Inequality for each $a, b, c, d$ being each area of four faces of a tetrahedron

Question

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.

Answer 1

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. :)

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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 $(1)$, $(2)$, $(3)$.

Answer 2

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.

Answer 3

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

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?

Answer 4

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 $a,b,c,d$.