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[Edit I've found a determinant that satisfies the letter of the previous version of this question, but not its "Cayley-Menger" spirit.]

A tetrahedron with face areas $w$, $x$, $y$, $z$ and "pseudo-face" areas $h$, $j$, $k$ has volume $V$ given by

$$\begin{eqnarray} 81 V^4 &=& 2 w^2 x^2 y^2 + 2 w^2 y^2 z^2 + 2 w^2 z^2 x^2 + 2 x^2 y^2 z^2 + h^2 j^2 k^2 \\[4pt] &-&h^2(w^2 x^2+y^2 z^2)-j^2(w^2 y^2+z^2 x^2)-k^2(w^2 z^2+x^2 y^2) \end{eqnarray}$$

All the cool kids in the world of tetrahedral formulas are the determinants or minors of some matrix or another, such as the Cayley-Menger or the Gram. I seek to express the above volume formula in such a form.

Note that, with $A$, $B$, $C$ the dihedral angles between respective face-pairs $(y,z)$, $(z,x)$, $(x,y)$, we have this formula for volume:

$$81 V^4 = 4 x^2 y^2 z^2 \left( 1 - 2 \cos{A} \cos{B} \cos{C} - \cos^2 A - \cos^2 B - \cos^2 C \right) = 4 x^2 y^2 z^2 \left|\begin{array}{ccc} 1 & -\cos C & -\cos B \\ -\cos C & 1 & -\cos A \\ -\cos B & -\cos A & 1 \end{array}\right|$$

We also have what I call the Second Law of Cosines ...

$$h^2 = y^2 + z^2 - 2 y z \cos A \qquad j^2 = z^2 + x^2 - 2 z x \cos B \qquad k^2 = x^2 + y^2 - 2 x y \cos C$$

... which gives rise to this determinant form of the volume formula:

$$81V^4 = \frac{1}{2} \left|\begin{array}{ccc} 2 x^2 & k^2-x^2-y^2 & j^2-z^2-x^2 \\ k^2-x^2-y^2 & 2 y^2 & h^2-y^2-z^2 \\ j^2-z^2-x^2 & h^2-y^2-z^2 & 2 z^2 \end{array}\right| $$

Verifying equality with my target volume formula requires the tetrahedral "Sum of Squares" identity:

$$w^2+x^2+y^2+z^2=h^2+j^2+k^2$$

Now, while I'm pleased to have a reasonably-straightforward determinant equation (thus answering my original question myself), that particular determinant doesn't quite seem a worthy peer of the Cayley-Menger ... perhaps because it doesn't treat all face areas equally.

Can we do better?

(Yes, "better" is subjective, but I think we know what I mean: it should have the Cayley-Menger spirit. I suppose my primary criterion is that the determinant be symmetric in the four face areas and the three pseudo-face areas; ideally, the individual entries are (subjectively) "uncomplicated".)

Here's a possibly-helpful discussion from the previous version of this question:


