Geogebra3

Let the extension of $PQ$ to meet $OT$ and $UR$ at $M$ and $N$ respectively. We note both $\triangle PRU$ and $\triangle OQT$ are both unit equilateral triangles. Therefore, $PN = \frac {\sqrt 3}2$ , $QN = 1 - \frac {\sqrt 3}2$ , and $PQ = PN - QN = \frac {\sqrt 3}2 - \left(1-\frac {\sqrt 3}2\right) = \sqrt 3 -1$ . Implying $n = \boxed 3$ .

As shown, extend $\overline{PQ}$ and call its intersections with $\overline{OT}$ and $\overline{UR}$ $E$ and $F$ , respectively. By symmetry, $OE=ET=UF=FR=\frac{1}{2}$ . Thus $\angle PUF = \cos^{-1}\frac{1}{2} = \frac{\pi}{3}$ and $PF=\frac{\sqrt{3}}{2}$ . Similarly, $QE = \frac{\sqrt{3}}{2}$ . Thus $PQ = EF - (QF) - (PE) = 1-2\left(1-\frac{\sqrt{3}}{2}\right) = \sqrt{3} - 1$ So $n=\boxed{3}$ .

Let's place a coordinate system such that $U=(0,0), R=(1,0)$ . Then, the two quarter circles with respective centers at $U$ and $R$ obviously intersect at $x=\frac 12$ .

The quarter circle around $U$ can be described by the equation $y=\sqrt{1-x^2} \Rightarrow y \left( \frac 12 \right) = \sqrt{1-\left( \frac 12 \right)^2} = \frac {\sqrt{3}}2$ .

If we define point $S=\left(\frac12,\frac12\right)$ to be the midpoint of the square, then $\left|\overline{PS}\right|=\frac {\sqrt{3}}2-\frac 12=\frac {\sqrt{3-1}}2$ .

Now, $\left|\overline{PQ}\right|$ is just twice as long as $\left|\overline{PS}\right|$ , so $\left|\overline{PQ}\right| = \boxed{\sqrt{3}-1}$ , which makes the answer 3 .