Solid Geometry Advanced – Problem 7: Find

A regular tetrahedron has edge length $a$. Let $\varphi$ be the dihedral angle between any two adjacent faces.

Find $\cos\varphi$.

(1) Use a coordinate model for a regular tetrahedron: $$A(0,0,0),\ B(a,0,0),\ C\left(\frac{a}{2},\frac{\sqrt3 a}{2},0\right),\ D\left(\frac{a}{2},\frac{\sqrt3 a}{6},\sqrt{\frac{2}{3}}a\right).$$ Then $\triangle ABC$ and $\triangle ABD$ are adjacent faces along edge $AB$.

(2) A normal to plane $ABC$ is $$\mathbf n_1=\overrightarrow{AB}\times\overrightarrow{AC}=(a,0,0)\times\left(\frac{a}{2},\frac{\sqrt3 a}{2},0\right)\sim(0,0,1).$$ (3) A normal to plane $ABD$ is $$\mathbf n_2=\overrightarrow{AB}\times\overrightarrow{AD}=(a,0,0)\times\left(\frac{a}{2},\frac{\sqrt3 a}{6},\sqrt{\frac{2}{3}}a\right)\sim\left(0,-\sqrt{\frac{2}{3}},\frac{\sqrt3}{6}\right).$$ (4) The dihedral angle $\varphi$ equals the angle between $\mathbf n_1$ and $\mathbf n_2$: $$\cos\varphi=\frac{|\mathbf n_1\cdot\mathbf n_2|}{|\mathbf n_1||\mathbf n_2|}.$$ Using the scaled vectors, this simplifies to $$\cos\varphi=\frac{1}{3}.$$ (Equivalently, it is a well-known invariant for the regular tetrahedron.) \]

For a regular tetrahedron, $\cos\varphi=\dfrac{1}{3}$.