If $a$, $b$, $c$, $d$, $e$, $f$ are edges of the tetrahedron with $w = \triangle def$, $x = \triangle dbc$, $y = \triangle aec$, and $z = \triangle abf$, then $$ 9 V^2 a^2 = [h,y,z] \qquad 9 V^2 \; b^2 = [j,z,x] \qquad 9 V^2 \; c^2 = [k,x,y]\\ 9 V^2 d^2 = [h,w,x] \qquad 9 V^2 \; e^2 = [j,w,y] \qquad 9 V^2 \; f^2 = [k,w,z] $$ where "$[\bullet]$" is the "Heronic product" $$\begin{eqnarray} [p,q,r] &:=& (p+q+r)(-p+q+r)(p-q+r)(p+q-r)\\ &=&-p^4-q^4-r^4+2p^2q^2+2q^2r^2+2r^2p^2 \end{eqnarray}$$ Consequently, we can substitute into the Cayley-Menger determinant formula to write $$\begin{eqnarray} 288 V^2 &=& \left| \begin{array}{ccccc} 0 & a^2 & b^2 & c^2 & 1 \\ a^2 & 0 & f^2 & e^2 & 1 \\ b^2 & f^2 & 0 & d^2 & 1 \\ c^2 & e^2 & d^2 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \end{array}\right| \\ &=& \frac{1}{(9V^2)^3}\left| \begin{array}{ccccc} 0 & [h,y,z] & [j,z,x] & [k,x,y] & 1 \\ \; [h,y,z] & 0 & [k,w,z] & [j,w,y] & 1 \\ \; [j,z,x] & [k,w,z] & 0 & [h,w,x] & 1 \\ \; [k,x,y] & [j,w,y] & [h,w,x] & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \end{array}\right|\\ &=:&\frac{1}{729V^6} \mathrm{det} M \end{eqnarray} $$ Since $\mathrm{det}M$ is proportional to the 8th-power of $V$ ---and therefore, the square of the target formula--- perhaps I should attempt to construct $\sqrt{M}$.


My investigations of matrix $M$ have been fruitless. (Note, though, that despite its complexity ---and the fact that it generates the square of what I want--- $M$ has the kind of symmetry I would expect from a properly Cayley-Menger-esque result.) I think it's time to add a bounty to this question. [Bounty expired.]


Edit Here's a tantalizing alternative "(negative-)square-of-what-I-want" matrix:

$$N := \left|\begin{array}{cccccc} 0 & z^2 & y^2 & h^2 & y^2 & z^2 \\ z^2 & 0 & x^2 & x^2 & j^2 & z^2 \\ y^2 & x^2 & 0 & x^2 & y^2 & k^2 \\ h^2 & x^2 & x^2 & 0 & w^2 & w^2 \\ y^2 & j^2 & y^2 & w^2 & 0 & w^2 \\ z^2 & z^2 & k^2 & w^2 & w^2 & 0 \end{array}\right| \qquad \mathrm{det}N = -\left(81 V^4\right)^2$$

The pattern in the entries can be described thusly:

Index the rows and columns with edges $(a,b,c,d,e,f)$. Then the $pq$-th element ($p\ne q$) corresponds to the face determined by edges $p$ and $q$. (Opposite edges $p$ and $q$ determine a pseudo-face.)

The minors of $N$ are very interesting. Writing $N_p$ for the matrix obtained by removing row and column $p$ from $N$:

$$\begin{array}{ccc} \mathrm{det} N_a = 2 w^2 x^2 \cdot 81 V^4 & \mathrm{det} N_b = 2 w^2 y^2 \cdot 81 V^4 & \mathrm{det} N_c = 2 w^2 z^2 \cdot 81 V^4 \\ \mathrm{det} N_d = 2 y^2 z^2 \cdot 81 V^4 & \mathrm{det} N_e = 2 z^2 x^2 \cdot 81 V^4 & \mathrm{det} N_f = 2 x^2 y^2 \cdot 81 V^4 \\ \end{array}$$

Note, for instance in the case of $N_a$, that faces $w$ and $x$ meet along edge $d$, which is opposite edge $a$.

Moreover, with $N_{pq}$ the matrix obtained by removing row $p$ and column $q$ (with $p\ne q$):

$$\mathrm{det} N_{ab} = - w^2 \left( k^2 - x^2 - y^2 \right) \cdot 81 V^4 = 2 w^2 x y \cos C \cdot 81 V^4 \qquad \text{etc}$$

I get the sense that this gets me closer to my goal.

