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Difference between revisions of "2003 AMC 12A Problems/Problem 11"
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== Solution ==
 
== Solution ==
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Suppose that the common perimeter is <math>P</math>
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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>
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The circle circumscriber about the square has a diameter equal to the diagonal of the square, which is <math>\frac{P\sqrt{2}}{4}</math>
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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>
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Therefore, the radius is <math>\frac{P\sqrt{2}}{8}</math> and the area os the circle is
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Therefore, the radius is <math>\frac{P\sqrt{2}}{8}</math> and the area of the circle is
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<math>\pi \cdot \left(\frac{P\sqrt{2}}{8}\right)^2 = \pi \cdot \frac{2P^2}{64}=\boxed{\frac{P^2 \pi}{32}=A}</math>
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<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 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.
+
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}=\boxed{\frac{P^2\pi}{27}=B}</math>
+
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 ==
 +
{{AMC12 box|year=2003|ab=A|num-b=10|num-a=12}}
 +
{{MAA Notice}}

Latest revision as of 12:31, 22 July 2017

Problem 11

A square and an equilateral triangle have the same perimeter. Let $A$ be the area of the circle circumscribed about the square and $B$ the area of the circle circumscribed around the triangle. Find $A/B$.

$\mathrm{(A) \ } \frac{9}{16}\qquad \mathrm{(B) \ } \frac{3}{4}\qquad \mathrm{(C) \ } \frac{27}{32}\qquad \mathrm{(D) \ } \frac{3\sqrt{6}}{8}\qquad \mathrm{(E) \ } 1$

Solution

Suppose that the common perimeter is $P$. Then, the side lengths of the square and triangle, respectively, are $\frac{P}{4}$ and $\frac{P}{3}$ The circle circumscribed about the square has a diameter equal to the diagonal of the square, which is $\frac{P\sqrt{2}}{4}$ Therefore, the radius is $\frac{P\sqrt{2}}{8}$ and the area of the circle is $\pi \cdot \left(\frac{P\sqrt{2}}{8}\right)^2 = \pi \cdot \frac{2P^2}{64}=\frac{P^2 \pi}{32}=A$

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 $\frac{2}{3}$ of an altitude. By $30-60-90$ right triangle properties, the altitude is $\frac{\sqrt{3}}{2} \cdot s$ where s is the side. So, the radius is $\frac{2}{3} \cdot \frac{\sqrt{3}}{2} \cdot \frac{P}{3} = \frac{P\sqrt{3}}{9}$ The area of the circle is $\pi \cdot \left(\frac{P\sqrt{3}}{9}\right)^2=\pi \cdot \frac{3P^2}{81}=\frac{P^2\pi}{27}=B$ So, $\frac{A}{B}=\frac{\frac{P^2 \pi}{32}}{\frac{P^2 \pi}{27}}=\frac{P^2 \pi}{32} \cdot \frac{27}{P^2\pi}=\boxed{\frac{27}{32} \implies \mathrm{(C) \ } \frac{27}{32}}$

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

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