China Western Mathematical Olympiad

Question

Let D be a point on the side BC of an acute triangle ABC. The circle with diameter BD meets the lines AB and AD respectively at the points X and P, which are different from the points B and D. The circle with diameter CD meets the lines AC and AD respectively at the points Y and Q, which are different from the points C and D. Through the point A draw two lines which are perpendicular to PX and QY with the feet of perpendicular M and N respectively. Prove that AMN ABC if and only if the line AD passes through the circumcenter of ABC. (Posed by Li Qiusheng) FIGURE_0 FIGURE_1

Accepted Answer

**Proof** Join the segments XY and DX. It follows from the given conditions that B, P, D, X are concyclic, and C, Y, Q, D are concyclic. Then AXM = BXP = BDP = QDC = AYN. FIGURE_1 And it follows from AMX = ANY = 90^ that AMC ANY, so MAX = NAY and AMAX = ANAY, and hence MAN = XAY. Combining the two results above, we have AMN AXY. So we get aligned AMN ABC & AXY ABC & XY BC DXY = XDB. aligned --- As A, X, D, Y are concyclic, we have DXY = DAY. As XDB = 90^ - ABC, DXY = XDB DAC = 90^ - ABC, which is equivalent to the fact that the line AD passes through the circumcenter of ABC. Hence, AMN ABC if and only if AD passes through the circumcenter of ABC.