Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2003 AMC 12A Problems/Problem 11"

(wrote solution)
 
Line 15: Line 15:
 
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}=\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 20:02, 14 August 2009

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 circumscriber 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 os the circle is $\pi \cdot \left(\frac{P\sqrt{2}}{8}\right)^2 = \pi \cdot \frac{2P^2}{64}=\boxed{\frac{P^2 \pi}{32}=A}$

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 $\frac{2}{3}$ 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 $\pi \cdot \left(\frac{P\sqrt{3}}{9}\right)^2=\pi \cdot \frac{3P^2}{81}=\boxed{\frac{P^2\pi}{27}=B}$ So, $\frac{A}{B}=\frac{\frac{P^2 \pi}{32}}{\frac{P^2 \pi}{27}}=\frac{P^2 \pi}{32} \cdot \frac{27}{P^2\pi}=\boxed{\frac{27}{32} \implies \mathrm{(C) \ } \frac{27}{32}}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_11&oldid=32666"