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Problem 1389

AIME late

Geometry Difficulty 5.7 Prove it Croatian Mathematical Competitions · Croatia

The angle at the vertex $B$ in a triangle $ABC$ is $120^\circ$ . Let $A_1$ , $B_1$ and $C_1$ be the points on the segments $\overline{BC}$ , $\overline{CA}$ and $\overline{AB}$ respectively, such that $AA_1$ , $BB_1$ and $CC_1$ are angle bisectors of the triangle $ABC$ . Determine the angle $\angle A_1B_1C_1$ . (Serbia 1997)

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $X$ be any point on the extension of the segment $\overline{AB}$ over the vertex $B$ . Note that $\angle ABB_1 = \angle B_1BC = \angle CBX = 60^\circ$ .

This implies that the point $A_1$ lies on the angle bisector of the angle $\angle B_1BX$ , and it also lies on the angle bisector of the angle $\angle BAC$ .
From this we conclude that $A_1$ is the centre of excircle of the triangle $ABB_1$ opposite to vertex $A$ and hence it lies on the angle bisector of the angle $\angle BB_1C$ .
So the line $B_1A_1$ is the angle bisector of the angle $\angle BB_1C$ . Analogously we prove that the line $B_1C_1$ is the angle bisector of the angle $\angle AB_1B$ .
Hence
$\angle A_1B_1C_1 = \angle A_1B_1B + \angle BB_1C_1 = \frac{1}{2}\angle CB_1B + \frac{1}{2}\angle BB_1A = \frac{1}{2} \cdot 180^\circ = 90^\circ.$

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