Difference between revisions of "2003 AMC 12A Problems/Problem 11"
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This distance is <math>\frac{2}{3}</math> of an altitude. By <math>30-60-90</math> right triangle properties, the altitude is <math>\frac{\sqrt{3}}{2} \cdot s</math> where s is the side. | This distance is <math>\frac{2}{3}</math> of an altitude. By <math>30-60-90</math> right triangle properties, the altitude is <math>\frac{\sqrt{3}}{2} \cdot s</math> where s is the side. | ||
So, the radius is <math>\frac{2}{3} \cdot \frac{\sqrt{3}}{2} \cdot \frac{P}{3} = \frac{P\sqrt{3}}{9}</math> | So, the radius is <math>\frac{2}{3} \cdot \frac{\sqrt{3}}{2} \cdot \frac{P}{3} = \frac{P\sqrt{3}}{9}</math> | ||
− | The area of the circle is <math>\pi \cdot \left(\frac{P\sqrt{3}}{9}\right | + | The area of the circle is <math>\pi \cdot \left(\frac{P\sqrt{3}}{9}\right)^2=\pi \cdot \frac{3P^2}{81}=\boxed{\frac{P^2\pi}{27}=B}</math> |
So, <math>\frac{A}{B}=\frac{\frac{P^2 \pi}{32}}{\frac{P^2 \pi}{27}}=\frac{P^2 \pi}{32} \cdot \frac{27}{P^2\pi}=\boxed{\frac{27}{32} \implies \mathrm{(C) \ } \frac{27}{32}}</math> | So, <math>\frac{A}{B}=\frac{\frac{P^2 \pi}{32}}{\frac{P^2 \pi}{27}}=\frac{P^2 \pi}{32} \cdot \frac{27}{P^2\pi}=\boxed{\frac{27}{32} \implies \mathrm{(C) \ } \frac{27}{32}}</math> |
Revision as of 21:02, 14 August 2009
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 triangle, respectively, are
and
The circle circumscriber about the square has a diameter equal to the diagonal of the square, which is
Therefore, the radius is
and the area os the circle is
Now consider the circle circumscriber around the equilateral triangle. Due to symmetry, the circle must share a center with the equilateral triangle. The radius of the circle is simply the distance from the center of the triangle to a vertex.
This distance is of an altitude. By
right triangle properties, the altitude is
where s is the side.
So, the radius is
The area of the circle is
So,