Library / Taiwan IMO Selection Camp /68 of 359

Taiwan IMO Selection Camp

Geometry Difficulty 5.2 AIME, harder Prove it Taiwan

In triangle $ABC$ , $AB = AC \ne BC$ , and point $I$ is its incenter. Line $BI$ meets $AC$ at point $D$ . It is known that the line through $D$ perpendicular to $AC$ meets $AI$ at point $E$ . Prove that: the reflection of $I$ across line $AC$ lies on the circumcircle of triangle $BDE$ .

Solution

Let $\Gamma$ be the circle centered at $E$ passing through the two points $B$ , $C$ . Since $DE \perp AC$ , the reflection $F$ of $C$ across $D$ lies on $\Gamma$ . From $\angle DCI = \angle ICB = \angle CBI$ , we know that line $DC$ is tangent to the circumcircle of triangle $IBC$ . Let $J$ be the reflection of $I$ across $D$ . Using directed segments, we know
$DC \cdot DF = -DC^2 = -DI \cdot DB = DJ \cdot DB,$
so we get that $J$ also lies on $\Gamma$ .

Let $I'$ be the reflection of $I$ across $AC$ . Since $IJ$ and $CF$ bisect each other, $CJFI$ is a parallelogram. From $\angle FI'C = \angle CIF = \angle FJC$ , we get that $I'$ lies on $\Gamma$ . From this we know $EI' = EB$ .

Note that $AC$ is the internal angle bisector of $\angle BDI'$ . Since $DE \perp AC$ , $DE$ is the external angle bisector of $\angle BDI'$ . Combined with $EI' = EB$ , we know that $E$ lies on the circumcircle of triangle $BDI'$ . This completes the proof.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge 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.