n$ be a positive integer, and let $a_0 \ge a_1 \ge \dots \ge a_n \ge 0$ be integers.
Prove that
\[ \sum_{i=0}^n i\binom{a_i}{2} \le \half \binom{a_0+a_1+\dots+a_n}{2}. \] | For $n=0$ (which we permit) there is nothing to prove.
Hence to prove by induction on $n$, it would be sufficient to verify
\[ 2n \binom{a_n}{2} \le \binom{a_0 + a_1 + \dots + a_n}{2}
- \binom{a_0 + a_1 + \dots + a_{n-1}}{2}. \]
Rearranging the terms around, that's equivalent to proving
\begin{align*}
\iff 2n(a_n... |
JMO-2025-notes_5 | Let $H$ be the orthocenter of an acute triangle $ABC$,
let $F$ be the foot of the altitude from $C$ to $AB$,
and let $P$ be the reflection of $H$ across $BC$.
Suppose that the circumcircle of triangle $AFP$ intersects line $BC$
at two distinct points $X$ and $Y$.
Prove that $CX = CY$. | Let $Q$ be the antipode of $B$.
\begin{claim*}
$AHQC$ is a parallelogram, and $APCQ$ is an isosceles trapezoid.
\end{claim*}
\begin{proof}
As $\ol{AH} \perp \ol{BC} \perp \ol{CQ}$ and $\ol{CF} \perp \ol{AB} \perp \ol{AQ}$.
\end{proof}
\begin{center}
\begin{asy}
pair A = dir(110);
pair B = dir(210);
pair C = dir(33... |
JMO-2025-notes_6 | Let $S$ be a set of integers with the following properties:
\begin{itemize}
\ii $\{ 1, 2, \dots, 2025 \} \subseteq S$.
\ii If $a, b \in S$ and $\gcd(a, b) = 1$, then $ab \in S$.
\ii If $s + 1$ is composite for some $s \in S$, then all positive divisors of $s + 1$ are in $S$.
\end{itemize}
Prove that $S$ contains all po... | We prove by induction on $N$ that $S$ contains $\{1, \dots, N\}$
with the base cases being $N = 1, \dots, 2025$ already given.
For the inductive step, to show $N+1 \in S$:
\begin{itemize}
\ii If $N+1$ is composite we're already done from the third bullet.
\ii Otherwise, assume $N+1 = p \ge 2025$ is an (odd) prime... |
USAMO-1996-notes_1 | Prove that the average of the numbers $n \sin n^{\circ}$ for
$n = 2,4,6,\dots,180$ is $\cot 1^{\circ}$. | Because
\[ n \sin n\dg + (180-n) \sin(180\dg-n\dg) = 180 \sin n\dg \]
So enough to show that
\[ \sum_{n=0}^{89} \sin (2n)\dg = \cot1\dg \]
Let $\zeta = \cos2\dg + i \sin 2 \dg$ be a primitive root.
Then
\begin{align*}
\sum_{n=0}^{89} \frac{\zeta^n - \zeta^{-n}}{2i}
&= \frac{1}{2i}
\left[ \frac{\zeta^{90}-1}{\zeta... |
USAMO-1996-notes_2 | For any nonempty set $S$ of real numbers,
let $\sigma(S)$ denote the sum of the elements of $S$.
Given a set $A$ of $n$ positive integers,
consider the collection of all distinct sums $\sigma(S)$ as $S$ ranges
over the nonempty subsets of $A$.
Prove that this collection of sums can be
partitioned into $n$ classes so th... | By induction on $n$ with $n=1$ being easy.
For the inductive step, assume
\[ A = \left\{ a_1 > a_2 > \dots > a_n \right\}. \]
Fix any index $k$ with the property that
\[ a_k > \frac{\sigma(A)}{2^k} \]
(which must exist since $\half+\frac14+\dots+\frac{1}{2^k} < 1$).
Then
\begin{itemize}
\ii We make $k$ classes for t... |
USAMO-1996-notes_3 | Let $ABC$ be a triangle. Prove that there is a line $\ell$ (in the plane
of triangle $ABC$) such that the intersection of the interior of
triangle $ABC$ and the interior of its reflection $A'B'C'$ in $\ell$ has
area more than $\frac23$ the area of triangle $ABC$. | All that's needed is:
\begin{claim*}
If $ABC$ is a triangle where $\half < \frac{AB}{AC} < 1$, then
the $\angle A$ bisector works.
\end{claim*}
\begin{proof}
Let the $\angle A$-bisector meet $BC$ at $D$.
The overlapped area is $2[ABD]$ and
\[ \frac{[ABD]}{[ABC]} = \frac{BD}{BC} = \frac{AB}{AB+AC} \]
by angl... |
USAMO-1996-notes_4 | An $n$-term sequence $(x_1, x_2, \dots, x_n)$
in which each term is either $0$ or $1$ is called a binary sequence of length $n$.
Let $a_n$ be the number of binary sequences of length $n$ containing
no three consecutive terms equal to $0$, $1$, $0$ in that order.
Let $b_n$ be the number of binary sequences of length $n$... | Consider the map from
sequences of the latter form to sequences of the first form by
\[ (y_1, \dots, y_{n+1}) \mapsto (y_1 + y_2, y_2 + y_3, \dots, y_n + y_{n+1}). \]
It is $2$-to-$1$. The end. |
USAMO-1996-notes_5 | Let $ABC$ be a triangle, and $M$ an interior point such that
$\angle MAB=10^\circ$, $\angle MBA=20^\circ$, $\angle MAC=40^\circ$ and $\angle MCA=30^\circ$.
Prove that the triangle is isosceles. | Let $\theta = \angle MBC < 80\dg$.
By trig Ceva, we get
\[
\frac{\sin 10\dg}{\sin 40\dg} \cdot \frac{\sin \theta}{\sin 20\dg}
\cdot \frac{\sin 30\dg}{\sin (80\dg-\theta)} = 1.
\]
This simplifies to
\[ \sin \theta = 4 \sin(80\dg- \theta) \sin 40\dg \cos 10\dg.\]
\begin{claim*}
We have $\theta=60\dg$.
\end{claim*}
... |
USAMO-1996-notes_6 | Determine with proof whether there is a subset $X \subseteq \ZZ$
with the following property: for any $n \in \ZZ$,
there is exactly one solution to $a+2b = n$, with $a,b \in X$.
\end{enumerate} | The idea is generating functions, but extra care is required
since exponents will be in $\ZZ$ rather than in $\ZZ_{\ge 0}$.
However, consider formally the limit
\[ f(x) = \prod_{k \ge 0} \left( 1 + x^{(-4)^k} \right). \]
For size reasons, this indeed converges formally to a power series,
in the sense that the coeffici... |
USAMO-1997-notes_1 | Let $p_1, p_2, p_3, \dots$ be the prime numbers
listed in increasing order,
and let $0 < x_0 < 1$ be a real number between 0 and 1.
For each positive integer $k$, define
\[ x_k = \begin{cases} 0 & \mbox{if} \; x_{k-1} = 0, \\[.1in]
{\displaystyle \left\{ \frac{p_k}{x_{k-1}} \right\}}
& \mbox{if} \; x_{k-1} \neq 0,
... | The answer is $x_0$ rational.
If $x_0$ is irrational, then all $x_i$ are irrational by induction.
