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?
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Best Answer
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 $Q S S'$.