Inspired by Calvin Lin 2 : An open challenge to all of Brilliant

Start with the circumcircle $\Gamma$ and the chord $AB$ .

• Find the perpendicular bisector of $AB$ . This meets $AB$ at $F$ , and $\Gamma$ at points $P,Q$ . Bisect the diameter $PQ$ to find the centre $O$ of the circumcircle.
• Extend the line $AB$ and find the point $X$ such that $FB = BX$ . Draw the circle $\Gamma_1$ which has centre $X$ and radius $BX\sqrt{3}$ . This is easy to do. For example, a circle of radius $BX$ and centre $B$ and a circle of radius $BX$ and centre $F$ meet at two points, both of which are a distance $BX\sqrt{3}$ from $X$ . Why do this? The circle $\Gamma_1$ is the locus of the points $P$ such that $AP = BP\sqrt{3}$ , and hence the midpoint $D$ of $BC$ must lie on $\Gamma_1$ .
• Since $O$ lies on the perpendicular bisector of $BC$ , the angle $\angle ODB = 90^\circ$ . If we construct the midpoint $Y$ of $OB$ and draw the circle $\Gamma_2$ with centre $Y$ that passes through $O$ and $B$ , then the point $D$ must lie on $\Gamma_2$ as well.
• Having identified the point $D$ , extend the line $BD$ to meet the circumcircle $\Gamma$ at $C$ , and we are done.

Note that since $\Gamma_1$ and $\Gamma_2$ meet at two points, there are two possible triangles that can be constructed for any particular circle $\Gamma$ and chord $AB$ . The diagram shows both triangles $ABC$ and $ABC'$ .

This construction holds for any chord $AB$ (if $AB$ is a diameter the triangles $ABC$ and $ABC'$ are congruent), and hence $ABC$ does not have to be equilateral.