So the sequence cannot become zero.
If $x_0$ is rational, then all are.
Now one simply observes that the denominators of $x_n$
are strictly decreasing, until we reach $0 = \frac 01$.
This concludes the proof.
\begin{rem... |
USAMO-1997-notes_2 | Let $ABC$ be a triangle.
Take noncollinear points $D$, $E$, $F$ on the perpendicular bisectors
of $BC$, $CA$, $AB$ respectively.
Show that the lines through $A$, $B$, $C$
perpendicular to $EF$, $FD$, $DE$ respectively are concurrent. | The three lines are the radical axii of the three circles
centered at $D$, $E$, $F$, so they concur. |
USAMO-1997-notes_3 | Prove that for any integer $n$,
there exists a unique polynomial $Q$ with coefficients
in $\{0,1,\dots,9\}$ such that $Q(-2) = Q(-5) = n$. | If we let
\[ Q(x) = \sum_{k \ge 0} a_k x^k \]
then $a_k$ is uniquely determined by $n \pmod{2^k}$ and $n \pmod{5^k}$.
Indeed, we can extract the coefficients of $Q$
exactly by the following algorithm:
\begin{itemize}
\ii Define $b_0 = c_0 = n$.
\ii For $i \ge 0$, let $a_i$ be the unique digit
satisfying $a_i \equ... |
USAMO-1997-notes_4 | To clip a convex $n$-gon means to choose a pair of consecutive sides $AB$, $BC$
and to replace them by the three segments $AM$, $MN$, and $NC$,
where $M$ is the midpoint of $AB$ and $N$ is the midpoint of $BC$.
In other words, one cuts off the triangle $MBN$ to obtain a convex $(n+1)$-gon.
A regular hexagon $\mathcal{P... | Call the original hexagon $ABCDEF$.
We show the area common to triangles $ACE$ and $BDF$ is in every
$\mathcal{P}_n$; this solves the problem since the area is $1/3$.
For every side of a clipped polygon,
we define its \emph{foundation} recursively as follows:
\begin{itemize}
\ii $AB$, $BC$, $CD$, $DE$, $EF$, $FA$
... |
USAMO-1997-notes_5 | If $a,b,c > 0$ prove that
\[
\frac{1}{a^3+b^3+abc}
+ \frac{1}{b^3+c^3+abc}
+ \frac{1}{c^3+a^3+abc}
\le \frac{1}{abc}.
\] | From $a^3 + b^3 \ge ab(a+b)$, the left-hand side becomes
\[
\sum_{\text{cyc}} \frac{1}{a^3+b^3+abc}
\le \sum_{\text{cyc}} \frac{1}{ab(a+b+c)}
= \frac{1}{abc} \sum_{\text{cyc}} \frac{c}{a+b+c}
= \frac{1}{abc}.
\] |
USAMO-1997-notes_6 | Suppose the sequence of nonnegative integers
$a_1, a_2, \dots, a_{1997}$ satisfies
\[ a_i + a_j \leq a_{i+j} \leq a_i + a_j + 1 \]
for all $i,j \geq 1$ with $i + j \leq 1997$.
Show that there exists a real number $x$ such that
$a_n = \lfloor nx \rfloor$ for all $1 \leq n \leq 1997$.
\end{enumerate} | We are trying to show there exists an $x \in \RR$
such that
\[ \frac{a_n}{n} \le x < \frac{a_n+1}{n} \qquad \forall n. \]
This means we need to show
\[ \max_i \frac{a_i}{i} < \min_j \frac{a_j+1}{j}. \]
Replace 1997 by $N$.
We will prove this by induction,
but we will need some extra hypotheses on the indices $i,j$
whic... |
USAMO-1998-notes_1 | Suppose that the set $\{1,2,\dots, 1998\}$
has been partitioned into disjoint pairs $\{a_i,b_i\}$ ($1\leq i\leq 999$)
so that for all $i$, $|a_i-b_i|$ equals $1$ or $6$.
Prove that the sum
\[ |a_1-b_1|+|a_2-b_2|+\dotsb +|a_{999}-b_{999}| \]
ends in the digit $9$. | Let $S$ be the sum.
Modulo $2$,
\[ S = \sum |a_i-b_i| \equiv \sum (a_i+b_i)
= 1 + 2 + \dots + 1998 \equiv 1 \pmod 2. \]
Modulo $5$,
\[ S = \sum |a_i-b_i| = 1 \cdot 999 \equiv 4 \pmod 5. \]
So $S \equiv 9 \pmod{10}$. |
USAMO-1998-notes_2 | Let $\mathcal C_1$ and $\mathcal C_2$ be concentric circles, with
$\mathcal C_2$ in the interior of $\mathcal C_1$.
From a point $A$ on $\mathcal C_1$ one draws the tangent $AB$
to $\mathcal C_2$ ($B\in \mathcal C_2$).
Let $C$ be the second point of intersection of ray $AB$ and
$\mathcal C_1$, and let $D$ be the midpoi... | By power of a point we have
\[ AE \cdot AF = AB^2
= \left( \half AB \right) \cdot \left( 2AB \right)
= AD \cdot AC \]
and hence $CDEF$ is cyclic.
Then $M$ is the circumcenter of quadrilateral $CDEF$.
\begin{center}
\begin{asy}
pair A = dir(170);
pair C = dir(40);
pair B = midpoint(A--C);
pair D = midpoint(A--B);
p... |
USAMO-1998-notes_3 | Let $a_0,a_1,\dots ,a_n$ be numbers from the interval $(0,\pi/2)$
such that $\tan (a_0-\frac{\pi}{4})+ \tan (a_1-\frac{\pi}{4})
+ \dotsb +\tan (a_n-\frac{\pi}{4}) \ge n-1$.
Prove that
\[ \tan a_0\tan a_1 \dotsm \tan a_n \ge n^{n+1}. \] | Let $x_i = \tan(a_i - \frac{\pi}{4})$.
Then we have that
\[ \tan a_i = \tan(a_i - 45\dg + 45\dg) = \frac{x_i + 1}{1 - x_i}. \]
If we further substitute $y_i = \frac{1 - x_i}{2} \in (0,1)$,
then we have to prove that the following statement:
\begin{claim*}
If $\sum_0^n y_i \le 1$ and $y_i \ge 0$, we have
\[ \prod_{i... |
USAMO-1998-notes_4 | A computer screen shows a $98 \times 98$ chessboard, colored in the usual way.
One can select with a mouse any rectangle with sides on the lines
of the chessboard and click the mouse button:
as a result, the colors in the selected rectangle switch
(black becomes white, white becomes black).
Find, with proof, the minimu... | The answer is $98$.
One of several possible constructions is to toggle
all columns and rows with even indices.
In the other direction, let $n = 98$ and
suppose that $k$ rectangles are used,
none of which are $n \times n$ (else we may delete it).
Then, for any two orthogonally adjacent cells,
the edge between them must... |
USAMO-1998-notes_5 | Prove that for each $n\geq 2$,
there is a set $S$ of $n$ integers
such that $(a-b)^2$ divides $ab$ for every distinct $a,b\in S$. | This is a direct corollary of the more difficult USA TST 2015/2, reproduced below.
\begin{quote}
Prove that for every positive integer $n$,
there exists a set $S$ of $n$ positive integers
such that for any two distinct $a,b \in S$, $a-b$ divides $a$ and $b$
but none of the other elements of $S$.
