We know 'triangle inequality'. I'm interested in the generalization of this inequality.

Here is my question.

Question: How can we represent a necessary and sufficient condition for each positive number $a, b, c, d$ being each area of four faces of a tetrahedron?

I've tried to get some kind of inequality, but I'm facing difficulty.

  • $\begingroup$ Have you taken a look at Heron's formula? With four (unknown) sides, and four (known) areas to express, I believe it would be close to sufficient. $\endgroup$
    – Arthur
    Aug 28, 2013 at 7:39

4 Answers 4


The Triangle Inequality is an aspect of the Law of Cosines.

$$\begin{align} a \le b + c \quad b \le c + a \quad c \le a + b \quad &\implies \qquad |b-c| \le a \le b+c\\ &\implies b^2 + c^2 - 2 b c \le a^2 \le b^2 + c^2 + 2 b c \\ &\implies \exists\;\theta, \; 0 \leq \theta \leq \pi \quad \text{s.t.} \quad a^2 = b^2 + c^2 - 2 b c \cos\theta \end{align}$$

where it turns out that "$\theta$" is exactly the angle that fits into the appropriate corner of the triangle. (Proof left to reader.)

For tetrahedra, we have this Law of Cosines involving face areas $W$, $X$, $Y$, $Z$ and dihedral angles $A$, $B$, $C$ (meeting at the vertex opposite face $W$) and $D$, $E$, $F$ (surrounding face $W$). For instance,

$$W^2 = X^2 + Y^2 + Z^2 - 2 Y Z \cos A - 2 Z X \cos B - 2 X Y \cos C$$

(with $A$ between faces $Y$ & $Z$, etc). Clearly, this gives the necessary condition $$W \leq X + Y + Z$$ and its kin, although these are not sufficient.

Interestingly, when you've come to know tetrahedra like I know them, you realize that there are in fact seven faces to each of these things: the four familiar ("standard") ones, and three that I call "pseudo-faces". A pseudo-face is the projection of the tetrahedron into a plane parallel to a pair of opposite edges. I denote the areas of these $H$, $J$, $K$.

More-interestingly, there's a Law of Cosines involving pseudo-faces: $$\begin{align} Y^2 + Z^2 - 2 Y Z \cos A \quad &= H^2 = \quad W^2 + X^2 - 2 W X \cos D \\ Z^2 + X^2 - 2 Z X \cos B \quad &= J^2 = \quad W^2 + Y^2 - 2 W Y \cos E \\ X^2 + Y^2 - 2 X Y \cos C \quad &= K^2 = \quad W^2 + Z^2 - 2 W Z \cos F \end{align}$$ which, together with the Law of Cosines above, proves this Sum-of-Squares identity: $$W^2 + X^2 + Y^2 + Z^2 = H^2 + J^2 + K^2 \qquad(1)$$

Now, given seven ostensible areas (four standard and three pseudo), the Law of Cosines leads to Triangle-Inequality-like conditions, such as $$|Y-Z| \leq H \leq Y+Z \qquad\qquad |W-X| \leq H \leq W + X \qquad (2)$$ (and likewise for $J$ and $K$). Of course, the areas must also satisfy the Sum-of-Squares identity $(1)$. But even this collection of conditions isn't sufficient to determine a tetrahedron. We need one more: $$\begin{align} 0 \quad \leq \quad &2 W^2 X^2 Y^2 + 2 W^2 Y^2 Z^2 + 2 W^2 Z^2 X^2 + 2 X^2 Y^2 Z^2 + H^2 J^2 K^2 \\ &-H^2\left(W^2 X^2+Y^2 Z^2\right) -J^2\left(W^2 Y^2+Z^2 X^2\right) -K^2\left(W^2 Z^2+X^2 Y^2\right) \qquad (3) \end{align}$$

When the right-hand side is in fact non-negative, it gives $81 V^4$, where $V$ is the volume of the tetrahedron.

Together, $(1)$, $(2)$, $(3)$ constitute my analogue of Menger's Theorem (which outlines conditions under which six edge-lengths can make a tetrahedron). For further information on this result, see my Bloog post "A Hedronometric Theorem of Menger".

