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 $A{V}^2=AP\cdot AQ$, $M{U}^2=MQ\cdot MP$. Also $AQ=MP$ as $AP=QM$, and so $A{V}^2=M{U}^2$, $AV=MU$. 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 $AB=\sqrt{A{C}^2+B{C}^2}=\sqrt{5}$. In addition triangle $AMC$ is right and isosceles, with $\angle AMC=\angle MAC=45^{\circ }$.



We employ the equality $A{V}^2=AP\cdot AQ$ 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 $A{V}^2=AP\cdot AQ$ 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 $x=\sqrt{\frac{\sqrt{5}-2}{2}}$, $PQ=2x=\sqrt{2\sqrt{5}-4}$.