Difference between revisions of "2023 USAJMO Problems"

Revision as of 14:23, 24 March 2023

Day 1

Problem 1

Find all triples of positive integers $(x,y,z)$ that satisfy the equation $\begin{align*} 2(x+y+z+2xyz)^2=(2xy+2yz+2zx+1)^2+2023 \end{align*}$

Problem 2

(Holden Mui) In an acute triangle $ABC$ , let $M$ be the midpoint of $\overline{BC}$ . 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 $\overline{AQ}$ . Prove that $NB=NC$ .

Problem 3

(Holden Mui) Consider an $n$ -by- $n$ board of unit squares for some odd positive integer $n$ . We say that a collection $C$ of identical dominoes is a maximal grid-aligned configuration on the board if $C$ consists of $(n^2-1)/2$ dominoes where each domino covers exactly two neighboring squares and the dominoes don't overlap: $C$ then covers all but one square on the board. We are allowed to slide (but not rotate) a domino on the board to cover the uncovered square, resulting in a new maximal grid-aligned configuration with another square uncovered. Let $k(C)$ be the number of distinct maximal grid-aligned configurations obtainable from $C$ by repeatedly sliding dominoes. Find the maximum value of $k(C)$ as a function of $n$ .

Day 2

Problem 4

(David Torres) Two players, $B$ and $R$ , play the following game on an infinite grid of unit squares, all initially colored white. The players take turns starting with $B$ . On $B$ 's turn, $B$ selects one white unit square and colors it blue. On $R$ 's turn, $R$ selects two white unit squares and colors them red. The players alternate until $B$ decides to end the game. At this point, $B$ gets a score, given by the number of unit squares in the largest (in terms of area) simple polygon containing only blue unit squares. What is the largest score $B$ can guarantee?

(A simple polygon is a polygon (not necessarily convex) that does not intersect itself and has no holes.

Problem 5

A positive integer $a$ is selected, and some positive integers are written on a board. Alice and Bob play the following game. On Alice's turn, she must replace some integer $n$ on the board with $n+a$ , and on Bob's turn he must replace some even integer $n$ on the board with $n/2$ . Alice goes first and they alternate turns. If on his turn Bob has no valid moves, the game ends.

After analyzing the integers on the board, Bob realizes that, regardless of what moves Alice makes, he will be able to force the game to end eventually. Show that, in fact, for this value of $a$ and these integers on the board, the game is guaranteed to end regardless of Alice's or Bob's moves.

Problem 6

(Anton Trygub) 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 $\angle DFE = 90^\circ$ . Line $FE$ intersects rays $AB$ and $AC$ at points $X$ and $Y$ , respectively. Prove that $\angle XDE = \angle EDY$ .

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	==Problem 6==		==Problem 6==
−	(''Anton Trygub'') Isosceles triangle <math>ABC</math>, with <math>AB=AC</math>, is inscribed in circle <math>\omega</math>. Let <math>D</math> be an arbitrary point inside <math>BC</math> such that <math>BD\neq DC</math>. Ray <math>AD</math> intersects <math>\omega</math> again at <math>E</math> (other than <math>A</math>). Point <math>F</math> (other than <math>E</math>) is chosen on <math>\omega</math> such that <math>\angle ~~DEF~~ = 90^\circ</math>. Line <math>FE</math> intersects rays <math>AB</math> and <math>AC</math> at points <math>X</math> and <math>Y</math>, respectively. Prove that <math>\angle XDE = \angle EDY</math>.	+	(''Anton Trygub'') Isosceles triangle <math>ABC</math>, with <math>AB=AC</math>, is inscribed in circle <math>\omega</math>. Let <math>D</math> be an arbitrary point inside <math>BC</math> such that <math>BD\neq DC</math>. Ray <math>AD</math> intersects <math>\omega</math> again at <math>E</math> (other than <math>A</math>). Point <math>F</math> (other than <math>E</math>) is chosen on <math>\omega</math> such that <math>\angle DFE = 90^\circ</math>. Line <math>FE</math> intersects rays <math>AB</math> and <math>AC</math> at points <math>X</math> and <math>Y</math>, respectively. Prove that <math>\angle XDE = \angle EDY</math>.

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