Baltic Way 2024 · Problem 13

Geometry

Let $A B C$ be an acute triangle with orthocentre $H$ . Let $D$ be a point outside the circumcircle of triangle $A B C$ such that $\angle A B D=\angle D C A$ . The reflection of $A B$ in $B D$ intersects $C D$ at $X$ . The reflection of $A C$ in $C D$ intersects $B D$ at $Y$ . The lines through $X$ and $Y$ perpendicular to $A C$ and $A B$ , respectively, intersect at $P$ . Prove that points $D, P$ and $H$ are collinear.

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Topics

Circles and tangency · Cyclic geometry · Triangles and centers

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Solution

From the reflections, we have

\angle D B X=180^{\circ}-\angle D B A=180^{\circ}-\angle D C A=\angle D C Y

(Fig. 15), so points $B, C, X, Y$ are concyclic. Define $Q=X P \cap B D$ and $R=Y P \cap C D$ (Fig. 16). Then due to the right angles, we find $\angle D Y R=\angle D X Q$ . Hence points $Q, R, X, Y$ are concyclic, too. Consequently, $\angle D Q R=\angle D X Y=\angle D B C$ , so $B C \| Q R$ . Since also $B H \| P Q$ and $C H \| P R$ , it follows that triangle $B H C$ is a homothetic image of triangle $Q P R$ with center $D$ . Hence $D, P$ and $H$ are collinear. Official solution diagram for Baltic Way 2024 Problem 13 (Figure 15).

Figure 15 Official solution diagram for Baltic Way 2024 Problem 13 (Figure 16).

Figure 16

Contest context

Results from Baltic Way 2024

11 teams

Mean score: 2.0 / 5
Scores of 4 or 5: 3 / 11
Estonia: 5 / 5

Score distribution

All team scores

Team	Score
Poland	5 / 5
Estonia	5 / 5
Germany	2 / 5
Ukraine	5 / 5
Latvia	1 / 5
Norway	1 / 5
Lithuania	1 / 5
Sweden	1 / 5
Denmark	0 / 5
Finland	1 / 5
Iceland	0 / 5

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