# “Japanese Theorem” on cyclic polygons: Higher-dimensional generalizations?

A beautiful theorem known as the Japanese Theorem (Wikipedia, MathWorld) says that, no matter how one triangulates a cyclic (inscribed in a circle) polygon, the sum of the radii of the incircles is the same: (Wikipedia image)

Q. Does this generalize to higher dimensions? In particular, if one partitions a convex polyhedron in $\mathbb{R}^3$, all of whose vertices lie on a sphere, into tetrahedra, is the sum of the radii of the inspheres independent of the tetrahedralization?

A bit of search has not resulted in an answer, which suggests that the answer may well be No...

Addendum. Following TMA's suggestion, I computed the radii sum for the five-tetrahedron partition—$1.334$, and the sum for the six-tetrahedron partition—$1.242$. Barring a computation error, this settles the question negatively. Too bad!  • I think the answer is no even for the unit cube (5 versus 6 decomposition), but you are in a better position to compute it than I. – The Masked Avenger May 4 '14 at 23:59
• Thanks for checking. I think there are at least two distinct ways to make a 6 part tetrahedral decomposition of the cube. If you take any statistic based on the inscribed spheres that naturally suggests itself as an invariant, say sum of surface areas, you can test it out on those 3 decompositions ( plus more suggested by prisms ) to see if you have an invariant. – The Masked Avenger May 5 '14 at 2:05
• What if you fix a triangulation of the polyhedron, or the number of tetrahedra in the triangulation? After all, in the 2-dimensional case all sides are already simplices. Of course, as The Masked Avenger suggests, another invariant you can consider is the sum of the areas. – Marco Golla May 5 '14 at 9:25
• I checked and $\sum r^2_i$ is not the same for the two partitions of the cube. – Joseph O'Rourke May 5 '14 at 10:29
• @JosephO'Rourke In a different angle, is this result used in Computer Science anywhere? – john mangual May 5 '14 at 11:54