# Volume of a tetrahedron, given areas of 6 triangles

Given an arbitrary tetrahedron, I always have an inscribed sphere inside, which contacts the tetrahedron at four new points (one on each of the four faces). Each face, which is itself a triangle, is now divided into $$3$$ small triangles, so I have $$12$$ such small triangles. It turns out that each pair of triangles with a common edge also has a common area; so I have $$6$$ different areas for these small triangles (one area corresponding to each edge).

The question is: If I start with the areas of those small triangles, is my tetrahedron fixed? If so, is there a simple expression for the volume of the tetrahedron in terms of those $$6$$ small areas?

Thank you so much for any attention to this problem.

• What do you mean by "each surface" - triangle? To create the three smaller triangles, do you "squeeze " down the tetrahedron? – Moti Apr 28 at 0:58
• You're looking for a generalization of Heron's formula. This question might be helpful, though not exactly what you asked for. – RobertTheTutor Apr 28 at 1:05
• I added a picture, showing those 12 triangles. – Intelligenti pauca Apr 28 at 14:00
• Observation: Knowing the three part areas of it, determines a face (including the position of the kissing point) up to shearing – Hagen von Eitzen Apr 28 at 14:36
• It's worth pointing out, to forestall an obvious attempt, that the volume of a tetrahedron is not determined just by the areas of the faces; e.g., see math.stackexchange.com/questions/160258/… . So the extra information given by the kissing points needs to be used somehow. Just counting constraints, you have $6$ parameters and want to determine the $12$ parameters of the vertex locations, up to translation ($3$ parameters) and rotation ($3$ parameters) of the whole tetrahedron... so the result should be unique. – mjqxxxx Apr 28 at 14:57

Introducing conventions familiar to me, I'll consider tetrahedron $$OABC$$ to have edges and face-angles about $$O$$, and face-areas

$$a := |OA| \qquad b := |OB| \qquad c := |OC| \\[6pt] \alpha := \angle BOC \qquad \beta := \angle COA \qquad \gamma := \angle AOB \\[6pt] W := |\triangle ABC| \qquad X := |\triangle OBC| \qquad Y := |\triangle OCA| \qquad Z:=|\triangle OAB|$$ Without fear of confusion, I'll also use $$A$$, $$B$$, $$C$$ to refer to the dihedral angles along edges $$a$$, $$b$$, $$c$$.

As OP notes, insphere-determined sub-triangles that share an edge have the same area. Let $$\sigma_a$$, $$\sigma_b$$, $$\sigma_c$$ be the areas of the triangles sharing respective edges $$a$$, $$b$$, $$c$$; further, let $$\sigma_d$$, $$\sigma_e$$, $$\sigma_f$$ be the areas of the triangles sharing edges $$|BC|$$, $$|CA|$$, $$|AB|$$. (I typically denote those edges $$d$$, $$e$$, $$f$$, but that's not important here.) Of course, the full face-areas are simple sums of these sub-areas:

$$W = \sigma_d + \sigma_e + \sigma_f \qquad X = \sigma_d+\sigma_b+\sigma_c \qquad \text{etc} \tag1$$

Below I show that $$\cos A$$, $$\cos B$$, $$\cos C$$ are also expressible in terms of the $$\sigma$$s, which in turn guarantees a unique volume, via $$81V^4 = 4X^2Y^2Z^2(1-2\cos A\cos B\cos C-\cos^2 A-\cos^2 B-\cos^2 C) \tag2$$

To get at the cosine relations, let's coordinatize: $$O=(0,0,0) \qquad A = (a,0,0) \qquad B = (b \cos\alpha,b\sin\alpha, 0) \\[6pt] C = (c\cos\beta,c\sin\beta\cos A,c\sin\beta\sin A)$$

It is "known" that the incenter of $$OABC$$ is given by $$I = \frac{W O + X A + Y B + Z C}{W+X+Y+Z} \tag3$$

