Silk Road Mathematics Competition

The lengths of the sides of $ABC$ denote by $a,b,c$. Let $I$ be the center of the incircle, $r$ its radius, $M$ the midpoint of $BC$. Denote by $r_a$ the radius of the excircle touching $BC$. Let $L$ be the point of tangency of excircle with $BC$. ${\omega }_0$ be the circle with center $O$, passing through $B$ and $C$, touching $\omega$. W.l.o.g. assume that $b\ge c$.

Firstly, $SK$ is a bisector of the angle $CSB$, since the homothety with center $S$, which transforms the incircle to ${\omega }_0$, transforms $BC$ to some line touching ${\omega }_0$ at $E$ the midpoint of $BC$.

It is well-known fact that $K$ and $L$ are symmetric with respect to $M$. So, if we prove that $ME=I_{a}L/2$ then we can immediately conclude that $K,E,I_a$ are collinear.

Let $OT\perp IK$, $OM=|x|$, where $x\ge 0$, if $O$ lies below $BC$, and $x<0$ otherwise. Let $R$ be the radius of ${\omega }_0$. Then $ME=R-x$. Consider the right triangle $OTI$, $OI=R-r$, $OT=(b-c)/2$, $IT=r-x$. By Pythagorean theorem

$$(1)(R-r{)}^2=(r-x{)}^2+(b-c{)}^2/4$$ From the right triangle $OMB$ we have $R^2-x^2=a^2/4$. Then using (1) we get $ME=R-x=(p-b)(p-c)/(2r)$, where $p$ is a semiperimeter of $ABC$. Then it is sufficient to prove that $$(2)r_a=\frac{(p-b)(p-c)}{r}$$ We have $r=S/p$. On the other hand, $S=r_a(p-a)$. Then (2) true iff $S^2=p(p-a)(p-b)(p-c)$, which is Heron's formula.