22. Let circles $\Gamma$ and $\Gamma'$ be two concentric circles with center $O$, with radii $20$ and $17$ respectively. An equilateral triangle $\triangle ABC$ is inside circle $\Gamma$ and outside circle $\Gamma'$, with point $A$ on circle $\Gamma$, and line $BC$ tangent to circle $\Gamma'$ at point $S$. Line $AO$ intersects $BC$ at point $P$, and $\frac{BP}{CP}=3$. If $AB$ can be expressed in the form $\frac{m}{\sqrt{n}}-\frac{p}{\sqrt{q}}$, where $m, n, p, q \in \mathbf{Z}_{+}$ and $(m, n)=(p, q)=1$, then $m+n+p+q=(\quad)$.
(A) 42
(B) 86
(C) 92
(D) 144
(E) 130
22. E.

In equilateral $\triangle ABC$, draw $AM \perp BC$ at point $M$.
Thus, $BM = MC$.
$$
\begin{array}{l}
\text{By } \frac{BP}{CP} = 3 \Rightarrow MP = PC \\
\Rightarrow \frac{AM}{MP} = \frac{\sqrt{3}}{\frac{1}{2}} = 2 \sqrt{3} .
\end{array}
$$

Since the line containing $BC$ is tangent to circle $\Gamma'$ at point $S$,
$$
\begin{array}{l}
\Rightarrow OS \perp BC \\
\Rightarrow \triangle AMP \backsim \triangle OSP \\
\Rightarrow \frac{AP}{AM} = \frac{OP}{OS} = \frac{\sqrt{13}}{2 \sqrt{3}} \\
\Rightarrow OP = \frac{\sqrt{13}}{2 \sqrt{3}} OS = \frac{17 \sqrt{13}}{2 \sqrt{3}} \\
\Rightarrow AP = OA - OP = 20 - \frac{17 \sqrt{13}}{2 \sqrt{3}} \\
\Rightarrow AB = \frac{2}{\sqrt{3}} AM = \frac{2}{\sqrt{3}} \left( \frac{2 \sqrt{3}}{\sqrt{13}} AP \right) \\
= \frac{4}{\sqrt{13}} AP = \frac{80}{\sqrt{13}} - \frac{34}{\sqrt{3}} \\
\Rightarrow m = 80, n = 13, p = 34, q = 3 \\
\Rightarrow m + n + p + q = 130 .
\end{array}
$$