2013 AMC 10B Problem 7

Label the points $P_0, \dots, P_5$ in order; consecutive points are $60^\circ$ apart. Up to rotation and reflection, three points can be spaced around the circle in only three ways: gaps of $2, 2, 2$ (equilateral), gaps of $1, 1, 4$ (isosceles), or gaps of $1, 2, 3$. So the scalene triangle has vertices like $P_0, P_1, P_3$.

Since $P_0$ and $P_3$ are diametrically opposite, $P_0P_3$ is a diameter of length $2$, and the angle at $P_1$ is a right angle (angle in a semicircle). The chord $P_0P_1$ spans a $60^\circ$ arc, so it equals the radius, $1$. The other leg is then $\sqrt{2^2 - 1^2} = \sqrt{3}$.

The area is
$$\frac{1}{2} \cdot 1 \cdot \sqrt{3} = \frac{\sqrt{3}}{2}.$$

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

Points equally spaced on a circle are a regular hexagon, whose side equals the radius. Any triangle using two opposite vertices contains a diameter and is therefore right-angled, so the side lengths fall out of the Pythagorean theorem. Classifying triangles by the gaps between chosen points is a quick way to see which shape is forced.

Coordinates: $P_0 = (1, 0)$, $P_1 = \left(\tfrac12, \tfrac{\sqrt3}{2}\right)$, $P_3 = (-1, 0)$. The base $P_0P_3$ has length $2$ along the $x$-axis and $P_1$ sits at height $\tfrac{\sqrt3}{2}$, so the area is $\tfrac12 \cdot 2 \cdot \tfrac{\sqrt3}{2} = \tfrac{\sqrt3}{2}$.

• Picking three points that form an isosceles triangle (like two apart and two apart), or forgetting the hypotenuse is the diameter 2, not the radius.
• Computing the area of the equilateral triangle $P_0P_2P_4$, which is $\frac{3\sqrt3}{4}$, or of the small triangle $P_0P_1P_2$, which is $\frac{\sqrt3}{4}$.