Baltic Way 2019

Segments $B{B}_1$ and $C{C}_1$ intersect inside triangle $ABC$, point $A$ is outside the circle, therefore ray $AO$ lies inside angle $BAC$. Assume that triangle $ABC$ is not isosceles. Then $\angle BAO\ne \angle OAC$. Without loss of generality $\angle BAO<\angle OAC$. Let $B_1^{\prime }$ and $C^{\prime }$ be points symmetrical with respect to line $AO$ to points $C_1$ and $B$ correspondingly. Then triangle $AB{C}^{\prime }$ is isosceles, and lines $C^{\prime }{C}_1$ and $B{B}_1^{\prime }$ intersect in some point $K$ on the line $AO$ (see fig.). Now point $B_1$ lies on the short arc $B_1^{\prime }{C}^{\prime }$, therefore the ray $B{B}_1$ is obtained from the ray $B{B}_1^{\prime }$ by rotation in clockwise direction, and therefore the intersection point of $B{B}_1$ and $AO$ lies below point $K$. By the analogous reasons the intersection point of $C_{1}C$ and $AO$ lies above point $K$. So for non-isosceles triangle the concurrence from the problem statement is impossible.



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Alternative solution.

Suppose that $B_1{C}_1$ intersects $BC$ at $X$. The polar line of $X$ is the line passing through $B{C}_1\cap B_{1}C=A$ and $B{B}_1\cap C{C}_1$. By assumption given in the problem statement, this line is $AO$. This is a contradiction since the polar line cannot pass through the center of the circle. It follows that $B_1{C}_1\parallel BC$ and it is clear now that $ABC$ is isosceles.