IMO 1997 SL 23

Origin: GBR

Problem

Let ABCD be a convex quadrilateral and O the intersection of its diagonals AC and BD. If OA sin \angleA + OC sin \angleC = OB sin \angleB + OD sin \angleD, prove that ABCD is cyclic.

Solution

Let K, L, M, and N be the projections of O onto the lines AB, BC, CD, and DA, and let \alpha1, \alpha2, \alpha3, \alpha4, \beta1, \beta2, \beta3, \beta4 denote the angles OAB, OBC, OCD, ODA, OAD, OBA, OCB, ODC, respectively. We start with the following observation: Since NK is a chord of the circle with diameter OA, we have OA sin \angleA = NK = ON cos \alpha1 + OK cos \beta1 (because \angleONK = \alpha1 and \angleOKN = \beta1). Analogous equalities also hold: OB sin \angleB = KL = OK cos \alpha2 + OL cos \beta2, OC sin \angleC = LM = OL cos \alpha3 + OM cos \beta3 and OD sin \angleD = MN = OM cos \alpha4 + ON cos \beta4. Now the condition in the problem can be restated as NK + LM = KL + MN (i.e., KLMN is circumscribed), i.e., OK(cos \beta1 −cos \alpha2) + OL(cos \alpha3 −cos \beta2) +OM(cos \beta3 −cos \alpha4) + ON(cos \alpha1 −cos \beta4) = 0. (1) To prove that ABCD is cyclic, it suffices to show that \alpha1 = \beta4. Assume the contrary, and let w.l.o.g. \alpha1 > \beta4. Then point A lies inside the circle BCD, which is further equivalent to \beta1 > \alpha2. On the other hand, from \alpha1 + \beta2 = \alpha3 + \beta4 we deduce \alpha3 > \beta2, and similarly \beta3 > \alpha4. Therefore,

since the cosine is strictly decreasing on (0, \pi), the left side of (1) is strictly negative, yielding a contradiction.