1971 AHSME Problems/Problem 35
Problem
Each circle in an infinite sequence with decreasing radii is tangent externally to the one following it and to both sides of a given right angle. The ratio of the area of the first circle to the sum of areas of all other circles in the sequence, is
Solution 1
WLOG, let the radius of the largest circle be (we can do this because scaling the diagram preserves the ratio of areas). Let the largest circle be and the second largest circle be . As in the diagram above, let have center with points of tangency at and . Let the tangents from these two points meet at . From the problem, we know that . Let the common tangent of and intersect and at and , respectively. Let intersect at .
By the Two Tangent Theorem, . Thus, is an isosceles right triangle, so . is a tangent-chord angle in , so . Thus, because has right angles and adjacent sides congruent, it is a square. Thus, .
is the incircle of , because it is tangent to all its sides. Let and be the semiperimeter of . Because , we see that and . Thus, is an isosceles right triangle, so . By the properties of an incircle's touchpoints, and . Thus, is the midpoint of , so .
Because is the common tangent of and , is on . Thus, by the Two Tangent Theorem, .
Now, we have the equation , which yields . We know that . Let the inradius of be . Now the formula in gives us the equation , which yields .
Because the ratio of the radii of and is , the ratio of their areas is . This ratio is the same as that of the area of to the third largest circle, the third to the fourth, etc., because removing the largest circle gives a scaled-down version of the same problem. Thus, the total area of the circles is given by the geometric sequence starting at (the area of ) with common ratio , which, by the infinite geometric series formula, has an infinite sum of . However, we are looking for the ratio of the area of to that of all of the circles excluding , so we have to subtract . This reveals that the area of the other circles is . Now, our desired answer is the ratio . Thus, we choose .
Solution 2 (Informed guess)
Draw a good diagram, and especially make sure that you are drawing a right angle. Then, you will realize that the big circle completely dwarfs all of the other circles. Answer choice (C), , is about , whereas the greatest of the other answer choices is about , which is unreasonably small based on our diagram. Thus, we choose answer .
See Also
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