Radius of a Sphere inscribed in a Convex Polyhedron

Question

My teacher gave me this problem in class as a challenge. It has stumped me for days, yet he refuses to give me the answer!

Let $PQRSTU$ and $PQR'S'T 'U'$ be two regular planar hexagons in three dimensional space, each having side length $1$, and such that $\angle TPT'=60^{\circ}$. Let $P$ be the convex polyhedron whose vertices are $P, Q, R, R', S, S', T, T', U, U'$. How would one go about determining:

a) The radius $r$ of the largest sphere that can be inscribed in the polyhedron?

b) Let $E$ be a sphere with radius $r$ enclosed in polyhedron $P$ (as derived in part a)). The set of all possible centers of $E$ is a line segment $\overline{XY}$. What is the length $XY$?

I have tried EVERYTHING in my arsenal... I even made a cutout polyhedron $P$! But i still cant solve it. Can someone help?

Answer 1

Hint to start: $P T T'$ is an equilateral triangle, and is the cross-section of your polyhedron through $P$ orthogonal to $PQ$. Same goes for $Q S S'$.

Answer 2

I have a very mediocre solution for the maximum radius $r$.

If we consider the cross section $SQS'$, we see that it is a equilateral triangle with side length $\sqrt{3}$. The maximum radius of a circle that can be inscribed in it is $\frac{1}{2}$, which is believe is the answer. This is however, far from a complete proof, as we fail to account for the restraint of the rest of the polyhedra, and so we might need to make a smaller value of $r$.

1) is this logic that the maximal radius is $\frac{1}{2}$ even correct?

2) How do we prove that this radius can actually be attained?

3) and what about part b) of this question... the movement of the sphere?

If anyone could give a complete solution of the three above questions, I would be sincerely grateful!