, 2012

Since $\triangle QUR$ and $\triangle SUR$ are equilateral, then $\angle QUR = \angle SUR = 60^\circ$. Since $QU=PU=TU=SU$ and $QP = PT=TS$, then $\triangle QUP$, $\triangle PUT$ and $\triangle TUS$ are congruent. Thus, $\angle QUP = \angle PUT = \angle TUS$. The angles around point $U$ add to $360^\circ$. Thus, $\angle SUR + \angle QUR + \angle QUP + \angle PUT + \angle TUS = 360^\circ$ and so $60^\circ + 60^\circ + 3\angle TUS = 360^\circ$ or $3\angle TUS = 240^\circ$ or $\angle TUS = 80^\circ$. Since $\triangle TUS$ is isosceles with $TU = SU$, then $\angle UST = \angle UTS$. Since the angles in $\triangle TUS$ add to $180^\circ$, then $\angle TUS + \angle UST + \angle UTS = 180^\circ$. Therefore, $80^\circ + 2\angle UST = 180^\circ$ and so $2\angle UST = 100^\circ$ or $\angle UST = 50^\circ$.