Construction of equilateral triangle with one vertex given inside an acute angle

Question

If I am given an acute angle in the plane and a vertex $A$ located inside that angle, how would I construct an equilateral triangle $ABC$ such that $B$ is on one side of the angle and $C$ is on the other side?

I thought of rotating one side of the angle by $60$ degrees around the point $A$, but I'm not sure if that would be the correct approach.

Answer 1

Let the angle have vertex $O$ and bounding rays $b$ and $c$.

• Construct exterior ray $p$ from $O$ that makes an angle of $30^\circ$ with $c$.
• Reflect $b$ in $c$ to get ray $b^\prime$.
• Construct the perpendicular from $A$ to $p$; let this perpendicular meet $b^\prime$ at $A^\prime$
• Construct a circle about $O$ through $A^\prime$; let this circle meet $b$ at $B$.
• Likewise, construct $C$ on $c$ (via $q$, $c^\prime$, and $A^{\prime\prime}$ in the figure).
• $\triangle ABC$ is equilateral. $\square$

If $B$ and $C$ lie on the angle's bounding lines (not merely the rays), then another solution triangle can be determined by taking $p$ and $q$ to be interior rays, and constructing points on rays opposite $b^\prime$, $c^\prime$, $b$, and $c$.

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Note. I got this construction by doing a preliminary run using coordinates and trig. Defining $\alpha := \angle BOC$, $\beta := \angle AOB$, $\gamma := \angle AOC$, I found that $$\frac{|\overline{OB}|}{|\overline{OA}|} = \frac{\cos\left(\gamma + 30^\circ\right)}{\cos\left(\alpha-30^\circ\right)} \qquad \frac{|\overline{OC}|}{|\overline{OA}|} = \frac{\cos\left(\beta + 30^\circ\right)}{\cos\left(\alpha-30^\circ\right)}$$

Observe that, writing $P$ and $Q$ for the feet of the perpendiculars from $A$ to $p$ and $q$, respectively, we have $$\angle AOP = \gamma + 30^\circ \qquad \angle AOQ = \beta + 30^\circ \qquad \angle A^\prime O P = \alpha - 30^\circ = \angle A^{\prime\prime} O Q $$ and the construction follows.