Difference between revisions of "2003 AMC 10A Problems/Problem 17"
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<math> \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 </math> | <math> \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 </math> | ||
− | == Solution == | + | == Solution 1 == |
Let <math>s</math> be the length of a side of the equilateral triangle and let <math>r</math> be the radius of the circle. | Let <math>s</math> be the length of a side of the equilateral triangle and let <math>r</math> be the radius of the circle. | ||
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<math>r=\frac{3\sqrt{3}}{\pi} \Rightarrow\boxed{\mathrm{(B)}\ \frac{3\sqrt{3}}{\pi}}</math> | <math>r=\frac{3\sqrt{3}}{\pi} \Rightarrow\boxed{\mathrm{(B)}\ \frac{3\sqrt{3}}{\pi}}</math> | ||
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+ | == Solution 2 == | ||
+ | As in Solution <math>1</math>, let <math>s</math> be the side length of the equilateral triangle and let <math>r</math> be the radius of the circumcircle. Remembering that the formula for radius of a circumcircle is <math>\frac{abc}{4A}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are side lengths of the triangle, and <math>A</math> is its area, we define each of the quantities in the formula. Since <math>a = b = c = s</math> and, by the formula for the area of an equilateral triangle, <math>A = \frac{s^2\sqrt{3}}{4}</math>, the circumradius is <math>\frac{s^3}{4\frac{s^2\sqrt{3}}{4}} = \frac{s}{\sqrt{3}}</math>. The area of the circumcircle is then <math>\pi(\frac{s}{\sqrt{3}})^2 = \frac{\pi s^2}{3}</math>. Because this is equal to the perimeter of the triangle, we equate this to <math>3s</math>. Upon solving, we note that the nonzero value of <math>s</math> is then <math>\frac{9}{\pi}</math>. We can now plug this back into the circumradius formula we found earlier to get that <math>r = \frac{\frac{9}{\pi}}{\sqrt{3}} = \boxed{\textbf{(B)} \frac{3 \sqrt{3}}{\pi}}</math>. | ||
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+ | ~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 12:48, 4 April 2024
Contents
[hide]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?
Solution 1
Let be the length of a side of the equilateral triangle and let be the radius of the circle.
In a circle with a radius , the side of an inscribed equilateral triangle is .
So .
The perimeter of the triangle is
The area of the circle is
So:
Solution 2
As in Solution , let be the side length of the equilateral triangle and let be the radius of the circumcircle. Remembering that the formula for radius of a circumcircle is , where , , and are side lengths of the triangle, and is its area, we define each of the quantities in the formula. Since and, by the formula for the area of an equilateral triangle, , the circumradius is . The area of the circumcircle is then . Because this is equal to the perimeter of the triangle, we equate this to . Upon solving, we note that the nonzero value of is then . We can now plug this back into the circumradius formula we found earlier to get that .
~ cxsmi
Video Solution
~savannahsolver
See Also
2003 AMC 10A (Problems • Answer Key • Resources) | ||
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.