1992 AMC 12 Problem 30

By symmetry, place the trapezoid with its bases centered on the same vertical axis and put the circle’s center at the midpoint of $AB$

Write the distance from the center to a leg and require the tangency point to lie on the leg segment; the minimum occurs at an endpoint case

Place $A=(-46,0),$ $B=(46,0),$ $D=(-\frac{19}{2},h),$ and $C=(\frac{19}{2},h).$ Symmetry forces the circle’s center to be $O=(0,0).$ The horizontal offset along each leg is $\frac{73}{2},$ so $$x^2=h^2+\left(\frac{73}{2}\right)^2.$$ The perpendicular from $O$ to a leg first has its foot on the segment when the foot reaches the upper endpoint; this boundary condition is $OD\perp AD.$ Thus $$\begin{aligned} (-\frac{19}{2},h)\cdot(\frac{73}{2},h)&=0,\\ h^2&=\frac{1387}{4}. \end{aligned}$$ Therefore $$m^2=\frac{1387+5329}{4}=1679.$$

Thus the correct answer is B.