1959 AHSME Problems/Problem 41
Problem
On the same side of a straight line three circles are drawn as follows: a circle with a radius of 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:
Solution
Let the radius of the two large circles be . Also, let the center of the small circle be and the centers of the two large circles be and , as in the diagram. Further, let the large circles intersect at , let the circle centered at be tangent to the line at , and let the circle centered at be tangent to the line at . Because and , . Because the two large circles are tangent at , is on the segment connecting their centers, . Thus, is a distance from . Because the small circle (with radius ) is tangent to the line, . Also, because the small circle and the circle centered at are tangent, . Because , by the Pythagorean Theorem, we have the following equation:
See also
1959 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 40 |
Followed by Problem 42 | |
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