1953 AMC 12 Problem 27

Attempt Problem 27 of the 1953 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 1953 AMC 12 solutions, or check the answer key.

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27.

The radius of the first circle is 11 inch, that of the second 12\dfrac12 inch, that of the third 14\dfrac14 inch and so on indefinitely. The sum of the areas of the circles is:
3π4\dfrac{3\pi}{4}
1.3π1.3\pi
2π2\pi
4π3\dfrac{4\pi}{3}
none of these
Answer: D
Concepts:circle areainfinite geometric series
Small hint:
Squaring each radius changes the common ratio
Big hint:
The areas form a geometric series with first term π\pi and ratio 14\frac{1}{4}
Solution:
The areas are π, π4, π16,… \pi,\ \frac{\pi}{4},\ \frac{\pi}{16},\ldots Therefore their sum is π1−14=4π3. \frac{\pi}{1-\frac{1}{4}}=\frac{4\pi}{3}.
Thus, the correct answer is D.
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