# BMO 2005 Problem 2

Problem 2 of the British Mathematical Olympiad 2005

In triangle ABC, ∠BAC = 120°. Let the angle bisectors of angles A, B and C meet the opposite sides in D, E and F respectively. Prove that the circle on diameter EF passes through D.

Solution by Steve Dinh, a.k.a. Võ Đức Diên (dedicated to my former professor Le Chi De)

Let BE meet CF at I, the incenter id ΔABC. Let J and S be the points of tangency of the inccircle with the sides BC and AC, respectively.

Link JS to meet BE at L. From E draw the perpendicular to CF to meet CF and BC at K and N, respectively. Also let CF meet ED at T, BE meet FD at V.

We have ∠BID = ∠ABI + ∠BAI = 90° - ½∠C = ∠JIC or ∠BIJ = ∠DIC

It’s easily seen that ∠EIK = ½ (∠B +∠C) = 30° or ∠BIC = 150° and ∠IEK = 90° - ∠EIK = 60° and since CI is the perpendicular bisector of EN, IE = IN and ∠INE = 60°. It follows that IEN is an equilateral triangle and ∠NIK = 30°.

We now have

∠IND = ∠NIK + ½∠C = 30° + ½∠C = 30° + (30° - ½∠B) = 60° - ½∠B and ∠DIN = ∠DIC - 30° = ∠BIJ - 30° = 90° - ½∠B - 30° = 60° - ½∠B

or ∠IND = ∠DIN and DE is bisector of ∠IEN or ∠DEN = 30°

From here, ∠ITE = 30° + 90° = 120°

Similarly on the other side, we have ∠IVF = 120°

But ∠ITE + ∠IVF = ∠BIC + ∠FDE or ∠FDE = ∠ITE + ∠IVF - ∠BIC = 90°

Therefore, the circle on diameter EF passes through D.

Note: The problem has been previously offered at the 56 Leningrad Mathematical Olympiad (1990). A different solution is available elsewhere.