National Olympiad of Argentina

One may assume $P$ between $A$ and $Q$. Let the incircle touch sides $BC$ and $CA$ at $U$ and $V$ respectively. By power of a point $AV^2 = AP \cdot AQ$, $MU^2 = MQ \cdot MP$. Also $AQ = MP$ as $AP = QM$, and so $AV^2 = MU^2$, $AV = MU$. On the other hand $CU = CV$ by equal tangents, hence $AC = AV + CV = MU + CU = MC$. Because $M$ is the midpoint of $BC$, it follows that $BC = 2AC = 2$. Therefore $AB = \sqrt{AC^2 + BC^2} = \sqrt{5}$. In addition triangle $AMC$ is right and isosceles, with $\angle AMC = \angle MAC = 45^\circ$.

We employ the equality $AV^2 = AP \cdot AQ$ again to compute $PQ$. First, $AV = \frac{1}{2}(AB+AC-BC) = \frac{\sqrt{5}-1}{2}$. (If the incircle touches $AB$ at $T$ then $AV = AT$, $BT = BU$, $CU = CV$ imply $AV+BU+CU = \frac{1}{2}(AB + BC + CA)$; on the other hand $BU + CU = BC$.) Second, $AM$ and $PQ$ have common midpoint $N$ because $AP = QM$. So if $PQ = 2x$ then $PN = NQ = x$, $AP = AN-x$, $AQ = AN+x$. Since $N$ is the midpoint of the hypotenuse $AM$ of the right triangle $AMC$ with $\angle MAC = 45^\circ$, we have $AN = \frac{AC}{\sqrt{2}} = \frac{1}{\sqrt{2}}$. Thus $AV^2 = AP \cdot AQ$ takes the form $\left(\frac{\sqrt{5}-1}{2}\right)^2 = \left(\frac{1}{\sqrt{2}} - x\right)\left(\frac{1}{\sqrt{2}} + x\right)$, or $\frac{3-\sqrt{5}}{2} = \frac{1}{2} - x^2$. Hence $x = \sqrt{\frac{\sqrt{5}-2}{2}}$, $PQ = 2x = \sqrt{2\sqrt{5}-4}$.