XXIX Rioplatense Mathematical Olympiad

First, let us notice that since $APB$ and $CDR$ are equilateral, we have $AB=AP$ and $CD=DR$. Hence, the difference between the perimeters of $APQRD$ and $ABCD$ equals $$(AD+DR+QR+PQ+AP)-(AD+CD+BC+AB)=PQ+QR-BC.$$ By hypothesis we know that $BC=24$ and $\text{per}(APQRD)-\text{per}(ABCD)=32$. Therefore, using the previous equality, we get $PQ+QR=56$.

On the other hand, we have that $P\hat{B}Q=360^{\circ }-A\hat{B}C-C\hat{B}Q-A\hat{B}P$, which gives $P\hat{B}Q=90^{\circ }$ since $A\hat{B}C=150^{\circ }$ and $A\hat{B}P=C\hat{B}Q=60^{\circ }$. Analogously, we can obtain that $Q\hat{C}R=90^{\circ }$. Using the Pythagorean theorem on the triangle $△PBQ$ we find that $PQ=\sqrt{18^2+24^2}=30$ and so, using that $PQ+QR=56$, it follows that $QR=26$. Finally, since $Q\hat{C}R=90^{\circ }$, we can apply the Pythagorean theorem again but now in the triangle $△CQR$ in order to get $Q{R}^2=C{Q}^2+C{R}^2$. From the fact that $QC=BC=24$ and $QR=26$ it follows that $CR=10$. Since $CR=CD$ because $△CDR$ is equilateral, it is $CD=10$.