Edit. Getting even closer. Augmenting matrix $N$ into more Menger-like form gives

$$\begin{align} P &:= \left|\begin{array}{ccccccc} 0 & z^2 & y^2 & h^2 & y^2 & z^2 & 1 \\ z^2 & 0 & x^2 & x^2 & j^2 & z^2 & 1 \\ y^2 & x^2 & 0 & x^2 & y^2 & k^2 & 1 \\ h^2 & x^2 & x^2 & 0 & w^2 & w^2 & 1 \\ y^2 & j^2 & y^2 & w^2 & 0 & w^2 & 1 \\ z^2 & z^2 & k^2 & w^2 & w^2 & 0 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 & 0 \end{array}\right| \\[6pt] \mathrm{det} P &= 2\left(w^4+x^4+y^4+z^4-h^2j^2-j^2k^2-k^2h^2\right) \cdot 81 V^4 \end{align}$$

This may be about as good as I can expect. I'd prefer, though, that the multiplied polynomial --if there must be one-- have an unambiguous sign (and vanish only trivially).

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Nonsensical thought: start with the trivial matrix you wrote $\begin{pmatrix}\cdot & 0 \\ 0 & 1\end{pmatrix}$ and apply determinant-preserving row and column operations? –  user2468 Mar 29 '12 at 19:55
    
What are '"pseudo-face" areas'? –  Salech Alhasov Mar 29 '12 at 20:03
2  
@Salech Pseudo-face areas are the areas of a tetrahedron's pseudo-faces! You can think of a pseudo-face as the shadow when projecting the tetrahedron into a plane parallel to a particular pair of opposite edges. The pseudo-face areas also "complete" a Law of Cosines for tetrahedra involving face areas and dihedral angles. For information, see here: daylateanddollarshort.com/math/pdfs/pseudofaces.pdf –  Blue Mar 29 '12 at 20:12
    
@Day Late Don, thank you for that information! –  Salech Alhasov Mar 29 '12 at 20:30
1  
@Salech: There's more where that came from! :) daylateanddollarshort.com/math/research-dron.html –  Blue Mar 29 '12 at 20:33
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2 Answers 2

up vote 3 down vote accepted

By Jove, I think I've got it!

I rectified the non-symmetricality of my original, non-symmetric determinant by adding an obvious $w$-specific row and column. This didn't work, but I thought, "How about doing the '1's thing, too?" Bingo!

$$\left|\begin{array}{ccccc} 2 w^2 & h^2 - w^2 - x^2 & j^2 - w^2 - y^2 & k^2 - w^2 - z^2 & 1 \\ h^2 - w^2 - x^2 & 2 x^2 & k^2 - x^2 - y^2 & j^2 - z^2 - x^2 & 1 \\ j^2 - w^2 - y^2 & k^2 - x^2 - y^2 & 2 y^2 & h^2 - y^2 - z^2 & 1 \\ k^2 - w^2 - z^2 & j^2 - x^2 - x^2 & h^2 - y^2 - z^2 & 2 z^2 & 1 \\ 1 & 1 & 1 & 1 & 0 \end{array}\right| = - 32 \cdot 81 V^4$$

The great thing about those "$1$"s ---especially with the "$0$" in the corner--- is that they facilitate quick-and-easy determinant-preserving row and column clean-up. I'm free, for instance, to add $w^2$ to the polynomials in the first row, and again in the first column; $x^2$ in the second row and second column; etc.

$$\Delta^{\star} := \left|\begin{array}{ccccc} 4 w^2 & h^2 & j^2 & k^2 & 1 \\ h^2 & 4 x^2 & k^2 & j^2 & 1 \\ j^2 & k^2 & 4 y^2 & h^2 & 1 \\ k^2 & j^2 & h^2 & 4 z^2 & 1 \\ 1 & 1 & 1 & 1 & 0 \end{array}\right| = - 32 \cdot 81 V^4$$

(I never suspected that the result I was after would be dominated by pseudo-face elements!)

I'm not sure this is the very-best counterpart to the matrix of the Cayley-Menger determinant (call it $\Delta$) ---the minors of my matrix aren't as interesting as the minors of the $\Delta$ matrix--- but at least I have a clean way to express a counterpart of Menger's Theorem:

Menger's Theorem. An ordered sextuple of edge lengths $(a,b,c,d,e,f)$ determines an actual, non-degenerate tetrahedron --with edges $a,b,c$ opposite respective edges $d,e,f$-- if and only if

  • The Heronic products $[d,e,f]$, $[d,b,c]$, $[a,e,c]$, $[a,b,f]$ are all positive.
  • The Cayley-Menger determinant, $\Delta$, is positive.

