jmc

The center of the semicircle is also the midpoint of $AB$. Let this point be O. Let $h$ be the length of $AD$. Rescale everything by 42, so $AU=2,AN=3,UB=4$. Then $AB=6$ so $OA=OB=3$. Since $ON$ is a radius of the semicircle, $ON=3$. Thus $OAN$ is an equilateral triangle. Let $X$, $Y$, and $Z$ be the areas of triangle $OUN$, sector $ONB$, and trapezoid $UBCT$ respectively. $X=\frac{1}{2}(UO)(NO)\sin O=\frac{1}{2}(1)(3)\sin 60^{\circ }=\frac{3}{4}\sqrt{3}$ $Y=\frac{1}{3}\pi (3{)}^2=3\pi$ To find $Z$ we have to find the length of $TC$. Project $T$ and $N$ onto $AB$ to get points $T^{\prime }$ and $N^{\prime }$. Notice that $UN{N}^{\prime }$ and $TU{T}^{\prime }$ are similar. Thus: $\frac{T{T}^{\prime }}{U{T}^{\prime }}=\frac{U{N}^{\prime }}{N{N}^{\prime }}⟹\frac{T{T}^{\prime }}{h}=\frac{1/2}{3\sqrt{3}/2}⟹T{T}^{\prime }=\frac{\sqrt{3}}{9}h$. Then $TC=T^{\prime }C-T^{\prime }T=UB-T{T}^{\prime }=4-\frac{\sqrt{3}}{9}h$. So: $Z=\frac{1}{2}(BU+TC)(CB)=\frac{1}{2}\left(8-\frac{\sqrt{3}}{9}h\right)h=4h-\frac{\sqrt{3}}{18}{h}^2$ Let $L$ be the area of the side of line $l$ containing regions $X,Y,Z$. Then $L=X+Y+Z=\frac{3}{4}\sqrt{3}+3\pi +4h-\frac{\sqrt{3}}{18}{h}^2$ Obviously, the $L$ is greater than the area on the other side of line $l$. This other area is equal to the total area minus $L$. Thus: $\frac{2}{1}=\frac{L}{6h+\frac{9}{2}\pi -L}⟹12h+9\pi =3L$. Now just solve for $h$. $$\begin{array}{rl}12h+9\pi & =\frac{9}{4}\sqrt{3}+9\pi +12h-\frac{\sqrt{3}}{6}{h}^2\\ 0& =\frac{9}{4}\sqrt{3}-\frac{\sqrt{3}}{6}{h}^2\\ h^2& =\frac{9}{4}(6)\\ h& =\frac{3}{2}\sqrt{6}\end{array}$$ Don't forget to un-rescale at the end to get $AD=\frac{3}{2}\sqrt{6}\cdot 42=63\sqrt{6}$. Finally, the answer is $63+6=\boxed{69}$.