AoPS - Problem

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Problem

Source: Belarusian olympiad 2023

Tags: geometry

nAalniaOMliO

13.05.2024 20:43

Point $D$ is the midpoint of $BC$ , where $ABC$ is an isosceles triangle ( $AB=AC$ ). On circle $(ABD)$ a point $P \neq A$ is chosen. $O$ is the circumcenter of $ACP$ , $Q$ is the foot of the perpendicular from $C$ onto $AO$ . Prove that the circumcenter of triangle $ABQ$ lies on the line $AP$

NO_SQUARES

13.05.2024 22:07

I think it is this problem.

PROF65

14.05.2024 01:15

Let $M$ midpoint of $AC$ ; $CQ$ cuts $AP$ at $R$ ; we have $\angle ACR=\angle AOM=\angle APC$ hence $AC$ is tangent to the circle $(PRC)$ at $C$ so $AP.AR=AC^2=AB^2$ thus $AB$ is tangent to the circle $(BRP)$ implies $\angle ABR=\angle APB =90^ {\circ}$ thus $AQBR$ is cyclic and then its center is on $AP$ . Best regards. RH HAS