OA=OB if <PAD = <ADP=< CBP =< PCB =< CPD

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Source: 2024 Czech and Slovak Olympiad III A p2

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#1 h May 18, 2024, 7:16 AM • 1 Y

Y by GeoKing

Let the interior point $P$ of the convex quadrilateral $ABCD$ be such that $|\angle PAD| = |\angle ADP| = |\angle CBP| = |\angle PCB| = |\angle CPD|.$ Let $O$ be the center of the circumcircle of the triangle $CPD$ . Prove that $|OA| = |OB|$ .

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gnoka

#2 h May 18, 2024, 2:38 PM • 2 Y

Y by GeoKing, soryn

From ${\angle}\mathrm{BPC} = {\angle}\mathrm{APD}$ , we obtain ${\angle}\mathrm{BPD} = {\angle}\mathrm{APC}$ . Additionally, since PB=PC and PD=PA, triangles PBD and PCA are congruent, leading to ${\angle}\mathrm{PBD} = {\angle}\mathrm{PCA}$ .

Observing that BC is parallel to PD, we have ${\angle}\mathrm{CBD} = {\angle}\mathrm{BDP}$ .So from $\angle PBD+\angle BDO$ $={\angle}\mathrm{PBC} - {\angle}\mathrm{CBD} + {\angle}\mathrm{BDP} + {\angle}\mathrm{PDO} = {\angle}\mathrm{PBC} + {\angle}\mathrm{PDO} = {\angle}\mathrm{CPD} + {\angle}\mathrm{DPO}$
$= {\angle}\mathrm{OPC} = {\angle}\mathrm{OCP}$ . And since ${\angle}\mathrm{OCP} = {\angle}\mathrm{OCA} + {\angle}\mathrm{PCA}$ , we find ${\angle}\mathrm{BDO} = {\angle}\mathrm{OCA}$ .
Given that BD=AC and OD=OC, triangle BDO is congruent to triangle ACO. Therefore, we have BO = AO.
Q.E.D.

This post has been edited 2 times. Last edited by gnoka, May 18, 2024, 2:39 PM

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#3 h Mar 30, 2025, 10:47 AM • 1 Y

Y by soryn

Problem is equivalent to the following problem.

Let $ABC$ be an acute triangle and $X$ and $Y$ are the points such that $CA=CX$ , $CB=CY$ and $AX \parallel CB$ , $BY \parallel AC$ . Let $O$ be the circumcenter of $(ABC)$ . Then prove that $OX=OY$ .

Let $D$ and $E$ be the second intersection points of $XA$ , $YB$ with $(ABC)$ . Since $\triangle XCD \sim \triangle CAB \sim \triangle YCE$ we have will have that $\frac{XD}{YE}=\frac{CB}{CA}$ . Also from $\triangle CXA \sim \triangle CYB$ we get that $\frac{XA}{BY}=\frac{CA}{CB}$ which implies that $pow(X, (ABC))=XA*XD=YE*YB=pow(Y,(ABC))$ $\implies$ $OX=OY$ .

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