\end{quote} |
USAMO-1998-notes_6 | Let $n \geq 5$ be an integer.
Find the largest integer $k$ (as a function of $n$)
such that there exists a convex $n$-gon $A_{1}A_{2}\dots A_{n}$
for which exactly $k$ of the quadrilaterals
$A_{i}A_{i+1}A_{i+2}A_{i+3}$ have an inscribed circle,
where indices are taken modulo $n$.
\end{enumerate} | The main claim is the following:
\begin{claim*}
We can't have both $A_1 A_2 A_3 A_4$
and $A_2 A_3 A_4 A_5$ be circumscribed.
\end{claim*}
\begin{proof}
If not, then we have the following diagram,
where $a = A_1A_2$, $b = A-2 A_3$, $c = A_3 A_4$, $d = A_4 A_5$.
\begin{center}
\begin{asy}
size(9cm);
p... |
USAMO-1999-notes_1 | Some checkers placed on an $n \times n$ checkerboard satisfy the following conditions:
\begin{enumerate}
\ii[(a)] every square that does not contain a checker shares a side with one that does;
\ii[(b)] given any pair of squares that contain checkers,
there is a sequence of squares containing checkers,
starting ... | Take a spanning tree on the set $V$ of checkers
where the $|V|-1$ edges of the tree are given by orthogonal adjacency.
By condition (a) we have
\[ \sum_{v \in V} (4-\deg v) \ge n^2 - |V| \]
and since $\sum_{v \in V} \deg v = 2(|V|-1)$ we get
\[ 4|V| - \left( 2|V|-2 \right) \ge n^2 - |V| \]
which implies $|V| \ge \frac{... |
USAMO-1999-notes_2 | Let $ABCD$ be a convex cyclic quadrilateral.
Prove that \[ |AB - CD| + |AD - BC| \geq 2|AC - BD|. \] | Let the diagonals meet at $P$,
and let $AP = pq$, $DP = pr$, $BP = qs$, $CP = rs$.
Then set $AB = qx$, $CD = rx$, $AD = py$, $BC = sy$.
In this way we compute
\[ \left\lvert AC-BD \right\rvert
= \left\lvert (p-s)(q-r) \right\rvert \]
and
\[ \left\lvert AB-CD \right\rvert
= \left\lvert q-r \right\rvert x. \]
By t... |
USAMO-1999-notes_3 | Let $p > 2$ be a prime and let $a$, $b$, $c$, $d$ be integers not divisible by $p$,
such that
\[ \left\{ \dfrac{ra}{p} \right\} + \left\{ \dfrac{rb}{p} \right\}
+ \left\{ \dfrac{rc}{p} \right\} + \left\{ \dfrac{rd}{p} \right\} = 2 \]
for any integer $r$ not divisible by $p$.
(Here, $\{t\} = t - \left\lfloor t \right... | First of all, we apparently have $r(a+b+c+d) \equiv 0 \pmod p$
for every prime $p$, so it automatically follows that $a+b+c+d \equiv 0 \pmod p$.
By scaling appropriately, and also replacing each number with its remainder modulo $p$,
we are going to assume that
\[ 1 = a \le b \le c \le d < p. \]
We are going to prove th... |
USAMO-1999-notes_4 | Let $a_1$, $a_2$, \dots, $a_n$ be a sequence of $n > 3$ real numbers
such that
\[ a_1 + \dots + a_n \ge n \quad\text{and}\quad
a_1^2 + \dots + a_n^2 \ge n^2. \]
Prove that $\max(a_1, \dots, a_n) \ge 2$. | Proceed by contradiction, assuming $a_i < 2$ for all $i$.
If all $a_i \ge 0$, then $ n^2 \le \sum_i a_i^2 < n \cdot 2^2$,
contradiction.
Otherwise, assume at least one $a_i$ is negative.
Note that if $-x$ and $-y$ are both present in the sequence ($x,y>0$),
then we can replace them with $-(x+y)$ and $0$.
So we may as... |
USAMO-1999-notes_5 | The Y2K Game is played on a $1 \times 2000$ grid as follows.
Two players in turn write either an S or an O in an empty square.
The first player who produces three consecutive boxes that spell SOS wins.
If all boxes are filled without producing SOS then the game is a draw.
Prove that the second player has a winning stra... | The main insight is that a construct of the form
\[ S \; \square \; \square \; S \]
(here the $\square$ is blank)
will kill any player which plays inside it.
We call this a \emph{trap} accordingly.
\begin{claim*}
The second player can force a trap to exist;
in this case the game will never end in a draw.
\end{clai... |
USAMO-1999-notes_6 | Let $ABCD$ be an isosceles trapezoid with $AB \parallel CD$.
The inscribed circle $\omega$ of triangle $BCD$ meets $CD$ at $E$.
Let $F$ be a point on the (internal) angle bisector of $\angle DAC$
such that $EF \perp CD$.
Let the circumscribed circle of triangle $ACF$
meet line $CD$ at $C$ and $G$.
Prove that the triang... | Note $E$ is contact point of $A$-excircle of $\triangle ACD$,
so $F$ is $A$-excenter.
Hence $CF$ is external angle bisector of $\angle ACG$
which implies $FA = FG$
(since $F$ is the arc midpoint on the circumcircle of $AFG$). |
USAMO-2000-notes_1 | Call a real-valued function $f$ \emph{very convex} if
\[ \frac{f(x)+f(y)}{2}
\ge f\left( \frac{x+y}{2} \right) + \left\lvert x-y \right\rvert \]
holds for all real numbers $x$ and $y$.
Prove that no very convex function exists. | For $C \ge 0$, we say a function $f$ is \emph{$C$-convex}
\[ \frac{f(x)+f(y)}{2}
\ge f\left( \frac{x+y}{2} \right) + C\left\lvert x-y \right\rvert. \]
Suppose $f$ is $C$-convex.
Let $a < b < c < d < e$ be any arithmetic progression,
such that $t = |e-a|$.
Observe that
\begin{align*}
f(a) + f(c) &\ge 2f(b) + C \cdo... |
USAMO-2000-notes_2 | Let $S$ be the set of all triangles $ABC$ for which
\[ 5 \left( \frac{1}{AP} + \frac{1}{BQ} + \frac{1}{CR} \right)
- \frac{3}{\min\{ AP, BQ, CR \}} = \frac{6}{r}, \]
where $r$ is the inradius and $P$, $Q$, $R$ are the points of tangency
of the incircle with sides $AB$, $BC$, $CA$ respectively.
Prove that all triangle... | We will prove the inequality
\[ \frac{2}{AP} + \frac{5}{BQ} + \frac{5}{CR} \ge \frac 6r \]
with equality when $AP : BQ : CR = 1 : 4 : 4$.
This implies the problem statement.
Letting $x= AP$, $y = BQ$, $z = CR$, the inequality becomes
\[ \frac2x + \frac5y + \frac5z \ge 6\sqrt{\frac{x+y+z}{xyz}}. \]
Squaring both sides ... |
USAMO-2000-notes_3 | A game of solitaire is played with $R$ red cards,
$W$ white cards, and $B$ blue cards.
A player plays all the cards one at a time.
With each play he accumulates a penalty.
