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All-Soviet-Union Mathematical Olympiad

Geometry Difficulty 4.6 AIME Prove it Soviet Union

Problem:
Given a circle $c$ and two fixed points $A$ , $B$ on it. $M$ is another point on $c$ , and $K$ is the midpoint of $BM$ . $P$ is the foot of the perpendicular from $K$ to $AM$ .

a. Prove that $KP$ passes through a fixed point (as $M$ varies).

b. Find the locus of $P$ .

Solution

Solution:

a.
Take $Y$ on the circle so that $\angle ABY = 90^\circ$ . Then $AY$ is a diameter and so $\angle AMY = 90^\circ$ . Take $X$ as the midpoint of $BY$ . Then triangles $BXK$ and $BYM$ are similar, so $XK$ is parallel to $YM$ . Hence $XK$ is perpendicular to $AM$ , and so $P$ is the intersection of $XK$ and $AM$ . In other words, $KP$ always passes through $X$ .

b.
$P$ must lie on the circle diameter $AX$ , and indeed all such points can be obtained (given a point $P$ on the circle, take $M$ as the intersection of $AP$ and the original circle). So the locus of $P$ is the circle diameter $AX$ .

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