# Bounding the total edge length of a polyhedron

For any convex polyhedron, let $L$ denote the total length of all its edges and let $D$ be its diameter (the max distance between any 2 points on the boundary). Find the largest possible $x$ and the smallest possible $y$ such that $yD \ge L \ge xD$ always holds.

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Why would we want to do that? – user53153 Feb 13 '13 at 5:37
A different way to ask the question - suppose you have a "limiting" ball and you put a convex polyhedron inside. Relative to a well known measure like the ball's diameter, how much total material can you embed? Conversely, what is the minimum amount where you don't waste any space? – Meina222 Feb 13 '13 at 6:06

There is no such $y$, since you can make an arbitrarily fine mesh that approximates a sphere; as you refine it, the total length of the edges grows without bound while the diameter stays the same.

To minimize $L$ for given $D$, I'd expect the infimum to be attained by a tetrahedron; the factor is lower for a degenerate tetrahedron with three points converging ($L=3D$) than for a regular tetrahedron ($L=6D$), so the largest possible $x$ should be $3$, but I don't know how to prove this.

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I agree with the $x$ = 3 bound. Intuitively one should be able to argue that 1st of all the diameter is attained at endpoints being vertices, and 2nd that one can cut out a tetrahedron out of the body, where the rest of the edges left out sum up to more than the edges of the tetrahedron. – Meina222 Feb 13 '13 at 16:00
Can you argue more formally why the fine mesh goes to infitnity as oppose to say converging like an integral might. As you get the number of faces to go to infinity, the perimeter of each faces goes to 0. Who wins the race? – Meina222 Feb 13 '13 at 16:02
My last question is actually silly - didn't spend too much time thinking on it. You're right - if you take the sphere, you can cut it in an ever increasing number of circles in parallel planes, which in turn can be approximated by regular polygons whose number of sides goes to infinity. The whole ensemble's total side sum does tend to infinity, and you can always connect parallel ciricle polygons in such a way as to form a tetrahedron. – Meina222 Feb 13 '13 at 16:19

Ultimately this question does not seem to be that interesting. As pointed out by joriki, $x = 3$, but is only attained in the limit by a degenerate tetrahedron, one of whose faces tends to being contained in an ifninitely small disk. And $y$ does not exist (see comments). It's maybe somewhat inetersting to formally prove $x = 3$ but that's about it.

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At least for convex polyhedron, $x = 3$.

Given any convex polyhedron $P$. Since $P$ is compact and the distance function $P \times P \ni (u,v) \mapsto |u-v| \in \mathbb{R}$ is continuous, there exists $u, v \in P$ such that $|u - v| = D(P)$, diameter of $P$.

If $u$ is not extremal, we can find $u_1, u_2 \in P$ s.t. $u = \frac{u_1 + u_2}{2}$. By parallelogram law of euclidean norm, we have:

$$2|u - v|^2 + 2 |\frac{u_1-u_2}{2}|^2 = |u_1-v|^2 + |u_2-v|^2$$

This leads to the contradiction that at least one of $|u_1-v|$ or $|u_2-v|$ is greater than $|u-v| = D(P)$. Same things happens to $v$. This means both $u$ and $v$ are extremal and hence are vertices of $P$.

Now by Steinitz's theorem, the undirected graph formed by the edges and vertices of $P$ is a 3-vertex connected planar graph. By Menger's theorem, there are at least 3 pairwise vertex-disjoint paths joining $u$ and $v$. So $L(P)$, the total length of edges of $P$, is at least $3 D(P)$. This means $x \ge 3$. joriki's example of degenerate tetrahedron shows that the limit $L(P)/D(P) = 3$ is attainable. From this we can conclude $x = 3$ for convex polyhedron.

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