66th Belarusian Mathematical Olympiad

Necessity. Let $B$, $S$, $C$ be collinear. Let $X$ be the midpoint of the segment $B_1{C}_1$. Then the bisector of $B_1{C}_1$ passes through $X$ and intersects the side $BC$ at $S$ since $S$ lies on $BC$ and from the condition it follows that $S{B}_1=S{C}_1$ (see Fig. 1). Let $O$ be the circumcenter of the triangle $ABC$. Since the triangle $ABC$ is isosceles, $AO$ is the bisector of the angle $BAC$. Since $OB=OA$, it follows that $\angle OB{B}_1=\angle OA{B}_1=\angle OA{C}_1$. By condition, $B{B}_1=A{C}_1$, hence the triangles $OB{B}_1$ and $OA{C}_1$ are equal. Then $\angle OB{B}_{1}B=\angle OA{C}_{1}A$, so $$\angle OB{B}_{1}A+\angle O{C}_{1}A=180^{\circ }-\angle OB{B}_{1}B+\angle O{C}_{1}A=180^{\circ },$$ which means that the quadrilateral $OB{B}_{1}A{C}_1$ is cyclic. Since $\angle B_{1}AO=\angle C_{1}AO$, we see that $O{B}_1=O{C}_1$ as the chords of the equal arcs. Hence, $OX$ is the median of the isosceles triangle $B_{1}O{C}_1$ and so $OX$ is the bisector of the angle $B_{1}O{C}_1$, i.e., $\angle XO{C}_1=0.5\angle B_{1}O{C}_1=0.5(180^{\circ }-\angle B_{1}A{C}_1)$. (Note that $OX$ is also the altitude of $B_{1}O{C}_1$ and since $SX\perp B_1{C}_1$, we see that $O$ lies on $SX$.) Therefore, $$\angle XO{C}_1=90^{\circ }-0.5\angle B_{1}A{C}_1=90^{\circ }-0.5\angle BAC=90^{\circ }-\angle OAC=$$ $$=[OA\perp BC]=\angle ACB=\angle C_{1}CS,$$ whence the quadrilateral $O{C}_{1}CS$ is cyclic. Therefore $\angle OS{C}_1=\angle OC{C}_1$. Then we have $$\frac{1}{2}\angle BAC=\angle OAC=[OA=OC\Rightarrow \angle OAC=\angle OCA]=\angle OCA=\angle OC{C}_1=\angle OS{C}_1=\angle XS{C}_1=90^{\circ }-\angle S{C}_{1}X=90^{\circ }-\angle S{C}_1{B}_1=90^{\circ }-\angle BAC.$$ Hence $\frac{3}{2}\angle BAC=90^{\circ }$, so $\angle BAC=60^{\circ }$, which means that the triangle $ABC$ is equilateral as required.



Sufficiency. It is easy to see that there exists a unique point $S$ satisfying the problem condition. Mark the point $A_1$ on the side $BC$ such that $A_{1}C=A{C}_1$ (see Fig. 2). Show that $A_1$ and $S$ coincide if the triangle $ABC$ is equilateral. Indeed, we have $$B{B}_1=A{C}_1=C{A}_1,A_{1}B=B_{1}A=C_{1}C,$$ $$\angle A_{1}B{B}_1=\angle B_{1}A{C}_1=\angle C_{1}C{A}_1=\angle BAC=60^{\circ }.$$ It follows that $△A_{1}B{B}_1\cong △B_{1}A{C}_1\cong △C_{1}C{A}_1$. Hence, $A_1{B}_1=B_1{C}_1=C_1{A}_1$, so the triangle $A_1{B}_1{C}_1$ is equilateral and $\angle A_1{B}_1{C}_1=\angle B_1{C}_1{A}_1=60^{\circ }=\angle BAC$, which means that $A_1$ and $S$ coincide. Therefore, if the triangle $ABC$ is equilateral, then the points $B$, $S$, $C$ are collinear.