Austria 2014

We denote the centroid of the triangle $ABC$ by $G$. As the centroid divides each median into parts in the ratio $2:1$, we have $$\overline{AG}=\frac{2}{3}\cdot \overline{AD}=12\text{and}\overline{BG}=\frac{2}{3}\cdot \overline{BE}=9.$$ By the Pythagorean theorem in the triangle $AGB$, we obtain $$\overline{AB}=\sqrt{{\overline{AG}}^2+{\overline{GB}}^2}=\sqrt{12^2+9^2}=15.$$

By Thales' theorem, $G$ lies on the circle with center $F$ and diameter $AB$. Therefore, we have $$\overline{GF}=\overline{FA}=\frac{1}{2}\cdot \overline{AB}=\frac{15}{2}.$$ As $\overline{FG}:\overline{GC}=1:2$, we obtain $$\overline{FC}=3\cdot \overline{FG}=\frac{45}{2}.$$