IMO 1982 SL 14

Problem

C2 (AUS 4) Let ABCD be a convex plane quadrilateral and let A1 denote the circumcenter of \triangleBCD. Define B1, C1, D1 in a corresponding way. (a) Prove that either all of A1, B1, C1, D1 coincide in one point, or they are all distinct. Assuming the latter case, show that A1, C1 are on opposite sides of the line B1D1, and similarly, B1, D1 are on opposite sides of the line A1C1. (This establishes the convexity of the quadrilateral A1B1C1D1.) (b) Denote by A2 the circumcenter of B1C1D1, and define B2, C2, D2 in an analogous way. Show that the quadrilateral A2B2C2D2 is similar to the quadrilateral ABCD.

Solution

(a) If any two of A1, B1, C1, D1 coincide, say A1 \equivB1, then ABCD is inscribed in a circle centered at A1 and hence all A1, B1, C1, D1 coin- cide. Assume now the opposite, and let w.l.o.g. \angleDAB + \angleDCB < 180◦. Then A is outside the circumcircle of \triangleBCD, so A1A > A1C. Simi- larly, C1C > C1A. Hence the perpendicular bisector lAC of AC sepa- rates points A1 and C1. Since B1, D1 lie on lAC, this means that A1 and C1 are on opposite sides B1D1. Similarly one can show that B1 and D1 are on opposite sides of A1C1. (b) Since A2B2 \perpC1D1 and C1D1 \perpAB, it follows that A2B2 \parallelAB. Similarly A2C2 \parallelAC, A2D2 \parallelAD, B2C2 \parallelBC, B2D2 \parallelBD, and C2D2 \parallelCD. Hence \triangleA2B2C2 ∼\triangleABC and \triangleA2D2C2 ∼\triangleADC, and the result follows.