# Artist needing to determine geometric angle for sculpture based on platonic solid

Dear Mathematicians I need your help for a new sculpture!

I will attach images but first imagine 2 hexagons - where one is rotated 30 deg. They are separated by 12 equilateral triangles. I need to confirm the angle between the 2 faces as show in the image. My computer render calculation makes it 145.222 deg but I need to confirm this is absolutely correct before proceeding to fabrication!

Your help would be greatly appreciated, thank you! Pete

This shows the angle I need to confirm: This is the computer design angle that I need to confirm: This is something like what the final sculpture will look like! This shows how I made it: • Looks correct. I find the angle you want equals to $\cos^{-1}\left(\frac{1-2\sqrt{3}}{3}\right) \sim 145.22189133^\circ$, matching your own estimate. Aug 20 at 10:20
• Thank you for for your generosity in helping me with the problem. It is greatly appreciated. You can see my art at petemoorhouse.co.uk if of interest. Have a great day! Aug 20 at 11:24
• I don't see what this question has to do with platonic solids. 21 hours ago

You are requesting a dihedral angle of a (regular) hexagonal antiprism. This is calculated as follows. Consider a hexagon with vertices $$v_0, \ldots, v_5$$, where $$v_k = \left(\cos \frac{\pi}{3} k, \sin \frac{\pi}{3} k, d \right),$$ and another hexagon with vertices $$w_0, \ldots, w_5$$ where $$w_k = \left(\cos \frac{\pi}{6} (2k+1), \sin \frac{\pi}{6} (2k+1), -d\right).$$
The distance these hexagons are separated by is $$2d$$. In order for the lateral triangles to be equilateral, the distance between, say, $$v_0$$ and $$w_0$$, must be $$1$$. This implies $$1 = \left(1 - \cos \frac{\pi}{6}\right)^2 + \left(0 - \sin \frac{\pi}{6}\right)^2 + (2d)^2,$$ hence $$d = \frac{\sqrt{\sqrt{3} - 1}}{2}.$$ The dihedral angle $$\theta$$ between triangular faces is therefore given by the equation $$\cos \theta = \frac{\vec a \cdot \vec b}{|\vec a| |\vec b|},$$ where $$\vec a = w_5 - \frac{v_0 + w_0}{2} = \left(\frac{\sqrt{3} - 2}{4}, -\frac{3}{4}, -d\right), \\ \vec b = v_1 - \frac{v_0 + w_0}{2} = \left(-\frac{\sqrt{3}}{4}, \frac{2\sqrt{3} - 1}{4}, d \right).$$ This gives $$\vec a \cdot \vec b = - \frac{\sqrt{3}}{4} - d^2, \\ |\vec a| |\vec b| = 1 - \frac{\sqrt{3}}{4} + d^2,$$ hence $$\theta = \arccos \frac{1 - 2 \sqrt{3}}{3} \approx 145.2218913319^\circ. \approx 145^\circ 13' 18.80879''.$$ The equivalent radian measure is approximately $$2.534600149715126$$ radians.
What you are looking for is the dihedral angle of a hexagonal antiprism. I found this site that contains the complicated formula: $$\cos \alpha = 1-\frac23\left(\frac{3-\tan^2{\frac{90}n}}{4} +\frac{\sin^2 \frac{270}{n}}{ \sin^2 \frac{180}{n}}\right)$$
The results table on that page only goes up to $$n=5$$, but putting in $$n=6$$ I get that the angle $$\alpha$$ is $$145.22189133...$$, so your software gives the correct angle rounded to three decimal places.