Let $A B C D$ be a unit square and let $P$ and $Q$ be points in the plane such that $Q$ is the circumcentre of triangle $B P C$ and $D$ is the circumcentre of triangle $P Q A$. Find all possible values of the length of segment $P Q$.
As $Q$ is the circumcentre of triangle $B P C$, we have $|P Q|=|Q C|$ and $Q$ lies on the perpendicular bisector $s$ of $B C$. On the other hand, as $D$ is the circumcentre of triangle $P Q A, Q$ lies on the circle centred at $D$ and passing through $A$. Thus $Q$ must be one of the two intersection points $Q_{1}$ and $Q_{2}$ of this circle and the line $s$. We may choose $Q_{1}$ to lie inside, and $Q_{2}$ outside of the square $A B C D$.

Let $E$ and $F$ be the midpoints of $A D$ and $B C$, respectively. We have $\left|A Q_{1}\right|=\left|D Q_{1}\right|=|D A|=1$. Hence $\left|E Q_{1}\right|=\frac{\sqrt{3}}{2}$ and $\left|F Q_{1}\right|=1-\frac{\sqrt{3}}{2}$. The Pythagorean theorem applied to the triangle $C F Q_{1}$ now yields

$$
\left|C Q_{1}\right|^{2}=|C F|^{2}+\left|F Q_{1}\right|^{2}=\left(\frac{1}{2}\right)^{2}+\left(1-\frac{\sqrt{3}}{2}\right)^{2}=2-\sqrt{3}
$$

and hence $\left|C Q_{1}\right|=\sqrt{2-\sqrt{3}}$. Similarly, $\left|Q_{2} E\right|=\frac{\sqrt{3}}{2}$, and the Pythagorean theorem applied to the triangle $C F Q_{2}$ now yields

$$
\left|C Q_{2}\right|^{2}=|C F|^{2}+\left|F Q_{2}\right|^{2}=\left(\frac{1}{2}\right)^{2}+\left(1+\frac{\sqrt{3}}{2}\right)^{2}=2+\sqrt{3}
$$

and hence $\left|C Q_{2}\right|=\sqrt{2+\sqrt{3}}$. Hence the possible values of the length of the segment $P Q$ are $\sqrt{2-\sqrt{3}}$ and $\sqrt{2+\sqrt{3}}$.

Remark. The actual location of the point $P$ is unimportant for us. Note however that the point $P$ exists because $P$ and $C$ are the two intersection points of the circle centred at $D$ passing through $A$ and the circle centred at $Q$ passing through $C$.