If $$P$$ is the point where the insphere touches face $$Z$$ (the $$xy$$-plane), then \begin{align} \sigma_a := |\triangle OAP| &= \frac12|OA|P_y = \frac12 a \frac{Yb\sin\gamma+ Z c \sin\beta \cos A}{W+X+Y+Z} \\[8pt] &= \frac{YZ(1+\cos A)}{W+X+Y+Z} \tag4 \end{align} Likewise, $$\sigma_b = \frac{ZX(1+\cos B)}{W+X+Y+Z} \qquad \sigma_c = \frac{X Y(1+\cos C)}{W+X+Y+Z} \tag5$$

Substituting $$X$$, $$Y$$, $$Z$$, $$\cos A$$, $$\cos B$$, $$\cos C$$ in terms of the $$\sigma$$s into $$(2)$$, we find

\begin{align} 81 V^4 =\, &16\,(\sigma_a + \sigma_b + \sigma_c + \sigma_d + \sigma_e + \sigma_f)^2 \\ &\cdot(p+q+r) (-p+q+r) (p-q+r) (p+q-r)\end{align} \tag{\star} where $$p := \sqrt{\sigma_a\sigma_d} \qquad q := \sqrt{\sigma_b\sigma_e} \qquad r := \sqrt{\sigma_c\sigma_f}$$

As a sanity check, a regular tetrahedron with side-length $$s$$ has equilateral faces of area $$W=X=Y=Z=\frac14s^2\sqrt{3}$$ and sub-faces of area $$\sigma_a=\cdots=\sigma_f = \frac1{12}s^2\sqrt{3}=p=q=r$$. As a result, $$(\star)$$ yields $$V^4 = \frac{16}{81} \left(6\cdot \frac1{12}s^2\sqrt{3}\right)^2\left(3\cdot \frac1{12}s^2\sqrt{3}\right)\left(\frac1{12}s^2\sqrt{3}\right)^3 = \left(\frac1{12} \sqrt{2} s^3\right)^4$$ as expected.

• This is so helpful! Thank you so much for the detailed answer! It helps me a lot. – Songbo Xie Apr 29 at 20:24
• Could you kindly explain how you obtain your equation (2)? I have no problems with other derivations. But by some simple numerical checks, it turns out that Eq.(2) sometimes gives correct answer for the volume, sometimes wrong, sometimes even imaginary. – Songbo Xie Apr 30 at 5:52
• @SongboXie: The most straightforward way to verify $(2)$ is to invoke the Spherical Law of Cosines to convert dihedral angle information to face-angle information; eg, $$\cos A=\frac{\cos\alpha-\cos\beta\cos\gamma}{\sin\beta\sin\gamma}$$ and, of course, the area formulas $X=\frac12bc\sin\alpha$, etc. Substituting and simplifying, $(2)$ becomes$$V^4=\frac1{6^4}a^4b^4c^4(1+2\cos\alpha\cos\beta\cos\gamma-\cos^2\alpha-\cos^2\beta-\cos^2\gamma)^2$$which agrees with the formula on Wikipedia. – Blue Apr 30 at 8:26
• @SongboXie: When doing test calculations be sure to use valid triples of dihedral angles. In particular, you must have $180^\circ \leq A+B+C \leq 540^\circ$. That's the angle-sum property in Spherical Geometry. (It (and the Spherical Law of Cosines) is relevant here because the faces at a vertex of a tetrahedron cut the surface of a sphere centered at that vertex to form a spherical triangle. The tetrahedron's dihedral and face-angles correspond to the triangle's "angles" and "sides".) – Blue Apr 30 at 8:34
• @SongboXie: It's worth noting that $$1-2\cos A\cos B\cos C-\cos^2 A-\cos^2 B-\cos^2 C \\ = -4\cos\tfrac12(A+B+C)\cos\tfrac12(-A+B+C)\cos\tfrac12(A-B+C)\cos\tfrac12(A+B-C)$$ By the angle-sum restriction, we have $90^\circ\leq\tfrac12(A+B+C)\leq 270^\circ$, so that $-\cos\tfrac12(A+B+C)$ is always non-negative. The product of the remaining factors must be non-negative, as well. – Blue Apr 30 at 8:39