IMO 1998 SL 7

Origin: GBR

Problem

Let ABC be a triangle such that \angleACB = 2\angleABC. Let D be the point on the side BC such that CD = 2BD. The segment AD is extended to E so that AD = DE. Prove that \angleECB + 180◦= 2\angleEBC.

Solution

We shall use the following result. Lemma. In a triangle ABC with BC = a, CA = b, and AB = c, i. \angleC = 2\angleB if and only if c2 = b2 + ab; ii. \angleC + 180◦= 2\angleB if and only if c2 = b2 −ab. Proof. i. Take a point D on the extension of BC over C such that CD = b. The condition \angleC = 2\angleB is equivalent to \angleADC = 1 2\angleC = \angleB, and thus to AD = AB = c. This is further equivalent to triangles CAD and ABD being similar, so CA/AD = AB/BD, i.e., c2 = b(a + b). ii. Take a point E on the ray CB such that CE = b. As above, \angleC + 180◦= 2\angleB if and only if \triangleCAE ∼\triangleABE, which is equivalent to EB/BA = EA/AC, or c2 = b(b −a). Let F, G be points on the ray CB such that CF = 1 3a and CG = 4 3a. Set BC = a, CA = b, AB = c, EC = b1, and EB = c1. By the lemma it follows that c2 = b2 + ab. Also b1 = AG and c1 = AF, so Stewart’s theorem gives us c2 1 = 3b2 + 1 3c2 −2 9a2 = b2 + 1 3ab −2 9a2 and b2 1 = −1 3b2 + 4 3c2 + 4 9a2 = b2 + 4 3ab + 4 9a2. It follows that b1 = 3a + b and c2 1 = b2 1 −
ab + 2 3a2 = b2 1 −ab1. The statement of the problem follows immediately by the lemma.