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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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    $PQRST$ can't be a hexagon, with only five corners...2011-09-15
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    Obviously $PQRSTU$ was meant. I edited it.2011-09-15
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    This is one of the problems from http://usamts.org and should not be discussed anywhere. (It is illegal to discuss)2011-09-29
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    @Kirthi: Thank you for pointing that out; I have upvoted your comment. Hopefully, this dissuades people from answering further. But it is the general policy here that we are not in the business of making sure other people behave ethically, and it is entirely up to alan to keep his integrity and refrain from submitting anything he is told here.2011-09-29
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    I would request the moderator to remove all the answers for a question from USAMTS contest.2011-09-30

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