If he plays a blue card, then he is charged a penalty
which is the number of white cards still in his hand.
If he plays a white card, then he is cha... | The minimum penalty is
\[ f(B,W,R) = \min (BW, 2WR, 3RB) \]
or equivalently, the natural guess of
``discard all cards of one color first''
is actually optimal (though not necessarily unique).
This can be proven directly by induction.
Indeed the base case $BWR = 0$
(in which case zero penalty is clearly achievable).
Th... |
USAMO-2000-notes_4 | Find the smallest positive integer $n$ such
that if $n$ squares of a $1000 \times 1000$ chessboard are colored,
then there will exist three colored squares
whose centers form a right triangle with sides
parallel to the edges of the board. | The answer is $n = 1999$.
For a construction with $n = 1998$,
take a punctured L as illustrated below (with $1000$ replaced by $4$):
\[ \begin{bmatrix} 1 \\ 1 \\ 1 \\ & 1 & 1 & 1 \end{bmatrix}. \]
We now show that if there is no right triangle,
there are at most $1998$ tokens (colored squares).
In every column with m... |
USAMO-2000-notes_5 | Let $A_1 A_2 A_3$ be a triangle,
and let $\omega_1$ be a circle in its plane
passing through $A_1$ and $A_2$.
Suppose there exists circles $\omega_2$, $\omega_3$, \dots, $\omega_7$
such that for $k = 2, 3, \dots, 7,$
circle $\omega_k$ is externally tangent to $\omega_{k-1}$ and passes
through $A_k$ and $A_{k+1}$ (indic... | The idea is to keep track of the subtended arc
$\widehat{A_i A_{i+1}}$ of $\omega_i$ for each $i$.
To this end, let $\beta = \dang A_1 A_2 A_3$,
$\gamma = \dang A_2 A_3 A_1$
and $\alpha = \dang A_1 A_2 A_3$.
\begin{center}
\begin{asy}
size(8cm);
defaultpen(fontsize(9pt));
pair A2 = (0,0);
pair O1 = (-8,0);
pair O2 = (... |
USAMO-2000-notes_6 | Let $a_1, b_1, a_2, b_2, \dots , a_n, b_n$ be nonnegative real numbers.
Prove that
\[ \sum_{i, j = 1}^{n} \min\{a_i a_j, b_i b_j\}
\le \sum_{i, j = 1}^{n} \min\{a_i b_j, a_j b_i\}. \]
\end{enumerate} | We present two solutions.
\paragraph{First solution by creating a single min (Vincent Huang and Ravi Boppana).}
Let $b_i = r_i a_i$ for each $i$,
and rewrite the inequality as
\[ \sum_{i,j} a_i a_j \left[
\min (r_i, r_j) - \min(1, r_i r_j) \right] \ge 0. \]
We now do the key manipulation to convert the double min
in... |
USAMO-2001-notes_1 | Each of eight boxes contains six balls.
Each ball has been colored with one of $n$ colors,
such that no two balls in the same box are the same color,
and no two colors occur together in more than one box.
Find with proof the smallest possible $n$. | The answer is $n = 23$.
Shown below is a construction using that many colors,
which we call $\{1,2,\dots,15,a,\dots,f,X,Y\}$.
\[
\begin{bmatrix}
X & X & X & 1 & 2 & 3 & 4 & 5 \\
1 & 6 & 11 & 6 & 7 & 8 & 9 & 10 \\
2 & 7 & 12 & 11 & 12 & 13 & 14 & 15 \\
3 & 8 & 13 & Y & Y & Y & a & b \\
4 & 9 & 14 &... |
USAMO-2001-notes_2 | Let $ABC$ be a triangle and let $\omega$ be its incircle.
Denote by $D_1$ and $E_1$ the points where $\omega$
is tangent to sides $BC$ and $AC$, respectively.
Denote by $D_2$ and $E_2$ the points on sides $BC$ and $AC$,
respectively, such that $CD_2=BD_1$ and $CE_2=AE_1$,
and denote by $P$ the point of intersection of ... | We have that $P$ is the Nagel point
\[ P = \left( s-a : s-b : s-c \right). \]
Therefore,
\[ \frac{PD_2}{AD_2} = \frac{s-a}{(s-a)+(s-b)+(s-c)} = \frac{s-a}{s}. \]
Meanwhile, $Q$ is the antipode of $D_1$.
The classical homothety at $A$ mapping $Q$ to $D_1$
(by mapping the incircle to the $A$-excircle)
has ratio $\frac{s-... |
USAMO-2001-notes_3 | Let $a, b, c$ be nonnegative real numbers
such that $a^2+b^2+c^2 + abc = 4$.
Show that \[ 0 \le ab + bc + ca - abc \le 2. \] | The left-hand side of the inequality is trivial;
just note that $\min \left\{ a,b,c \right\} \le 1$.
Hence, we focus on the right side.
We use Lagrange Multipliers.
Define \[ U = \left\{ (a,b,c) \mid a,b,c > 0
\text{ and } a^2+b^2+c^2 < 1000 \right\}. \]
This is an intersection of open sets, so it is open.
Its closu... |
USAMO-2001-notes_4 | Let $ABC$ be a triangle and $P$ any point
such that $PA$, $PB$, $PC$ are the sides of an obtuse triangle,
with $PA$ the longest side.
Prove that $\angle BAC$ is acute. | Using Ptolemy's inequality and Cauchy-Schwarz,
\begin{align*}
PA \cdot BC
&\le PB \cdot AC + PC \cdot AB \\
&\le \sqrt{(PB^2+PC^2)(AB^2+AC^2)} \\
&< \sqrt{PA^2 \cdot (AB^2+AC)^2} = PA \cdot \sqrt{AB^2+AC^2}
\end{align*}
meaning $BC^2 < AB^2+AC^2$, so $\angle BAC$ is acute.
\begin{remark*}
[Lokman G\"{o}k\c{c... |
USAMO-2001-notes_5 | Let $S \subseteq \ZZ$ be such that:
\begin{enumerate}
\ii[(a)] there exist $a,b \in S$ with
$\gcd(a,b) = \gcd(a-2, b-2) = 1$;
\ii[(b)] if $x$ and $y$ are elements of $S$ (possibly equal),
then $x^2-y$ also belongs to $S$.
\end{enumerate}
Prove that $S = \ZZ$. | Call an integer $d > 0$ \emph{shifty} if
$S = S+d$ (meaning $S$ is invariant under shifting by $d$).
First, note that if $u, v \in S$, then for any $x \in S$,
\[ v^2 - (u^2-x) = (v^2-u^2) + x \in S. \]
Since we can easily check that $|S| > 1$
and $S \neq \{n, -n\}$ we conclude there exists a shifty integer.
We claim ... |
USAMO-2001-notes_6 | Each point in the plane is assigned a real number.
Suppose that for any nondegenerate triangle, the number at its incenter
is the arithmetic mean of the three numbers at its vertices.
Prove that all points in the plane were assigned the same number.
\end{enumerate} | In this solution, we denote points by capital letters
and the associated values by lowercase letters
(e.g.\ $p$ denotes the value of our function at $P$).
\begin{claim*}
If $ABCD$ is an isosceles trapezoid, then $a+c=b+d$.
\end{claim*}
\begin{proof}
Indeed, suppose WLOG that rays $BA$ and $CD$ meet at $X$.
