2009 AMC 10A Problems/Problem 21

Problem

Many Gothic cathedrals have windows with portions containing a ring of congruent circles that are circumscribed by a larger circle, In the figure shown, the number of smaller circles is four. What is the ratio of the sum of the areas of the four smaller circles to the area of the larger circle?

$[asy] unitsize(6mm); defaultpen(linewidth(.8pt)); draw(Circle((0,0),1+sqrt(2))); draw(Circle((sqrt(2),0),1)); draw(Circle((0,sqrt(2)),1)); draw(Circle((-sqrt(2),0),1)); draw(Circle((0,-sqrt(2)),1)); [/asy]$

$\mathrm{(A)}\ 3-2\sqrt2 \qquad \mathrm{(B)}\ 2-\sqrt2 \qquad \mathrm{(C)}\ 4(3-2\sqrt2) \qquad \mathrm{(D)}\ \frac12(3-\sqrt2) \qquad \mathrm{(E)}\ 2\sqrt2-2$

Solution

Draw some of the radii of the small circles as in the picture below.

$[asy] unitsize(12mm); defaultpen(linewidth(.8pt)); draw(Circle((0,0),1+sqrt(2))); draw(Circle((sqrt(2),0),1)); draw(Circle((0,sqrt(2)),1)); draw(Circle((-sqrt(2),0),1)); draw(Circle((0,-sqrt(2)),1)); draw( (sqrt(2),0) -- (0,sqrt(2)) -- (-sqrt(2),0) -- (0,-sqrt(2)) -- cycle ); draw( (0,sqrt(2)) -- (0,1+sqrt(2)) ); draw( (0,-sqrt(2)) -- (0,-1-sqrt(2)) ); [/asy]$

It is now obvious that the radius $R$ of the large circle can be expressed using the radius $r$ of the small circles as $R=2r + 2r\sqrt 2$ . Then the area of the large circle is $L = \pi R^2 = \pi (2r)^2 (1+\sqrt 2)^2 = 4\pi r^2 (3+2\sqrt 2)$ . The area of four small circles is $S = 4\pi r^2$ . Hence their ratio is:

\begin{align*}
\frac SL 
& = \frac{4\pi r^2}{4\pi r^2 (3+2\sqrt 2)} \\
& = \frac 1{3+2\sqrt 2} \\
& = \frac 1{3+2\sqrt 2} \cdot \frac{3-2\sqrt 2}{3 - 2\sqrt 2} \\
& = \frac{3 - 2\sqrt 2}{3^2 - (2\sqrt 2)^2} \\
& = \frac{3 - 2\sqrt 2}1 \\
& = \boxed{3 - 2\sqrt 2}


But the answer is C.
\end{align*} (Error compiling LaTeX. Unknown error_msg)

2009 AMC 10A (Problems • Answer Key • Resources)
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2009 AMC 10A Problems/Problem 21

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