HMMT-Feb 2018 guts, Problem 12

$\triangle P N R$ 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 $\frac{P A}{A N}$.

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 $\frac{P A}{A N}=\frac{P R}{R N}=\frac{19}{18}$ $\fbox{\frac{19}{18}}$.

Answer: \frac{19}{18}