Then ... |
USAMO-2002-notes_1 | Let $S$ be a set with $2002$ elements, and let $N$ be an integer with $0 \leq N \leq 2^{2002}$.
Prove that it is possible to color every subset of $S$
either black or white so that the following conditions hold:
\begin{enumerate}
\ii[(a)] the union of any two white subsets is white;
\ii[(b)] the union of any two bl... | We will solve the problem with $2002$
replaced by an arbitrary integer $n \ge 0$.
In other words, we prove:
\begin{claim*}
For any nonnegative integers $n$ and $N$ with $0 \le N \le 2^n$,
it is possible to color the $2^n$ subsets
of $\{1, \dots, n\}$ black and white
satisfying the conditions of the problem.
\en... |
USAMO-2002-notes_2 | Let $ABC$ be a triangle such that
\[
\left( \cot \frac{A}{2} \right)^2
+ \left( 2\cot \frac{B}{2} \right)^2
+ \left( 3\cot \frac{C}{2} \right)^2
= \left( \frac{6s}{7r} \right)^2,
\]
where $s$ and $r$ denote its semiperimeter and its inradius, respectively.
Prove that triangle $ABC$ is similar to a triangle $T$ ... | Let $x = s-a$, $y = s-b$, $z = s-c$ in the usual fashion, then the equation reads
\[ x^2 + 4y^2 + 9z^2 = \left( \frac67(x+y+z) \right)^2. \]
However, by Cauchy-Schwarz, we have
\[ \left( 1 + \tfrac14 + \tfrac19 \right)\left( x^2 + 4y^2 + 9z^2 \right) \ge \left( x+y+z \right)^2 \]
with equality if and only if $1 : \tfr... |
USAMO-2002-notes_3 | Prove that any monic polynomial (a polynomial with leading coefficient $1$)
of degree $n$ with real coefficients is the average
of two monic polynomials of degree $n$ with $n$ real roots. | First,
\begin{lemma*}
If $p$ is a monic polynomial of degree $n$,
and $p(1)p(2) < 0$, $p(2)p(3) < 0$, \dots, $p(n-1)p(n) < 0$
then $p$ has $n$ real roots.
\end{lemma*}
\begin{proof}
The intermediate value theorem already guarantees
the existence of $n-1$ real roots.
The last root is obtained by considering... |
USAMO-2002-notes_4 | Determine all functions $f \colon \RR \to \RR$ such that
\[ f(x^2 - y^2) = x f(x) - y f(y) \]
for all pairs of real numbers $x$ and $y$. | The answer is $f(x) = cx$, $c \in \RR$ (these obviously work).
First, by putting $x=0$ and $y=0$ respectively
we have \[ f(x^2) = xf(x) \quad\text{and}\quad f(-y^2) = -yf(y). \]
From this we deduce that $f$ is odd, in particular $f(0) = 0$.
Then, we can rewrite the given as $f(x^2-y^2) + f(y^2) = f(x^2)$.
Combined wit... |
USAMO-2002-notes_5 | Let $a$, $b$ be integers greater than $2$.
Prove that there exists a positive integer $k$
and a finite sequence $n_1, n_2, \dots, n_k$ of
positive integers such that $n_1 = a$, $n_k = b$,
and $n_i n_{i+1}$ is divisible by $n_i + n_{i+1}$
for each $i$ ($1 \leq i < k$). | Consider a graph $G$ on the vertex set $\{3, 4, \dots\}$
and with edges between $v$, $w$ if $v + w \mid vw$;
the problem is equivalent to showing that $G$ is connected.
First, note that $n$ is connected to $n(n-1)$, $n(n-1)(n-2)$, etc.\
up to $n!$.
But for $n > 2$, $n!$ is connected to $(n+1)!$ too:
\begin{itemize}
... |
USAMO-2002-notes_6 | I have an $n \times n$ sheet of stamps, from which I've been asked
to tear out blocks of three adjacent stamps in a single row or column.
(I can only tear along the perforations separating adjacent stamps,
and each block must come out of the sheet in one piece.)
Let $b(n)$ be the smallest number of blocks I can tear ou... | For the lower bound: there are $2n(n-2)$ places one could put a block.
Note that each block eliminates at most $14$ such places.
For the upper bound, the construction of $\frac15$ is easy to build.
Here is an illustration of one possible construction
for $n=9$ which generalizes readily,
using only vertical blocks.
\[
... |
USAMO-2003-notes_1 | Prove that for every positive integer $n$
there exists an $n$-digit number divisible by $5^n$
all of whose digits are odd. | This is immediate by induction on $n$.
For $n = 1$ we take $5$;
moving forward if $M$ is a working $n$-digit number then exactly one of
\begin{align*}
N_1 &= 10^n + M \\
N_3 &= 3 \cdot 10^n + M \\
N_5 &= 5 \cdot 10^n + M \\
N_7 &= 7 \cdot 10^n + M \\
N_9 &= 9 \cdot 10^n + M
\end{align*}
is divisible by $5^{n+... |
USAMO-2003-notes_2 | A convex polygon $\mathcal{P}$ in the plane
is dissected into smaller convex polygons by drawing all of its diagonals.
The lengths of all sides and all diagonals of the polygon $\mathcal{P}$ are rational numbers.
Prove that the lengths of all sides of all polygons
in the dissection are also rational numbers. | Suppose $AB$ is a side of a polygon in the dissection,
lying on diagonal $XY$, with $X$, $A$, $B$, $Y$ in that order.
Then \[ AB = XY - XA - YB. \]
In this way, we see that it actually just suffices to
prove the result for a quadrilateral.
We present two approaches to this end.
\paragraph{First approach (trig).}
Cons... |
USAMO-2003-notes_3 | Let $n$ be a positive integer.
For every sequence of integers
\[ A = (a_0, \; a_1, \; a_2, \; \dots, a_n) \]
satisfying $0 \le a_i \le i$, for $i=0,\dots,n$,
we define another sequence
\[ t(A)= (t(a_0), \; t(a_1), \; t(a_2), \; \dots, \; t(a_n)) \]
by setting $t(a_i)$ to be the number of terms in the
sequence $A$ that ... | We go by strong induction on $n$ with the base cases $n=1$ and $n=2$ done by hand.
Consider two cases:
\begin{itemize}
\ii If $a_0 = 0$ and $a_1 = 1$, then $1 \le t(a_i) \le i$ for $i \ge 1$;
now apply induction to
\[ \left(t(a_1)-1, \; t(a_2)-1, \; \dots, \; t(a_n)-1\right). \]
\ii Otherwise, assume that $a_0 = a_1 =... |
USAMO-2003-notes_4 | Let $ABC$ be a triangle.
A circle passing through $A$ and $B$
intersects segments $AC$ and $BC$ at $D$ and $E$, respectively.
Lines $AB$ and $DE$ intersect at $F$,
while lines $BD$ and $CF$ intersect at $M$.
Prove that $MF = MC$ if and only if $MB \cdot MD = MC^2$. | Ceva theorem plus the similar triangles.
\begin{center}
\begin{asy}
size(7.5cm);
pair C = Drawing("C", (3,0), dir(-45));
pair D = Drawing("D", (0,0));
pair M = Drawing("M", (1,-0.7));
pair F = Drawing("F", 2*M-C);
pair B = Drawing("B", conj(C/M) * (M-C)+C, dir(130));
pair E = Drawing("E", extension(D,F,B... |
USAMO-2003-notes_5 | Let $a$, $b$, $c$ be positive real numbers.
