IMO 1985/B2 solution

IMO 1985

Problem B2

A circle center O passes through the vertices A and C of the triangle ABC and intersects the segments AB and BC again at distinct points K and N respectively. The circumcircles of ABC and KBN intersect at exactly two distinct points B and M. Prove that angle OMB is a right angle.

Solution

The three radical axes of the three circles taken in pairs, BM, NK and AC are concurrent. Let X be the point of intersection. [They cannot all be parallel or B and M would coincide.] The first step is to show that XMNC is cyclic. The argument depends slightly on how the points are arranged. We may have: ∠XMN = 180^o - ∠BMN = ∠BKN = 180^o - ∠AKN = ∠ACN = 180^o - ∠XCN, or we may have ∠XMN = 180^o - ∠BMN = 180^o - ∠BKN = ∠AKN = 180^o - ∠ACN = 180^o - ∠XCN.

Now XM.XB = XK.XN = XO² - ON². BM·BX = BN·BC = BO² - ON², so XM·XB - BM·BX = XO² - BO². But XM·XB - BM·BX = XB(XM - BM) = (XM + BM)(XM - BM) = XM² - BM². So XO² - BO² = XM² - BM². Hence OM is perpendicular to XB, or ∠OMB = 90^o.

Solutions are also available in Murray S Klamkin, International Mathematical Olympiads 1978-1985, MAA 1986, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.

26th IMO 1985