Rice university vigre: edge length minimizing polyhedra  (Page 3/4)

When θ is the angle between the old top and the new one, then for the variation P _θ we have

which has a minimum at $\theta =0.$

Next, we varied the cube by fixing a vertex, lifting the diagonal vertex a certain height and the two adjacent vertices by half that height:

In this case, when θ was the angle between the old top and the new one, then for the variation P _θ we have:

which has minimum at $\theta =0.$

The next variation of the cube is collapsing two of the sides together, so as to form a shape approaching the Regular Triangular Prism:

When θ is the angle of collapse, then

which has minimum at $\theta =0,$ when the figure is a cube. Observe that although the figure with $\theta =\frac{\pi }{6}$ is the Regular Triangular Prism, $\mathcal{F}\left(P_{\theta }\right){|}_{\frac{\pi }{6}}$ with the formula given above is not $\mathcal{F}\left(RTP\right).$ This is because when $\theta =\frac{\pi }{6},$ two vertical edges have collapsed into one, and our formula above double counts these edges at $\theta =\frac{\pi }{6}.$

Symmetrization of polyhedra

The next method we considered is derived from the Isoperimetric Problem, which asks to find the figure in the plane with unit area and whose boundary curve has the smallest length. The solution, of course, is the disk of unit area. An argument for this is via symmetrization. That is, take a region A in the plane and draw a line ℓ which cuts A into two connected regions $A_1,A_2$ both with area $1/2.$ Suppose then that the part of the boundary curve of A corresponding to A ₁ has smaller length than that of $A_2,$ that is

Consider the new region $A_1\cup R_{\ell }\left(A_1\right)$ where $R_{\ell }\left(A_1\right)$ is the reflection of A ₁ across the line $\ell .$ Then $A_1\cup R_{\ell }\left(A_1\right)$ has unit area, but the length of the boundary curve of $A_1\cup R_{\ell }\left(A_1\right)$ is less than or equal to that of $A.$ Hence, if A is a solution to the Isoperimetric Problem, then the claim is that A must be symmetric across every line which slices A into two regions of equal area, and so A must be a disk.

Although the solution to the Isoperimetric Problem is far more subtle than the argument given above, we nonetheless take the argument above as an analogy. That is, given a polyhedra $P,$ we slice P via a plane H which cuts P into two polyhedra $P_1,P_2$ with equal volume. Then letting P ₁ have the smaller total edge length, where we only sum the edges which correspond to edges of $P,$ then the polyhedron $P_1\cup R_H\left(P_1\right),$ where $R_H\left(P_1\right)$ is the reflection of P ₁ across $H,$ will have $V(P_1\cup R_H\left(P_1\right))=V\left(P\right)$ and $E(P_1\cup R_H\left(P_1\right))\le E\left(P\right),$ assuming that no new edges have been introduced when we took the reflection of P ₁ across H and merged it with $P_1.$

The difficulty is finding planes given a polyhedron P which halve the volume and which produce new polyhedra with no new edges introduced at the plane H when we reflect the two halves across $H.$ We call such a plane a bisecting plane of $P.$ If P and H intersect along a face of P which is not perpendicular to H , then a new edge will be introduced when we reflect. For the Regular Triangular Prism, there are only two such planes, and the Regular Triangular Prism is symmetric under the reflection across these planes:

There are only two such planes for the cube