Prove that
\[ \frac{(2a+b+c)^2}{2a^2+(b+c)^2}
+ \frac{(2b+c+a)^2}{2b^2+(c+a)^2}
+ \frac{(2c+a+b)^2}{2c^2+(a+b)^2} \le 8. \] | This is a canonical example of tangent line trick.
Homogenize so that $a + b + c = 3$.
The desired inequality reads
\[ \sum_{\text{cyc}} \frac{(a+3)^2}{2a^2+(3-a)^2} \le 8. \]
This follows from
\[ f(x) = \frac{(x+3)^2}{2x^2+(3-x)^2}
\le \frac{1}{3} (4x + 4) \]
which can be checked as
$\frac 13 (4x+4)(2x^2+(3-x)^2) - ... |
USAMO-2003-notes_6 | At the vertices of a regular hexagon are written six nonnegative integers
whose sum is $2003^{2003}$.
Bert is allowed to make moves of the following form:
he may pick a vertex and replace the number written
there by the absolute value of the difference between the numbers
written at the two neighboring vertices.
Prove ... | If $a \le b \le c$ are \emph{odd} integers,
the configuration which has $(a,b-a,b,c-b,c,c-a)$ around the hexagon
in some order (up to cyclic permutation and reflection)
is said to be \emph{great} (picture below).
\begin{claim*}
We can reach a great configuration
from any configuration with odd sum.
\end{claim*}
\b... |
USAMO-2004-notes_1 | Let $ABCD$ be a quadrilateral circumscribed about a circle,
whose interior and exterior angles are at least $60$ degrees.
Prove that
\[ \frac{1}{3}|AB^3 - AD^3|
\le |BC^3 - CD^3| \le 3|AB^3 - AD^3|. \]
When does equality hold? | Clearly it suffices to show the left inequality.
Since $AB+CD = BC+AD \implies |AB-AD| = |BC-CD|$, it suffices to prove
\[ \frac13(AB^2 + AB \cdot AD + AD^2)
\le BC^2 + BC \cdot CD + CD^2. \]
This follows by noting that
\begin{align*}
BC^2 + BC \cdot CD + CD^2
&\ge BC^2 + CD^2 - 2(BC)(CD)\cos(\angle BCD) \\
&= BD... |
USAMO-2004-notes_2 | Let $a_1$, $a_2$, \dots, $a_n$ be integers whose greatest common divisor is $1$.
Let $S$ be a set of integers with the following properties:
\begin{enumerate}
\ii[(a)] $a_i \in S$ for $i = 1, \dots, n$.
\ii[(b)] $a_i - a_j \in S$ for $i, j = 1, \dots, n$, not necessarily distinct.
\ii[(c)] If $x, y \in S$ and $x+... | The idea is to show any linear combination of the $a_i$ are in $S$,
which implies (by Bezout) that $S = \ZZ$.
This is pretty intuitive, but the details require some care
(in particular there is a parity obstruction at the second lemma).
First, we make the following simple observations:
\begin{itemize}
\ii $0 \in S$,... |
USAMO-2004-notes_3 | For what real values of $k > 0$ is it possible to
dissect a $1 \times k$ rectangle into two similar but noncongruent polygons? | Answer: the dissection is possible for every $k > 0$ except for $k = 1$.
\paragraph{Construction.}
By symmetry it suffices to give a construction for $k > 1$
(since otherwise we replace $k$ by $k\inv$).
For every integer $n \ge 2$ and real number $r > 1$,
we define a shape $\mathcal R(n,r)$ as follows.
\begin{itemize}... |
USAMO-2004-notes_4 | Alice and Bob play a game on a $6$ by $6$ grid.
On his turn, a player chooses a rational number not yet appearing in the grid
and writes it in an empty square of the grid.
Alice goes first and then the players alternate.
When all squares have numbers written in them, in each row,
the square with the greatest number in ... | Bob can win.
Label the first two rows as follows:
\[
\begin{bmatrix}
a & b & c & d & e & f \\
d' & e' & f' & a' & b' & c'
\end{bmatrix}
\]
These twelve boxes thus come in six \emph{pairs}, $(a, a')$, $(b, b')$ and so on.
\begin{claim*}
Bob can ensure that the order relation
of the labels is the same be... |
USAMO-2004-notes_5 | Let $a$, $b$, $c$ be positive reals.
Prove that
\[ ( a^5-a^2+3 )( b^5-b^2+3 )( c^5-c^2+3 )
\ge \left( a+b+c \right)^3. \] | Observe that for all real numbers $a$, the inequality
\[ a^5-a^2+3 \ge a^3+2 \]
holds.
Then the problem follows by H\"{o}lder in the form
\[ (a^3+1+1)(1+b^3+1)(1+1+c^3) \ge (a+b+c)^3. \] |
USAMO-2004-notes_6 | A circle $\omega$ is inscribed in a quadrilateral $ABCD$.
Let $I$ be the center of $\omega$.
Suppose that \[ (AI + DI)^2 + (BI + CI)^2 = (AB + CD)^2. \]
Prove that $ABCD$ is an isosceles trapezoid.
\end{enumerate} | Here's a completely algebraic solution.
WLOG $\omega$ has radius $1$, and let $a$, $b$, $c$, $d$ be the
lengths of the tangents from $A$, $B$, $C$, $D$ to $\omega$.
It is known that
\[ a+b+c+d = abc+bcd+cda+dab \qquad (\star) \]
which can be proved by, say $\tan$-addition formula.
Then, the content of the problem is to... |
USAMO-2005-notes_1 | Determine all composite positive integers $n$ for which
it is possible to arrange all divisors of $n$ that are greater than $1$
in a circle so that no two adjacent divisors are relatively prime. | The only bad ones are $n = pq$, products of two distinct primes.
Clearly they can't be so arranged, so we show all others work.
\begin{itemize}
\ii If $n$ is a power of a prime, the result is obvious; any arrangement works.
\ii If $n = p_1^{e_1} \dots p_k^{e_k}$ for some $k \ge 3$,
then first situate $p_1p_2$, $p_2p_3... |
USAMO-2005-notes_2 | Prove that the system of equations
\begin{align*}
x^6 + x^3 + x^3y + y &= 147^{157} \\
x^3 + x^3y + y^2 + y + z^9 &= 157^{147}
\end{align*}
has no integer solutions. | Sum the equations and add $1$ to both sides to get
\[ (x^3+y+1)^2 + z^9 = 147^{157} + 157^{147} + 1
\equiv 14 \pmod{19} \]
But $a^2 + b^9 \not\equiv 14 \pmod{19}$
for any integers $a$ and $b$,
since the ninth powers modulo $19$ are $0$, $\pm 1$
and none of $\{13, 14, 15\}$ are squares modulo $19$.
Therefore, there ar... |
USAMO-2005-notes_3 | Let $ABC$ be an acute-angled triangle,
and let $P$ and $Q$ be two points on side $BC$.
Construct a point $C_1$ in such a way that the convex quadrilateral $APBC_1$ is cyclic,
$\ol{QC_1} \parallel \ol{CA}$, and $C_1$ and $Q$ lie on opposite sides of line $AB$.
