Track / Stage 4 / 305 of 340 #565 of 2000

Problem 565

AMC 12 late, AIME early

Geometry Difficulty 5.0 Find the answer

12. [7] $\triangle P N R$ 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

Official solution

Answer: $\square$
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 $\frac{P A}{A N}=\frac{P R}{R N}=\frac{19}{18}$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.