2015 IMO Problems/Problem 3

Let $ABC$ be an acute triangle with $AB>AC$ . Let $\Gamma$ be its circumcircle, $H$ its orthocenter, and $F$ the foot of the altitude from $A$ . Let $M$ be the midpoint of $BC$ . Let $Q$ be the point on $\Gamma$ such that $\angle HKQ=90^\circ$ . Assume that the points $A$ , $B$ , $C$ , $K$ , and $Q$ are all different, and lie on $\Gamma$ in this order.

Prove that the circumcircles of triangles $KQH$ and $FKM$ are tangent to each other.

The Actual Problem

Let $ABC$ be an acute triangle with $AB > AC$ . Let $\gamma$ be its circumcircle, $H$ its orthocenter, and $F$ the foot of the altitude from $A$ . Let $M$ be the midpoint of $BC$ . Let $Q$ be the point on $\gamma$ such that $\angle HQA = 90◦$ and let $K$ be the point on $\gamma$ such that $\angle HKQ = 90◦$ . Assume that the points $A$ , $B$ , $C$ , $K$ and $Q$ are all different and lie on $\gamma$ in this order. Prove that the circumcircles of triangles $KQH$ and $FKM$ are tangent to each other.

Solution

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2015 IMO (Problems) • Resources
Preceded by Problem 2	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 4
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2015 IMO Problems/Problem 3

The Actual Problem

Solution

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