Baltic Way 2010 · Problem 11

Geometry

Let $A B C D$ be a square and let $S$ be the point of intersection of its diagonals $A C$ and $B D$ . Two circles $k, k^{\prime}$ go through $A, C$ and $B, D$ ; respectively. Furthermore, $k$ and $k^{\prime}$ intersect in exactly two different points $P$ and $Q$ . Prove that $S$ lies on $P Q$ .

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Topics

Cyclic geometry

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Solution

It is clear that $PQ$ is the radical axis of $k$ and $k'$ . The power of $S$ with respect to $k$ is $-|AS| \cdot |CS|$ and the power of $S$ with respect to $k'$ is $-|BS| \cdot |DS|$ . Because $ABCD$ is a square, these two numbers are clearly the same. Thus, $S$ has the same power with respect to $k$ and $k'$ and lies on the radical axis $PQ$ of $k$ and $k'$ .

Contest context

Results from Baltic Way 2010

10 teams

Mean score: 4.9 / 5
Scores of 4 or 5: 10 / 10
Estonia: 5 / 5

Score distribution

All team scores

Team	Score
Poland	5 / 5
Lithuania	5 / 5
Germany	5 / 5
Latvia	5 / 5
Denmark	4 / 5
Sweden	5 / 5
Estonia	5 / 5
Norway	5 / 5
Finland	5 / 5
Iceland	5 / 5

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