1997 Fermat Problem 25

In triangle ABC, R is the mid-point of BC, CS = 3SA, and \frac{AT}{TB} = \frac{p}{q}. If w is the area of \triangle CRS, x is the area of \triangle RBT, z is the area of \triangle ATS, and x^2 = wz, then the value of \frac{p}{q} is

\textbf{(A)}\ \frac{\sqrt{21}-3}{2}\quad \textbf{(B)}\ \frac{\sqrt{21}+3}{2}\quad \textbf{(C)}\ \frac{\sqrt{21}-3}{6}\quad \textbf{(D)}\ \frac{\sqrt{105}+3}{6}\quad \textbf{(E)}\ \frac{\sqrt{105}-3}{6}

If there are no answer choices shown, enter a numerical answer.

Full credit to this problem is given to the CEMC, you may view all Fermat contests here.