Difference between revisions of "2003 AMC 12A Problems/Problem 11"
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== Solution == | == Solution == | ||
− | Suppose that the common perimeter is <math>P</math> | + | Suppose that the common perimeter is <math>P</math>. |
Then, the side lengths of the square and triangle, respectively, are <math>\frac{P}{4}</math> and <math>\frac{P}{3}</math> | Then, the side lengths of the square and triangle, respectively, are <math>\frac{P}{4}</math> and <math>\frac{P}{3}</math> | ||
The circle circumscribed about the square has a diameter equal to the diagonal of the square, which is <math>\frac{P\sqrt{2}}{4}</math> | The circle circumscribed about the square has a diameter equal to the diagonal of the square, which is <math>\frac{P\sqrt{2}}{4}</math> | ||
Therefore, the radius is <math>\frac{P\sqrt{2}}{8}</math> and the area of the circle is | Therefore, the radius is <math>\frac{P\sqrt{2}}{8}</math> and the area of the circle is | ||
− | <math>\pi \cdot \left(\frac{P\sqrt{2}}{8}\right)^2 = \pi \cdot \frac{2P^2}{64}= | + | <math>\pi \cdot \left(\frac{P\sqrt{2}}{8}\right)^2 = \pi \cdot \frac{2P^2}{64}=\frac{P^2 \pi}{32}=A</math> |
Now consider the circle circumscribed 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. | Now consider the circle circumscribed 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 <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)^2=\pi \cdot \frac{3P^2}{81}= | + | The area of the circle is <math>\pi \cdot \left(\frac{P\sqrt{3}}{9}\right)^2=\pi \cdot \frac{3P^2}{81}=\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> | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2003|ab=A|num-b=10|num-a=12}} | {{AMC12 box|year=2003|ab=A|num-b=10|num-a=12}} | ||
+ | {{MAA Notice}} |
Latest revision as of 13:31, 22 July 2017
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 circumscribed about the square has a diameter equal to the diagonal of the square, which is Therefore, the radius is and the area of the circle is
Now consider the circle circumscribed 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,
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.