2013 AMC 10B Problems/Problem 7

Problem

Six points are equally spaced around a circle of radius 1. Three of these points are the vertices of a triangle that is neither equilateral nor isosceles. What is the area of this triangle?

$\textbf{(A)}\ \frac{\sqrt{3}}{3}\qquad\textbf{(B)}\ \frac{\sqrt{3}}{2}\qquad\textbf{(C)}\ \textbf{1}\qquad\textbf{(D)}\ \sqrt{2}\qquad\textbf{(E)}\ \text{2}$

Solution

If there are no two points on the circle that are adjacent, then the triangle would be equilateral. If the three points are all adjacent, it would be isosceles. Thus, the only possibility is two adjacent points and one point two away. Because one of the sides of this triangle is the diameter, the opposite angle is a right angle. Also, because the two adjacent angles are one sixth of the circle apart, the angle opposite them is thirty degrees. This is a $30-60-90$ triangle. If the original six points are connected, a regular hexagon is created. This hexagon consists of six equilateral triangles, so the radius is equal to one of its side lengths. The radius is $1$ , so the side opposite the thirty degree angle in the triangle is also $1$ . From rules with $30-60-90$ triangles, the area is $1\cdot\sqrt3/2$ = $\boxed{\textbf{(B) } \frac{\sqrt3}{2}}$

2013 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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2013 AMC 10B Problems/Problem 7

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