Balkan Mathematical Olympiad

Let the lines $BE$ and $ST$ intersect at point $L$. It is enough to prove that $AC$ passes through $L$, that is, the points $A,L,C$ are collinear (Figure 1).



We observe that the circles $c(O,R)$ and $c_1(K,KC)$ are homothetic with respect to homothety $H(C,m)$, that is, homothety with center $C$ and ratio $$m=\frac{R}{KC}.$$ Then the extension of $CS$ meets the circle ($c$) at a point $S_1$ homothetic to $S$ with respect to $H(C,m)$. The extension of $CT$ meets the circle ($c$) at a point $T_1$ homothetic to $T$ with respect to $H(C,m)$. Therefore the segment $S_1{T}_1$ is homothetic to the segment $ST$ and hence $ST\parallel S_1{T}_1$ (1)

Since $ES$ and $ET$ are tangents to the circle ($c_1$), $EK$ is the perpendicular bisector of the segment $ST$. Hence $L$ is the midpoint of $ST$.

Moreover, $\angle AEB=90^{\circ }$, and hence $AE\perp EB$. Since $EB$ is the perpendicular bisector of $ST$, we have that: $ST\parallel AE$ (2)

From relations (1) and (2) we conclude that the quadrilateral $AE{S}_1{T}_1$ is isosceles trapezium and hence $E{S}_1=A{T}_1$ and thus $\angle EC{S}_1=\angle AC{T}_1$. (3)

Since $ES$ and $ET$ are tangents of the circumcircle of the triangle $CST$ (at $S$, $T$), $CE$ is the symmedian corresponding to the vertex $C$. $$\text{Hence }\angle EC{S}_1=\angle LC{T}_1.(4)$$ From relations (3) and (4) we conclude that the points $A,L,C$ are collinear.