Construct a point $B_1$ in such a way that the convex quadri... | It is enough to prove that $A$, $B_1$, and $C_1$ are collinear,
since then $\dang C_1QP = \dang ACP = \dang AB_1P = \dang C_1B_1P$.
\begin{center}
\begin{asy}
pair A = dir(110);
pair B = dir(210);
pair C = dir(330);
pair P = 0.7*C+0.3*B;
draw(A--B--C--cycle);
pair O_1 = circumcenter(A, P, C);
pair B_1 = 2*O_1-C;
pair... |
USAMO-2005-notes_4 | Legs $L_1$, $L_2$, $L_3$, $L_4$ of a square table each have length $n$,
where $n$ is a positive integer.
For how many ordered $4$-tuples $(k_1, k_2, k_3, k_4)$ of nonnegative integers
can we cut a piece of length $k_i$ from the end of leg $L_i$
and still have a stable table?
(The table is \emph{stable} if it can be pl... | Flip the table upside-down so that that the table's surface
rests on the floor.
Then, we see that we want the truncated legs
to have endpoints $A$, $B$, $C$, $D$ which are coplanar (say).
\begin{claim*}
This occurs if and only if $ABCD$ is a parallelogram.
\end{claim*}
\begin{proof}
Obviously $ABCD$ being a paralle... |
USAMO-2005-notes_5 | Let $n > 1$ be an integer.
Suppose $2n$ points are given in the plane, no three of which are collinear.
Suppose $n$ of the given $2n$ points are colored blue and the other $n$ colored red.
A line in the plane is called a \emph{balancing line}
if it passes through one blue and one red point
and, for each side of the lin... | Consider the convex hull $\mathcal H$ of the polygon. There are two cases.
\paragraph{Easy case: the convex hull has both colors.}
If the convex hull $\mathcal H$ is not all the same color,
there exist two edges of $\mathcal H$ (at least) which have differently colored endpoints.
The extensions of those sides form bal... |
USAMO-2005-notes_6 | For a positive integer $m$,
let $s(m)$ denote the sum of the decimal digits of $m$.
A set $S$ positive integers is \emph{$k$-stable}
if $s(\sum_{x \in X} x) = k$ for any nonempty subset $X \subseteq S$.
For each integer $n \ge 2$ let $f(n)$ be the minimal $k$
for which there exists a $k$-stable set with $n$ integers.
... | \paragraph{Construction showing $f(n) \le 9 \left\lceil \log_{10} \binom{n+1}{2} \right\rceil$.}
Let $n \ge 1$ and $e \ge 1$ be integers satisfying $1 + 2 + \dots + n < 10^e$.
Consider the set
\[ S = \left\{ 10^e-1, \; 2(10^e-1), \; \dots, \; n(10^e-1) \right\}. \]
For example, if $n = 6$ and $e = 3$,
we have $S = \{99... |
USAMO-2006-notes_1 | Let $p$ be a prime number and let $s$ be an integer with $0 < s < p$.
Prove that there exist integers $m$ and $n$ with $0 < m < n < p$ and
\[ \left \{\frac{sm}{p} \right\} < \left \{\frac{sn}{p} \right \} < \frac{s}{p} \]
if and only if $s$ is not a divisor of $p-1$. | It's equivalent to $ms \bmod p < ns \bmod p < s$,
where $x \bmod p$ means the remainder when $x$ is divided by $p$,
by slight abuse of notation.
We will assume $s \ge 2$ for simplicity,
since the case $s = 1$ is clear.
For any $x \in \{1, 2, \dots, s-1\}$
we define $f(x)$ to be the unique number in $\{1, \dots, p-1\}$... |
USAMO-2006-notes_2 | Let $k > 0$ be a fixed integer.
Compute the minimum integer $N$ (in terms of $k$) for which
there exists a set of $2k+1$ distinct positive integers
that has sum greater than $N$,
but for which every subset of size $k$ has sum at most $N/2$. | The answer is $N = k(2k^2+3k+3)$ given by
\[ S = \left\{ k^2+1, k^2+2, \dots, k^2+2k+1 \right\}. \]
To show this is best possible,
let the set be $S = \{ a_0 < a_1 < \dots < a_{2k} \}$
so that the hypothesis becomes
\begin{align*}
N + 1 &\le a_0 + a_1 + \dots + a_{2k} \\
N/2 &\ge a_{k+1} + \dots + a_{2k}.
\end{al... |
USAMO-2006-notes_3 | For integral $m$, let $p(m)$ be the greatest prime divisor of $m$.
By convention, we set $p(\pm 1) = 1$ and $p(0) = \infty$.
Find all polynomials $f$ with integer coefficients
such that the sequence
\[ \{ p(f(n^2)) - 2n \}_{n \geq 0} \]
is bounded above.
(In particular, this requires $f(n^2) \neq 0$ for $n \ge 0$.) | If $f$ is the (possibly empty)
product of linear factors of the form $4n-a^2$,
then it satisfies the condition.
We will prove no other polynomials work.
In what follows, assume $f$ is irreducible and nonconstant.
It suffices to show for every positive integer $c$,
there exists a prime $p$ and a nonnegative integer $n$... |
USAMO-2006-notes_4 | Find all positive integers $n$ for which there exist an integer $k \ge 2$
and positive rational numbers $a_1$, \dots, $a_k$ satisfying
$a_1 + a_2 + \dots + a_k = a_1 a_2 \dots a_k = n$. | The answer is all $n$ other than $1,2,3,5$.
\begin{claim*}
The only solution with $n \le 5$ is $n = 4$.
\end{claim*}
\begin{proof}
The case $n=4$ works since $2+2 = 2\cdot2=4$. So assume $n > 4$.
We now contend that $k > 2$.
Indeed, if $a_1 + a_2 = a_1 a_2 = n$ then
$(a_1-a_2)^2 = (a_1+a_2)^2 - 4a_1a_2 = n^... |
USAMO-2006-notes_5 | A mathematical frog jumps along the number line.
The frog starts at $1$, and jumps according to the following rule:
if the frog is at integer $n$,
then it can jump either to $n+1$ or to $n + 2^{m_n+1}$
where $2^{m_n}$ is the largest power of $2$ that is a factor of $n$.
Show that if $k \ge 2$ is a positive integer
and ... | We will think about the problem in terms of
finite sequences of jumps $(s_1, s_2, \dots, s_\ell)$,
which we draw as
\[ 1 = x_0
\taking{s_1} x_1
\taking{s_2} x_2
\taking{s_3} \dots
\taking{s_\ell} x_\ell \]
where $s_k = x_k - x_{k-1}$ is the length of some hop.
We say the sequence is \emph{valid} if it has the
p... |
USAMO-2006-notes_6 | Let $ABCD$ be a quadrilateral,
and let $E$ and $F$ be points on sides $AD$ and $BC$,
respectively, such that $\frac{AE}{ED} = \frac{BF}{FC}$.
Ray $FE$ meets rays $BA$ and $CD$ at $S$ and $T$, respectively.
Prove that the circumcircles of triangles $SAE$, $SBF$, $TCF$, and $TDE$
pass through a common point.
