IMO 1995 SL G1

Origin: BUL | Category: Geometry

Problem

Let A, B, C, and D be distinct points on a line, in that order. The circles with diameters AC and BD intersect at X and Y . O is an arbitrary point on the line XY but not on AD. CO intersects the circle with diameter AC again at M, and BO intersects the other circle again at N. Prove that the lines AM, DN, and XY are concurrent.

Solution

The result is trivial if O coincides with X or Y , so let us assume it does not. From OB \cdot ON = OC \cdot OM = OX \cdot OY we deduce that BCMN is a cyclic quadrilateral. Further, if O lies between X and Y , then \angleMAD + \angleMND = \angleMAD +\angleMNB +\angleBND = \angleMAD +\angleMCA+\angleAMC = 180◦. Similarly, we also have \angleMAD + \angleMND = 180◦if O is not on the segment XY . Therefore ADNM is cyclic. Now let AM and DN intersect at Z and let the line ZX intersect the two circles at Y1 and Y2. Then ZX \cdot ZY1 = ZM \cdot ZA = ZN \cdot ZD = ZX \cdot ZY2. Hence Y1 = Y2 = Y , implying that Z lies on XY .

Second solution. Let Z1, Z2 be the points in which AM, DN respectively meet XY , and P = BC \capXY . Then, from \triangleOPC ∼\triangleAPZ1, we have PZ1 = PA\cdotPC PO = PX2 PO and analogously PZ2 = PX2 PO . Hence, we conclude that Z1 \equivZ2.