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I am making a dodecahedron that needs to fit inside of a sphere. The sphere has a diameter of 56mm. What is largest possible measurement of one segment of a pentagon side of a dodecahedron that would fit inside the sphere? How do I determine this?

unfolded dodecahedron

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up vote 3 down vote accepted

If you know the side $\ell$ of the pentagons, the radius of the circumcribing sphere is $r=\tfrac{\ell}{4}(\sqrt{15}+\sqrt{3})$.This is given in Wikipedia (which is, I guess, where you got the picture from :) ) and I recall having read it being deduced in a book written by Coxeter.

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and so it is possible to solve it backwards as well, if you know the radius you can find l - correct? – cwd Nov 21 '11 at 15:38
Yes. ${}{}{}{}$ – Mariano Suárez-Alvarez Nov 21 '11 at 15:44
$\ell=4(\sqrt{15}+\sqrt{3})^{-1}r$. – Mariano Suárez-Alvarez Nov 21 '11 at 19:45

The radius $R$ of the sphere circumscribing the dodecahedron, having edge length $a$, is given here as follows $$\color{blue}{R=\frac{\sqrt{3}(\sqrt{5}+1)a}{4}}$$ $$\implies a=\frac{(\sqrt{5}-1)R}{\sqrt{3}}$$ Now, substituting the radius $R=\frac{56}{2}=28\space mm$, the edge length of the dodecahedron $$a=\frac{(\sqrt{5}-1)(28)}{\sqrt{3}}\approx 19.98203703 \space mm$$

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The answers provided here are correct. Let me explain the formula.

By joining together the diagonals of the pentagons on the dodecahedron we build a cube. For reference use the following figure:

cube in dodecahedron

Let us focus in the blue pentagon. The diagonal separates the pentagon into a triangle and a trapezoid. The other 11 diagonals (chosen here. Note that you could form up to 5 cubes by selecting different sets of diagonals) form a cube. The vertices of the cube lie in a sphere which is the same sphere circumscribing the dodecahedron. So, to find the radius of that sphere is just half of the diagonal of the cube.

It is a well known fact (see here Pentagon ) that the length of a diagonal $d$ of a pentagon of side length $\ell$ is given by \begin{equation} d = \phi \ell \end{equation}

where $\phi=(1+ \sqrt{5})/2$ is the golden ratio.

Now, the diagonal of the cube is \begin{equation} D = \sqrt{ d^2 + d^2 + d^2} = \sqrt{3} d = \sqrt{3} \phi \ell. \end{equation}

So the radius of the sphere is \begin{equation} r = \frac{D}{2} = \frac{\sqrt{3} \, \phi \ell}{2} = \frac{\sqrt{3} (1 + \sqrt{5}) \ell }{4}. \end{equation}

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