Randomly plotting the angle again

The locus of points giving the same $\theta$ value divides the square into two regions; one (green) where the angle $APB$ is bigger than $\theta$ , and one where it's smaller (blue). The probability of being in one of these regions is equal to the area of that region (since the probability space has unit area).

At the median $\theta$ value, these two regions will each have area $\frac12$ (by definition of the median). So if we can find the area, we get an equation for $\theta$ .

To find the locus, let the coordinates of $P$ be $(x,y)$ ; then $\theta=\tan^{-1} \frac{y}{x}+\tan^{-1} \frac{1-y}{x}$

Taking the tangent of both sides, $\begin{aligned} \tan \theta &=\frac{ \frac{y}{x}+\frac{1-y}{x}}{1-\frac{y}{x} \cdot \frac{1-y}{x}} \\ \tan \theta &=\frac{x}{x^2-y+y^2} \\ \left(x^2-y+y^2 \right) \tan \theta &=x \end{aligned}$

which we recognise as the equation of a circle. In fact, it's the equation of a circle with centre $(\frac12 \cot \theta,\frac12)$ and radius $\frac12 \csc \theta$ . As you might intuitively expect, this circle passes through $(0,0)$ and $(0,1)$ as well.

The region enclosed by the locus and the square is made up of a rectangular portion and a circular segment. Its area is given by $\cot \theta+\frac{2\theta-\sin 2\theta}{8\sin^2 \theta}$

We want this equal to $\frac12$ . Solving numerically, we find $\theta=1.3841\ldots$ for the median.

Alternatively, we can rearrange to the form in the question: $\begin{aligned} \cot \theta+\frac{2\theta-\sin 2\theta}{8\sin^2 \theta} &=\frac12 \\ 8\cos \theta \sin \theta + 2\theta-\sin 2\theta &=4\sin^2 \theta \\ 3\sin 2\theta+2\cos 2\theta + 2\theta &=2 \end{aligned}$

giving the answer $\boxed7$ .

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There are cases when the locus doesn't look exactly like the image above; but none of these are near the median.