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Combinatorial Geometry!!

This is an interesting problem I saw in an NMTC paper.... A solid cube is chopped off at each of it's 8 corners to create an equilateral triangle with 3 new corners. All possible diagonals are drawn. Find the number of these diagonals which completely lie inside the new 3-D body.

Note by Samarth M.O.
3 years, 10 months ago

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The truncated cube has \(3\times8=24\) vertices. Each vertex \(v\) belongs to two of the octagonal faces and one triangular face. Thus there are \(7+7-1=13\) vertices to which \(v\) can be joined which result in a nontrivial diagonal which lies on the surface of the truncated cube. Thus there are \(10\) vertices to which \(v\) can be joined forming a nontrivial diagonal lying inside the truncated cube.

Thus there are \(\tfrac12 \times 24 \times 10 = 120\) nontrivial diagonals lying inside the interior of the truncated cube. Mark Hennings · 3 years, 10 months ago

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Note that no diagonal lies outside the truncated cube. Therefore the total number of diagonals inside the truncated cube is \(\binom{24}{2}=300\). But we counted edges and the diagonal of each octagonal face with that too. We subtract the \(36\) edges and the \(120\) diagonals of all the octagonal faces to get \(300-36-120=\boxed{144}\). For some reason I got a different answer than Mark, so who made the mistake? Daniel Liu · 3 years, 10 months ago

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@Daniel Liu \({24 \choose 2}=276\) Mark Hennings · 3 years, 10 months ago

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