And now, my "hedronometric" (face-based) form ...

Hedronometric Menger's Theorem. An ordered septuple of areas $(w,x,y,z;h,j,k)$ determines an actual, non-degenerate tetrahedron --with pseudo-faces $h,j,k$ associated with respective face-pair pairs $(w,x;y,z)$, $(w,y;z,x)$, $(w,z;x,y)$-- if and only if

  • $w^2+x^2+y^2+z^2 = h^2 + j^2 + k^2$
  • The Heronic products $[h,w,x]$, $[h,y,z]$, $[j,w,y]$, $[j,z,x]$, $[k,w,z]$, $[k,x,y]$ are all positive.
  • The determinant $\Delta^{\star}$ is negative.

Time to update my bloog post on the subject!

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This may be related: math.stackexchange.com/questions/394031/… –  Daved May 17 '13 at 0:41
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Ok, so starting with that interesting matrix you found: $$N := \left|\begin{array}{cccccc} 0 & z^2 & y^2 & h^2 & y^2 & z^2 \\ z^2 & 0 & x^2 & x^2 & j^2 & z^2 \\ y^2 & x^2 & 0 & x^2 & y^2 & k^2 \\ h^2 & x^2 & x^2 & 0 & w^2 & w^2 \\ y^2 & j^2 & y^2 & w^2 & 0 & w^2 \\ z^2 & z^2 & k^2 & w^2 & w^2 & 0 \end{array}\right| \qquad \mathrm{det}N = -\left(81 V^4\right)^2$$

You can start to remove the negative part of the determinant, if you ...

Reverse the order of columns (three interchanges) $$N := \left|\begin{array}{cccccc} z^2 & y^2 & h^2 & y^2 & z^2 & 0 \\ z^2 & j^2 & x^2 & x^2 & 0 & z^2 \\ k^2 & y^2 & x^2 & 0 & x^2 & y^2 \\ w^2 & w^2 & 0 & x^2 & x^2 & h^2 \\ w^2 & 0 & w^2 & y^2 & j^2 & y^2 \\ 0 & w^2 & w^2 & k^2 & z^2 & z^2 \end{array}\right| \qquad \mathrm{det}N = \left(81 V^4\right)^2$$

Or reverse the order of rows (three interchanges) $$N := \left|\begin{array}{cccccc} z^2 & z^2 & k^2 & w^2 & w^2 & 0 \\ y^2 & j^2 & y^2 & w^2 & 0 & w^2 \\ h^2 & x^2 & x^2 & 0 & w^2 & w^2 \\ y^2 & x^2 & 0 & x^2 & y^2 & k^2 \\ z^2 & 0 & x^2 & x^2 & j^2 & z^2 \\ 0 & z^2 & y^2 & h^2 & y^2 & z^2 \end{array}\right| \qquad \mathrm{det}N = \left(81 V^4\right)^2$$

For every interchange of rows or columns, the determinant multiplies by $-1$.

And... That's all I got. If I put either matrix into mathematica to find the square root, the evaluation continues indefinitely. So now the quest is now to find the square root. Good Luck!

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Thanks for the input, but the negative sign wasn't the roadblock (and fixing it isn't worthy of the bounty). –  Blue Nov 18 '12 at 1:44
    
I didn't put this up for the bounty... :| –  Zchpyvr Nov 18 '12 at 1:49
    
Rather, I wanted to help others find a possible solution based on my initial approach. And if my end result isn't helpful or conducive, then oh well. I conceded my inability to solve it at the end of my answer. So I agree, this answer isn't worthy of the bounty, and shouldn't be selected. –  Zchpyvr Nov 18 '12 at 1:52
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