$20 \cdot 19$ Let $C_{1}, C_{2}$, and $C_{3}$ be three parallel chords on the same side of the center of a circle. The distance between $C_{1}$ and $C_{2}$ is equal to the distance between $C_{2}$ and $C_{3}$. The lengths of the three chords are 20, 16, and 8. The radius of the circle is 
(A) 12.
(B) $4 \sqrt{7}$.
(C) $\frac{5 \sqrt{65}}{3}$.
(D) $\frac{5 \sqrt{22}}{2}$.
(E) There is no unique solution according to the given conditions.
(31st American High School Mathematics Examination, 1980)
[Solution] According to the problem, we draw the right figure. Let $r$ be the radius, $x$ be the distance from the center $O$ to the nearest chord, and $y$ be the common distance between the chords. By the Pythagorean theorem, we have
$$
\left\{\begin{array}{l}
r^{2}=x^{2}+10^{2}, \\
r^{2}=(x+y)^{2}+8^{2}, \\
r^{2}=(x+2 y)^{2}+4^{2} .
\end{array}\right.
$$

From (2) - (1) we get $0=2 x y+y^{2}-36$,
From (3) - (2) we get $0=2 x y+3 y^{2}-48$.
From (5) - (4) we get $2 y^{2}=12$, so $y=\sqrt{6}$.
Substituting $y=\sqrt{6}$ into the above equations, we get $r=\frac{5 \sqrt{22}}{2}$. Therefore, the answer is $(D)$.