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 A square and an equilateral triangle have the same perimeter. Let <math>A</math> be the area of the circle circumscribed about the square and <math>B</math> the area of the circle circumscribed around the triangle. Find <math>A/B</math>.   A square and an equilateral triangle have the same perimeter. Let <math>A</math> be the area of the circle circumscribed about the square and <math>B</math> the area of the circle circumscribed around the triangle. Find <math>A/B</math>. 
   
−  <math> \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 </math>
 +  $ \mathrm{(A) \ 
−   
−  == Solution ==
 
−  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>
 
−  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
 
−  <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>
 
−   
−  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>306090</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>
 
−  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>
 
−   
−  == See Also ==
 
−  {{AMC12 boxyear=2003ab=Anumb=10numa=12}}
 
−  {{MAA Notice}}
 
Revision as of 21:12, 21 May 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 .
$ \mathrm{(A) \