Problem 1161

Let $A$ be the vertex from which the three notable lines originate. We know that $F$ bisects the arc $BC$ not containing $A$. If we draw the perpendicular bisector of $BC$, it passes through points $F$ and $O$, where $O$ is the center of the circumcircle, and $O F \perp B C$ and $A M \perp B C$, that is, $O F \| A M$. Furthermore, we know that the lines $O F$ and $A S$ have a common point on the side $BC$, since both bisect it.
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Based on this, the construction proceeds as follows: We draw the line $O F$ and draw a line parallel to it through the point $M$. This intersects the circle at the vertex $A$. We draw a perpendicular to $A M$ through the intersection point of the lines $A S$ and $O F$, and the intersection points of this perpendicular with the circle are the vertices $B$ and $C$ of the triangle.

From the construction, it follows that the obtained triangle, if it exists, satisfies the condition. If the problem has a solution, there is clearly only one. We claim that if the points $M, F, S$ are on an arc smaller than a semicircle, then the problem has a solution. An important step in the construction is the construction of the intersection point of the lines $A S$ and $O F$. This intersection point is only an interior point of the circle if $A$ and $S$ are on different sides of the diameter $O F$, and since $A M \| O F$, $M$ and $S$ are also on different sides of the line $O F$, that is, $F$ indeed lies between $M$ and $S$. We know that the diameter $O F$ separates the points $A$ and $S$, similarly, the line $B C$ does as well. Reflect the point $A$ over $O F$, since $A A^{\prime} \| B C$, it follows that $B C$ also separates the points $A^{\prime}$ and $S$, that is, $A^{\prime}$ is on one of the arcs defined by the points $B$ and $C$, while $S$ is on the other. The angle $M A A^{\prime}$ is a right angle, from which it follows that $M, F$, and $S$ are indeed on an arc smaller than a semicircle.