Larry Riddle, Agnes Scott College

Below is the graph of the self-contacting scaling factor as a function of θ, along with the horizontal line at \(1/\phi = 0.61803\ldots\). We see that there are exactly four intersections. We claim that these occur at θ = 60°, 108°, 120°, and 144°.

If θ = 108°, then
\[
\begin{align}
\cos(108^\circ) &= \frac{1-\sqrt{5}}{4} = -\frac{1}{2\phi} = -\frac{1}{2}(\phi-1) \\ \\
\cos(216^\circ) &= 2\cos^2(108^\circ) - 1 = \frac{1}{2\phi^2}-1 = -\frac{\phi}{2}
\end{align}
\]
Substituting these into (*)
\[
2\cos(2\theta)r^2 + 2\cos(\theta)r + 1 = -\phi r^2 - (\phi-1)r + 1 = -(\phi r - 1)(r+1) = 0
\]
Therefore \(r_{sc} = 1/\phi\) is the solution when θ = 108°.
#### θ = 120°

We can still use (*) with θ = 120°. This gives
\[
\begin{align}
2\cos(2\theta)r^2 + 2\cos(\theta)r + 1 &= 2\cos(240^\circ)r^2 + 2\cos(120^\circ)r + 1 \\ \\
&= -r^2 - r + 1 = r^2+r-1
\end{align}
\]
and so \(r_{sc} = 1/\phi\) is the solution when θ = 120°.
#### θ = 144°

For 135° < θ < 180°, the self-contacting value is [Proof]
\[r_{sc} = -\frac{1}{2\cos(\theta)}\]

Therefore when θ = 144°, we have
\[
r_{sc} = -\frac{1}{2\cos(144^\circ)} = -\frac{1}{-2\cos(36^\circ)} = \frac{1}{2\frac{1+\sqrt{5}}{4}} = \frac{2}{1+\sqrt{5}} = \frac{1}{\phi}.
\]