[Geometry]
Problem: Let $T=8$. Let $A=(1,5)$ and $B=(T-1,17)$. Compute the value of $x$ such that $(x, 3)$ lies on the perpendicular bisector of $\overline{A B}$.

Solution: The midpoint of $\overline{A B}$ is $\left(\frac{T}{2}, 11\right)$, and the slope of $\overleftrightarrow{A B}$ is $\frac{12}{T-2}$. Thus the perpendicular bisector of $\overline{A B}$ has slope $\frac{2-T}{12}$ and passes through the point $\left(\frac{T}{2}, 11\right)$. Thus the equation of the perpendicular bisector of $\overline{A B}$ is $y=\left(\frac{2-T}{12}\right) x+\left(11-\frac{2 T-T^{2}}{24}\right)$. Plugging $y=3$ into this equation and solving for $x$ yields $x=\frac{96}{T-2}+\frac{T}{2}$. With $T=8$, it follows that $x=\frac{96}{6}+\frac{8}{2}=16+4=\mathbf{2 0}$.

Final answer: 20