1990 USAMO Problems/Problem 5


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 $90^{\circ}$, $A'$ 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.

Solution 3 (Radical Lemma)

Let $\omega_1$ be the circumcircle with diameter $AB$ and $\omega_2$ be the circumcircle with diameter $AC$. We claim that the second intersection of $\omega_1$ and $\omega_2$ other than $A$ is $A'$, where $A'$ is the feet of the perpendicular from $A$ to segment $BC$. Note that \[\angle AA'B=90^{\circ}=\angle AB'B\] so $A'$ lies on $\omega_1.$ Similarly, $A'$ lies on $\omega_2$. Hence, $AA'$ is the radical axis of $\omega_1$ and $\omega_2$. By the Radical Lemma, it suffices to prove that the intersection of lines $MN$ and $PQ$ lie on $AA'$. But, $MN$ is the same line as $CC'$ and $PQ$ is the same line as $BB'$. Since $AA', BB'$, and $CC'$ intersect at the orthocenter $H$, $H$ lies on the radical axis $AA'$ and we are done. $\blacksquare$

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

See Also

1990 USAMO (ProblemsResources)
Preceded by
Problem 4
Followed by
Last Question
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All USAMO Problems and Solutions

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