<s>[INST] Area of the shadow of a regular polygon inscribed in a sphere. Consider the situation given below:
Let a regular polygon be inscribed in a sphere such that its circumcentre is at a distance $r$ from the centre of the sphere of radius $R$. A point source of light is kept at the centre of the sphere. How can we calculate the area of the shadow made on the surface of the sphere.
I tried to use the relation: $ \Omega = \frac{S}{R^2} $
But of course that is the case when a circle would be inscribed. So can I somehow relate it for any general polygon?
 [/INST] Let $\theta = \cos^{-1}\frac{r}{R}$ and $\phi = \frac{2\pi}{n}$ where
$n$ is the number of sides of the regular polygon.
Choose the coordinate system so that the vertices of the regular polygon are located at
$$\vec{v}_k = (R\sin\theta\cos(k\phi), R\sin\theta\sin(k\phi), R\cos\theta) \quad\text{ for } k = 0, 1, \ldots, n - 1.$$
Let $\hat{v}_k = \frac{\vec{v}_k}{|\vec{v}_k|}$ be the corresponding unit vectors
and $\hat{z} = (1,0,0)$.
Recall following formula by Oosterom and Strackee on solid angle,

Given any 3 unit vectors, $\vec{a}, \vec{b}, \vec{c}$, the solid angle $\Omega$ spanned by these 3 vectors satisfy
  $$\tan\frac{\Omega}{2} = \frac{ \vec{a} \cdot (\vec{b} \times \vec{c}) }{1 + \vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}}$$

Apply this to the unit vectors $\hat{v}_0, \hat{v}_1$ and $\hat{z}$, we find the corresponding solid angle $\Omega$ satisfy
$$\tan\frac{\Omega}{2} = \frac{\sin^2\theta\sin\phi}{1 + 2\cos\theta + \cos^2\theta + \sin^2\theta\cos\phi}
= \frac{\sin\phi}{
\frac{1+\cos\theta}{1-\cos\theta} + \cos\phi}
= \frac{\sin\phi}{\frac{R+r}{R-r}+\cos\phi}
$$
So the desired area is
$$\verb/Area/ = nR^2\Omega = 2nR^2\tan^{-1}\left[\frac{\sin\left(\frac{2\pi}{n}\right)}{\frac{R+r}{R-r}+\cos\left(\frac{2\pi}{n}\right)}\right]$$
Update
Let $s = \sin\frac{\pi}{n}$, $c = \cos\frac{\pi}{n}$ and $D = \frac{R+r}{R-r}$, we 
can simplify above expression as
$$\begin{align}\tan^{-1}\left[\frac{2sc}{D+c^2-s^2}\right]
&= \frac{\pi}{n} - \tan^{-1}\left[\frac{\frac{s}{c} - \frac{2sc}{D+c^2-s^2}}{1 + \frac{s}{c}\frac{2sc}{D + c^2 - s^2}}\right]
= \frac{\pi}{n} - \tan^{-1}\left[\frac{s(D-1)}{c(D+1)}\right]\\
&= \frac{\pi}{n} - \tan^{-1}\left[\frac{sr}{cR}\right]
\end{align}$$
So a simpler version of the area is
$$\verb/Area/ = 2R^2\left(\pi - n\tan^{-1}\left(\frac{r}{R}\tan\frac{\pi}{n}\right)\right)$$
It is easy to see this equals to the expression in Harish Chandra Rajpoot's answer.
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