\end{enumer... | \begin{center}
\begin{asy}
size(8cm);
defaultpen(fontsize(9pt));
pair A = (-1,3);
pair B = (-3,-2);
pair C = (6,-2);
pair D = (3,5);
pair M = (A*C-B*D)/(A+C-B-D);
pair E = 0.7*A+0.3*D;
pair F = 0.7*B+0.3*C;
pair S = extension(E, F, A, B);
pair T = extension(C, D, E, F);
... |
USAMO-2007-notes_1 | Let $n$ be a positive integer.
Define a sequence by setting $a_1 = n$ and, for each $k > 1$,
letting $a_k$ be the unique integer in the range $0 \leq a_k \leq k-1$
for which $a_1 + a_2 + \dots + a_k$ is divisible by $k$.
(For instance, when $n = 9$
the obtained sequence is $9,1,2,0,3,3,3,\dots$.)
Prove that for any $n$... | For each $k$, the number
\[ b_k \coloneq \frac 1k (a_1 + \dots + a_k) \]
is a nonnegative integer.
\begin{claim*}
The sequence $(b_k)$ is eventually constant.
\end{claim*}
\begin{proof}
Since
\[ b_{k+1}
= \frac{a_1 + \dots + a_k + a_{k+1}}{k+1}
\le \frac{k b_k + k}{k+1} < b_k + 1 \]
and therefore $b_{k+... |
USAMO-2007-notes_2 | Decide whether it possible to cover all lattice points in $\RR^2$
by an (infinite) family of disks whose interiors are disjoint
such that the radius of each disk is at least $5$. | The answer is no.
Assume not.
Take a disk $\odot O$ not touching any member of the family,
and then enlarge it until it is maximal.
Then, it must be tangent to at least three other disks,
say $\odot A$, $\odot B$, $\odot C$.
Suppose WLOG that $\angle AOB \le 120\dg$.
Denote the radii of $\odot O$, $\odot A$, $\odot B$... |
USAMO-2007-notes_3 | Let $S$ be a set containing $n^2+n-1$ elements.
Suppose that the $n$-element subsets of $S$ are partitioned into two classes.
Prove that there are at least $n$ pairwise disjoint sets in the same class. | We present two solutions which are really equivalent, but phrased differently.
We refer to the two classes as ``red'' and ``blue'', respectively.
\paragraph{First solution (Grant Yu).}
We define a set of $n+1$ elements to be \emph{useful}
if it has $n$-element subsets in each class.
Consider a \textbf{maximal collect... |
USAMO-2007-notes_4 | An \emph{animal} with $n$ \emph{cells}
is a connected figure consisting of $n$ equal-sized square cells
(equivalently, a polyomino with $n$ cells).
A \emph{dinosaur} is an animal with at least $2007$ cells.
It is said to be \emph{primitive}
it its cells cannot be partitioned into two or more dinosaurs.
Find with proof ... | In fact it's true for any tree with maximum degree $\le 4$.
Here is the solution of Andrew Geng.
Let $T$ be such a tree (a spanning tree of the dinosaur graph)
which corresponds to a primitive dinosaur.
\begin{claim*}
There exists a vertex $v$ such that when $v$ is deleted, no dinosaurs result.
\end{claim*}
\begin{... |
USAMO-2007-notes_5 | Prove that for every nonnegative integer $n$,
the number $7^{7^n}+1$ is the product
of at least $2n+3$ (not necessarily distinct) primes. | We prove this by induction on $n$ by showing that
\[ \frac{X^7+1}{X+1} = X^6 - X^5 + \dots + 1 \]
is never prime for $X = 7^{7^n}$,
hence we gain at least two additional prime factors
whenever we increase $n$ by one.
Indeed, the quotient may be written as
\[ \left( X+1 \right)^6 - 7X \cdot (X^2+X+1)^2 \]
which becomes... |
USAMO-2007-notes_6 | Let $ABC$ be an acute triangle
with $\omega$, $S$, and $R$ being its incircle,
circumcircle, and circumradius, respectively.
Circle $\omega_{A}$ is tangent internally to $S$
at $A$ and tangent externally to $\omega$.
Circle $S_{A}$ is tangent internally to $S$
at $A$ and tangent internally to $\omega$.
Let $P_A$ and $... | It turns out we can compute $P_AQ_A$ explicitly.
Let us invert around $A$ with radius $s-a$
(hence fixing the incircle) and then compose this
with a reflection around the angle bisector of $\angle BAC$.
We denote the image of the composed map via
\[ \bullet \mapsto \bullet^\ast \mapsto \bullet^+. \]
We overlay this inv... |
USAMO-2008-notes_1 | Prove that for each positive integer $n$,
there are pairwise relatively prime integers $k_0$, \dots, $k_n$,
all strictly greater than $1$, such that
$k_0k_1 \dots k_n - 1$ is the product of two consecutive integers. | In other words, if we let
\[ P(x) = x(x+1) + 1 \]
then we would like there to be infinitely many primes
dividing some $P(t)$ for some integer $t$.
In fact, this result is true in much greater generality.
We first state:
\begin{theorem}
[Schur's theorem]
If $P(x) \in \ZZ[x]$ is nonconstant and $P(0) = 1$,
then th... |
USAMO-2008-notes_2 | Let $ABC$ be an acute, scalene triangle,
and let $M$, $N$, and $P$ be the midpoints of
$\ol{BC}$, $\ol{CA}$, and $\ol{AB}$, respectively.
Let the perpendicular bisectors of $\ol{AB}$ and $\ol{AC}$
intersect ray $AM$ in points $D$ and $E$ respectively,
and let lines $BD$ and $CE$ intersect in point $F$, inside triangle ... | We present four solutions.
\paragraph{Barycentric solution.}
First, we find the coordinates of $D$.
As $D$ lies on $\ol{AM}$, we know $D=(t:1:1)$ for some $t$.
Now by perpendicular bisector formula, we find
\[ 0 = b^2(t-1) + (a^2-c^2) \implies t = \frac{c^2+b^2-a^2}{b^2}. \]
Thus we obtain \[ D = \left( 2S_A : c^2 : c... |
USAMO-2008-notes_3 | Let $n$ be a positive integer. Denote by $S_n$ the set of points $(x, y)$
with integer coordinates such that
\[ \left\lvert x\right\rvert + \left\lvert y + \half \right\rvert < n. \]
A path is a sequence of distinct points
$(x_1 , y_1), (x_2, y_2), \dots, (x_\ell, y_\ell)$ in $S_n$
such that, for $i = 2, \dots, \ell$,... | \paragraph{First solution (local).}
We proceed by induction on $n$.
The base case $n=1$ is clear, so suppose $n > 1$.
Let $S$ denote the set of points
\[ S = \left\{ (x,y) : x + \left\lvert y+\frac12 \right\rvert \ge n - 2 \right\}. \]
An example when $n=4$ is displayed below.
\begin{center}
\begin{asy}
int n = 4;
for... |
USAMO-2008-notes_4 | For which integers $n \ge 3$ can one find a triangulation
of regular $n$-gon consisting only of isosceles triangles? | The answer is $n$ of the form $2^a(2^b+1)$
where $a$ and $b$ are nonnegative integers not both zero.
Call the polygon $A_1 \dots A_n$
with indices taken modulo $n$.
We refer to segments $A_1 A_2$, $A_2 A_3$, \dots, $A_n A_1$
as \emph{short sides}.
Each of these must be in the triangulation.
Note that
\begin{itemize}
... |
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