Dynamic Geometry: P131

Let the diameter of the circle be $AB$ , the moving triangle vertex be $C$ , the square vertex on $AB$ be $D$ , and the side length of the square be $a$ . Then

$\begin{aligned} AD + DB & = AB \\ a \csc \alpha + a \sec \alpha & = 2 \\ \implies a & = \frac 2{\csc \alpha + \sec \alpha} = \frac {2\sin \alpha \cos \alpha}{\sin \alpha + \cos \alpha} = \frac {\sin 2\alpha}{\sin \alpha + \cos \alpha} \end{aligned}$

When the area of the square,

$\begin{aligned} a^2 & = \frac {400}{1421} \\ \frac {\sin^2 2\alpha}{(\sin \alpha + \cos \alpha)^2} & = \frac {400}{1421} \\ \frac {\sin^2 2\alpha}{\sin 2 \alpha + 1} & = \frac {400}{1421} \\ 1421 \sin^2 2\alpha - 400 \sin 2 \alpha - 400 & = 0 \\ (29 \sin 2 \alpha - 20)(49 \sin 2 \alpha + 20) & = 0 & \small \blue{\text{Since }\alpha \text{ is acute,}} \\ \sin 2 \alpha & = \frac {20}{29} \\ \frac {2\tan \alpha}{1 + \tan^2 \alpha} & = \frac {20}{29} \\ \frac 1{\cot \alpha + \tan \alpha} & = \frac {10}{29} \\ \frac 1{\tan (\blue{90^\circ - \alpha}) + \tan \alpha} & = \frac {10}{29} & \small \blue{\text{Note that }\beta = 90^\circ - \alpha} \\ \implies \tan \alpha + \tan \blue \beta & = \frac {29}{10} \end{aligned}$

Therefore $p+q = 10 + 29 = \boxed{39}$ .