# insphere/circumsphere ratio of a polyhedron the same as its dual polyhedron?

Is the $r/R$ ratio for any polyhedron always the same as the $r/R$ ratio of the dual of that polyhedron?

Given any polyhedron, we can find the biggest sphere that fits inside it (its insphere) and the smallest sphere outside it (its circumsphere). I'm interested in $r/R$, which is the ratio of the radius of the insphere to the radius of the circumsphere.

Motivation: the worst-case linear distortion of a polyhedral global map projection is roughly (at least) $r/R$. The $r/R$ ratio is one of many ways to measure how close a polyhedron is to a sphere. Some of those measures say the dodecahedron is the closest Platonic solid to the sphere. Others say the icosahedron is the closest Platonic solid to the sphere. Polyhedrons with larger r/R ratio are better, in the sense that they (usually) produce maps with less linear distortion. I'm surprised to discover that that ratio seem to be equal between the dodecahedron and the icosahedron (both ~ $0.7946$), and a (somewhat smaller) ratio is the same between the cube and the octahedron (both ~ $0.5773$). (I'm not so surprised that an (even smaller) ratio is the same between the remaining Platonic solid and its dual polyhedron: $1/3$). Can this surprising-to-me fact be generalized to some larger group of polyhedrons? What other polyhedron have a $r/R$ ratio the same as the $r/R$ ratio of its dual polyhedron? Do any polyhedron have a $r/R$ ratio different from the $r/R$ ratio of its dual polyhedron?

(In this answer, I do as the questioner did, calling the smallest containing sphere the circumsphere and its center the circumcenter, even if this sphere does not touch all vertices. Similarly, I call the largest enclosed sphere the insphere and its center the incenter, even if this sphere does not touch all faces.)

Assuming that the circumcenters and incenters of both the original polyhedron and the dual all coincide, these ratios should be the same.

A convex polyhedron which contains the origin in its interior can be described either as the convex hull of a set $\{a_i\}\subseteq {\Bbb R}^3$ of vertices or as the intersection of a set of bounding half-spaces $\{S_j\}$, where each bounding half-space is of the form $$S_j=\{x\mid x\cdot b_j\le 1\}, \qquad b_j\in {\Bbb R}^3.$$

Letting the circumsphere be centered at the origin, the circumradius $R$ will equal the maximum value of any $|a_i|$. Also, since the distance between the origin and the plane $\{x\mid x\cdot b_j=1\}$ is $|b_j|^{-1}$, if the insphere is centered at the origin, the inradius will equal the minimum value of any $|b_j|^{-1}$.

The polar dual of a given polyhedron can be constructed by interchanging the $a_i$s and $b_j$s. So, if the original polyhedron had circumradius $R$ and inradius $r$, and if the circumcenter and incenter of the polar dual are still at the origin, then the polar dual must have circumradius $1/r$ and inradius $1/R$, and the ratio $R/r$ of the circumradius to the inradius remains the same.

For an example, take a cuboctahedron with vertices $$(\pm 1, \pm 1, 0), \ (\pm 1, 0, \pm 1), \ (0, \pm 1, \pm 1).$$ In this case, the circumsphere and insphere are both centered at the origin. The radius $\sqrt{2}$ circumsphere touches all vertices; the insphere has radius $1$ and touches only the square faces; and the ratio $R/r$ is $\sqrt{2}$. The dual of this polyhedron is the rhombic dodecahedron, with vertices $$(\pm 1,0,0), \ (0,\pm 1,0), \ (0,0,\pm 1),\ (\pm\frac12, \pm\frac12, \pm\frac12).$$ The circumsphere and the insphere are both still centered at the origin. The circumsphere has radius $1$ and touches only those vertices where four faces meet (the first six above); the insphere has radius $1/\sqrt{2}$ and touches all faces; and the ratio $R/r$ is still $\sqrt{2}$.

• Beautiful answer! – nbubis Jan 22 '13 at 8:36
• Isn't there some circular reasoning here? You're trying to prove the assertion "Assuming that the circumcenters and incenters of both the original polyhedron and the dual all coincide, these ratios are the same.". Then you choose a coordinate space in which the circumcenter is the origin; ok. Then you say the duality in question is reciprocation about a sphere centered at that origin. That doesn't seem justified; for example, mightn't it be that the original polyhedron and dual were spherical-reciprocation duals with respect to a sphere centered at some other origin? – Don Hatch Nov 2 '20 at 3:54
• It seems the language of this proof breaks down in the case of, say, a non-cube rectangular box, for which there is no "insphere" or "incenter" according to the definition proposed. However, in that case there is still a unique "inradius", so the question is still well-defined. I wonder, could the language of this proof be fixed up so that that this case is covered (and proved)? I suspect a well-chosen change in the statement to be proved might resolve both my previous concern and this one. – Don Hatch Nov 2 '20 at 4:06

Yes, it is true for dual solids. For dual solids cube & octahedron $$\color{blue}{\frac{r}{R}=\frac{1}{\sqrt{3}}\approx 0.577350269}$$ and for dual solids dodecahedron & icosahedron $$\color{blue}{\frac{r}{R}=\sqrt{\frac{5+2\sqrt{5}}{15}}\approx 0.794654472}$$

For derivation of radii $r$ & $R$ for all platonic solids, you may go through HCR's Formula for regular polyhedrons

For table of radii $r$ & $R$ for all platonic solids, you may go through Table of platonic solids by HCR

• The original question already pointed out that the ratios seemed to be equal for the duals of the 5 Platonic solids. This answer shows that the ratios actually are exactly equal, which is nice to know. What about other dual solids, such as the triangular prism and its dual the triangular bipyramid, the cuboctahedron and its dual the rhombic dodecahedron, the Archimedean solids and their duals the Catalan solids, etc.? – David Cary Jun 18 '15 at 4:39
• Of course your question has concern about many duals. Please give me time I will summarize them & update as soon as possible. – Harish Chandra Rajpoot Jun 18 '15 at 4:51
• You may want to look at David Moews answer, which already covers all those shapes. Do you have anything to add that isn't already covered by that answer? – David Cary Jun 18 '15 at 14:51

the ratio is not the same between the cube and the octahedron they are not (both ~ 0.5773) octahedron 0.577350269 cube 0.471404521

• The ratio for the cube is $1/\sqrt3$, which seems to be the number that you gave for the octahedron. The number you gave for the cube seems to be $\sqrt{2/9}$; where did that come from? – Andreas Blass Mar 14 '13 at 12:32