381. Given a triangle $A B C$ and a point $O$ inside it. Construct a segment with its midpoint at point $O$, the ends of which lie on the boundary of triangle $A B C$. What is the maximum number of solutions this problem can have?
$\triangleright$ We need to find a point $X$ on the boundary of the triangle, for which the point $X^{\prime}$ symmetric to it (with respect to $O$) also lies on the boundary of the triangle.

To do this, let's ask ourselves: where can $X^{\prime}$, the point symmetric to $X$ on the boundary of triangle $ABC$, be? In this form, the answer is clear: on the boundary of the symmetric triangle $A^{\prime} B^{\prime} C^{\prime}$ (points $A^{\prime}, B^{\prime}, C^{\prime}$ are symmetric to points $A, B, C$ with respect to point $O$). Therefore, we need to construct this symmetric triangle and take as $X^{\prime}$ one of the points of its intersection with the original triangle. There can be a maximum of 6 such intersection points, which group into three pairs, meaning the problem can have a maximum of three solutions. $\triangleleft$

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