1954 AMC 12 Problem 26
Attempt Problem 26 of the 1954 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 1954 AMC 12 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
26.
The straight line is divided at so that Circles are described on and as diameters and a common tangent meets produced at Then equals:
the diameter of the smaller circle
the radius of the smaller circle
the radius of the larger circle
the difference of the two radii
Answer: B
Small hint:
The point is the external center of similitude of the two circles
Big hint:
Let and compare the distances from to the two centers in the ratio
Solution:
Let so and Measured from the circle centers are at and and their radii are in the ratio If the common external tangent makes the external center of similitude, so Hence and which is the radius of the smaller circle.
Thus, the correct answer is B.
Problem 26 in other years
- 1950 AMC 12
- 1951 AMC 12
- 1952 AMC 12
- 1953 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
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- 1978 AMC 12
- 1979 AMC 12
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- 1989 AMC 12
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- 1999 AMC 12