Baltic Way 2025 · Problem 14

Geometry

In an acute triangle $ABC$ with $AB<AC$ , the circumcircle is $\omega$ and the orthocentre is $H$ . The point $P$ on major arc $BC$ of $\omega$ satisfies $\angle PCB+\angle ACB=90^\circ$ . The point $R$ on segment $AC$ satisfies $BR=CR$ . Prove that $\angle HPR=\angle ACB$ .

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Topics

Circles and tangency · Cyclic geometry · Triangles and centers

Solutions

Solution

Let $BH$ meet $\omega$ again at $P'$ . Then

\angle CBP'=90^\circ-\angle ACB=\angle PCB,

so $PBCP'$ is an isosceles trapezoid.

By symmetry, triangles $BPR$ and $CP'R$ are congruent, while triangles $CRH$ and $CRP'$ are mirror images with respect to $AC$ . Therefore there is a rotation sending $\triangle BRP$ to $\triangle CRH$ . Hence $\triangle PRH\sim\triangle CBR$ , and so

\angle HPR=\angle ACB.

Contest context

Results from Baltic Way 2025

11 teams

Mean score: 1.8 / 5
Scores of 4 or 5: 3 / 11
Estonia: 5 / 5

Score distribution

All team scores

Team	Score
Germany	5 / 5
Estonia	5 / 5
Poland	0 / 5
Lithuania	1 / 5
Norway	1 / 5
Latvia	0 / 5
Finland	5 / 5
Denmark	1 / 5
Sweden	0 / 5
Ukraine	1 / 5
Iceland	1 / 5

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