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Folding Polygons into Pyramids

- Posted in Math by

When I find incorrect answers in, which is way more often than it should be for only one month of activity, I try to report the problem.


Which of the following regular polygons can be folded into a pyramid without cutting or overlapping any area? regular polyhedra The multiple choice approach of the problem is in error, because TWO of the regular shapes can each be folded into (non-flat) triangular pyramids.

TRIANGLE: Let its verteces be A, B, and C. Create Midpoints for AB, BC, and CA, called D, E, and F respectively. The triangular base is triangle DEF. This solution has the added bonus of creating a regular tetrahedron, but that was not a requirement.

SQUARE: Let its verteces be A, B, C, and D. Create Midpoints for AB and DA called E and F respectively. The triangular base is triangle ECF. Points A, B, and D meet above the base to form the pyramidal apex. It is an irregular tetrahedron of positive volume \(V = \frac{L \times W \times H}{3} = \frac{1}{3} \times\frac{1}{2} unit \times \frac{1}{2} unit \times 1 unit\ = \frac{1}{12}unit^3\) *, but is a valid solution. (Sides of the base measure \(EC=\frac{\sqrt{5}}{2}, CF=\frac{\sqrt{5}}{2}\) , and \(EF=\frac{\sqrt{2}}{2}\)). A square post-it note thus folded proves it to be true.

* CORRECTION: That would be the volume of a square base, but this base is triangular, so it has half that area. \(V = \frac{h \times A}{3} = \frac{1}{3} \times ( \frac{1}{2} \times \frac{1}{2} unit \times \frac{1}{2} unit ) \times 1 unit\ = \frac{1}{24}unit^3\) Square folded into irregular tetrahedron

Note: The problem on has since been reworded so that it asks only for the polygon that can form a regular pyramid. But as we can see in my next post, that eliminates the irregular pyramids made by this square and with a regular pentagon, but doesn't necessarily exclude regular tetrahedrons made from any regular polygon leaving uncut, non-overlapping flaps.