Zig, say hello to Zag .....

Without loss of generality let the corners of the square be $A(0,2), B(2,2), C(2,0)$ and $D(0,0)$ . Now let point $P$ have coordinates $(a,b)$ ; then by symmetry the coordinates of point $Q$ must be $(2 - a, 2 - b)$ . Then since $DP = PQ$ , we have that

$a^{2} + b^{2} = (2 - 2a)^{2} + (2 - 2b)^{2} \Longrightarrow 3a^{2} + 3b^{2} - 8a - 8b + 8 = 0$

$\Longrightarrow (a - \frac{4}{3})^{2} + (b - \frac{4}{3})^{2} = \frac{8}{9}$ .

This means that $P$ lies on a circle centered at $O(\frac{4}{3}, \frac{4}{3})$ with radius $r = \frac{2}{3}\sqrt{2}$ . Therefore, $\angle ADP$ will be minimized when $DP$ is tangent to this circle.

Now by symmetry $OD$ makes an angle of $45$ degrees with the $x$ -axis and has length $\frac{4}{3}\sqrt{2}$ . Thus $\angle PDO = 45 - \angle ADP$ , and so

$\sin(\angle PDO) = \sin(45 - \angle ADP) = \dfrac{OP}{OD} = \dfrac{\frac{2}{3}\sqrt{2}}{\frac{4}{3}\sqrt{2}} = \dfrac{1}{2}$ .

This implies that $45 - \angle ADP = \sin^{-1}(\frac{1}{2}) = 30$ , and so the minimum value for $\angle ADP$ is $45 - 30 = \boxed{15}$ degrees.

Imgur

Let DP = PQ = QB = x in the unit square shown.

By symmetry, the midpoint M of PQ would also be the midpoint of diagonal DB, giving $DM = \frac{1}{\sqrt{2}}$ Using cosine rule in triangle DPM, $\cos \theta = \frac{x^2 + \frac{1}{2} - \frac{x^2}{4}}{2 \frac {x}{\sqrt{2}}}$ To minimize angle ADP, we must find x that maximizes angle theta or minimize cos(theta), so differentiating the expression and equating to zero gives $x = \sqrt{\frac{2}{3}} and \cos \theta = \frac{\sqrt{3}}{2}$ Thus theta = 30° and angle ADP = 45° - 30° = 15°

P.S. After solving the problem, plotted it to scale and discovered that angle PMD = 90°. There must be an interesting explanation for this, which can lead to a quicker solution. Hope someone interprets that.