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anonymous
 5 years ago
A cube with volume of 1 cm^3 has a total Edge Length of 12 (because 12 edges of length 1). Which polyhedra with volume 1 have less total Edge Length than the cube? (A polyhedron is just any flatfaced 3dimensional object, like a tetrahedron, Egyptian pyramid, pup tent, shoebox, parallelipiped, hexagonal prism, etc.)
anonymous
 5 years ago
A cube with volume of 1 cm^3 has a total Edge Length of 12 (because 12 edges of length 1). Which polyhedra with volume 1 have less total Edge Length than the cube? (A polyhedron is just any flatfaced 3dimensional object, like a tetrahedron, Egyptian pyramid, pup tent, shoebox, parallelipiped, hexagonal prism, etc.)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I believe the answer to the question is that no polyhedrons of volume one have a total edge length of less than 12. I don't have a proof, but here is a sketch of my ideas. First, note that cubes minimize total edge length for rectangular prisms. The proof of this is easy. Let a, b, and c be the sides of the prism. Then clearly abc=1. Furthermore, total edge length is 4a+4b+4c. But by AMGM: \[4a+4b+4c \ge 3 \big((4a)(4b)(4c) \big)^\frac{1}{3}=12(abc)^\frac13=12\] Now, and here's the sketchy part, consider your general polyhedron of volume 1 to consist of n^3 cubes of side length 1/n as n approaches infinity. Calculus tells us that we can create all solids in this manner. For some notation, let A and B be two figures composed of these smaller cubes in some manner. Let A+B be ANY way of joining these two figures together (including even not having them touch). Let M be the polyhedron you are making. Let E(A) be the total edge length of A and let V(A) be the volume of A. Now convince yourself, preferably through a proof of some sort, that the following is true: \[E(A+B)\ge \frac{E(A)+E(B)}{V(A)+V(B)}\] Inducting on this, we get \[E(\sum A_i)\ge\frac{\sum E(A_i)}{\sum V(A_i)}\] Then, taking A_i to be each small cube and taking the limit, you get that \[E(M)\ge\frac{\sum E(A_i)}{\sum V(A_i)}=\frac{n^3\frac{12}{n^3}}{1}=12\] This is just a sketch, but filling in the details should give a full proof.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0awesome effort! unfortunately the thirdtolast equation fails in cases where E(A+B) << E(A)+E(B) (which happens when interior edging vanishes because of a gluing together) first check out the edge length of a few pyramids and triangular prisms before making too many sweeping claims...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Aha, touche. I tried looking at at several other cases and didn't find anything, same with my claim that the formula was true. But alas, here we are: Consider an equilateral triangular prism of side length (of the triangle) a and length of the entire prism d. Then the volume of the figure is \[\frac{a^2d\sqrt3}{4}=1 \implies a^2d=\frac{4}{\sqrt3}\] Since the total edge length is 6a+3d, we have, again by AMGM, that \[6a+3d=3a+3a+3d \ge 3\big((3a)(3a)(3d)\big)^\frac13\] \[=9(a^2d)^\frac13=9\left(\frac{4}{\sqrt3}\right)^\frac13 \approx 11.896\] Since AMGM always attains equality, there must, at the very least, be a prism of this shape with total edge length less than 12. Cool problem
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