Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Imagine that I have a truncated icosahedron consisting of 60 identical vertices, each of degree $deg(v) = 3$, and fixed edge length $L$. I'd like to assign some constant curvature or bending angle $\theta$ to each edge s.t. I can deform the truncated icosahedron into its circumscribing sphere.

As a function of the edge length $L$, what value of $\theta$ allows me to properly perform this deformation?

share|cite|improve this question
I don't understand how you're using the "bending angle" to deform the icosahedron. Are you looking for the angle $\theta$ that each edge makes with the circumscribing sphere? – anon Jul 17 '11 at 23:49
@anon, sorry for the confusion! No I'm looking for the bending angle that places each edge on the surface of the sphere. – R.H. Jul 18 '11 at 3:04
Oh, you mean the angle formed by the arc which results from a radial projection of an edge onto the circumscribing sphere. – anon Jul 19 '11 at 12:15
up vote 2 down vote accepted

Fix $L=1$. Then the radius of the circumscribed sphere is $r=\frac{1}{4} \sqrt{58+18 \sqrt{5}} \approx 2.478$. Now look at the isosceles triangle formed by the center of the sphere and one edge. It has sides of length $r$ and base length 1. So the angles at either end of the base are $\cos^{-1} (1/(2r)) \approx 78.3593^\circ$. The angle between the tangent to the sphere at one endpoint and the edge is then about $11.6407^\circ$.
           Soccer Ball

share|cite|improve this answer

In general the circumscribed radius ($R$) of a truncated icosahedron, having 12 congruent regular pentagonal faces & 20 congruent regular hexagonal faces each with edge length $L$, is given by the generalized expression (derived in Mathematical analysis of truncated icosahedron by HCR) $$\bbox[4pt, border: 1px solid blue;]{\color{blue}{R}=\color{red}{\frac{L\sqrt{58+18\sqrt{5}}}{4}}\color{purple}{\approx 2.478018659\space L}}$$ Now, join one of end-points & the mid-point of the edge to the center of truncated icosahedron, we get a right triangle with base $\color{blue}{\frac{L}{2}}$ & hypotenuse $\color{blue}{\frac{L\sqrt{58+18\sqrt{5}}}{4}}$

Hence, the angle $\theta$ between the edge & the radius is given as $$\cos \theta=\frac{\frac{L}{2}}{\frac{L\sqrt{58+18\sqrt{5}}}{4}}=\frac{2}{\sqrt{58+18\sqrt{5}}}=\sqrt{\frac{29-9\sqrt{5}}{218}}$$ $$\implies \color{blue}{ \theta=\cos^{-1}\sqrt{\frac{29-9\sqrt{5}}{218}}\approx 78.35927686^{o}}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.