IMO 2000 SL G5

Origin: GBR | Category: Geometry

Problem

The tangents at B and A to the circumcircle of an acute- angled triangle ABC meet the tangent at C at T and U respectively. AT meets BC at P, and Q is the midpoint of AP; BU meets CA at R, and S is the midpoint of BR. Prove that \angleABQ = \angleBAS. Determine, in terms of ratios of side lengths, the triangles for which this angle is a maximum.

Solution

Since \angleABT = 180◦−\gamma and \angleACT = 180◦−\beta, the law of sines gives BP PC = SABT SACT = AB\cdotBT \cdotsin \gamma AB\cdotBT \cdotsin \beta = AB sin \gamma AC sin \beta = c2 b2 , which implies BP = c2a b2+c2 . Denote by M and N the feet of perpendiculars from P and Q on AB. We have cot \angleABQ = BN NQ = 2BN PM = BA+BM BP sin \beta = c+BP cos \beta BP sin \beta = b2+c2+ac cos \beta ca sin \beta

2(b2+c2)+a2+c2−b2 2ca sin \beta

a2+b2+3c2 4SABC = 2 cot \alpha + 2 cot \beta + cot \gamma. Similarly, cot \angleBAS = 2 cot \alpha + 2 cot \beta + cot \gamma; hence \angleABQ = \angleBAS. Now put p = cot \alpha and q = cot \beta. Since p+q \geq0, the A-G mean inequality gives us cot \angleABQ = 2p + 2q + 1−pq p+q \geq2p + 2q + 1−(p+q)2/4 p+q = 7 4(p + q) + p+q \geq2

4 = \sqrt 7. Hence \angleABQ \leqarctan \sqrt 7. Equality holds if and only if cot \alpha = cot \beta = \sqrt 7, i.e., when a : b : c = 1 : 1 : \sqrt 2.