IMO 1992 SL 11

Origin: JAP

Problem

In a triangle ABC, let D and E be the intersections of the bisec- tors of \angleABC and \angleACB with the sides AC, AB, respectively. Determine the angles \angleA, \angleB, \angleC if ∡BDE = 24◦, ∡CED = 18◦.

Solution

Let I be the incenter of \triangleABC. Since 90◦+ \alpha/2 = \angleBIC = \angleDIE = 138◦, we obtain that \angleA = 96◦. A B C D E I E′ D′ S Let D′ and E′ be the points symmetric to D and E with respect to CE and BD respectively, and let S be the intersection point of ED′ and BD. Then \angleBDE′ = 24◦and \angleD′DE′ = \angleD′DE −\angleE′DE = 24◦,which means that DE′ bisects the angle SDD′. Moreover, \angleE′SB = \angleESB = \angleEDS + \angleDES = 60◦and hence SE′ bisects the angle D′SB. It follows that E′ is the excenter of \triangleD′DS and consequently \angleD′DC = \angleDD′C = \angleSD′E′ = (180◦−72◦)/2 = 54◦. Finally, \angleC = 180◦−2 \cdot 54◦= 72◦and \angleB = 12◦.