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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... – J. M. Sep 15 '11 at 2:49
Obviously $PQRSTU$ was meant. I edited it. – Robert Israel Sep 15 '11 at 3:08
This is one of the problems from and should not be discussed anywhere. (It is illegal to discuss) – user16885 Sep 29 '11 at 19:34
@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. – Zev Chonoles Sep 29 '11 at 21:49
I would request the moderator to remove all the answers for a question from USAMTS contest. – user16885 Sep 30 '11 at 1:20

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 $Q S S'$.

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Robert, I have become curious in this question as well, and have looked at this hint for guidance. I dont see, however, how this helps. I understand that $PQSS'TT'$ is a "wedge" with a 60 degree angle... does that mean the maxiumum radius is 1/2? Probably not... can you please expand? – user14044 Sep 15 '11 at 10:51
I get a maximum radius of $1/2$ and $XY$ of length $(5 - \sqrt{13})/3$ – Robert Israel Sep 15 '11 at 16:08
Basic method: find coordinates for all points, normals to the planes through $PQR$, $PQR'$, $QRR'$, $RSR'$, $STS'$, $TUT'$, $UPU'$. Find equation of line through $(P+T+T')/3$ and $Q+S+S')/3$. Look at inequalities that say distance from point on this line to each plane $\ge 1/2$... – Robert Israel Sep 15 '11 at 18:01
I would really appreciate it if you could expand on those ideas, perhaps with a complete solution. Thank you :) – user14044 Sep 15 '11 at 18:43
@Robert: See Kirthi Raman's comment on the question. – Zev Chonoles Sep 29 '11 at 21:50

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!

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See Kirthi Raman's comment on the question. – Zev Chonoles Sep 29 '11 at 21:50

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