20.115 A sequence of circles with decreasing radii is formed such that each circle is tangent to the next circle and to the two sides of a given right angle. The ratio of the area of the first circle to the sum of the areas of all other circles in the sequence is
(A) $(4+3 \sqrt{2}): 4$.
(B) $9 \sqrt{2}: 2$.
(C) $(16+12 \sqrt{2}): 1$.
(D) $(2+2 \sqrt{2}): 1$.
(E) $(3+2 \sqrt{2}): 1$.
(22nd American High School Mathematics Examination, 1971)
[Solution] Let $O$ denote the vertex of the right angle, $C$ and $C^{\prime}$ the centers of the circles, and $r$ and $r^{\prime}$ the radii of any two adjacent circles $\left(r>r^{\prime}\right)$. If $T$ is the point of tangency of the two circles, then

and
$$
O T=O C^{\prime}+r^{\prime}=(\sqrt{2}+1) r^{\prime},
$$

From these two equations, we get the ratio of the radii of adjacent circles
$$
\frac{r^{\prime}}{r}=\frac{(\sqrt{2}-1)}{(\sqrt{2}+1)}=(\sqrt{2}-1)^{2} \text {. }
$$

If $r$ is the radius of the first circle in the sequence, then its area is $\pi r^{2}$, and the sum of the areas of all other circles, which form a geometric series, is
$$
\begin{aligned}
& \pi r^{2}\left[(\sqrt{2}-1)^{4}+(\sqrt{2}-1)^{8}+\cdots\right] \\
= & \frac{\pi(\sqrt{2}-1)^{4} r^{2}}{1-(\sqrt{2}-1)^{4}}=\frac{\pi r^{2}}{(\sqrt{2}+1)^{4}-1} .
\end{aligned}
$$
$\therefore$ The required ratio of the areas is
$$
\left[(\sqrt{2}+1)^{4}-1\right]: 1=(16+12 \sqrt{2}): 1 .
$$

Therefore, the answer is (C).