The following diagram represents three faces of a regular icosahedron, flattened out for ease of illustration. If they are folded appropriately, line $ab$, in blue, passes through the interior of the icosahedron. How does one find the angle between the line $ab$ and the plane containing the face $A$?
(Incidentally, the practical application of this problem is that I need to figure out what angle to set my miter saw to to cut the legs for a decorative table. It would make me happy if I could get the ends of the legs to exactly match the positions of the vertices of a regular icosahedron.)
The result is:
$$\theta=\text{atan}\left(\frac{2}{\Phi+1}\right)\frac{180}{\pi}\approx37.38 \ \text{degrees}$$
where $\Phi=\dfrac{1+\sqrt{5}}{2}\approx 1.618$ is the Golden Ratio, $atan$ is the "arc tan" function (known under the name $\tan^{-1}$ on pocket computers), and factor $\frac{180}{\pi}$ is there for the conversion from radians to degrees.
My computation is based on the following figure which represents a 2D projection of the icosahedron without angle distortion.
Proof: Let $\theta=$ angle $(\vec{AM},\vec{AI})$ and $\alpha=$ angle $(\vec{AB},\vec{AP}).$
I have used the 3D coordinates of the icosahedron that can be found in (http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/phi3DGeom.html#section3.5 ).
In 2D, coordinates of points $A,B,I$ are $A(0,\Phi), B(\Phi,1), I(1,0)$. Formula (1) results from the following computation:
$$\tan(\theta)=\tan((\alpha+\theta)-\alpha)=\dfrac{\tan(\alpha+\theta)-tan(\alpha)}{1+\tan(\alpha+\theta)tan(\alpha)}$$
$$=\dfrac{(1+\tfrac{1}{\Phi})-(1-\tfrac{1}{\Phi})}{1+(1+\tfrac{1}{\Phi})(1-\tfrac{1}{\Phi})}=\dfrac{\tfrac{2}{\Phi}}{2-\tfrac{1}{\Phi^2}}$$
and the application of relationship $\Phi^2=\Phi+1 \ \iff \ \tfrac{1}{\Phi}=\Phi-1$.
The second picture brings an understanding of the 3D corresponding points.