67th Czech and Slovak Mathematical Olympiad

It suffices to prove that if $AB<AC$ then $S_b$ lies inside the circumcircle $k$ of triangle $B{S}_{c}C$. The midline $S_b{S}_c$ is parallel to $BC$ (Fig. 1). Let line $S_b{S}_c$ intersect $k$ for the second time at $P$. We will show that $S_b$ lies on the segment $S_{c}P$ (as opposed to lying on the ray opposite to $P{S}_c$). To that end, it suffices to prove $\angle BCA<\angle BCP$. By symmetry about the perpendicular bisector of $BC$ we have $\angle BCP=\angle CBA$, so we need to prove $\angle BCA<\angle CBA$ which is in fact clearly equivalent to the given $AB<AC$.

Fig. 1

Fig. 2

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Alternative solution.

By power of $A$ with respect to $k$, there exists a point $Q$ on the ray $AC$ such that $AQ\cdot AC=AB\cdot A{S}_c=\frac{1}{2}A{B}^2$. Then (Fig. 2) $$AQ=\frac{A{B}^2}{2\cdot AC}<\frac{A{C}^2}{2\cdot AC}=\frac{1}{2}AC=A{S}_b,$$ hence $Q$ lies on segment $A{S}_b$. As before we conclude that $S_b$ lies inside the circumcircle of triangle $B{S}_{c}C$.

Fig. 2