FYI: The Bloog also has a number of other notes on "Hedronometry" ---my name for the dimensionally-enhanced trigonometry of tetrahedra--- both Euclidean and non-. (The earliest notes need some editing love. I was just using them for TeX practice waaaay-back-when. :)

So, one way to answer your question is this:

$W$, $X$, $Y$, $Z$ can be areas of faces of a tetrahedron if and only if there exist non-negative $H$, $J$, $K$ satisfying $(1)$, $(2)$, $(3)$.

  • $\begingroup$ Thank you. I guess your idea would be one possible answer to my question. $\endgroup$
    – mathlove
    Aug 28, 2013 at 12:06
  • $\begingroup$ As @achillehui's answer demonstrates, the collection of inequalities such as $W \leq X + Y + Z$ are necessary and sufficient for $W$, $X$, $Y$, $Z$ are areas of faces of some tetrahedra. On the sufficiency side, the inequalities serve to justify the existence of the pseudo-face areas $H$, $J$, $K$. (But, note that, in general, these pseudo-areas are not ---and cannot be--- uniquely determined by the face areas.) I should adjust my answer to make the same connection. $\endgroup$
    – Blue
    Aug 28, 2013 at 14:17

Lemma Given 4 positive numbers $A_1, A_2, A_3, A_4$, in order for them to be realizable as the four face areas of a non-degenerate tetrahedron, a necessary and sufficient condition is the existence of 4 unit vectors $\hat{n}_1, \hat{n}_2, \hat{n}_3, \hat{n}_4$ ( no 2 lies on the same line, no 3 lies on the same plane) such that

$$A_1 \hat{n}_1 + A_2 \hat{n}_2 + A_3 \hat{n}_3 + A_4 \hat{n}_4 = \vec{0}$$

The necessary part is standard vector analysis. One can use the unit normal vectors for the faces as $\hat{n}_i$. I will skip this part of proof. For the sufficient part, let's say we indeed have 4 such unit vectors. Now define 4 planes $P_i, i = 1,\ldots, 4$ by

$$P_i = \Big\{\; \vec{x} \in \mathbb{R}^3 : \vec{x} \cdot \vec{n}_i = \frac{1}{A_i}\;\Big\}$$ and consider the intersections of any 3 of the planes:

$$\vec{x}_1 = P_2 \cap P_3 \cap P_4,\;\; \vec{x}_2 = P_3 \cap P_4 \cap P_1,\;\; \vec{x}_3 = P_4 \cap P_1 \cap P_2,\;\; \vec{x}_4 = P_1 \cap P_2 \cap P_3$$

It is clear $\vec{x}_2, \vec{x}_3, \vec{x}_4$ lies on the plane $P_1$ and hence satisfy $\vec{x}_i \cdot \hat{n}_1 = \frac{1}{A_1}$ for $i \ne 1$. In general, we have $\vec{x}_i \cdot \hat{n}_j = \frac{1}{A_j}$ whenever $i \ne j$. Notice $$\vec{x_1} \cdot \hat{n}_1 = -\frac{1}{A_1} \vec{x_1} \cdot ( A_2 \hat{n}_2 + A_3 \hat{n}_3 + A_4 \hat{n}_4 ) = -\frac{1}{A_1} \left( \frac{A_2}{A_2} + \frac{A_3}{A_3} + \frac{A_4}{A_4} \right) = -\frac{3}{A_1}$$

We can conclude the distance of $\vec{x}_1$ from the plane $P_1$ is $\frac{4}{A_1}$. One way to interpret this result is if we from a tetrahedron $X$ from $\vec{x}_1, \vec{x}_2, \vec{x}_3$ and $\vec{x}_4$. The height of the tetrahedron with respect to the face opposite to $\vec{x}_1$ (i.e. the one contained in $P_1$ ) is $\frac{4}{A_1}$. Similar conclusions can be drawn for other vertices $\vec{x}_2, \vec{x}_3$ and $\vec{x}_4$.

