Difference between revisions of "1967 AHSME Problems/Problem 20"
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== Solution == | == Solution == | ||
− | <math>\fbox{A}</math> | + | |
+ | All answers correctly grow as <math>m^2</math>, so we let <math>m=1</math>. | ||
+ | |||
+ | The radius of the first circle is <math>\frac{1}{2}</math>, so its area is <math>\frac{\pi}{4}</math>. | ||
+ | |||
+ | The diagonal of the second square is the diameter of the first circle, which is <math>1</math>. Therefore, the side length of the square is <math>\frac{\sqrt{2}}{2}</math>. | ||
+ | |||
+ | 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 <math>R</math>, 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 <math>R^2</math>. | ||
+ | |||
+ | Since the first square has side <math>1</math> and the second square has side <math>\frac{\sqrt{2}}{2}</math>, we know <math>R = \frac{\sqrt{2}}{2}</math>, and <math>R^2 = \frac{1}{2}</math>. | ||
+ | |||
+ | Since the area of the first circle is <math>\frac{\pi}{4}</math>, and the common ratio of areas is <math>\frac{1}{2}</math>, the sum of all the areas of the circles is the sum of the infinite geometric sequence, which is <math>\frac{A_1}{1 - r} = \frac{\frac{\pi}{4}}{1 - \frac{1}{2}} = \frac{\pi}{2}</math>. This is for side length <math>m=1</math>, and as noted before, there should be an <math>m^2</math> factor in addition to this number to generalize it from the unit square. This gives answer <math>\fbox{A}</math>. | ||
== See also == | == See also == |
Latest revision as of 19:06, 12 July 2019
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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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