Library / Saudi Arabian Mathematical Competitions /25 of 395

Saudi Arabian Mathematical Competitions

Geometry Difficulty 4.6 AIME Prove it Saudi Arabia

Let $ABC$ be a triangle with $AB \neq AC$ . The incircle of triangle $ABC$ is tangent to $BC$ , $CA$ , $AB$ at $D$ , $E$ , $F$ , respectively. The perpendicular line from $D$ to $EF$ intersects $AB$ at $X$ . The second intersection point of circumcircles of triangles $AEF$ and $ABC$ is $T$ . Prove that $TX \perp TF$ .

Solution

Let $G$ be the intersection point of $DX$ and $EF$ .

Consider the inversion $I_{I}^{ID^{2}}$ , then the circumcircle of $\triangle ABC$ is sent to the nine-point circle of $\triangle DEF$ and the line $EF$ is sent to the circumcircle of $\triangle AEF$ . Hence $I$ , $T$ , and $G$ are collinear.

Then $AI \parallel GX$ since they are both perpendicular to $EF$ . We have
$\angle GTF = \angle ITF = \angle IFG = \angle IAE = \angle FAI = \angle FXG$
so, quadrilateral $TXGF$ is cyclic, this implies that
$\angle XTF = \angle XGF = 90^{\circ}.$
$\square$

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 reproduced verbatim; metadata (topic, difficulty) added by this project.