Recall the four face areas of a tetrahedron is inversely proportional to the corresponding heights, we find the four face areas of $X$ has the right ratios:

$$ \text{Area}(X\cap P_1) : \text{Area}(X\cap P_2) : \text{Area}(X\cap P_3) : \text{Area}(X\cap P_4) = A_1 : A_2 : A_3 : A_4$$

By scaling $X$ with right amount, we can realize $A_1, A_2, A_3, A_4$ as the four face areas of a tetrahedron.

A trivial corollary of the lemma is if $A_i$ are realizable face areas, then we have inequalities like: $$A_1 = |A_1 \hat{n}_1 | = |A_2 \hat{n}_2 + A_3 \hat{n}_3 + A_4 \hat{n}_4| < A_2 + A_3 + A_4$$ The inequality is strict because no two $\hat{n}_i$ is lying on the same line. If we group these inequalities together, we find

Another set of necessary condition for $A_i$ to be realizable as face areas: $$\begin{cases} A_1 < A_2 + A_3 + A_4\\ A_2 < A_3 + A_4 + A_1\\ A_3 < A_4 + A_1 + A_2\\ A_4 < A_1 + A_2 + A_3\tag{*1} \end{cases}$$ It turns out $(*1)$ is also a sufficient condition.

From $(*1)$, we can deduce two inequalities: $$(A_1 + A_2)^2 - (A_3 - A_4)^2 > 0\quad\text{ and }\quad(A_1 - A_2)^2 - (A_3 + A_4)^2 < 0$$ For $\theta \in [0,\pi]$, if we define a function $f(\theta)$ by:

$$f(\theta) = (A_1^2 + 2A_1A_2\cos\theta + A_2^2) - (A_3^2 - 2A_3A_4\cos\theta + A_4^2)$$

The above two inequalities implies $f(0) > 0$ and $f(\pi) < 0$. This means we can find a $\theta \in (0,\pi)$ such that $f(\theta) = 0$. Now start with any two unit vectors $\hat{n}_1$ and $\hat{n}_3$. If one rotate $\hat{n}_1$ for an angle $\theta$ to get a new unit vector $\hat{n}_2$ and rotate $\hat{n}_3$ for an angle $\pi - \theta$ to get another new vector $\hat{n}_4$, we will have:

$$\begin{align}|A_1 \hat{n}_1 + A_2 \hat{n}_2|^2 = & (A_1^2 + 2A_1A_2\cos\theta + A_2^2) = (A_3^2 - 2A_3A_4\cos\theta + A_4^2)\\ = & |A_3 \hat{n}_3 + A_4 \hat{n}_4|^2\end{align}$$ It is then clear a further rotation of $\hat{n}_3$ and $\hat{n}_4$ will allow us to produce 4 unit vectors satisfying the condition in Lemma and hence $A_1, A_2, A_3, A_4$ are realizable as face areas of a tetrahedron.

  • 1
    $\begingroup$ +1. "It turns out $(*1)$ is also a sufficient condition." Indeed. In then end, you're declaring that a pair of opposing dihedral angles will be supplementary to lock-down a fifth degree of freedom, and then showing you can get the sixth. (Importantly, there are other choices; eg, one could declare that the three dihedral angles opposite the largest face are equal, use the Law of Cosines to determine their size, and then argue that the inequalities allow that another angle can be chosen.) So, the inequalities $(*1)$ provide that there are tetrahedra, but they don't uniquely determine one. $\endgroup$
    – Blue
    Aug 28, 2013 at 14:08
  • 1
    $\begingroup$ @Blue, interesting. I can see how your "three dihedral angles opposite the largest face are equal" translate to the normal vectors I use as another way of constructing the desired tetrahedron. $\endgroup$ Aug 28, 2013 at 14:17

$$a<b+c+d$$ $$b<c+d+a$$ $$c<d+a+b$$ $$d<a+b+c$$

Is that what you want?

This condition is necessary. You can't have sufficient condition with areas alone.

Let's investigate if a tetrahedron with triangle areas $a,b,c,d$ is possible if the conditions are true.

Let $a>b, a>c, a>d$. Make $a$ a regular triangle with side $A$. Now you must construct one more point on position $(x,y,z)$. Triangles $b,c,d$ must have heights $H_b,H_c,H_d$ such that $AH_b/2=b$ etc. Now we have three constraints and three variables, can you solve it?

