Larry Riddle, Agnes Scott College

The Golden Self-Contracting Angles

The golden ratio is $\phi = \frac{1+\sqrt{5}}{2}.$ It is the unique positive solution to the equation $$x^2-x-1=0$$, and its reciprocal $$1/\phi$$ is the unique positive solution to $$x^2+x-1=0$$. Two useful properties of $$\phi$$ are that \begin{align} \frac{1}{\phi} &= \phi-1 \\ \frac{1}{\phi^2} &= 2-\phi \end{align} See the Wikipedia article for more information.

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°.

θ = 60°

When 0° < θ < 90°, the self-contacting scaling factor r is the unique solution [Proof] to the equation $\sum_{k=0}^{N-1} r^{k+2}\cos(k\theta) = \frac{1}{2}$ where N is the smallest integer such that Nθ ≥ 90°. If θ = 60°, then N = 2, and the equation for r is $\frac{1}{2} = r^2 + r^3\cos(60^\circ) = r^2 + \frac{r^3}{2} \\ \\ \Rightarrow 0 = r^3 + 2r^2 - 1 = (r^2 + r - 1)(r+1)$ Therefore r satisfies $$r^2+r-1=0$$, and hence $$r = 1/\phi$$.

θ = 108°

For 90° < θ < 135°, the self-contacting factor r satisfies the equation [Proof] $(*) \quad 2\cos(2\theta)r^2 + 2\cos(\theta)r + 1 = 0$
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 $r_{sc} = -\frac{1}{2\cos(144^\circ)} = -\frac{1}{-2\cos(36^\circ)} = \frac{1}{2\frac{1+\sqrt{5}}{4}} = \frac{2}{1+\sqrt{5}} = \frac{1}{\phi}.$