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Saudi Arabian IMO Booklet

Geometry Difficulty 8.0 Shortlist Prove it Saudi Arabia

Let $BB'$ , $CC'$ be the altitudes of an acute-angled triangle $ABC$ . Two circles passing through $A$ and $C'$ are tangent to $BC$ at points $P$ and $Q$ . Prove that $A$ , $B'$ , $P$ , $Q$ are concyclic.

Solution

Since $BP^2 = BQ^2 = BA \cdot BC'$ and the quadrilaterals $AC'A'C$ , $AB'A'B$ are cyclic ( $AA'$ is the altitude) we have
$CP \cdot CQ = CB^2 - BP^2 = CB^2 - BA \cdot BC' = BC^2 - BC \cdot BA' = BC \cdot CA' = CA \cdot CB'$
Clearly this is equivalent to the required assertion. $\square$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.