Belorusija 2012

Question

Given a right-angled triangle ABC with C = 90^. FIGURE_0 Triangle AMN equal to ABC is constructed on the hypotenuse of ABC, ANM = 90^, AN = BC (see the fig.). The incircle _1 of triangle AMN touches the hypotenuse AM at point P, and the incircle _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 _1 and _2 are concurrent. FIGURE_1 FIGURE_2

Accepted Answer

Note that MAN = 90^ - BAC = ABC. So MN = AC, FIGURE_1 Fig. 1 FIGURE_2 Fig. 2 AN = BC and, moreover, AP = BQ. Let R be the midpoint of AB. Then APR = BQR (AP = BQ, AR = BR and PAR = QBR). So PRA = 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 _1 and _2. Then O_1AT = 0.5 MAN = 0.5 ABC = TBO_2, which gives AO_1 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.