HMMT February

Geometry Difficulty 5.3 AIME, harder Find the answer

Triangle $\triangle P N R$ has side lengths $P N=20, N R=18$ , and $P R=19$ . Consider a point $A$ on $P N$ . $\triangle N R A$ is rotated about $R$ to $\triangle N^{\prime} R A^{\prime}$ so that $R, N^{\prime}$ , and $P$ lie on the same line and $A A^{\prime}$ is perpendicular to $P R$ . Find $\frac{P A}{A N}$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Denote the intersection of $P R$ and $A A^{\prime}$ be $D$ . Note $R A^{\prime}=R A$ , so $D$ , being the altitude of an isosceles triangle, is the midpoint of $A A^{\prime}$ . Thus, $\angle A R D=\angle A^{\prime} R D=\angle N R A$ so $R A$ is the angle bisector of $P N R$ through $R$ . By the angle bisector theorem, we have $\frac{P A}{A N}=\frac{P R}{R N}=\frac{19}{18}$ .

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