Problem 51

First, we note that $\triangle P Q S$ and $\triangle R Q S$ are equilateral. Join $P$ to $R$. Since $P Q R S$ is a rhombus, then $P R$ and $Q S$ bisect each other at their point of intersection, $M$, and are perpendicular. Note that $Q M=M S=\frac{1}{2} Q S=3$. Since $\angle P S Q=60^{\circ}$, then $P M=P S \sin (\angle P S M)=6 \sin \left(60^{\circ}\right)=6\left(\frac{\sqrt{3}}{2}\right)=3 \sqrt{3}$. Since $P T=T R$, then $\triangle P R T$ is isosceles. Since $M$ is the midpoint of $P R$, then $T M$ is perpendicular to $P R$. Since $S M$ is also perpendicular to $P R$, then $S$ lies on $T M$. By the Pythagorean Theorem in $\triangle P M T$, since $M T>0$, we have $M T=\sqrt{P T^{2}-P M^{2}}=\sqrt{14^{2}-(3 \sqrt{3})^{2}}=\sqrt{196-27}=\sqrt{169}=13$. Therefore, $S T=M T-M S=13-3=10$.