Track / Stage 5 / 336 of 400 #936 of 2000

Problem 936

AIME late

Geometry Difficulty 5.9 Prove it

\section*{Problem 14}

$\mathrm{ABCD}$ is a parallelogram. The excircle of $\mathrm{ABC}$ opposite $\mathrm{A}$ has center $\mathrm{E}$ and touches the line $\mathrm{AB}$ at $\mathrm{X}$ . The excircle of ADC opposite $\mathrm{A}$ has center $\mathrm{F}$ and touches the line $\mathrm{AD}$ at $\mathrm{Y}$ . The line $\mathrm{FC}$ meets the line $\mathrm{AB}$ at $\mathrm{W}$ , and the line $\mathrm{EC}$ meets the line $\mathrm{AD}$ at $\mathrm{Z}$ . Show that $\mathrm{WX}=\mathrm{YZ}$ .

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

\section*{Solution}

We have the familiar result that AY is perimeter ADC (chase round using the fact that the two tangents from the same point have the same length). Similarly, $\mathrm{AX}=$ perimeter $\mathrm{ABC}=$ perimeter $\mathrm{ADC}$ . So $\mathrm{AX}=\mathrm{AY}(*)$

$\mathrm{AE}$ is parallel to the bisector of $\mathrm{ACD}$ , which is perpendicular to $\mathrm{CF}$ . So $\mathrm{CW}$ is perpendicular to AE. Hence AW = AC. Similarly AZ = AC. Hence AW = AZ. Subtracting from $\left({ }^{*}\right)$ gives result.

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