1959 AMC 12 Problem 41

Attempt Problem 41 of the 1959 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 1959 AMC 12 solutions, or check the answer key.

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41.

On the same side of a straight line three circles are drawn as follows: a circle with a radius of 44 inches is tangent to the line, the other two circles are equal, and each is tangent to the line and to the other two circles. The radius of the equal circles is:

2424

2020

1818

1616

1212

Answer: D
Concepts:tangent circlesdistance formula
Difficulty rating: 1590
Small Hint:

The small circle lies symmetrically between the two equal circles

Big Hint:

If an equal circle has radius R,R, compare the center distance R2+(R4)2\sqrt{R^2+(R-4)^2} with R+4R+4

Solution:

Let the equal circles have radius R.R. Their centers are 2R2R apart, so the center of the radius-44 circle lies midway between them. The horizontal and vertical separations between its center and either large center are RR and R4.R-4. Tangency gives R2+(R4)2=(R+4)2. R^2+(R-4)^2=(R+4)^2. Simplifying yields R216R=0,R^2-16R=0, and positivity gives R=16.R=16.

Thus, the correct answer is D.

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Problem 41 in Other Years

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