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Everyone knows Heron formula for the area of a triangle as a function of its sides. Moving to three dimensions, what is the maximum volume of a tetrahedron with given four face areas $a$, $b$, $c$, and $d$?

Obviously, this is an optimization problem. For example, with $a=b=c=d=1$ one can construct a tetrahedron with any volume between zero and $\frac{2\sqrt2}{3\sqrt{3\sqrt3}}$. The latter is the volume of the regular tetrahedron with side 1, and it is this tetrahedron that gives the maximum volume (from symmetry considerations). The zero volume is presented by any degenerate tetrahedron (such as the "tetrahedrons" with vertices given by $$ A=(0,0,0),\quad B=(2,0,0),\quad C=(1+e,1,0),\quad D=(1-e,-1,0), $$ for any real $e$, all have four faces 1 and volume zero).

A degenerate tetrahedron can not be constructed for given face areas when $S_1\not=S_2+S_3+S_4$ and $S_1+S_2\not=S_3+S_4$ for any assignment of $a,b,c,d$ to $S_1,S_2,S_3,S_4$. Then a complementary problem is what is the minimum volume of a tetrahedron with given four faces $a$, $b$, $c$, and $d$?

The "maximum" problem somewhat resembles "What is the maximum area of the quadrilateral given by its sides $a$, $b$, $c$, and $d$?", for which Brahmagupta's formula gives the solution. See also this question, and you can find numerous other pieces of study of this planar problem in the Internet. In this problem one can intuitively guess that the four vertices of the maximum area quadrilateral should live on a circle, and then proceed with the derivation and proof of the formula. For the maximum volume of tetrahedron, what intuitive guess can be made?

I suspect the maximum volume is obtained together with the maximum sum of the six digedral angles at the six edges of the tetrahedron. In the above example, the sum is $2\pi\approx6.28$ for the degenerate tetrahedra and $6\cdot2\arcsin\frac1{\sqrt3}\approx7.39$ for the regular tetrahedron. To support this intuition, the volume is given by any of the six identities like this one: $$ 3V\cdot AB=2c d\sin AB_\angle, $$ where $c$ is the area of the face opposing vertex $C$, and $AB_\angle$ is the dihedral angle at edge $AB$, so all the dihedral angles should be maximized.

From this guess on, the radius of the circumscribed sphere should be minimized, perhaps. I can not prove this guess, but it concords with the expression of the area of a triangle given by its sides: $S=4abc/R$ (if finding the area of the triangle by its sides were an optimization problem, then you would minimize the circumcircle).

The problem was originally given to me by Ernst D. Krupnikov, who claims its closed form solution is cumbersome and difficult to derive without a help of a computer algebra system. Here is a particular case of the problem you may find amusing to solve: what is the maximum volume of the tetrahedron with face areas 120, 150, 160, 250?

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  • $\begingroup$ I suppose that by "side" you mean "face", don't you? But in that case I don't understand your zero-volume example, as the four faces don't have the same area. $\endgroup$ – Aretino Jan 5 '18 at 11:25
  • $\begingroup$ For the particular case, it is a right tetrahedron with edge of lengths $15,16,20$ attached at the apex and volume $800$. It is indeed amusing to solve but the math I use involve finding the root of some quartic polynomial, That means there is a closed form in terms of radicals but it is something very ugly to write it down... $\endgroup$ – achille hui Jan 5 '18 at 12:52
  • $\begingroup$ Isn't this just a maximization problem of the Cayley-Menger determinant? mathworld.wolfram.com/Cayley-MengerDeterminant.html $\endgroup$ – Jack D'Aurizio Jan 5 '18 at 14:13
  • $\begingroup$ I indeed mistaken "side" for "face", thank you. And I also confused you by using same letters to denote face areas and vertices, sorry. I shall fix the text $\endgroup$ – Dmitry Baksheev Jan 6 '18 at 3:03
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Given four face-areas, the tetrahedron of maximum volume is the one in which opposite edges are perpendicular.

