2003 AMC 12A Problems/Problem 11
Problem 11
A square and an equilateral triangle have the same perimeter. Let 
be the area of the circle circumscribed about the square and 
the area of the circle circumscribed around the triangle. Find 
.
Solution
Suppose that the common perimeter is 
. Then the side lengths of the square and the triangle are 
and 
, respectively.
The circle circumscribed about the square has a diameter equal to the diagonal of the square, which is therefore 
, and so its radius is 
. Hence the area of the circle is ![\[A = \pi\left(\frac{P\sqrt{2}}{8}\right)^2 = \pi \cdot \frac{2P^2}{64} = \frac{\pi P^2}{32}.\]](http://latex.artofproblemsolving.com/8/3/4/834aad0ddc2e59e9237c5d28b90067f67e216489.png)
Now consider the circle circumscribed around the equilateral triangle. By symmetry, its center must be the same as that of the triangle, so its radius is simply the distance from the center of the triangle to a vertex. Recalling that the centroid of any triangle divides its medians in the ratio 
, and that the medians of an equilateral triangle are the same as its altitudes, we deduce that the radius is 
of the total length of an altitude. Since the side length of this triangle is , the length of an altitude is 
, so finally the radius is ![\[\frac{2}{3} \cdot \frac{P\sqrt{3}}{6} = \frac{P\sqrt{3}}{9},\]](http://latex.artofproblemsolving.com/c/c/0/cc0b0e7da5184457c4f94d029a3f76e9fb99c801.png)
and thus the area of this circle is ![\[B = \pi\left(\frac{P\sqrt{3}}{9}\right)^2 = \pi \cdot \frac{3P^2}{81} = \frac{\pi P^2}{27}.\]](http://latex.artofproblemsolving.com/8/f/9/8f9269089d2a28eef4a19c93757fa5ecbcf027e6.png)
This gives ![\[\frac{A}{B}=\frac{\left(\frac{\pi P^2}{32}\right)}{\left(\frac{\pi P^2}{27}\right)} = \frac{\pi P^2}{32} \cdot \frac{27}{\pi P^2}= \boxed{\mathrm{(C) \ } \frac{27}{32}}.\]](http://latex.artofproblemsolving.com/8/4/a/84a3328d03f7189a4ebf9899cdae32976ebb75a0.png)
See Also
| 2003 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 10 |
Followed by Problem 12 |
| 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 12 Problems and Solutions | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
