1967 AHSME Problems/Problem 20
Problem
A circle is inscribed in a square of side , then a square is inscribed in that circle, then a circle is inscribed in the latter square, and so on. If is the sum of the areas of the first circles so inscribed, then, as grows beyond all bounds, approaches:
Solution
All answers correctly grow as , so we let .
The radius of the first circle is , so its area is .
The diagonal of the second square is the diameter of the first circle, which is . Therefore, the side length of the square is .
Now we note that the picture is self-similar; if we erase the outer square, erase the outer circle, rotate the picture, and dilate the picture from the center in both the x- and y-directions by an equal scaling factor, we will get the original picture. Therefore, the side lengths of successive squares form a geometric sequence with common ratio , as do the radii of the circles. On the other hand, the areas of the squares (and areas of the circles) form a geometric sequence with common ratio .
Since the first square has side and the second square has side , we know , and .
Since the area of the first circle is , and the common ratio of areas is , the sum of all the areas of the circles is the sum of the infinite geometric sequence, which is . This is for side length , and as noted before, there should be an factor in addition to this number to generalize it from the unit square. This gives answer .
See also
1967 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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