IMO 2001 SL G1

Origin: UKR | Category: Geometry

Problem

Let A1 be the center of the square inscribed in acute triangle ABC with two vertices of the square on side BC. Thus one of the two re- maining vertices of the square is on side AB and the other is on AC. Points B1, C1 are defined in a similar way for inscribed squares with two vertices on sides AC and AB, respectively. Prove that lines AA1, BB1, CC1 are concurrent.

Solution

Let MNPQ be the square inscribed in \triangleABC with M \inAB, N \inAC, P, Q \inBC, and let AA1 meet MN, PQ at K, X respectively. Put MK = PX = m, NK = QX = n, and MN = d. Then BX XC = m n = BX + m XC + n = BP CQ = d cot \beta + d d cot \gamma + d = cot \beta + 1 cot \gamma + 1. Similarly, if BB1 and CC1 meet AC and BC at Y, Z respectively then CY Y A = cot \gamma+1 cot \alpha+1 and AZ ZB = cot \alpha+1 cot \beta+1. Therefore BX XC CY Y A AZ ZB = 1, so by Ceva’s theorem, AX, BY, CZ have a common point.

Second solution. Let A2 be the center of the square constructed over BC outside \triangleABC. Since this square and the inscribed square correspond- ing to the side BC are homothetic, A, A1, and A2 are collinear. Points B2, C2 are analogously defined. Denote the angles BAA2, A2AC, CBB2, B2BA, ACC2, C2CB by \alpha1, \alpha2, \beta1, \beta2, \gamma1, \gamma2. By the law of sines we have sin \alpha1 sin \alpha2 = sin(\beta + 45◦) sin(\gamma + 45◦), sin \beta1 sin \beta2 = sin(\gamma + 45◦) sin(\alpha + 45◦), sin \gamma1 sin \gamma2 = sin(\alpha + 45◦) sin(\beta + 45◦). Since the product of these ratios is 1, by the trigonometric Ceva’s theorem AA2, BB2, CC2 are concurrent.