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, 1997

Geometry Difficulty 7.9 National Olympiad, round 2 Prove it Hong Kong

Let $\theta$ be a fixed angle. Let two circles touch each other internally at point $A$ . Let $AB$ be a chord of the outer circle and such that $AB$ is not a diameter. Let $M$ be a variable point on the major arc $AB$ of the outer circle. $MA$ meets the inner circle at $N$ . $P$ is a point on the segment $MB$ such that $\angle MPN = \theta$ . Determine the locus of $P$ while $M$ varies.

Solution

Let $AB$ meet the inner circle at $C$ . Let $D$ be the point on the tangent at $B$ to the outer circle such that $\angle BDC = \theta$ , and $D$ lies on opposite side of $AB$ as $M$ . We claim that the locus of $P$ is $\text{ABD}$ .

Firstly, since $A$ is the homothetic centre of the two circles, we have $NC \parallel MB$ . It follows that $\frac{AN}{NM} = \frac{AC}{CB}$ . Secondly, since $\angle NMP = \angle CBD$ by the tangent and $\angle MPN = \angle BDC = \theta$ , we have $\triangle MNP \sim \triangle BCD$ . Note that the triangles are directly similar. Therefore, $MNAP$ is similar to $BCAD$ . Thus, $A$ is the centre of spiral similarity mapping $MP$ to $BD$ . It follows that $MB \cap PD = P$ lies on $(ABD)$ , which is fixed.

It remains to notice that $P$ approaches $A$ when $M$ approaches $A$ , and $P$ approaches $D$ when $M$ approaches $B$ . Due to continuity, the locus of $P$ is $\text{ABD}$ from $A$ to $D$ .

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