Consider two concentric circles of radius $17$ and $19.$ The larger circle has a chord, half of which lies inside the smaller circle. What is the length of the chord in the larger circle?
$\textbf{(A)}\ 12\sqrt{2} \qquad\textbf{(B)}\ 10\sqrt{3} \qquad\textbf{(C)}\ \sqrt{17 \cdot 19} \qquad\textbf{(D)}\ 18 \qquad\textbf{(E)}\ 8\sqrt{6}$
Label the center of both circles $O$. Label the chord in the larger circle as $\overline{ABCD}$, where $A$ and $D$ are on the larger circle and $B$ and $C$ are on the smaller circle. Construct the radius perpendicular to the chord and label their intersection as $M$. Because a radius that is perpendicular to a chord bisects the chord, $M$ is the midpoint of the chord. 
Construct segments $\overline{AO}$ and $\overline{BO}$. These are radii with lengths 17 and 19 respectively. 
Then, use the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem). In $\triangle OMA$, we have
\begin{align*} OM^2 & = OA^2 - AM^2 \\ & = OA^2 - \left( \frac{AD}{2} \right)^2 \\ & = 19^2 - \frac{AD^2}{4} . \end{align*}
In $\triangle OMB$, we have
\begin{align*} OM^2 & = OB^2 - BM^2 \\ & = OB^2 - \left( \frac{BC}{2} \right)^2 \\ & = OB^2 - \left( \frac{AD}{4} \right)^2 \\ & = 17^2 - \frac{AD^2}{16} . \end{align*}
Equating these two expressions, we get 
\[19^2 - \frac{AD^2}{4} = 17^2 - \frac{AD^2}{16}\]
and $AD=\boxed{\textbf{(E) }8 \sqrt{6}}$.
~eisthefifthletter and [Steven Chen](https://artofproblemsolving.comhttps://www.professorchenedu.com)