I'm asking and answering this question to provide a partial answer to this question and a comment on this answer at MO.

The isoperimetric ratio $\mu$ of a solid is the ratio $A^3/V^2$, where $A$ is the surface area of the solid and $V$ is its volume. By the isoperimetric inequality, the minimal isoperimetric ratio achieveable is that of the sphere,

$$ \mu_\infty=\frac{(4\pi)^3}{(4\pi/3)^2}=36\pi\approx113.1\;. $$

For any number $n$ of vertices, we can ask for the minimal isoperimetric ratio $\mu_n$ of a polyhedron with $n$ vertices, and we'll have $\mu_m\gt\mu_n\gt\mu_\infty$ for $m\lt n$. The present question asks for $\mu_5$.

  • $\begingroup$ @Asaf: It does -- though I've been picking up speed again these days after being too busy with other things... $\endgroup$ – joriki May 7 '13 at 8:09

Without loss of generality, we can assume that the faces are triangles; the case in which some of the triangles are coplanar and form a larger face is included as a special case.

With $k$ triangles, there are $3k/2$ edges, so by Euler's polyhedron formula $5-3k/2+k=2$ and thus $k=6$. Thus there are $9$ edges, so all pairs of vertices are connected by an edge except one. We can consider the triangle formed by the three vertices not contained in that pair as the base, and the two vertices in the pair are on different sides of this triangular base and form two tetrahedra with it.

Let the volume $V$ be given, and also let the base and a partition of the volume into volumes of the two tetrahedra be given. This constrains the other two points to planes parallel to the base, and we can ask where in those planes they must be to minimize the surface area.

Consider one of the points placed such that its projection onto the base is the incentre of the base. Then all three triangles it forms have the same altitude and form the same angle with the base. Thus if we move the point in the plane, the rate of change in the total area of these three triangles will be proportional to the sum $\sum_il_i\cos\theta_i$, with $l_i$ the side lengths of the base and $\theta_i$ the angles between the inward normals of the base and the direction in which the point is moved. But this is just the total movement along a direction perpendicular to that direction as we move around the base, so this sum vanishes. Thus the position above the incentre is a stationary point for the total area of the three triangles, and by writing the total area out in coordinates with the base in the $xy$ plane it can be shown that it's in fact the only stationary point and thus the minimum.

The radius of the incircle is $ r=B/s $, where $B$ is the area of the base and $s$ is its semiperimeter. Given the height $d$ of the point above the base, the altitudes of all three circles are $h=\sqrt{d^2+r^2}$, and thus the total area of the three triangles is $sh=s\sqrt{d^2+(B/s)^2}=\sqrt{(ds)^2+B^2}$. For given $B$ and $d$, and thus given volume of the tetrahedron, this is minimal for minimal $s$, and this is achieved for an equilateral triangle. Thus we may conclude that to achieve the minimal isoperimetric ratio the base must be an equilateral triangle and the other two points must lie above its centre.

A certain partition of the total volume into two volumes for the two tetrahedra corresponds to a partition of the total height of the two tetrahedra into their individual heights. The total area $\sqrt{(d_1s)^2+B^2}+\sqrt{(d_2s)^2+B^2}$ of the six faces of the polyhedron is minimal in the symmetric case $d_1=d_2=d$.

Thus, the polyhedron is determined by the side length $a$ of the equilateral triangle and the height $d$ of the two tetrahedra. In terms of these two lengths, we have

$$ \begin{align} s&=\frac32a\;,\\ B&=\frac12a\frac{\sqrt3}2a=\frac{\sqrt3}4a^2\;,\\ A&=2\sqrt{(ds)^2+B^2}\;,\\ V&=2\cdot\frac13dB\;. \end{align} $$

Thus, with $\lambda:=d/a$ the isoperimetric ratio is

$$ \begin{align} \frac{A^3}{V^2} &= \frac{\left(9\lambda^2+\frac34\right)^{3/2}}{\left(\frac23\frac{\sqrt3}4\lambda\right)^2}\\ &= \frac32\left(36\lambda^{2/3}+3\lambda^{-4/3}\right)^{3/2}\;. \end{align} $$

The minimum is attained at $\lambda=1/\sqrt6$, leading to $\mu_5=3^5=243$, as conjectured in the MO answer.

Here's a table comparing the result to the isoperimetric ratios of a regular tetrahedron, a regular octahedron and a sphere:

$$ \begin{array}{c|rl} n&\mu_n\\\hline 4&2^3\cdot3^{7/2}&\approx374.1\\ 5&3^5&=243\\ 6&2^2\cdot3^{7/2}&\approx187.1\\ \infty&36\pi&\approx113.1 \end{array} $$

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.