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Berkeley Math Circle: Monthly Contest 6

Geometry Difficulty 4.8 AIME Prove it United States

Problem:

Let ABCABC be a triangle, and let X,Y,ZX, Y, Z be the excenters opposite A,B,CA, B, C. The incircle of triangle ABCABC touches BC,CA,ABBC, CA, AB at points D,E,FD, E, F. Finally, let II and OO denote the incenter and circumcenter of triangle ABCABC.
Prove that lines DX,EY,FZ,IODX, EY, FZ, IO are concurrent.

Solution

Solution:

The fact that DX,EY,FZDX, EY, FZ are concurrent follows from the fact that triangles DEFDEF and XYZXYZ are homothetic; indeed, note that EFEF and YZYZ are both perpendicular to the internal angle bisector of ∠BAC\angle BAC.

Now, to see that the concurrence point lies on IOIO, note that point II is the orthocenter of triangle XYZXYZ, and OO is the nine-point center of triangle XYZXYZ. Thus line IOIO is the Euler line of triangle XYZXYZ and thus passes through the circumcenter SS of triangle XYZXYZ. But II is the circumcenter of triangle DEFDEF, hence line SISI passes through the concurrency point.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.