1. **Problem Statement:** We have triangle $\triangle XYZ$ with circumcenter $S$, where $S$ is the intersection of the perpendicular bisectors $\overline{TS}$, $\overline{US}$, and $\overline{VS}$. Given $VS=66$, $YZ=82$, and $ZS=110$, find $XS$, $UZ$, and $VY$. 2. **Key Concept:** The circumcenter $S$ is equidistant from all vertices of the triangle. This means: $$XS = YS = ZS$$ 3. **Given:** - $VS = 66$ - $YZ = 82$ - $ZS = 110$ 4. **Find $XS$:** Since $S$ is the circumcenter, $XS = ZS = 110$. 5. **Find $UZ$ and $VY$:** Points $U$ and $V$ lie on the sides of the triangle where the perpendicular bisectors intersect. - $U$ lies on side $YZ$, so $UZ$ is half of $YZ$ because $U$ is the midpoint of $YZ$ (property of perpendicular bisector). - $V$ lies on side $XY$, so $VY$ is half of $XY$. But $XY$ is not given directly. 6. **Calculate $UZ$:** $$UZ = \frac{YZ}{2} = \frac{82}{2} = 41$$ 7. **Calculate $VY$:** Since $V$ is the midpoint of $XY$, $VY = VX$. 8. **Find $XY$ using the circumradius:** Since $S$ is the circumcenter, $XS = YS = ZS = 110$. Using the triangle side $YZ=82$, and $S$ lies on the perpendicular bisector of $YZ$, $S$ is equidistant from $Y$ and $Z$. Similarly, $V$ is the midpoint of $XY$, so $VY = VX = \frac{XY}{2}$. But $VS=66$ is given, and $VS$ is the distance from $V$ to $S$. Since $S$ is the circumcenter, $S$ lies on the perpendicular bisector of $XY$ at $V$. Using the right triangle $VYS$ with $VS=66$ and $YS=110$, by Pythagoras: $$XY = 2 \times VY$$ Calculate $VY$: $$VY = \sqrt{YS^2 - VS^2} = \sqrt{110^2 - 66^2} = \sqrt{12100 - 4356} = \sqrt{7744} = 88$$ Therefore, $$XY = 2 \times 88 = 176$$ So, $$VY = 88$$ **Final answers:** $$XS = 110$$ $$UZ = 41$$ $$VY = 88$$