# Finding the dihedral angle of a octahedron

I am trying to find the dihedral angle of a regular octahedron. All the triangles on a octahedron are equilateral triangles. So, when I drop the perpendicular from the top of the octahedron to the square base, the point it hits should be the midpoint of any of the triangles. Not sure if I am going the right direction here.

Choose the vertices for the octahedron to be the six points $$\begin{array}{c} (\pm 1, 0, 0) \\ (0, \pm 1, 0) \\ (0, 0, \pm 1) \end{array}$$.

Write down the vectors for a pair of edges of a given face. For example, for the face in the first octant, you could calculate

\begin{align} \vec{v}_1 &= (0, 1, 0) - (1, 0, 0) = (-1, 1, 0) \\ \vec{w}_1 &= (0, 0, 1) - (1, 0, 0) = (-1, 0, 1) \end{align}

Now, find a vector normal to the face by calculating the cross product $\vec{n}_1 = \vec{v}_1 \times \vec{w}_1 = (1, 1, 1)$.

Do the analogous calculation for an adjacent face to find its normal $\vec{n}_2$. Now, calculate the angle between these normal vectors, using

$$\cos \theta = \frac{\vec{n}_1 \cdot \vec{n}_2}{||\,\vec{n}_1\,||\;||\,\vec{n}_2\,||}$$

Convince yourself (really do it!) that the dihedral angle is the supplement to this angle. Namely, $\pi - \theta$.

Yes, your direction is correct but you left it off too soon.

Dihedral angle can be next found directly by drawing a 3D diagram of the top half of the Octahedron. Take equilateral triangle $$ABV$$ to find altitude $$h$$ by the Pythagoras thm:

From center point $$M$$ of $$AB$$ the vertex $$V$$ and center of base $$O$$ draw a right triangle (yellow).

$$h^2 = \big( \sqrt3\big)^2 -1^2 \rightarrow h=\sqrt2$$

Semi-dihedral angle is

$$\angle OMV = \tan^{-1}\sqrt2= \cos^{-1}\dfrac{1}{\sqrt3} =\sin^{-1}\sqrt{\dfrac{2}{3}}$$

and the dihedral is double this.

Fix a vertex $v$ in octahedron $O$ of edge length $l$ where $l$ is suitably large. If we place $v$ at origin, then consider intersection of unit sphere $S$ and $O$ That is we have spherical four polygon $P=ABCD$ on $S$ : Since face in $O$ is regular triangle so edge length of $P$ is $\frac{\pi}{3}$ And note that internal angle of $P$ is dihedral angle $\theta$.

If two diagonals $AC,\ BD$ intersect at $E$ then consider a spherical triangle $EAB$. $\angle AEB=\frac{\pi}{2}$ and $L:=|AE|=|BE|$ Hence cosine law implies that $$\cos\ \frac{\pi}{3}=\cos^2L +\sin^2L\cos\ \frac{\pi}{2}$$ $$\cos\ L=\cos\ L\cos\ \frac{\pi}{3} +\sin\ L\sin\ \frac{\pi}{3}\cos\ \angle BAE$$

$$L=\frac{\pi}{4},\ \theta=2\angle BAE=2\cos^{-1}\frac{1}{\sqrt{3}}$$