USA Tstst

Geometry Difficulty 7.3 National Olympiad, round 2 Prove it United States

Let $ABC$ be a scalene triangle with incenter $I$ . The incircle of $ABC$ touches $BC$ , $CA$ , $AB$ at points $D$ , $E$ , $F$ , respectively. Let $P$ be the foot of the altitude from $D$ to $EF$ , and let $M$ be the midpoint of $BC$ . The rays $AP$ and $IP$ intersect the circumcircle of triangle $ABC$ again at points $G$ and $Q$ , respectively. Show that the incenter of triangle $GQM$ coincides with $D$ .

Solution

Refer to the figure below.

Claim — The point $Q$ is the Miquel point of $BFEC$ . Also, $\overline{QD}$ bisects $\angle BQC$ .
Proof. Inversion around the incircle maps line $EF$ to $(AIEF)$ and the nine-point circle of $\triangle DEF$ to the circumcircle of $\triangle ABC$ (as the midpoint of $EF$ maps to $A$ , etc.). This implies $P$ maps to $Q$ ; that is, $Q$ coincides with the second intersection of $(AFIE)$ with $(ABC)$ . This is the claimed Miquel point.
The spiral similarity mentioned then gives $\frac{QB}{BF} = \frac{QC}{CE}$ , so $\overline{QD}$ bisects $\angle BQC$ . $\square$

Claim — We have $(QG; BC) = -1$ , so in particular $\overline{GD}$ bisects $\angle BGC$ .
Proof. Note that
$-1 = (AI; EF) \stackrel{Q}{=} (\overline{AQ} \cap \overline{EF}, P; E, F) \stackrel{A}{=} (QG; BC).$
The last statement follows from Apollonian circle, or more bluntly $\frac{GB}{GC} = \frac{QB}{QC} = \frac{BD}{DC}$ . $\square$

Hence $\overline{QD}$ and $\overline{GD}$ are angle bisectors of $\angle BQC$ and $\angle BGC$ . However, $\overline{QM}$ and $\overline{QG}$ are isogonal in $\angle BQC$ (as median and symmedian), and similarly for $\angle BGC$ , as desired.

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