1990 USAMO Problems/Problem 5

Problem

An acute-angled triangle $ABC$ is given in the plane. The circle with diameter $\, AB \,$ intersects altitude $\, CC' \,$ and its extension at points $\, M \,$ and $\, N \,$ , and the circle with diameter $\, AC \,$ intersects altitude $\, BB' \,$ and its extensions at $\, P \,$ and $\, Q \,$ . Prove that the points $\, M, N, P, Q \,$ lie on a common circle.

Solution

Let $A'$ be the intersection of the two circles. $AA'$ is perpendicular to both $BA'$ , $CA'$ implying $B$ , $C$ , $A'$ are collinear. Since $A'$ is the foot of the altitude from $A$ : $A$ , $H$ , $A'$ are concurrent, where $H$ is the orthocentre.

Now, $H$ is also the intersection of $BB'$ , $CC'$ which means that $AA'$ , $MN$ , $PQ$ are concurrent. Since $A$ , $M$ , $N$ , $A'$ and $A$ , $P$ , $Q$ , $A'$ are cyclic, $M$ , $N$ , $P$ , $Q$ are cyclic by the radical axis theorem.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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1990 USAMO (Problems • Resources)
Preceded by Problem 4	Followed by Final Question
1 • 2 • 3 • 4 • 5
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1990 USAMO Problems/Problem 5

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