Let $I$ be the incentre of a triangle $ABC$. Let $F$ and $G$ be the projections of $A$ onto the lines $BI$ and $CI$, respectively. Rays $AF$ and $AG$ intersect the circumcircles of the triangles $CFI$ and $BGI$ for the second time at points $K$ and $L$, respectively. Prove that the line $AI$ bisects the segment $KL$.
Problem
Source: Baltic Way 2021, Problem 12
Tags: geometry, geometry proposed, incenter
MP8148
16.11.2021 04:15
Note that $\angle ICK = 180^\circ - \angle IFK = 90^\circ$ and similarly $\angle IBL = 90^\circ$, so $\overline{CK}$ and $\overline{BL}$ meet at the $A$-excenter, $I_A$. Also note that $\overline{AG} \parallel \overline{CK}$ and $\overline{AF} \parallel \overline{BL}$. Thus since $A$, $I$, $I_A$ are collinear and $ALI_AK$ is a parallelogram, the conclusion follows.
Orestis_Lignos
18.11.2021 10:13
Note that $\angle ALB=\angle BIC=90^\circ+\angle A/2$ and $\angle LBA=90^\circ-\angle ABI=90^\circ-\angle B/2$, hence $\frac{AL}{AB}=\frac{\cos \angle B/2}{\cos \angle A/2}$. Similarly, $\frac{AK}{AC}=\frac{\cos \angle C/2}{\cos \angle A/2}$, so by dividing these two relations and using $\frac{AB}{AC}=\frac{\sin \angle C}{\sin \angle B}$, we infer that $$\frac{AL}{AK}=\frac{\cos \angle B/2}{\cos \angle C/2} \cdot \frac{\sin \angle C}{\sin \angle B}=\frac{\sin \angle C/2}{\sin \angle B/2}$$ In addition, $\frac{\sin \angle LAI}{\sin \angle IAK}=\frac{\sin \angle B/2}{\sin \angle C/2}$, hence $\frac{AL}{AK} \cdot \frac{\sin \angle LAI}{\sin \angle IAK}=1$. To conclude, let $M \equiv AI \cap LK$ and note that $\frac{LM}{MK}=\frac{AL}{AK} \cdot \frac{\sin \angle LAI}{\sin \angle IAK}=1,$ and so we are done.