Logo for LIVE by Po-Shen Loh

LIVE

English

1971 AMC 12 Problem 35

Attempt Problem 35 of the 1971 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1971 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

35.

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+32):4(4+3\sqrt2):4

92:29\sqrt2:2

(16+122):1(16+12\sqrt2):1

(2+22):1(2+2\sqrt2):1

(3+22):1(3+2\sqrt2):1

Answer: C
Concepts:tangent circlesgeometric sequencearea ratioradical
Difficulty rating: 2380
Small Hint:

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

Big Hint:

The radius ratio is 3−223-2\sqrt2, so the area ratio is its square

Solution:

If consecutive radii are r>r′,r\gt r', their centers lie on the angle bisector at distances r2r\sqrt2 and r′2r'\sqrt2 from the vertex. External tangency gives 2(r−r′)=r+r′, \sqrt2(r-r')=r+r', so r′r=3−22.\frac{r'}{r}=3-2\sqrt2. The ratio of successive areas is q=(3−22)2. q=(3-2\sqrt2)^2. Therefore the first area divided by the sum of all later areas is 1−qq=1q−1=(3+22)2−1=16+122. \begin{aligned} \frac{1-q}{q} &=\frac1q-1\\ &=(3+2\sqrt2)^2-1\\ &=16+12\sqrt2. \end{aligned}

Therefore, the correct answer is C.

← Problem 34#34
Full Exam

Problem 35 in Other Years

1950 AMC 12 · 1951 AMC 12 · 1952 AMC 12 · 1953 AMC 12 · 1954 AMC 12 · 1955 AMC 12 · 1956 AMC 12 · 1957 AMC 12 · 1958 AMC 12 · 1959 AMC 12 · 1960 AMC 12 · 1961 AMC 12 · 1962 AMC 12 · 1963 AMC 12 · 1964 AMC 12 · 1965 AMC 12 · 1966 AMC 12 · 1967 AMC 12 · 1968 AMC 12 · 1969 AMC 12 · 1970 AMC 12 · 1972 AMC 12 · 1973 AMC 12