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Belarus — Selection and Training Session

Geometry Difficulty 4.5 AIME Prove it Belarus

Let the incircle of the triangle $ABC$ touch the side $AB$ at point $Q$ ; the incircles of the triangles $QAC$ and $QBC$ touch $AQ$ , $AC$ and $BQ$ , $BC$ at points $P$ , $T$ and $D$ , $F$ , respectively.
Prove that $PDFT$ is a cyclic quadrilateral.

Solution

(Solution by A. Gaponenko, D. Voynov.) First, note that incircles of the triangles $QAC$ and $QBC$ touch $CQ$ at the same point $X$ (well-known fact). Hence $CF = CX = CT$ . Also $AP = AT$ , $BF = BD$ . Now we have
$\angle TFD = 180^\circ - \angle TFC - \angle BFD =$
$= 180^\circ - (90^\circ - \frac{1}{2} \angle C) - (90^\circ - \frac{1}{2} \angle B) =$
$= \frac{1}{2}(\angle C + \angle B) = 90^\circ - \frac{1}{2} \angle A = \angle APT.$
It follows that
$\angle TFD + \angle TPD = \angle TFD + (180^\circ - \angle APT) = 180^\circ$
which finishes the proof.

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