Baltic Way 2021 · Problem 12

Geometry

Let $I$ be the incentre of a triangle $A B C$ . Let $F$ and $G$ be the projections of $A$ onto the lines $B I$ and $C I$ , respectively. Rays $A F$ and $A G$ intersect the circumcircles of the triangles $C F I$ and $B G I$ for the second time at points $K$ and $L$ , respectively. Prove that the line $A I$ bisects the segment $K L$ .

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Topics

Angles and distances · Triangles and centers

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Solution

Since $\angle IFK = 90^\circ$ , then $IK$ is the diameter of the circumcircle of $CFI$ , hence also $\angle ICK = 90^\circ$ . Similarly is $IL$ the diameter of the circumcircle of $BGI$ and $\angle IBL = 90^\circ$ . Therefore are the lines $CK$ and $GL$ parallel, also $BL$ and $FK$ are parallel.

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Results from Baltic Way 2021

12 teams

Mean score: 2.6 / 5
Scores of 4 or 5: 5 / 12
Estonia: 5 / 5

Score distribution

All team scores

Team	Score
St. Petersburg	5 / 5
Estonia	5 / 5
Germany	2 / 5
Latvia	5 / 5
Lithuania	2 / 5
Poland	5 / 5
Denmark	0 / 5
Norway	5 / 5
Finland	0 / 5
Sweden	0 / 5
Iceland	2 / 5
Ireland	0 / 5

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