THE 68th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD

We have $\frac{AI}{ID}=\frac{AB}{BD}=\frac{EC}{DE}=\frac{BE}{DE}=\frac{IE}{DE}$, which proves the first relation.

Let $G=EF\cap BC$. We know that $GD$ is the bisector of the angle $\angle EGI$, hence $\frac{GI}{GE}=\frac{ID}{DE}=\frac{AI}{IE}$. (1)

Triangles $FBE$ and $BGE$ are similar (AA), therefore $E{B}^2=EF\cdot EG=E{I}^2$. This means that triangles $FIE$ and $IGE$ are also similar (SAS), which leads to $\frac{FI}{IE}=\frac{GI}{GE}$. (2)

From (1) and (2) we obtain $FI=AI$, and the conclusion.



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Alternative solution.

Let $O$ be the circumcenter of triangle $ABC$. We show that triangles $AIO$ and $I{I}^{\prime }E$ are similar, which proves that the points $I,O,F,E$ are co-cyclic.

$OE$ and $I{I}^{\prime }$ are both perpendicular to $BC$, hence they are parallel. It follows that $\angle I^{\prime }IE\equiv \angle OEI\equiv \angle OAI$.

We prove that $\frac{AI}{I{I}^{\prime }}=\frac{AO}{IE}$. From the angle bisector theorem we obtain $\frac{AI}{ID}=\frac{AB}{BD}=\frac{AC}{CD}=\frac{AB+AC}{BC}=\frac{b+c}{a}$, hence $\frac{AI}{AD}=\frac{b+c}{a+b+c}$. As $AD=l_a=\frac{2bc\cos \frac{A}{2}}{b+c}$, we obtain $AI=\frac{2bc\cos \frac{A}{2}}{a+b+c}$.

But $IE=EB$ and $\frac{EB}{BC}=\frac{\sin \frac{A}{2}}{\sin A}=\frac{1}{2\cos \frac{A}{2}}$. Moreover, $EI=EB=\frac{a}{2\cos \frac{A}{2}}$.

It follows that, $AI\cdot IE=\frac{abc}{a+b+c}$.



On the other hand, $AO\cdot I{I}^{\prime }=R\cdot 2r=\frac{2RS}{p}=\frac{abc}{a+b+c}$, hence $AO\cdot I{I}^{\prime }=AI\cdot IE$, which, together with $\angle I^{\prime }IE\equiv \angle OAI$, proves the similarity of the triangles $AIO$ and $I{I}^{\prime }E$.

We deduce that $\angle EIO\equiv \angle I{I}^{\prime }F\equiv \angle OEF\equiv \angle OFE$, hence $IOEF$ is cyclic (the order of the points and the arguments depend slightly on the configuration).

It follows that $\angle IOF\equiv \angle IE{I}^{\prime }\equiv \angle AOI$. We also have $\angle IAO\equiv \angle IEO\equiv \angle IFO$, which means that triangles $AIO$ and $IFO$ are congruent (SAA), therefore $IF=IA$.