Midradius

The Radius of the Midsphere of a Polyhedron, also called the Interradius. For a Regular Polyhedron with Schläfli Symbol $\{q, p\}$, the Dual Polyhedron is $\{p, q\}$. Denote the Inradius $r$, midradius $\rho$, and Circumradius $R$, and let the side length be $a$. Then

$$r^2 = \left[{a \csc\left({\pi\over p}\right)}\right]^2+R^2=a^2+\rho^2$$ (1)

$$\rho^2 = \left[{a\cot\left({\pi\over p}\right)}\right]^2+R^2.$$ (2)

For Regular Polyhedra and Uniform Polyhedra, the Dual Polyhedron has Circumradius $\rho^2/r$ and Inradius $\rho^2/R$. Let $\theta$ be the Angle subtended by the Edge of an Archimedean Solid. Then

$$r = {\textstyle{1\over 2}}a\cos({\textstyle{1\over 2}}\theta)\cot({\textstyle{1\over 2}}\theta)$$ (3)

$$\rho = {\textstyle{1\over 2}}a\cot({\textstyle{1\over 2}}\theta)$$ (4)

$$R = {\textstyle{1\over 2}}a\csc({\textstyle{1\over 2}}\theta),$$ (5)

so

$$r:\rho:R = \cos({\textstyle{1\over 2}}\theta):1:\sec({\textstyle{1\over 2}}\theta)$$ (6)

(Cundy and Rollett 1989). Expressing the midradius in terms of the Inradius $r$ and Circumradius $R$ gives

$$\rho = {\textstyle{1\over 2}}\sqrt{2}\sqrt{r^2+r\sqrt{r^2+a^2}}$$

$$= \sqrt{R^2-{\textstyle{1\over 4}}a^2}$$ (7)

for an Archimedean Solid.

References

Cundy, H. and Rollett, A. Mathematical Models, 3rd ed. Stradbroke, England: Tarquin Pub., pp. 126-127, 1989.