Library / /492 of 1394

, 2018

Geometry Difficulty 5.1 AIME, harder Prove it United States

Problem:

$\triangle P N R$ 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 $\triangle N' R A'$ , so that $R$ , $N'$ , and $P$ lie on the same line and $A A'$ is perpendicular to $P R$ . Find $\frac{P A}{A N}$ .

Solution

Solution:

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

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.