# Intersecting Circles II

The app Euclidea, which I have mentioned in the last post, has an interesting level 4.2. I say interesting, because I was close to beaking my head while trying to solve it for the first time, although its solution is surprisingly easy.

Suppose that we’re given a straight line and an external point $P$. The exercise is to construct a second line through $P$ which cuts the given one at $\delta = \displaystyle\frac{\pi}{3} = 60^°$. $\mbox{Construct an angle of}~60°$ So we draw two circles. The first is centered at $P$ with a radius large enough to cut the line in two points, one of them labeled $Q$. The second is then constructed to be centered in $Q$ with radius $\overline{PQ}$. This defines a triangle $\bigtriangleup_{RPQ}$ which of course is equilateral and therefore has an inner angle of $\alpha = \pi / 3$.

With the circle $\bigcirc_Q$ cutting the given line in $S$, we draw a line through $R$ and $S$, thereby cutting $\bigcirc_P$ in $T$. The solution is then provided by the line going through $T$ and $P$. $\mbox{Solution}$

Proof: The so constructed and hatched triangles $\bigtriangleup_{RQS}$ and $\bigtriangleup_{TPR}$ both are isosceles, so $$\angle_R(\bigtriangleup_{RQS}) = \gamma \quad \mbox{and} \quad \angle_R(\bigtriangleup_{TPR}) = \beta~.$$ Since the outer angles of $\bigtriangleup_{RPQ}$ in $R$ with respect to $\overline{TS}$ plus $\alpha$ sum up to $\pi$, we therefore have $$\beta + \gamma = \pi - \frac{\pi}{3}$$ and because $\delta + \beta + \gamma = \pi$ as well, we get $\delta = \displaystyle\frac{\pi}{3}$ as requested.