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Problem 884

AIME late

Geometry Difficulty 5.7 Prove it

7.7 From the sides of an acute-angled $\triangle ABC$ , construct three triangles $\triangle BCA_{1}, \triangle CAB_{1}, \triangle ABC_{1}$ , such that $\angle CAB_{1}=\angle C_{1}AB=\alpha, \angle ABC_{1}=\angle A_{1}BC=\beta, \angle BCA_{1}=\angle B_{1}CA=\gamma$ , where $\alpha, \beta, \gamma$ are all acute angles. Prove that $AA_{1}, BB_{1}, OC_{1}$ are concurrent. $(\alpha+\beta+\gamma=180^{\circ}$ when, it is a former Soviet MO 7)

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

7.7 Let $A A_{1}$ , $B B_{1}$ , and $O C_{1}$ intersect $B C$ , $C A$ , and $A B$ at points $D$ , $E$ , and $F$ , respectively, then
$\frac{B D}{D C}=\frac{S_{\triangle A B A_{1}}}{S_{\triangle A C A_{1}}}=\frac{A B \cdot B A_{1} \sin (B+\beta)}{C A \cdot A_{1} C \sin (C+\gamma)}=\frac{A B \cdot \sin \gamma \cdot \sin (B+\beta)}{C A \cdot \sin \beta \cdot \sin (C+\gamma)} ;$

Similarly, we get two other similar equations. Multiplying the three equations yields 1. By the converse of Ceva's theorem, the proof is complete.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.