Maths Olympiad Prep

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Problem 1302

AIME late
Geometry Difficulty 5.1 Find the answer HMMT February

A semicircle with radius 2021 has diameter ABAB and center OO. Points CC and DD lie on the semicircle such that ∠AOC<∠AOD=90∘\angle AOC < \angle AOD = 90^{\circ}. A circle of radius rr is inscribed in the sector bounded by OAOA and OCOC and is tangent to the semicircle at EE. If CD=CECD=CE, compute ⌊r⌋\lfloor r \rfloor.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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Official solution

We are given m∠EOC=m∠CODm \angle EOC = m \angle COD and m∠AOC+m∠COD=2m∠EOC+m∠COD=90∘m \angle AOC + m \angle COD = 2m \angle EOC + m \angle COD = 90^{\circ} So m∠EOC=30∘m \angle EOC = 30^{\circ} and m∠AOC=60∘m \angle AOC = 60^{\circ}. Letting the radius of the semicircle be RR, we have (R−r)sin⁡∠AOC=r⇒r=13R(R-r) \sin \angle AOC = r \Rightarrow r = \frac{1}{3} R so ⌊r⌋=⌊20213⌋=673\lfloor r \rfloor = \left\lfloor\frac{2021}{3}\right\rfloor = 673

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.