Mathematics Asked by David Dong on November 22, 2020

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. (USAMO, 1990)

$MPNQ$ is cyclic if and only if $CBC’B’$ is cyclic. This is because $MPNQ$ is cyclic if and only if $MXcdot XN=PXcdot XQ$ by the power of $X$ on the circle whose existence is to be proven. Notice that $B’$ is on the circle with diameter $AB$, because $angle AB’B=90^{circ}$, thus $B’Xcdot XB=MXcdot XN.$

Also, $PXcdot XQ=CXcdot XC’,$ because $C’$ must be on the circle with diameter $AC$. Therefore, $MXcdot NX=PXcdot QXLongleftrightarrow B’Xcdot XB=CXcdot XC’$, which happens if and only if $CBC’B’$ is cyclic.

$CBC’B’$ is cyclic. This is because $angle B’CC’=angle C’BB’$, since $triangle ACC’simtriangle ABB’.$ Therefore, $MPNQ$ is also cyclic.

I think my solution is likely correct, but my handout gives a different solution, and I’m not very familiar with geometry. Is there anything that I’m missing?

Your solution is correct though it could be made neater.

Let circle with diameter AC be $omega_C$ and circle with diameter AB be $omega_B$. As the diameter subtends a right angle on the semicircle, $C'$ lies on $omega_C$ and $B'$ lies on $omega_B$.

Now

- Power of orthocenter $X$ wrt $omega_C$ is $$CXcdot C'X = PXcdot QX$$
- Power of orthocenter $X$ wrt $omega_B$ is $$BXcdot B'X = MXcdot NX$$
- Since $$angle BB'C = 90^{circ} = angle BC'C$$ $BC'B'C$ is cyclic.

Power of X wrt circumcircle of $BC'B'C$ is $$CXcdot C'X = BXcdot B'X$$

from which follows $$PXcdot QX = MXcdot NX$$

Hence proved, $MPNQ$ is cyclic.

Answered by cosmo5 on November 22, 2020

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