2003 AMC 10A Problem 17

Below is the professionally curated solution for Problem 17 of the 2003 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 10A solutions, or check the answer key.

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Concepts:equilateral triangle circumcircle, circumcenter, and circumradius special right triangle

Difficulty rating: 1600

17.

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?

$\dfrac{3\sqrt{2}}{\pi}$

$\dfrac{3\sqrt{3}}{\pi}$

$\sqrt{3}$

$\dfrac{6}{\pi}$

$\sqrt{3}\pi$

Solution:

Let the side length be $s$ and the circumradius be $R.$ From a $30$ - $60$ - $90$ triangle formed by the center and a side, $R = \dfrac{s}{\sqrt{3}},$ so $s = R\sqrt{3}.$

The perimeter is $3s = 3R\sqrt{3}$ and the circle's area is $\pi R^2.$

Setting them equal, $3R\sqrt{3} = \pi R^2,$ so $R = \dfrac{3\sqrt{3}}{\pi}.$

Thus, the correct answer is B.

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