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"

(Repaired Vandalism made by Wesleythang)
(Improved explanation, formatting, grammar, and LaTeX)
 
(One intermediate revision by one other user not shown)
Line 6: Line 6:
  
 
== 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 the triangle are <math>\frac{P}{4}</math> and <math>\frac{P}{3}</math>, respectively.
−
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.
+
The circle circumscribed about the square has a diameter equal to the diagonal of the square, which is therefore <math>\frac{P}{4} \cdot \sqrt{2} = \frac{P\sqrt{2}}{4}</math>, and so its radius is <math>\frac{\left(\frac{P\sqrt{2}}{4}\right)}{2} = \frac{P\sqrt{2}}{8}</math>. Hence the area of the circle is <cmath>A = \pi\left(\frac{P\sqrt{2}}{8}\right)^2 = \pi \cdot \frac{2P^2}{64} = \frac{\pi P^2}{32}.</cmath>
−
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>
+
Now consider the circle circumscribed around the equilateral triangle. By symmetry, its center must be the same as that of the triangle, so its radius is simply the distance from the center of the triangle to a vertex. Recalling that the centroid of any triangle divides its medians in the ratio <math>2:1</math>, and that the medians of an equilateral triangle are the same as its altitudes, we deduce that the radius is <math>\frac{2}{3}</math> of the total length of an altitude. Since the side length of this triangle is <math>\frac{P}{3}</math>, the length of an altitude is <math>\frac{P}{3}\sin\left(60^{\circ}\right) = \frac{P}{3} \cdot \frac{\sqrt{3}}{2} = \frac{P\sqrt{3}}{6}</math>, so finally the radius is <cmath>\frac{2}{3} \cdot \frac{P\sqrt{3}}{6} = \frac{P\sqrt{3}}{9},</cmath>
−
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>
+
and thus the area of this circle is <cmath>B = \pi\left(\frac{P\sqrt{3}}{9}\right)^2 = \pi \cdot \frac{3P^2}{81} = \frac{\pi P^2}{27}.</cmath>
−
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>
+
 
 +
This gives <cmath>\frac{A}{B}=\frac{\left(\frac{\pi P^2}{32}\right)}{\left(\frac{\pi P^2}{27}\right)} = \frac{\pi P^2}{32} \cdot \frac{27}{\pi P^2}= \boxed{\mathrm{(C) \ } \frac{27}{32}}.</cmath>
  
 
== 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}}
 
{{MAA Notice}}

Latest revision as of 16:33, 20 June 2025

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 the triangle are $\frac{P}{4}$ and $\frac{P}{3}$, respectively.

The circle circumscribed about the square has a diameter equal to the diagonal of the square, which is therefore $\frac{P}{4} \cdot \sqrt{2} = \frac{P\sqrt{2}}{4}$, and so its radius is $\frac{\left(\frac{P\sqrt{2}}{4}\right)}{2} = \frac{P\sqrt{2}}{8}$. Hence the area of the circle is \[A = \pi\left(\frac{P\sqrt{2}}{8}\right)^2 = \pi \cdot \frac{2P^2}{64} = \frac{\pi P^2}{32}.\]

Now consider the circle circumscribed around the equilateral triangle. By symmetry, its center must be the same as that of the triangle, so its radius is simply the distance from the center of the triangle to a vertex. Recalling that the centroid of any triangle divides its medians in the ratio $2:1$, and that the medians of an equilateral triangle are the same as its altitudes, we deduce that the radius is $\frac{2}{3}$ of the total length of an altitude. Since the side length of this triangle is $\frac{P}{3}$, the length of an altitude is $\frac{P}{3}\sin\left(60^{\circ}\right) = \frac{P}{3} \cdot \frac{\sqrt{3}}{2} = \frac{P\sqrt{3}}{6}$, so finally the radius is \[\frac{2}{3} \cdot \frac{P\sqrt{3}}{6} = \frac{P\sqrt{3}}{9},\] and thus the area of this circle is \[B = \pi\left(\frac{P\sqrt{3}}{9}\right)^2 = \pi \cdot \frac{3P^2}{81} = \frac{\pi P^2}{27}.\]

This gives \[\frac{A}{B}=\frac{\left(\frac{\pi P^2}{32}\right)}{\left(\frac{\pi P^2}{27}\right)} = \frac{\pi P^2}{32} \cdot \frac{27}{\pi P^2}= \boxed{\mathrm{(C) \ } \frac{27}{32}}.\]

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_11&oldid=251698"