Maximum Angle of Sight

Let the position of the slower runner be $Q$ and that of the faster runner be $R$ , and let $|SQ| = x \gt 0$ . Then $|SR| = 3x$ and

$\theta = \angle RPS - \angle QPS = \arctan(|SR| / |SP|) - \arctan(|SQ| / |SP|) = \arctan(3x) - \arctan(x)$ .

Differentiating $\theta$ with respect to $x$ we see that

$\dfrac{d\theta}{dx} = \dfrac{3}{1 + 9x^{2}} - \dfrac{1}{1 + x^{2}} = \dfrac{3(1 + x^{2}) - (1 + 9x^{2})}{(1 + 9x^{2})(1 + x^{2})} = \dfrac{2 - 6x^{2}}{(1 + 9x^{2})(1 + x^{2})}$ ,

which equals $0$ when $2 - 6x^{2} = 0 \Longrightarrow x^{2} = \dfrac{1}{3} \Longrightarrow x = \dfrac{1}{\sqrt{3}}$ .

Now since $\theta \ge 0 \space \forall \space x \ge 0$ , $\theta(0) = 0$ and $\displaystyle \lim_{x \to \infty} \theta(x) = 0$ , we know that the critical point found above will represent a maximum, and so the desired maximum is

$\theta_{\text{max}} = \arctan(\sqrt{3}) - \arctan(1/\sqrt{3}) = 60^{\circ} - 30^{\circ} = \boxed{30^{\circ}}$ .

$\tan{y}=x, \tan(\theta+y)=3x$

$\tan{\theta+y}=\dfrac{\tan{\theta}+\tan{y}}{1-\tan{y}\tan{\theta}}=3x$

$\tan{\theta}=\dfrac{2x}{3x^2+1}=\dfrac{2}{3x+1/x}\geq\dfrac{1}{\sqrt{3}}$ by AM-GM, when $\theta=\dfrac{\pi}{6}$