Agnes Scott College
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°.

goldenSCgraph

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] \[ (*) \quad 2\cos(2\theta)r^2 + 2\cos(\theta)r + 1 = 0 \]
 
 
cos(108°)
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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] \[r_{sc} = -\frac{1}{2\cos(\theta)}\]
 
 
cos(36°)
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Therefore when θ = 144°, we have \[ r_{sc} = -\frac{1}{2\cos(144^\circ)} = -\frac{1}{-2\cos(36^\circ)} = \frac{1}{2\frac{1+\sqrt{5}}{4}} = \frac{2}{1+\sqrt{5}} = \frac{1}{\phi}. \]