1983 AHSME Problems/Problem 23
Problem
In the adjoining figure the five circles are tangent to one another consecutively and to the lines and . If the radius of the largest circle is and that of the smallest one is , then the radius of the middle circle is
Solution
Consider three consecutive circles, as shown in the diagram above; observe that their centres , , and are collinear by symmetry. Let , , and be the points of tangency, and let and be segments parallel to the upper tangent (i.e. ), as also shown. Since is parallel to (the three points are collinear), is parallel to (as both are parallel to ), and is parallel to (as both are perpendicular to , due to the tangent being perpendicular to the radius), we have .
Now, if we let , and be the radii of the three circles (from smallest to largest), then and . Thus, from the similarity that we just proved, (where e.g. because of collinearity). This equation reduces to , i.e. , so the ratio of consecutive radii is constant, forming a geometric sequence. In this case, as the first radius is and, four radii later, the radius is , this constant ratio is . Therefore the middle radius is , which is choice .
See Also
1983 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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