33rd IMO 1992 shortlist Problem 11 BD and CE are angle bisectors of the triangle ABC. ∠BDE = 24 o and ∠CED = 18 o . Find angles A, B, C.
∠SBC + ∠SCB = 180 o - ∠BSC = ∠SED + ∠SDE = 42 o . Hence ∠B + ∠C = 84 o , so ∠A = 96 o . Take D' on BC such that CD' = CD, and E' on BC such that BE' = BE. Let ED' meet BD at S. ∠ESB = ∠SED + ∠SDE = 2·18 o + 24 o = 60 o . Hence ∠E'SB = 60 o . Hence ∠D'SE' = 60 o . So E'S is the external bisector of ∠D'SD. Now EC is perpendicular to DD', so ∠EDD' = 90 o - ∠CED = 72 o . ∠EDE' = 2·24 o , so ∠E'DD' = 72 o - 48 o = 24 o . But ∠E'DS = ∠EDS = 24 o , so E'D is the bisector of ∠SDD'. Hence E' must also lie on the external bisector of ∠DD'S. In other words, ∠E'D'S = ∠CD'D. But ∠ED'D = 180 o - ∠D'ED - ∠D'DE = 180 o - 2·18 o - 72 o = 72 o . Hence ∠CD'D = (180 o - 72 o )/2 = 54 o . Hence ∠C = 180 o - 2·54 o = 72 o . Hence ∠B = 12 o . 33rd IMO shortlist 1992 © John Scholes jscholes@kalva.demon.co.uk 25 Nov 2003 Last updated/corrected 25 Nov 2003