# A challenging straightedge and compass construction Three points $A,O,B$ are given, and $0<\theta=\widehat{AOB}<\frac{\pi}{3}$.
It is known that there are two points $A',B'$ on the segments $OA,OB$ such that $$BB'=B'A'=A'A$$ holds. How to find them with straightedge and compass?

The problem is straightforward to solve through trigonometry: if we set $$OA=A,\;OB=b,\;\cos\theta=c,\; AA'=x$$ it boils down to solving the second-degree equation: $$(a-x)^2+(b-x)^2 - 2(a-x)(b-x)c = x^2,$$ but I wasn't able to find an elegant solution through straightedge and compass only.

• Is it possible that the triangles $\Delta AOB,\Delta A'OB'$ are similar? If so, then a simple construction seems immediate. – Semiclassical Jun 3 '16 at 17:41
• @Semiclassical: that cannot hold in general. If $AOB$ and $A'OB'$ are similar then $\frac{a-x}{b-x}=\frac{a}{b}$, but we know that $x$ is the solution of an irreducible quadratic equation. – Jack D'Aurizio Jun 3 '16 at 17:43
• You're right, and it's actually easy to construct a counterexample: If $A'B'$ is perpendicular to $OA$, then there's no way that $AB$ and $A'B'$ can be parallel. – Semiclassical Jun 3 '16 at 17:50

Thanks to Xaver, a simple solution (but a not-so-trivial one). Lemma 1. If $$P\in OA$$ and $$Q\in OB$$ fulfills $$PA=QB$$, then $$PB\cap QA$$ lies on a line
that is parallel to the angle bisector of $$\widehat{AOB}$$.

Lemma 2. $$C=AB'\cap A'B$$ lies on a fixed circle $$\Gamma$$ through the incenter of $$AOB$$,
since $$\widehat{ACB}=\frac{\pi+\theta}{2}$$.

So, let $$P$$ be a point of $$OA$$ and $$Q$$ be the corresponding $$Q$$-point on $$OB$$, as in Lemma 1.

Let $$D=BP\cap AQ$$ and $$\ell$$ be the line through $$D$$ that is parallel to the angle bisector of $$\widehat{AOB}$$.

Let $$I$$ be the incenter of $$AOB$$ and $$\Gamma$$ the circumcircle of $$AIB$$.

Then $$\color{blue}{C=\ell \cap \Gamma}$$ and $$A',B'$$ are easily found.

Let $S$ be the intersection of $AB'$ and $BA'$. Let $\varphi:=\angle ASA'$. Then $\theta+2\varphi=180°$ always holds.

I haven't really worked out the details, but I'm sure that this helps to construct the segments $AB'$ and $BA'$ (and therefore $A'$ and $B'$).