IMO 1992 LL SPA66

Origin: SPA

Problem

A circle of radius \rho is tangent to the sides AB and AC of the triangle ABC, and its center K is at a distance p from BC. (a) Prove that a(p −\rho) = 2s(r −\rho), where r is the inradius and 2s the perimeter of ABC. (b) Prove that if the circle intersect BC at D and E, then DE = 4 rr1(\rho −r)(r1 −\rho) (r1 −r) , where r1 is the exradius corresponding to the vertex A. 7 The statement of the problem is obviously wrong, and the authors couldn’t de- termine a suitable alteration of the formulation which would make the problem correct. We put it here only for completeness of the problem set.