Multiple tetrahedra?
⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ x + y + z = 1 x + y + z = 6 x − y − z = 1 x − y − z = 6 y − z − x = 1 y − z − x = 6 z − y − x = 1 z − y − x = 6
The above equations define planes in 3 space.
How many ways can you choose four of them so that the enclosed region defines a regular tetrahedron?
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1 solution
I think a much better way of presenting this solution is as follows:
- Claim: A (regular) tetrahedron is formed by 4 planes if and only if {insert condition here}
- From the conditions, this implies that we must have X.
For example, in your solution, it doesn't (yet) explain why no other combination of 4 planes would work.
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There I've cleaned it up a bit.
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Great! You can now use "regular tetrahedron" everywhere in the solution.
Four planes enclose a regular tetrahedron if:
For the planes above, the first two are parallel, the third and fourth are parallel, the fifth and sixth are parallel, and the seventh and eighth are parallel, and, no two parallel planes can be part of of four that form the boundary of a regular tetrahedron.
However, it just so happens, that for the planes above, if you choose any two that aren't parallel, the dihedral angle between them is arccos ( 3 1 ) .
So, you will get a regular tetrahedron if you pick the four planes as follows:
Therefore, there are 2 ⋅ 2 ⋅ 2 ⋅ 2 = 2 4 ways of choosing these planes which will enclose a regular tetrahedron.