2013 AMC 10B Problem 7

1. Six equally spaced points on a circle are the corners of a regular hexagon. Number them $0, 1, 2, 3, 4, 5$ going around. Lines from the center to the corners cut the hexagon into six equilateral triangles, so neighboring points are exactly $1$ apart (the same as the radius).
2. Pick three points and count how many steps around the circle you move from each chosen point to the next. The three step counts add up to $6$, so the only options are:
- $2, 2, 2$ (points $0, 2, 4$): all sides equal, equilateral.
- $1, 1, 4$ (points $0, 1, 2$): two sides equal, isosceles.
- $1, 2, 3$ (points $0, 1, 3$): all three sides different. This is our triangle.
3. Points $0$ and $3$ are straight across the circle from each other, so side $0$–$3$ is a diameter, of length $2$.
4. A triangle drawn inside a circle with a diameter as one side has a right angle at the third corner. So the angle at point $1$ is $90^\circ$, and the diameter is the hypotenuse.
5. Side $0$–$1$ joins neighbors, so it has length $1$. By the Pythagorean theorem, the third side is $\sqrt{2^2 - 1^2} = \sqrt{3}$.
6. The area of a right triangle is half the product of the two legs:
$$\frac{1}{2} \cdot 1 \cdot \sqrt{3} = \frac{\sqrt{3}}{2}.$$

The answer is $\boxed{\textbf{(B)}\ \frac{\sqrt{3}}{2}}$.

Equally spaced points on a circle form a regular polygon, and for a hexagon the side equals the radius. Any triangle that uses two opposite points contains a diameter, so it has a right angle, and the Pythagorean theorem finishes the job. Counting the steps between chosen points is a quick way to list every possible triangle shape.

Use coordinates with the center at $(0, 0)$. Put point $0$ at $(1, 0)$ and point $3$ at $(-1, 0)$. Point $1$ is $60^\circ$ around the circle, at $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$. Take the diameter as the base, length $2$; the height is how far point $1$ is above it, $\frac{\sqrt{3}}{2}$. The area is $\frac{1}{2} \cdot 2 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$.