Solution:

The members of $\mathcal{F}$ are the segments from $(\cos\theta, 0)$ to $(0, \sin\theta)$ for $0 \lt \theta \lt 90^\circ,$ lying on the lines $\frac{x}{\cos\theta} + \frac{y}{\sin\theta} = 1;$ the segment $\overline{AB}$ is the member with $\theta = 60^\circ.$ For a point $(x, y)$ of $\overline{AB}$ with $x, y \gt 0,$ let $$g(\theta) = \frac{x}{\cos\theta} + \frac{y}{\sin\theta} - 1,$$ so the point lies on the member for angle $\theta$ exactly when $g(\theta) = 0.$ Note $g \to +\infty$ at both endpoints of $(0^\circ, 90^\circ)$ and $g(60^\circ) = 0.$ If $g'(60^\circ) \neq 0,$ then $g$ is negative on one side of $60^\circ,$ and the intermediate value theorem produces another zero on that side — the point is covered by another segment. So $C$ must satisfy $g'(60^\circ) = 0,$ and for that point $60^\circ$ is the strict global minimum of $g,$ so no other segment contains it.

Now $g'(\theta) = \frac{x \sin\theta}{\cos^2\theta} - \frac{y \cos\theta}{\sin^2\theta},$ and $g'(60^\circ) = 0$ gives $x \sin^3 60^\circ = y \cos^3 60^\circ,$ i.e. $y = 3\sqrt{3}\,x.$ Intersecting with $\overline{AB}:$ $y = \frac{\sqrt{3}}{2} - \sqrt{3}\,x$ gives $3x = \frac{1}{2} - x,$ so $x = \frac{1}{8}$ and $y = \frac{3\sqrt{3}}{8},$ an interior point of $\overline{AB}.$

Therefore $OC^2 = \frac{1}{64} + \frac{27}{64} = \frac{28}{64} = \frac{7}{16},$ and $p + q = 7 + 16 = 23.$