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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.
4 years 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.

- 4 years ago

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?

- 4 years ago

$${24 \choose 2}=276$$

- 4 years ago