2003 AMC 10A Problems/Problem 17

Problem

The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle?

$\mathrm{(A) \ } \frac{3\sqrt{2}}{\pi}\qquad \mathrm{(B) \ }  \frac{3\sqrt{3}}{\pi}\qquad \mathrm{(C) \ } \sqrt{3}\qquad \mathrm{(D) \ } \frac{6}{\pi}\qquad \mathrm{(E) \ } \sqrt{3}\pi$

Solution 1

Let $s$ be the length of a side of the equilateral triangle and let $r$ be the radius of the circle.

In a circle with a radius $r$, the side of an inscribed equilateral triangle is $r\sqrt{3}$.

So $s=r\sqrt{3}$.

The perimeter of the triangle is $3s=3r\sqrt{3}$

The area of the circle is $\pi r^{2}$

So: $\pi r^{2} = 3r\sqrt{3}$

$\pi r=3\sqrt{3}$

$r=\frac{3\sqrt{3}}{\pi} \Rightarrow\boxed{\mathrm{(B)}\ \frac{3\sqrt{3}}{\pi}}$

Solution 2

As in Solution $1$, let $s$ be the side length of the equilateral triangle and let $r$ be the radius of the circumcircle. Remembering that the formula for radius of a circumcircle is $\frac{abc}{4A}$, where $a$, $b$, and $c$ are side lengths of the triangle, and $A$ is its area, we define each of the quantities in the formula. Since $a = b = c = s$ and, by the formula for the area of an equilateral triangle, $A = \frac{s^2\sqrt{3}}{4}$, the circumradius is $\frac{s^3}{4\frac{s^2\sqrt{3}}{4}} = \frac{s}{\sqrt{3}}$. The area of the circumcircle is then $\pi(\frac{s}{\sqrt{3}})^2 = \frac{\pi s^2}{3}$. Because this is equal to the perimeter of the triangle, we equate this to $3s$. Upon solving, we note that the nonzero value of $s$ is then $\frac{9}{\pi}$. We can now plug this back into the circumradius formula we found earlier to get that $r = \frac{\frac{9}{\pi}}{\sqrt{3}} = \boxed{\textbf{(B)} \frac{3 \sqrt{3}}{\pi}}$.

~cxsmi

Video Solution

https://youtu.be/El9eQJmDO78

~savannahsolver

See Also

2003 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png