1971 AMC 12 Problem 35

Question

Each circle in an infinite sequence with decreasing radii is tangent externally to the one following it and to both sides of a given right angle. The ratio of the area of the first circle to the sum of the areas of all the other circles in the sequence is:

(4+3\sqrt2):4

9\sqrt2:2

(16+12\sqrt2):1

(2+2\sqrt2):1

(3+2\sqrt2):1

Answer: C

Concepts:tangent circles geometric sequence area ratio radical

Difficulty rating: 2380

Small hint:

The centers lie on the angle bisector; find the ratio of consecutive radii

Big hint:

The radius ratio is

3-2\sqrt2

, so the area ratio is its square

Solution:

If consecutive radii are

r\gt r',

their centers lie on the angle bisector at distances

r\sqrt2

and

r'\sqrt2

from the vertex. External tangency gives

\sqrt2(r-r')=r+r',

so

\frac{r'}{r}=3-2\sqrt2.

The ratio of successive areas is

q=(3-2\sqrt2)^2.

Therefore the first area divided by the sum of all later areas is

\begin{aligned} \frac{1-q}{q} &=\frac1q-1\\ &=(3+2\sqrt2)^2-1\\ &=16+12\sqrt2. \end{aligned}

Therefore, the correct answer is C.