40th Hellenic Mathematical Olympiad

Since $\angle CI{I}_a=\angle BI{I}_a=90^{\circ }$ the excenter $I_a$ belongs to the circumcircle of the triangle $BIC$. Moreover in the triangle $CDA$, $CI,C{I}_a$ are the internal and external bisector, respectively, and therefore the points $A,D$ are the conjugate harmonics of the points $I,I_a$. Since $M$ is the midpoint of the segment $AD$, from the relation of Newton we have: $$M{A}^2=MI\cdot M{I}_a.(1)$$ Taking the power of $M$ with respect to the circle $(I,B,C)$ we have: $$MI\cdot M{I}_a=MF\cdot MB.(2)$$ From (1) and (2) it follows that: $$M{A}^2=MF\cdot MB,$$ Which means that $MA$ is tangent of the circumcircle of the triangle $AFB$. Therefore $$\angle MBA=\angle FAM.(3)$$ and finally $$\begin{array}{rl}\angle FAC+\angle FCA& =\angle FAM+\frac{\angle A}{2}+\angle FCI+\frac{\angle C}{2}=\\ & (3)\angle MBA+\frac{\angle A}{2}+\angle FBI+\frac{\angle C}{2}=\frac{\angle A}{2}+\frac{\angle B}{2}+\frac{\angle C}{2}=90^{\circ }.\end{array}$$ Hence: $AF\perp FC$.

Figure 2

We draw from $C$ the parallel line to the bisector $AD$ which meets the line $AB$ at point $N$ and the line $AM$ at point $Z$. Taking angle equalities we have: $$\angle ACN=\angle DAC=\frac{\angle A}{2},\angle ANC=\angle BAD=\frac{\angle A}{2}$$ And so the triangle $ACN$ is isosceles. Since $M$ is the midpoint of $AD$ it follows that $Z$ is the midpoint of $CN$. Hence $AZ\perp CN\Rightarrow \angle AZC=90^{\circ }$. Therefore, in order to have $\angle AFC=90^{\circ }$, it is enough to prove that the quadrilateral $CFAZ$ is cyclic. In fact, we have $$\begin{gathered} \angle CFZ=180^{\circ }-\angle BFC=180^{\circ }-\angle BIC \\ =180^{\circ }-\left(90^{\circ }+\frac{\angle A}{2}\right)=90^{\circ }-\frac{\angle A}{2}=\angle CAZ. \end{gathered}$$

Figure 3