The points M and N and the circle k are given. Construct an equilateral triangle ABC inscribed in circle k such that |AM|=|BN|.

Question

I chose an arbitrary length r and constructed two circles with centers in M ​​and N of radius r and marked their intersections with the given circle. Then I looked at the angle formed by the center of the given circle and those two intersections. I found the bisector of that angle and marked the intersection of the bisector and the circle with S. Finally, I found points A and B on the given circle such that triangles ASO and BS0 are equilateral, where O is the center of the given circle. However, I get that at such construction, |AM| is only approximately equal to |BN|. How can I solve this problem?

Answer 1

Particular case where there is only two solutions:

As can be seen in figure, in this case center O of the circle and vertex A are on perpendicular bisector of MN . ABC is one solution and it's mirror about the diameter parallel with MN , triangle A'B'C' is the second one.

Answer 2

Rotate point $N$ by $120°$ counterclockwise about the center $O$ of the circle, to $N'$ (this can be done by constructing three circles, as in figure below). If the perpendicular bisector of $MN'$ intersects the given circle at $A$, then $AM=AN'$. Finally, rotate $A$ by $120°$ clockwise about $O$, to $B$: we have then $AM=AN'=BN$ (by the congruence of triangles $OAN'$ and $OBN$), as requested.

Both intersections of the perpendicular bisector give a solution, and the initial rotation can also be clockwise: hence we can have up to four distinct solutions.