Baltic Way 1996 · Problem 3

Solution:

As $Q$ is the circumcentre of triangle $BPC$, we have $|PQ| = |QC|$ and $Q$ lies on the perpendicular bisector $s$ of $BC$. On the other hand, as $D$ is the circumcentre of triangle $PQA$, $Q$ 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 $ABCD$.

Let $E$ and $F$ be the midpoints of $AD$ and $BC$, respectively. We have $|AQ_1| = |DQ_1| = |DA| = 1$. Hence $|EQ_1| = \frac{\sqrt{3}}{2}$ and $|FQ_1| = 1 - \frac{\sqrt{3}}{2}$. The Pythagorean theorem applied to the triangle $CFQ_1$ now yields

$$|CQ_1|^2 = |CF|^2 + |FQ_1|^2 = \left(\frac{1}{2}\right)^2 + \left(1 - \frac{\sqrt{3}}{2}\right)^2 = 2 - \sqrt{3}$$

and hence $|CQ_1| = \sqrt{2 - \sqrt{3}}$.

Similarly, $|Q_2E| = \frac{\sqrt{3}}{2}$, and the Pythagorean theorem applied to the triangle $CFQ_2$ now yields

$$|CQ_2|^2 = |CF|^2 + |FQ_2|^2 = \left(\frac{1}{2}\right)^2 + \left(1 + \frac{\sqrt{3}}{2}\right)^2 = 2 + \sqrt{3}$$

and hence $|CQ_2| = \sqrt{2 + \sqrt{3}}$.

Hence the possible values of the length of the segment $PQ$ are $\sqrt{2 - \sqrt{3}}$ and $\sqrt{2 + \sqrt{3}}$.