Silk Road Mathematics Competition

Let $BC=a$, $AC=b$, $AB=c$ and $\angle CAB=\alpha$, $\angle ABC=\beta$, $\angle BCA=\gamma$. Let $D$ be the intersection point of $BL$ and $CK$. Note that $△PAQ$ is isosceles. $$\angle BKL=\angle APK-\angle ABK=\frac{\pi -\alpha }{2}-\frac{\beta }{2}=\frac{\alpha +\beta +\gamma -\alpha -\gamma }{2}=\frac{\gamma }{2}=\frac{\angle ACB}{2}.$$ Since $\angle IKL=\angle BKL=\angle ABC/2=\angle ACI$, the points $I,K,Q,C$ are concyclic, and hence, $\angle IKC=\angle IQC=\pi /2$. Similarly, $\angle ILB=\pi /2$. Therefore, the points $B,L,K,C$ are concyclic, and the points $I,L,D,K$ are also. Particularly, $BC$ is a diameter of the circumscribed circle of $△CLK$ and $ID$ is a diameter of the circumscribed circle of $△ILK$.

Using the sines law in $△ILK$ we have $$ID=\frac{IK}{\sin \angle LDK}=\frac{LK}{\cos \angle LCK},$$ and in $△CLK$ we have $a=\frac{LK}{\sin \angle LCK}$. Therefore, $ID=a\tan \angle LCK$. Since $\sin \angle LCK=\sin \angle IQK=\sin (\alpha /2)$, we also can write that $ID=\tan (\alpha /2)$. On the other hand, $r=AQ\tan (\alpha /2)$ and $AQ=(b+c-a)/2$, where $r$ is a radius of incircle of $△ABC$.

Circumcircle of $△ILK$ touches the incircle of $△ABC⟺$ diameter of the circumcircle of $△ILK$ is equal to the radius of incircle of $△ABC$.