Tuesday, June 16, 2015

Cloudy Puzzle: Solution

We want to establish a mathematical relationship between the key variables of the problem: \(h, R, \theta\) and \(\phi\).

Consider the triangle PRQ; we can write, \[\tan\left(\frac{\pi}{2} - \theta \right) = \frac{QP}{RQ} = \frac{QP}{OQ - OR}.\] 
Recognizing that the length of the segment OP is \(R+ h\), we can further write:
\[\tan\left(\frac{\pi}{2} - \theta \right) = \frac{(R+h) \cos \phi}{(R+h) \sin \phi - R}.\]
Using the variable \(a = h/R\), we can try to solve for \(\phi\), which in this case leads to a quadratic equation. Let us limit \(\theta < \pi/2\), since all other cases are trivial extensions.

If we let,  \[A = a^{2} \tan^{2} \theta + a^{2} + 2 a \tan^{2}\theta + 2 a + \tan^{2} \theta,\] and \[B = a \tan \theta + \tan \theta - 1,\] then the two solutions are: \[\phi = -2 \tan^{-1} \left( \frac{a + 1 \pm \sqrt{A}}{B} \right).\]
Due to the geometry, acceptable values of \(\theta\) and \(\phi\) lie in the region \([0, \pi/2]\). This constraint may be used to cull out the extraneous root.

It is useful to consider the limit of \(a = h/R \ll 1.\) If we take the limit as it goes to zero, then  \[\phi = -2 \tan^{-1} \left( \frac{ \pm|\tan \theta| + 1}{\tan \theta - 1} \right).\]

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