AoPS - Problem

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Problem

Source: Indonesian TST for IMO 2023 stage 1: Test 3 - Geometry

Tags: projective geometry, geometry, Indonesia, Indonesia TST

DavyDuf

27.02.2023 10:08

Given an acute triangle $ABC$ with circumcenter $O$ . The circumcircle of $BCH$ and a circle with diameter of $AC$ intersect at $P (P \neq C)$ . A point $Q$ on segment of $PC$ such that $PB = PQ$ . Prove that $\angle ABC = \angle AQP$

anphang

27.02.2023 10:34

What is to be proven is equivalent to $A,Q,B,C$ concyclic. I'll just name $Q$ the intersection of $(ABC)$ and $(CP)$ instead, and prove that $PBQ$ is isosceles in $B$ .

$AQBC$ is cyclic, so

$\widehat{PQB}=\widehat{CQB}=\widehat{CAB}$ Also, we have

$\widehat{CAB}=\widehat{CAH}+\widehat{HAB}=\widehat{HBC}+\widehat{HCB}$ since there are right angles and concyclic points everywhere Finally, with cyclic quadrilateral

$CHPB$ we have

$\widehat{HBC}+\widehat{HCB}=180-\widehat{CHB}=180-\widehat{CPB}=\widehat{BPQ}\quad\square$