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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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2 Answers 2

up vote 2 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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