HMMT February

Construct $\overline{A C}, \overline{A Q}, \overline{B Q}, \overline{B D}$, and let $R$ denote the intersection of $\overline{A C}$ and $\overline{B D}$. Because $A B C D$ is cyclic, we have that $\triangle A B R \sim \triangle D C R$ and $\triangle A D R \sim \triangle B C R$. Thus, we may write $A R=4 x, B R=2 x, C R=6 x, D R=12 x$. Now, Ptolemy applied to $A B C D$ yields $140 x^{2}=1 \cdot 3+2 \cdot 4=11$. Now $\overline{B Q}$ is a median in triangle $A B D$. Hence, $B Q^{2}=\frac{2 B A^{2}+2 B D^{2}-A D^{2}}{4}$. Likewise, $C Q^{2}=\frac{2 C A^{2}+2 C D^{2}-D A^{2}}{4}$. But $P Q$ is a median in triangle $B Q C$, so $P Q^{2}=\frac{2 B Q^{2}+2 C Q^{2}-B C^{2}}{4}=\frac{A B^{2}+B D^{2}+C D^{2}+C A^{2}-B C^{2}-A D^{2}}{4}=$ $\frac{(196+100) x^{2}+1^{2}+3^{2}-2^{2}-4^{2}}{4}=\frac{148 x^{2}-5}{2}=\frac{148 \cdot \frac{11}{140}-5}{2}=\frac{116}{35}$.