Radioactive Target Practice

Set up a coordinate axis system $Oxyz$ with the radioactive source at the origin and the photographic plate occupying the plane $z=1$ . Introduce spherical polar angles $\theta, \phi$ with respect to these axes, so that $\theta$ is the polar angle from the positive $z$ -axis, and $\phi$ is the azimuthal angle measured from the $x$ -axis.

Then a radioactive particle emitted by the source in the direction defined by the angles $\theta$ and $\phi$ will hit the target provided that the distance $r$ of the point of impact from the point $P$ on the the plate that is closest to the source (the centre of the target square) satisfies $r \; \le \; \tfrac12\mathrm{min}\left(|\sec\phi|,|\mathrm{cosec}\,\phi|\right)$ (assuming that the target square has its sides parallel to the $x$ - and $y$ -axes). Thus an emitted radioactive particle will hit the target provided that $0 \; \le \; \tan\theta \; \le \; \tfrac12\mathrm{min}\left(|\sec\phi|,|\mathrm{cosec}\,\phi|\right)$ On the other hand, an emitted particle will hit the plate provided that $0 \le \theta < \tfrac12\pi$ . Thus the probability that a particle that hits the photographic plate in fact hits the target is $\begin{array}{rcl} p & = & \displaystyle \mathbb{P}\left[0 \le \tan\theta \le \tfrac12\mathrm{min}\left(|\sec\phi|,|\mathrm{cosec}\,\phi|\right) \; \Big| \; 0 \le \theta < \tfrac12\pi \right] \\ & = & \displaystyle \frac{\mathbb{P}\left[0 \le \tan\theta \le \tfrac12\mathrm{min}\left(|\sec\phi|,|\mathrm{cosec}\,\phi|\right) \right] }{\mathbb{P}\left[0 \le \theta < \frac12\pi\right]} \\ & = & \displaystyle 2\mathbb{P}\left[0 \le \tan\theta \le \tfrac12\mathrm{min}\left(|\sec\phi|,|\mathrm{cosec}\,\phi|\right)\right] \end{array}$ By symmetry this is equal to $\begin{array}{rcl} p & = & \displaystyle 8\mathbb{P}\left[ 0 \le \tan\theta \le \tfrac12\sec\phi \,,\, -\tfrac14\pi \le \phi \le \tfrac14\pi\right] \\ & = & \displaystyle 8 \times \frac{1}{4\pi} \int_{-\frac14\pi}^{\frac14\pi}\,d\phi \int_0^{\tan^{-1}(\frac12\sec\phi)} \sin\theta\,d\theta \\ & = & \displaystyle\frac{2}{\pi} \int_{-\frac14\pi}^{\frac14\pi}\,d\phi \Big[-\cos\theta\Big]_0^{\tan^{-1}(\frac12\sec\phi)} \\ & = & \displaystyle 1 - \frac{2}{\pi}\int_{-\frac14\pi}^{\frac14\pi} \frac{2\cos\phi}{\sqrt{1 + 4\cos^2\phi}}\,d\phi \; = \; 1 - \frac{4}{\pi}\int_{-\frac14\pi}^{\frac14\pi} \frac{\cos\phi}{\sqrt{5 - 4\sin^2\phi}}\,d\phi \\ & = & \displaystyle 1 - \frac{4}{\pi}\int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} \frac{du}{\sqrt{5-4u^2}} \; = \; \displaystyle 1 - \tfrac{4}{\pi}\sin^{-1}\sqrt{\tfrac25} \; = \; 1 - \tfrac{4}{\pi} \tan^{-1}\sqrt{\tfrac23} \end{array}$ which makes the answer $4+1+2+3 = \boxed{10}$ .

Outline of solution without calculus (more tedious)

$P(\text{Hit the target} = \frac{\text{Area of intersection between the sphere's surface and pyramid ABCDE}}{\text{Area of hemisphere}}=\frac{A}{2\pi}$

For simplicity, the radius of the sphere would be 1.

$A$ would simply be the area of a spherical square. This area can be found through spherical cosine rules and area of spherical triangle.