2018 AMC 10B Problem 12

Let $O$ be the center of the circle; it is the midpoint of $AB$, so $CO$ is the median from $C$. The centroid $G$ lies on this median, two thirds of the way from $C$ to $O$, which puts it one third of the way from $O$ to $C$:
$$OG = \frac{1}{3} OC = \frac{12}{3} = 4.$$

As $C$ runs around the circle, $G$ therefore stays at distance $4$ from the fixed point $O$ and sweeps out a circle of radius $4$ (missing the two points corresponding to $C = A$ and $C = B$, which are excluded).

The enclosed area is $\pi \cdot 4^2 = 16\pi \approx 50.27$, which rounds to $50$.

The answer is $\boxed{\textbf{(C)}\ 50}$.

With $A$ and $B$ fixed, the centroid is the image of $C$ under a dilation centered at the midpoint of $AB$ with ratio $\frac{1}{3}$ (since $G = \frac{A + B + C}{3}$ and $A + B = 2O$). A dilation sends a circle to a circle and scales area by the ratio squared, so the locus is a circle of one ninth the area.

Coordinates: $O = (0,0)$, $A = (-12, 0)$, $B = (12, 0)$, $C = (12\cos\theta, 12\sin\theta)$. Then $G = \left(\frac{12\cos\theta}{3}, \frac{12\sin\theta}{3}\right) = (4\cos\theta, 4\sin\theta)$, a circle of radius $4$.

• Using the full radius 12 (area 144 pi) or the half-way point (radius 6, area 36 pi, choice B) instead of the 1/3 ratio for the centroid.
• Measuring the centroid's distance from $C$ (which is $8$) rather than from the fixed point $O$.