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Three equilateral triangles

Source: Centroamerican Olympiad 2014, problem 5

June 11, 2014

geometryincentergeometric transformationreflectiontrigonometrycircumcircleangle bisector

Problem Statement

Points

A

B

C

and

D

are chosen on a line in that order, with

AB

and

CD

greater than

BC

. Equilateral triangles

APB

BCQ

and

CDR

are constructed so that

P

Q

and

R

are on the same side with respect to

AD

. If

\angle PQR=120^\circ

, show that

\frac{1}{AB}+\frac{1}{CD}=\frac{1}{BC}.