  • $\begingroup$ I think no. I need a necessary and sufficient condition, so I would like you to prove that every $(a,b,c,d)$ can be areas of a tetrahedron if it satisfies your inequalities. In other words, I think you just get a necessary condition though I would like you to write more details. $\endgroup$
    – mathlove
    Aug 28, 2013 at 7:26
  • 1
    $\begingroup$ Lay out a triangle for the base with the largest of the four areas. Given the areas of the other sides, and the fact that the area of a triangle is determined by base and height, choose a side of the base for each triangle. The third point of each triangle lies on a cylinder. The question is whether there is enough freedom in the choices (eg the original triangle) to enable a configuration where the three cylinders meet at a single point. It is possible that the intermediate value theorem could be used if a continuous deformation could be defined. $\endgroup$ Aug 28, 2013 at 7:36
  • $\begingroup$ @mathlove I added some more info, can you make the described system of equations and check if it's solvable? $\endgroup$
    – Džuris
    Aug 28, 2013 at 7:51

Just an idea without writing any proof.

The volume $V$ of a tetrahedron $O-PQR$ can be represented by six edges $OP=p, OQ=q, OR=r, QR=l, RP=m, PQ=n$ as the following: $$144V^2=p^2l^2(-p^2+q^2+r^2-l^2+m^2+n^2)+q^2m^2(p^2-q^2+rc^2+l^2-m^2+n^2)+r^2n^2(p^2+q^2-r^2+l^2+m^2-n^2)-l^2q^2r^2-p^2m^2r^2-p^2q^2n^2-l^2m^2n^2.$$

Let the right side of this expression be $F(p, q, r; l, m, n)$.

On the other hand, the necessary and sufficient condition for given four triangles $OPQ, ORQ, PRQ, OPR$ being the faces of a tetrahedron is that $$F(p, q, r; l, m, n)>0.$$

By Heron's formula, $$4a=\sqrt{(p^2+q^2+n^2)^2-2(p^4+q^4+n^4)}, 4b=\sqrt{(q^2+r^2+l^2)^2-2(q^4+r^4+l^4)},$$ $$4c=\sqrt{(l^2+m^2+n^2)^2-2(l^4+m^4+n^4)}, 4d=\sqrt{(p^2+r^2+m^2)^2-2(p^4+r^4+m^4)}.$$

The problem is that it seems impossible to represent $F$ only by $a,b,c,d$.

  • $\begingroup$ "The problem is that it seems impossible to represent $F$ only by $a$, $b$, $c$, $d$." Correct. Since the six edge-lengths are independent (albeit subject to certain relations), there are six degrees of freedom in determining a tetrahedron. Four face-areas alone cannot account for these. $\endgroup$
    – Blue
    Aug 28, 2013 at 12:48
  • $\begingroup$ As for accounting-for the extra degrees of freedom, one must be careful. My approach is to seek pseudo-faces. (Although there are three such faces, the Sum-of-Squares identity reduces the added degrees of freedom to just two.) Alternatively, you can restrict your universe of tetrahedra to, say, those whose opposing edges are orthogonal; this universe includes regular tetrahedra as well as "right-corner" tetrahedra (for which the Law of Cosines reduces to the Pythagorean theorem $W^2 = X^2 + Y^2 + Z^2$). These approaches work fine. (continued) $\endgroup$
    – Blue
    Aug 28, 2013 at 13:15
  • $\begingroup$ (part 2) In 1999, Marcin Mazur wondered if volume and circumradius, along with face areas, would determine a tetrahedra; they do not. Non-uniqueness was first shown by M.SE user @RobertIsrael in 2000; five years later, Yang and Zeng provided a continuum of tetrahedra sharing the same six parameters. (I happen to have a Bloog post about this, too, wherein I apply hedronometry to the analysis.) $\endgroup$
    – Blue
    Aug 28, 2013 at 13:18
  • $\begingroup$ @Blue:Thank you for nice information. $\endgroup$
    – mathlove
    Aug 29, 2013 at 14:45

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