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Geometry Difficulty 6.4 National olympiad Prove it Belarus

Given a right-angled triangle $ABC$ with $\angle C = 90^\circ$ .

Triangle $AMN$ equal to $ABC$ is constructed on the hypotenuse of $ABC$ , $\angle ANM = 90^\circ$ , $AN = BC$ (see the fig.). The incircle $\Gamma_1$ of triangle $AMN$ touches the hypotenuse $AM$ at point $P$ , and the incircle $\Gamma_2$ of triangle $ABC$ touches the side $BC$ at point $Q$ .
Prove that the segment $PQ$ , the hypotenuse $AB$ , and the segment connecting the centers of $\Gamma_1$ and $\Gamma_2$ are concurrent.

Solution

Note that $\angle MAN = 90^\circ - \angle BAC = \angle ABC$ . So $MN = AC$ ,

Fig. 1

Fig. 2

$AN = BC$ and, moreover, $AP = BQ$ . Let $R$ be the midpoint of $AB$ . Then $\triangle APR = \triangle BQR$ ( $AP = BQ$ , $AR = BR$ and $\angle PAR = \angle QBR$ ). So

$\angle PRA = \angle BRQ$ , hence the angles $PRA$ and $BRQ$ are vertical. Therefore, $PQ$ meets $AB$ at point $R$ (see fig. 1). On the other hand, let $T$ be an intersection point of $AB$ and the segment connecting the centers $Q_1$ and $Q_2$ of $\Gamma_1$ and $\Gamma_2$ . Then $\angle O_1AT = 0.5\angle MAN = 0.5\angle ABC = \angle TBO_2$ , which gives $AO_1 \parallel BO_2$ . Since, moreover, $AO_1 = BO_2$ , we conclude that $AO_1BO_2$ is a parallelogram. Since $T$ is a point of intersection of its diagonals we have $AT = TB$ . It follows that $T$ is the midpoint of $AB$ , i.e. points $T$ and $R$ coincide (see fig. 2). Thus, all three segments $AB$ , $PQ$ , $O_1O_2$ meet at the midpoint of $AB$ .

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