IMO 2001 SL G7

Origin: BUL | Category: Geometry

Problem

Let O be an interior point of acute triangle ABC. Let A1 lie on BC with OA1 perpendicular to BC. Define B1 on CA and C1 on AB similarly. Prove that O is the circumcenter of ABC if and only if the perimeter of A1B1C1 is not less than any one of the perimeters of AB1C1, BC1A1, and CA1B1.

Solution

We denote by p(XY Z) the perimeter of a triangle XY Z. If O is the circumcenter of \triangleABC, then A1, B1, C1 are the midpoints of the corresponding sides of the triangle, and hence p(A1B1C1) = p(AB1C1) = p(A1BC1) = p(A1B1C). Conversely, suppose that p(A1B1C1) \geqp(AB1C1), p(A1BC1), p(A1B1C). Let \alpha1, \alpha2, \beta1, \beta2, \gamma1, \gamma2 denote \angleB1A1C, \angleC1A1B, \angleC1B1A, \angleA1B1C, \angleA1C1B, \angleB1C1A. Suppose that \gamma1, \beta2 \geq\alpha. If A2 is the fourth vertex of the parallelogram B1AC1A2, then these conditions imply that A1 is in the interior or on the border of \triangleB1C1A2, and there- fore p(A1B1C1) \leqp(A2B1C1) = p(AB1C1). Moreover, if one of the inequalities \gamma1 \geq\alpha, \beta2 \geq\alpha is strict, then p(A1B1C1) is strictly less than A B C O A1 B1 C1 A2 \alpha1 \beta1 \gamma1 \alpha2 \beta2 \gamma2 p(AB1C1), contrary to the assumption. Hence \beta2 \geq\alpha =⇒\gamma1 \leq\alpha, \gamma2 \geq\beta =⇒\alpha1 \leq\beta, \alpha2 \geq\gamma =⇒\beta1 \leq\gamma, (1) the last two inequalities being obtained analogously to the first one. Be- cause of the symmetry, there is no loss of generality in assuming that \gamma1 \leq\alpha. Then since \gamma1 + \alpha2 = 180◦−\beta = \alpha + \gamma, it follows that \alpha2 \geq\gamma. From (1) we deduce \beta1 \leq\gamma, which further implies \gamma2 \geq\beta. Similarly, this leads to \alpha1 \leq\beta and \beta2 \geq\alpha. To sum up, \gamma1 \leq\alpha \leq\beta2, \alpha1 \leq\beta \leq\gamma2, \beta1 \leq\gamma \leq\alpha2. Since OA1BC1 and OB1CA1 are cyclic, we have \angleA1OB = \gamma1 and \angleA1OC = \beta2. Hence BO : CO = cos \beta2 : cos \gamma1, hence BO \leqCO. Analogously, CO \leqAO and AO \leqBO. Therefore AO = BO = CO, i.e., O is the circumcenter of ABC.