Volume of a tetrahedron

(Edited 6/20/23: I lost the images for this post, and they’re in 3D. I haven’t reconstructed them, so there’s an additional challenge for you!)

This is a challenging one: Given all the information at one corner of a tetrahedron (all three surface angles and all three edge lengths), what is the volume of the tetrahedron? [Image lost, not reconstructed]

The volume of any pyramid is equal to the area of its base times its height, divided by three: \[V = \frac{Ah}{3}\]

Since a tetrahedron is a pyramid from all of its faces, we would expect the final formula to be symmetrical. I’ll use \(\Delta BCD\) as the base, but we should come up with the same formula using \(\Delta ACD\) or \(\Delta ABD\).

The area of a triangle given two sides and the shared angle is given by \[A = \frac{bc \sin \beta}{2}\]

This can be easily seen by drawing \(\overline{BZ} \perp \overline{CD}\). The area of \(\Delta BCD\) is \(\frac{b \cdot BZ}{2}\); \(\sin \beta = \frac{BZ}{c}\), so \(BZ = c \sin\beta\) and thus \(A = \frac{bc \sin \beta}{2}\).

[Image lost, not reconstructed]Next, calculate the height of this tetrahedron. I’ll draw in two line segments. \(h \, (\overline{EA})\) is the height, and is perpendicular to \(\Delta BCD\). \(\Delta AED\) is a right triangle, so \(h^2 + d^2 = a^2\) and \(h^2 = a^2 – d^2\). We are given \(a\), so if we can calculate \(d\), we can calculate \(h\).

[Image lost, not reconstructed]Consider the base, \(\Delta BCD\). I’ve added three more segments: \(\overline{EG}\perp\overline{CD}\), \(\overline{EF}\perp\overline{BD}\), and \(g \, (\overline{FG})\). This gives us the quadrilateral \(EFDG\), where \(\angle F\) and \(\angle G\) are right angles. Because we have a quadrilateral with opposite angles that are supplementary, we know that the four vertices fall on a circle.

[Image lost, not reconstructed] In the next diagram, I’ve drawn the circle around the quadrilateral and added the center point and two radii. Because \(\angle FDG\) is an inscribed angle, it is half the size of central angle \(\angle FHG\). By the law of cosines, \(g^2 = r^2 + r^2 – 2\cdot r \cdot r \cdot \cos{2\beta} = 2r^2 (1 – \cos{2\beta})\). We use the trigonometric identity \(\cos{2\beta} = 1 – 2\sin^2\beta\) to determine that \(g^2 = 4r^2\sin^2\beta\) and \(r = \frac{g}{2\sin\beta}\). Thus, \(d = 2r = \frac{g}{\sin\beta}\).

From the law of cosines, we know that \(g^2 = DG^2 + DF^2 – 2 \cdot DG \cdot DF \cdot \cos\beta\).

[Image lost, not reconstructed] Since \(\Delta AFD\) is a right triangle, \(DF = a \cos\alpha\). Likewise, \(DG = a \cos\gamma\). Hence, \(g^2 = a^2 \cos^2\alpha + a^2 \cos^2\gamma – 2 \cdot a \cos\alpha \cdot a \cos\gamma \cdot \cos\beta\) \(=a^2 (\cos^2\alpha + \cos^2\gamma – 2\cos\alpha\cos\beta\cos\gamma)\).

We can now calculate \(h\). \(d^2 = \frac{g^2}{\sin^2\beta} = \frac{a^2 (\cos^2\alpha + \cos^2\gamma – 2\cos\alpha\cos\beta\cos\gamma)}{\sin^2\beta}\), so \(h^2 = a^2 – \frac{a^2 (\cos^2\alpha + \cos^2\gamma – 2\cos\alpha\cos\beta\cos\gamma)}{\sin^2\beta}\) \(= \frac{a^2(\sin^2\beta – \cos^2\alpha – \cos^2\gamma + 2\cos\alpha\cos\beta\cos\gamma)}{\sin^2\beta}\) \(=\frac{a^2(1-\cos^2\alpha – \cos^2\beta – \cos^2\gamma + 2\cos\alpha\cos\beta\cos\gamma)}{\sin^2\beta}\). Thus, \(h = \frac{a}{\sin\beta}\sqrt{1-\cos^2\alpha – \cos^2\beta – \cos^2\gamma + 2\cos\alpha\cos\beta\cos\gamma}\).

Finally, \(V = \frac{Ah}{3} = \frac{b\cdot c\cdot \sin\beta \cdot a}{3 \cdot 2 \cdot \sin\beta} \sqrt{1-\cos^2\alpha – \cos^2\beta – \cos^2\gamma + 2\cos\alpha\cos\beta\cos\gamma}\) \(=\frac{abc}{6} \sqrt{1-\cos^2\alpha – \cos^2\beta – \cos^2\gamma + 2\cos\alpha\cos\beta\cos\gamma}\).

So, given the three surface angles and three edge lengths from a given vertex of a tetrahedron, the volume of the tetrahedron is:

\[V=\frac{abc}{6} \sqrt{1-\cos^2\alpha – \cos^2\beta – \cos^2\gamma + 2\cos\alpha\cos\beta\cos\gamma}\]