IMO 1991 SL 7

Origin: CHN

Problem

Let O be the center of the circumsphere of a tetrahedron ABCD. Let L, M, N be the midpoints of BC, CA, AB respectively, and assume that AB+BC = AD+CD, BC+CA = BD+AD, and CA+AB = CD + BD. Prove that \angleLOM = \angleMON = \angleNOL.

Solution

The given equations imply AB = CD, AC = BD, AD = BC. Let L1, M1, N1 be the midpoints of AD, BD, CD respectively. Then the above

equalities yield L1M1 = AB/2 = LM, L1M1 \parallelAB \parallelLM; L1M = CD/2 = LM1, L1M \parallelCD \parallelLM1. Thus L, M, L1, M1 are coplanar and LML1M1 is a rhombus as well as MNM1N1 and LNL1N1. Then the A B C D M N L M1 N1 L1 Q segments LL1, MM1, NN1 have the common midpoint Q and QL \perpQM, QL \perpQN, QM \perpQN. We also infer that the line NN1 is perpendicular to the plane LML1M1 and hence to the line AB. Thus QA = QB, and similarly, QB = QC = QD, hence Q is just the center O, and \angleLOM = \angleMON = \angleNOL = 90◦.