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All-Soviet-Union Mathematical Olympiad

Geometry Difficulty 5.4 AIME, harder Prove it Soviet Union

Problem:

In the triangle $ABC$ , the angle $C$ is obtuse and $D$ is a fixed point on the side $BC$ , different from $B$ and $C$ . For any point $M$ on the side $BC$ , different from $D$ , the ray $AM$ intersects the circumcircle $S$ of $ABC$ at $N$ . The circle through $M$ , $D$ and $N$ meets $S$ again at $P$ , different from $N$ . Find the location of the point $M$ which minimises $MP$ .

Solution

Solution:

Take $A'$ on the circle $S$ such that $AA'$ is parallel to $BC$ . Let the ray $AD$ meet $S$ again at $P'$ . Then $\angle MNP' = \angle ANP'$ (same angle) $= \angle AA'P'$ ( $A'AP'N$ cyclic) $= \angle A'DB$ ( $BC$ parallel to $AA'$ ) $= \angle MDP$ (opposite angles). So $MDNP'$ is cyclic, so $P$ must be $P'$ . Since $P$ is a fixed point, independent of $M$ , we minimise $MP$ by taking $M$ as the foot of the perpendicular from $P$ to $BC$ .

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