Say that two polyhedra in $\mathbb{R}^3$ have isomeasures (my terminology) if they have: the same volume, the same surface area, the same sum of all edge lengths, and the same number of vertices. The generalization to $\mathbb{R}^d$ requires the identical sums of the $k$-dimensional measures of all $k$-faces on the exterior of the polytope.

For example, a unit cube has volume $1$, surface area $6$, edge-length sum $12$, and $8$ vertices.

Q. Are there incongruent but isomeasure simplicies in dimensions $d \ge 3$?

There are isomeasure triangles:

Here the blue triangle has base length $2$ and height $1$, while the red triangle has base $\frac{1}{2} \left(-1+\sqrt{2}+\sqrt{6 \sqrt{2}-5}\right)$ and height $\frac{1}{2} \left(-1+\sqrt{6 \sqrt{2}-5}+\sqrt{2 \left(6 \sqrt{2}-5\right)}\right)$, approximately $1.14055$ and $1.75354$ respectively.

Gerhard Paseman pointed out (in a comment to a now deleted question) that there are isomeasure prisms, but I have been unsuccessful in constructing isomeasure tetrahedra.

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    $\begingroup$ Now that this question has appeared, I can add my second comment: fix some scalene triangle and sum of edge lengths L, and consider the compact figure of all points (x,y,z) such that the suggested simplex has sum of edge lengths L. Now to each such point, associate the sum of areas and volume. There will be many cases when two points on the figure have the same ordered pair (a,v): the trick is to find those when v is nonzero. Fixing a v, one finds level curves on the surface at the same height with constant volume. Now, ... Gerhard "Leaves The Rest To You" Paseman, 2014.07.13 $\endgroup$ – Gerhard Paseman Jul 14 '14 at 0:48
  • $\begingroup$ @GerhardPaseman: That approach seems promising... $\endgroup$ – Joseph O'Rourke Jul 14 '14 at 1:08
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    $\begingroup$ I assume you are familiar with Schanuel's excellent paper "What is the length of a potato?", but your notion of isomeasure reminded me of it, so in the off chance you don't know that paper, I thought I'd mention it. $\endgroup$ – Theo Johnson-Freyd Jul 14 '14 at 3:02
  • $\begingroup$ Schanuel: Springer link. $\endgroup$ – Joseph O'Rourke Jul 14 '14 at 10:36

There are $6$ degrees of freedom for a tetrahedron ($6$ edge lengths that can vary independently in some open set). The surface area, volume and sum of edge lengths are just $3$ parameters. So for a generic tetrahedron there should be a $3$-parameter family of tetrahedra with the same surface area, volume and sum of edge lengths.


For example, let's fix the base as a triangle with edges of length $d_{12} = 2$, $d_{23} = 3$ and $d_{13} = 4$ (and thus area $\sqrt{135}/4$). If the other three edge lengths are $d_{14} = a$, $d_{24}=b$ and $d_{34} = c$, then we have sum of edge lengths, total area and volume $$ \eqalign{ E(a,b,c) = &9 + a + b + c\cr A(a,b,c) =& \dfrac{1}{4} \left(\sqrt {135}+\sqrt {-{b}^{4}+2\,{b}^{2}{c}^{2}-{c}^{4}+32\,{b}^{2}+32\, {c}^{2}-256}\right.\cr &+\sqrt {-{a}^{4}+2\,{a}^{2}{c}^{2}-{c}^{4}+18\,{a}^{2}+18 \,{c}^{2}-81}\cr &\left.+\sqrt {-{a}^{4}+2\,{a}^{2}{b}^{2}-{b}^{4}+8\,{a}^{2}+8\, {b}^{2}-16}\right)\cr V(a,b,c) =& \dfrac{1}{12} \sqrt{-16\,{a}^{4}+21\,{b}^{2}{a}^{2}+11\,{a}^{2}{c}^{2}-9\,{b}^{4}-3\,{b}^{ 2}{c}^{2}-4\,{c}^{4}-48\,{a}^{2}+99\,{b}^{2}+84\,{c}^{2}-576} \cr}$$ If for example we take $a_0 = 3$, $b_0 = 4$, $c_0 = 5$, then in the $ab$ plane (with $c=12-a-b$ to get $E(a,b,c) = E(a_0,b_0,c_0)$) we get the curves plotted here ($A(a,b,c) = A(a_0,b_0,c_0)$ in blue, $V(a,b,c) = V(a_0,b_0,c_0)$ in red):

enter image description here

Evidently there are four intersections, and thus four values of $(a,b,c)$ giving the same edge length sum, total area and volume with this base. One is $(a,b,c) = (3,4,5)$, the others are numerically close to $ (4.119343882, 2.962425391, 4.918230727)$, $ (3.309158209, 4.827583483, 3.863258308)$ and $ (4.431953503, 4.534086354, 3.033960143)$.

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  • $\begingroup$ Thanks, Robert, for this convincing high-level viewpoint! $\endgroup$ – Joseph O'Rourke Jul 14 '14 at 9:57
  • $\begingroup$ So, if the base triangle face is fixed, generically there will be just a finite number of isomeasure tetrahedra. $\endgroup$ – Joseph O'Rourke Jul 14 '14 at 10:35
  • $\begingroup$ I think generically there will be more. Fix the base triangle B, a height h, and a normal direction. Points in a plane at height h from B determine a fixed volume, and that plane is divided into many smooth compact level sets, each set corresponding to a constant sum of edge lengths. For each nontrivial level set, there are often for uncountably many values of the sum of areas at least two and perhaps more points on the level set which share that value for sum of areas. I think you get uncountably many which aren't congruent. Gerhard "Ask Me About Twisty Thinking" Paseman, 2014.07.14. $\endgroup$ – Gerhard Paseman Jul 15 '14 at 0:30
  • $\begingroup$ If you mean, for a given tuple of base, volume, edge sum and area sum, yes, there will be finitely many. There will be uncountably many classes of such isomeasure objects, even if you fix volume and (say) sum of edge lengths. Gerhard "Hopes He Got It Right" Paseman, 2014.07.14 $\endgroup$ – Gerhard Paseman Jul 15 '14 at 0:38

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