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More precisely, can an orientable closed compact 2-manifold embedded in $\mathbb{R}^3$ be, up to an isometry, uniquely identified by its surface area and the volume it encloses? Conversely, can a counter example be given to disproof this claim?

In the answer to an earlier question related to the isoperimetric inequality a proof for the statement that "sphere is the only closed surface in $\mathbb{R}^3$ that minimizes the surface area to volume ratio" has been outlined. That and the beginning of this post on Brunn-Minkowski inequality on Terence Tao's What's New made me curious about the existence (or lack thereof) more stringent constraints akin to the isoperimetric inequality.

As a first step towards constructing a counterexample, I examined the simple case of rectangular cube having the same surface area and volume as a cube. I ended up with the following equation: $$ (\underbrace{a b c}_{volume})^\frac{2}{3} = \frac{1}{6}\times\underbrace{(2ab+2bc+2ac)}_{\text{surface area}}\,, $$ where $a$, $b$, and $c$ are the dimensions of the rectangular cube. (For a cube of the sidelength $d$, $(d^3)^{2/3}=\frac{1}{6}\,6\,d^2$; hence the above equation.) I have no idea how to proceed with the search for any positive definite solutions to the above problem. I suppose it can be simplified by invoking (isotropic) scaling to set one of the dimensions of the rectangular cube to $1$ and the following equation: $$ (a b)^2 = \left(\frac{2}{6}\right)^3(ab+b+a)^3\,. $$

S.G.
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  • Since a cube of side length $d$ has volume $d^3$ and surface area $6d^2$, shouldn't your first equation be just $(abc)^{2/3} = 2ab + 2bc + 2ac$? Simplifying your second equation to $(ab)^2 = 8(ab + b + a)^3$. – Reese Johnston Nov 19 '17 at 05:08
  • One dimension down, note that an equilateral triangle with side $a$ will have the same perimeter and area as any number of rectangles. You get a quadratic to solve and I believe the problem will have have infinitely many solutions. – Ted Shifrin Nov 19 '17 at 05:10
  • OK, in your final formula, notice that $(ab + b + a)^3$ has all positive coefficients, and it has an $a^2b^2$ term in it with coefficient at least $1$. That means that the right-hand side is strictly larger than $(ab)^2$. So there's no solution with these particular shapes. Inspired by Ted's suggestion, maybe try comparing a regular tetrahedron with a cube or rectangular cube? – Reese Johnston Nov 19 '17 at 05:15
  • @Reese, Thank you for the comment! Just corrected it. – S.G. Nov 19 '17 at 15:53
  • @Ted Shifrin, good point. I should have thought about that. – S.G. Nov 19 '17 at 16:00

2 Answers2

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Your simple example is in a good direction but too simple. We know the cube is the most efficient parallepiped in terms of volume to surface ratio so you won't get another parallelepiped to match both. But we should be able to find two parallelepipeds, one a flat pancake and one a long needle that have the same surface and volume. Let one be $1 \times 1 \times 100$ and the other $a \times a \times b$ with $a \gt b$. Then we can write two equations and solve them(thanks Alpha) $$a^2b=100\\2a^2+4ab=402\\2a^2+\frac {400}a=402\\a=\frac 12(3\sqrt{89}-1)\approx 13.651$$

Ross Millikan
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Here is a counterexample:

Attach 3 tetrahedrons from face to face. There is only one way of doing this and this surface has a volume $3V$ and surface area $8A$ where $V$ is the volume of a tetrahedron and $A$ is the area of one of its faces. Notice there will be a "middle" tetrahedron touching two others and 2 "lone" tetrahedrons. If you attach a 4th tetrahedron to the middle or one of the lone tetrahedrons you get volume $4V$ and area $10A$ in both cases yet the two shapes would not be isometric.

Ezzaddin
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