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 1

Let $A'$ be the intersection of the two circles (other than $A$ ). $AA'$ 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.

Solution 2

Define $A'$ as the foot of the altitude from $A$ to $BC$ . Then, $AA' \cap BB' \cap CC'$ is the orthocenter. We will denote this point as $H$ Since $\angle AA'C$ and $\angle AA'B$ are both $90^{\Circle}$ , $A'$ lies on the circles with diameters $AC$ and $AB$ .

Now we use the Power of a Point theorem with respect to point $H$ . From the circle with diameter $AB$ we get $AH \cdot A'H = MH \cdot NH$ . From the circle with diameter $AC$ we get $AH \cdot A'H = PH \cdot QH$ . Thus, we conclude that $PH \cdot QH = MH \cdot NH$ , which implies that $P$ , $Q$ , $M$ , and $N$ all lie on a circle.

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

1990 USAMO (Problems • Resources)
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1990 USAMO Problems/Problem 5

Contents

Problem

Solution 1

Solution 2

See Also