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Berkeley Math Circle Monthly Contest 3

Geometry Difficulty 4.7 AIME Prove it United States

Problem:

The inscribed circle of a triangle $ABC$ touches the sides $BC$ , $CA$ , $AB$ at $D$ , $E$ , and $F$ respectively. Let $X$ , $Y$ , and $Z$ be the incenters of triangles $AEF$ , $BFD$ , and $CDE$ , respectively. Prove that $DX$ , $EY$ , and $CZ$ meet at one point.

Solution

Solution:

Consider the midpoint $M$ of arc $EF$ on the incircle of $\triangle ABC$ . Angles $AFM$ and $MFE$ are equal since they intercept equal arcs $FM$ and $ME$ , and so $M$ is on the bisector of $\angle AFE$ . Similarly, $M$ is on the bisector of $\angle FEA$ , and therefore $M$ coincides with $X$ . Moreover, angles $FDX$ and $XDE$ are equal since they intercept equal arcs $FX$ and $XE$ , and so $DX$ is the angle bisector of $\angle D$ in $\triangle DEF$ . Similarly, $EY$ and $FZ$ are the other two angle bisectors in $\triangle DEF$ . But the three angle bisectors in a triangle always meet!

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