IMO 1997 SL 8

Origin: GBR

Problem

Four different points A, B, C, D are chosen on a circle \Gamma such that the triangle BCD is not right-angled. Prove that: (a) The perpendicular bisectors of AB and AC meet the line AD at cer- tain points W and V , respectively, and that the lines CV and BW meet at a certain point T . (b) The length of one of the line segments AD, BT, and CT is the sum of the lengths of the other two. Original formulation. In triangle ABC the angle at A is the smallest. A line through A meets the circumcircle again at the point U lying on the arc BC opposite to A. The perpendicular bisectors of CA and AB meet AU at V and W, respectively, and the lines CV, BW meet at T . Show that AU = TB + TC.

Solution

(a) Denote by b and c the perpendicular bisectors of AB and AC re- spectively. If w.l.o.g. b and AD do not intersect (are parallel), then \angleBCD = \angleBAD = 90◦, a contradiction. Hence V, W are well-defined. Now, \angleDWB = 2\angleDAB and \angleDV C = 2\angleDAC as oriented an- gles, and therefore \angle(WB, V C) = 2(\angleDV C −\angleDWB) = 2\angleBAC = 2\angleBCD is not equal to 0. Consequently CV and BW meet at some T with \angleBTC = 2\angleBAC.

(b) Let B′ be the second point of intersection of BW with \Gamma. Clearly AD = BB′. But we also have \angleBTC = 2\angleBAC = 2\angleBB′C, which implies that CT = TB′. It follows that AD = BB′ = |BT \pm TB′| = |BT \pm CT|. Remark. This problem is also solved easily using trigonometry.