$20 \cdot 114$ As shown in the figure, five circles are each tangent to lines $L_{1}, L_{2}$, and are tangent to each other in sequence. If the radius of the largest circle is 18, and the radius of the smallest circle is 8, then the radius of the middle circle is
(A) 12.
(B) 12.5.
(C) 13.
(D) 13.5.
(E) 14.
(34th American High School Mathematics Examination, 1983)
［Solution］Consider the case of three circles; the case of five circles is similar.
As shown in the figure, the centers $P, Q, R$ of the three circles are collinear, and this line bisects the angle formed by the two tangents.

Let $A, B, C$ be the points of tangency, and let $P S, Q T$ be line segments parallel to the tangents. Then $\triangle P Q S \backsim \triangle Q R T$.
If $x, y, z$ are the radii of the small, medium, and large circles, respectively, then
$$
Q S=y-x, \quad R T=z-y .
$$

Thus, $\quad \frac{Q S}{P Q}=\frac{R T}{Q R}$ becomes $\frac{y-x}{x+y}=\frac{z-y}{y+z}$.
This simplifies to $y^{2}=x z$, or $\frac{y}{x}=\frac{z}{y}$. This shows that the ratio of consecutive radii is constant.

Applying this conclusion to the original problem, the consecutive radii of the five circles form a geometric sequence.
Given the problem, the common ratio can be calculated as $\sqrt[4]{\frac{18}{8}}$, so the radius of the middle circle is $8 \sqrt{\frac{18}{8}}=12$. Therefore, the answer is $(A)$.