(I call these "perfect" tetrahedra, although they are called "orthocentric" in the literature.)

You can find a proof of this in the 1975 Pacific Journal of Mathematics article "The Orthocentric Simplex as an Extreme Simplex (PDF)" by Leon Gerber. (As the title suggests, the results apply to arbitrary-dimensional figures, not merely three-dimensional tetrahedra.) The argument takes a few pages, but involves fairly elementary vector techniques, with a touch of calculus.

In my own note, "Heron-like Hedronometric Results for Tetrahedral Volume (PDF)", I show that the volume of a perfect tetrahedron with face areas $W$, $X$, $Y$, $Z$ is given by what I call the "Heron Quartic": $$\begin{align} 0 = 27 U^4 &+ U^3 \left( s_\star s_1 + 8(4 s_1 s_2 - 27 s_3) \right) \\ &- U^2 \left( s_\star^2 s_2 + 12 s_\star(3 s_1 s_3-28 s_4) - 48(9 s_3^2-16 s_2 s_4) \right) \\ &+ U\phantom{^3} \left(s_\star^3 s_3+40 s_\star s_1 s_4-576 s_\star s_3 s_4 + 1536 s_1 s_4^2 \right) \\ &- \phantom{U^3} s_4 ( s_\star^2 - 64 s_4)^2 \end{align}$$

where $U = 81 V^4$, and the $s_i$ are symmetric polynomials in the squares of the face-areas ( $w=W^2$, etc): $$s_1 := w+x+y+z \qquad s_2 := w x + w y + w z + x y + x z + y z$$ $$s_3 := w x y+w x z+w y z+ x y z \qquad s_4 := w x y z$$ $$s_\star := 4s_2 - s^2 = - w^2-x^2-y^2-z^2+2wx+2wy+2wz+2xy+2xz+2yz$$

In the case where $W^2 = X^2 + Y^2 + Z^2$, the polynomial factors as

$$0 = (U - 4 X^2 Y^2 Z^2) (\ldots\text{extraneous cubic}\ldots)$$

so that

$$9 V^2 = 2 XYZ$$ as mentioned in this answer. The maximizing figure is a "right-corner tetrahedron", with $\cos A = \cos B = \cos C = 0$. (Note: In general, a tetrahedron satisfying $W^2=X^2+Y^2+Z^2$ isn't necessarily a right-corner tetrahedron, but a perfect one must be.) Your "amusing" example falls into this category (with $W = 250$), so that $$9 V^2 = 2\cdot 120\cdot 150 \cdot 160 \qquad\to\qquad V = 800$$


BTW: Although the Heron Quartic is guaranteed (by the Descartes Rule of Signs) to have a positive root $U$, it's actually possible for it to have as many as three of them. I haven't been able to prove that there's an unambiguous choice corresponding to the given tetrahedron's volume. Even so, it's possible to prove that perfect volume is at least "locally" maximal using a few hedronometric relations, giving an alternative to Gerber's vector-based approach. We simply optimize the hedronometric volume formula ...

$$\begin{align} 81 V^4 &= H^2 J^2 K^2 - 2 ( W X - Y Z )( W Y - Z X )( W Z - X Y ) \\ &-H^2 ( W X - Y Z)^2 - J^2 ( W Y - Z X)^2 - K^2 ( W Z - X Y )^2 \end{align}$$

(where $H$, $J$, $K$ are the tetrahedron's pseudofaces), subject to the sum-of-squares identity ... $$W^2 + X^2 + Y^2 + Z^2 = H^2 + J^2 + K^2$$

... finding that stationary points (more-specifically, local maxima) must correspond to perfect tetrahedra. See my "Heron-like Results" note for details.

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Ernst D. Krupnikov suggests the following lemma:

If the four faces $a$, $b$, $c$, $d$, of a tetrahedron satisfy identity $a^2+b^2+c^2=d^2$, then the volume $V$ of the tetrahedron satisfies identity $9V^2=2abc$.

Can you prove the lemma? Does the reverse implication hold?

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