Track / Stage 5 / 35 of 400 #635 of 1964

Problem 635

AIME late

Geometry Difficulty 5.0 Prove it Autumn Tournament · Bulgaria

Given is an acute triangle $ABC$ with incenter $I$ and the incircle touches $BC$ , $CA$ , $AB$ at $D$ , $E$ , $F$ . The circle with center $C$ and radius $CE$ meets $EF$ for the second time at $K$ . If $X$ is the $C$ -excircle touchpoint with $AB$ , show that $CX$ , $KD$ , $IF$ concur.
(Kristyan Vasilev)

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

We claim the concurrency point is the $F$ -antipode $F'$ . It is well-known that this is $\overline{CX} \cap \overline{IF}$ .
Let $P = \overline{EF} \cap \overline{DF'}$ and $Q = \overline{DF} \cap \overline{EF'}$ . Then since $\angle PEQ = \angle PDQ = 90^\circ$ , $DEPQ$ is cyclic.
Now, we have
$\begin{aligned} \angle EFD &= \angle EFF' = 90^\circ - \angle PF'E = 90^\circ - \angle DFE \\ &= -90^\circ + \left( 90^\circ - \frac{\angle A}{2} \right) + \left( 90^\circ - \frac{\angle B}{2} \right) = \frac{\angle C}{2}, \end{aligned}$
so the center of ( $DEPQ$ ) lies on ( $CDE$ ) (and is on the same side of $\overline{DE}$ as $C$ ). On the other hand, it also lies on the perpendicular bisector of $\overline{DE}$ , so it must be $C$ itself. This gives us $P = K$ , and the conclusion follows. $\square$

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