# Problem A. 540. (September 2011)

**A. 540.** Let *A*_{1}*A*_{2}*A*_{3} be a non-equilateral triangle, let point *M* be its orthocenter, let *F* be its Feuerbach point, and let *k* be the circumcircle of the triangle. For *i*=1,2,3 denote by *k*_{i} the circle that is internally tangent to *k* and tangent to the sides *A*_{i}*A*_{i+1} and *A*_{i}*A*_{i+2}. (The indices are considered modulo 3, i.e. *A*_{4}=*A*_{1} and *A*_{5}=*A*_{2}.) Let *T*_{i} be the point of tangency between *k* and *k*_{i}. Prove that the lines *A*_{1}*T*_{1}, *A*_{2}*T*_{2}, *A*_{3}*T*_{3} and *MF* are concurrent.

Proposed by: *Gábor Damásdi* and *Márton Mester, *Budapest

(5 pont)

**Deadline expired on October 10, 2011.**

**Solution. **Denote by *b* the incircle and by *f* the 9-point circle and let *H* be the external homothety center between *f* and *k*.

Apply Monge's circle theorem to the circles *b*, *k* and *k*_{i}. The three external homothety centers between *b* and *k*_{i}, *k* and *k*_{i}, *b* and *k* are *A*_{i}, *T*_{i} and *H*, respectively; by the theorem these points are collinear. Therefore the line *A*_{i}*T*_{i} passes through *H* (*i*=1,2,3).

Now apply Monge's theorem to the circles *b*, *f* and *k*. The three homotety centers between them are *M*, *F* and *H*. Therefore the line *FM* also passes through *H*.

### Statistics:

6 students sent a solution. 5 points: Ágoston Tamás, Gyarmati Máté, Mester Márton, Omer Cerrahoglu, Szabó 928 Attila. 0 point: 1 student.

Problems in Mathematics of KöMaL, September 2011