IMO 1992 SL 3

Origin: CHN

Problem

The diagonals of a quadrilateral ABCD are perpendicular: AC\perpBD. Four squares, ABEF, BCGH, CDIJ, DAKL, are erected ex- ternally on its sides. The intersection points of the pairs of straight lines CL, DF; DF, AH; AH, BJ; BJ, CL are denoted by P1, Q1, R1, S1, respec- tively, and the intersection points of the pairs of straight lines AI, BK; BK, CE; CE, DG; DG, AI are denoted by P2, Q2, R2, S2, respectively. Prove that P1Q1R1S1 ∼= P2Q2R2S2.

Solution

Consider two squares AB′CD′ and A′BC′D. Since AC \perpBD, these two squares are homothetic, which implies that the lines AA′, BB′, CC′, DD′ are concurrent at a certain point O. Since the rotation about A by 90◦ takes ∆ABK into ∆AFD, it fol- lows that BK \perpDF. Denote by T the intersection of BK and DF. The rotation about some point X by 90◦maps BK into DF if and only if TX bisects an angle between BK and DF. Therefore \angleFTA = \angleATK = 45◦. Moreover, the quad- A B D C E F L K T A′ rilateral BA′DT is cyclic, which implies that \angleBTA′ = BDA′ = 45◦ and consequently that the points A, T, A′ are collinear. It follows that the

point O lies on a bisector of \angleBTD and therefore the rotation R about O by 90◦takes BK into DF. Analogously, R maps the lines CE, DG, AI into AH, BJ, CL. Hence the quadrilateral P1Q1R1S1 is the image of the quadrilateral P2Q2R2S2, and the result follows.