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Difference between revisions of "1967 AHSME Problems/Problem 20"

(Created page with "== Problem == A circle is inscribed in a square of side <math>m</math>, then a square is inscribed in that circle, then a circle is inscribed in the latter square, and so on. I...")
 
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== Solution ==
 
== Solution ==
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<math>\fbox{A}</math>
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All answers correctly grow as <math>m^2</math>, so we let <math>m=1</math>.
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The radius of the first circle is <math>\frac{1}{2}</math>, so its area is <math>\frac{\pi}{4}</math>.
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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>.
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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>.
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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>. 
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 +
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 ==

Revision as of 19:06, 12 July 2019

Problem

A circle is inscribed in a square of side $m$, then a square is inscribed in that circle, then a circle is inscribed in the latter square, and so on. If $S_n$ is the sum of the areas of the first $n$ circles so inscribed, then, as $n$ grows beyond all bounds, $S_n$ approaches:

$\textbf{(A)}\ \frac{\pi m^2}{2}\qquad \textbf{(B)}\ \frac{3\pi m^2}{8}\qquad \textbf{(C)}\ \frac{\pi m^2}{3}\qquad \textbf{(D)}\ \frac{\pi m^2}{4}\qquad \textbf{(E)}\ \frac{\pi m^2}{8}$

Solution

All answers correctly grow as $m^2$, so we let $m=1$.

The radius of the first circle is $\frac{1}{2}$, so its area is $\frac{\pi}{4}$.

The diagonal of the second square is the diameter of the first circle, which is $1$. Therefore, the side length of the square is $\frac{\sqrt{2}}{2}$.

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 $R$, 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 $R^2$.

Since the first square has side $1$ and the second square has side $\frac{\sqrt{2}}{2}$, we know $R = \frac{\sqrt{2}}{2}$, and $R^2 = \frac{1}{2}$.

Since the area of the first circle is $\frac{\pi}{4}$, and the common ratio of areas is $\frac{1}{2}$, the sum of all the areas of the circles is the sum of the infinite geometric sequence, which is $\frac{A_1}{1 - r} = \frac{\frac{\pi}{4}}{1 - \frac{1}{2}} = \frac{\pi}{2}$. This is for side length $m=1$, and as noted before, there should be an $m^2$ factor in addition to this number to generalize it from the unit square. This gives answer $\fbox{A}$.

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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