Library / /6 of 397

, 2021

Geometry Difficulty 4.4 AIME Prove it Taiwan

Let $I$ be the incenter of triangle $ABC$ , and let $D$ be the foot of the perpendicular from $I$ to side $BC$ . Let $D'$ be the reflection of $D$ with respect to $I$ , and suppose $\overline{AD'} = \overline{ID'}$ . With $D'$ as center, draw a circle $\Gamma$ passing through $A$ , $I$ , meeting $AB$ , $AC$ at $X$ , $Y$ . Let $Z$ be a point on $\Gamma$ such that $AZ$ is perpendicular to $BC$ .
Prove that $AD$ , $D'Z$ , $XY$ are concurrent.

Solution

Solution. Note that $AI$ bisects $\angle XAY$ , so $D'I$ is perpendicular to $XY$ , hence $\triangle AXY$ is similar (homothetic) to $\triangle ABC$ . Let $E$ be the foot of the perpendicular from $I$ to $CA$ , and let $I'$ be the incenter of $\triangle AXY$ ; note that $AD$ meets $XY$ at the point $T$ , the foot of the perpendicular from $I'$ to $XY$ , so it suffices to show that $D'$ , $T$ , $Z$ are collinear. From $\overline{IY} = \overline{II'}$ , $\overline{IE} = \overline{ID'}$ , and
$\angle YIE = \frac{1}{2} |\angle B - \angle C| = \angle D'II'$
we get $\triangle YIE \sim \triangle I'ID'$ , so $I'D' \parallel XY$ . Let $M$ , $M'$ be the midpoints of $\overline{BC}$ , $\overline{XY}$ respectively; then $I'TM'D'$ is a rectangle, and it is well known that $AD' \parallel IM \parallel I'M'$ , from which we get
$\angle ITD' = \angle IM'D' = \angle D'AZ = \angle AZD'$
Combined with $IT \parallel AZ$ , this gives that $D'$ , $T$ , $Z$ are collinear.

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 translated into English from zh; metadata (topic, difficulty) added by this project.