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!