olymmath

USAJMO 2024

United States of America Junior Mathematical Olympiad 2024.

Day 1

Problem 1

  • Day 1

Let ABCDABCD be a cyclic quadrilateral with AB=7AB = 7 and CD=8CD = 8 . Point PP and QQ are selected on segment ABAB such that AP=BQ=3AP = BQ = 3 . Points RR and SS are selected on segment CDCD such that CR=DS=2CR = DS = 2 . Prove that PQRSPQRS is a cyclic quadrilateral.

Proposed by Evan O'Dorney

Answer, solutions and source

Problem 2

  • Day 1

Let mm and nn be positive integers. Let SS be the set of integer points (x,y)(x,y) with 1≤x≤2m1\leq x\leq 2m and 1≤y≤2n1\leq y\leq 2n . A configuration of mnmn rectangles is called happy if each point in SS is a vertex of exactly one rectangle, and all rectangles have sides parallel to the coordinate axes. Prove that the number of happy configurations is odd.

Proposed by Serena An and Claire Zhang

Answer, solutions and source

Problem 3

  • Day 1

A sequence a1,a2,…a_{1}, a_{2}, \ldots of positive integers is defined recursively by a1=2a_{1}=2 and an+1=ann+1−1 for n≥1a_{n+1}=a_{n}^{n+1}-1 \text { for } n \geq 1 Prove that for every odd prime pp and integer kk, some term of the sequence is divisible by pkp^{k}.

Answer, solutions and source

Day 2

Problem 4

  • Day 2

Let n≥3n \geq 3 be an integer. Rowan and Colin play a game on an n×nn \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:

  • no matter how Rowan permutes the rows of the coloring, Colin can then permute the columns to restore the original grid coloring; and
  • no matter how Colin permutes the columns of the coloring, Rowan can then permute the rows to restore the original grid coloring. In terms of nn , how many orderly colorings are there?

Proposed by Alec Sun

Answer, solutions and source

Problem 6 · Also USAMO 2024 Problem 5

  • Day 2

Point DD is selected inside acute △ABC\triangle ABC so that ∠DAC=∠ACB\angle DAC = \angle ACB and ∠BDC=90∘+∠BAC\angle BDC = 90^{\circ} + \angle BAC . Point EE is chosen on ray BDBD so that AE=ECAE = EC . Let MM be the midpoint of BCBC .

Show that line ABAB is tangent to the circumcircle of triangle BEMBEM .

Proposed by Anton Trygub

Answer, solutions and source