IMO 1984 SL 15

Origin: LUX

Problem

Angles of a given triangle ABC are all smaller than 120◦. Equilateral triangles AFB, BDC and CEA are constructed in the exterior of \triangleABC. (a) Prove that the lines AD, BE, and CF pass through one point S. (b) Prove that SD + SE + SF = 2(SA + SB + SC).

Solution

(a) Since rotation by 60◦around A transforms the triangle CAF into \triangleEAB, it follows that ∡(CF, EB) = 60◦. We similarly deduce that

∡(EB, AD) = ∡(AD, FC) = 60◦. Let S be the intersection point of BE and AD. Since ∡CSE = ∡CAE = 60◦, it follows that EASC is cyclic. Therefore ∡(AS, SC) = 60◦= ∡(AD, FC), which implies that S lies on CF as well. (b) A rotation of EASC around E by 60◦transforms A into C and S into a point T for which SE = ST = SC + CT = SC + SA. Summing the equality SE = SC +SA and the analogous equalities SD = SB +SC and SF = SA + SB yields the result.