Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2009 AMC 10A Problems/Problem 21"

(fixed a bug in the solution)
(→‎See Also)
Line 65: Line 65:
 
== See Also ==
 
== See Also ==
  
−
{{AMC10 box|year=2009|ab=A|num-b=|num-a=}}
+
{{AMC10 box|year=2009|ab=A|num-b=20|num-a=22}}

Revision as of 11:42, 2 July 2009

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)) ); draw( (0,sqrt(2)) -- (0,-sqrt(2)), dashed ); [/asy]

Out of symmetry, the quadrilateral in the center must be a square. Its side is obviously $2r$, and therefore its diagonal is $2r\sqrt{2}$. We can now compute the length of the vertical diameter of the large circle as $2r + 2r\sqrt{2}$. Hence $2R=2r + 2r\sqrt{2}$, and thus $R=r+r\sqrt{2}=r(1+\sqrt{2})$.

Then the area of the large circle is $L = \pi R^2 = \pi r^2 (1+\sqrt 2)^2 = \pi r^2 (3+2\sqrt 2)$. 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}{\pi r^2 (3+2\sqrt 2)} \\ & = \frac 4{3+2\sqrt 2} \\ & = \frac 4{3+2\sqrt 2} \cdot \frac{3-2\sqrt 2}{3 - 2\sqrt 2} \\ & = \frac{4(3 - 2\sqrt 2)}{3^2 - (2\sqrt 2)^2} \\ & = \frac{4(3 - 2\sqrt 2)}1 \\ & = \boxed{4(3 - 2\sqrt 2)} \end{align*}

See Also

2009 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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
All AMC 10 Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_21&oldid=32228"