2018 AMC 10B Problem 12

The centroid is where the three medians of a triangle meet. It lies on each median, two thirds of the way from the corner to the midpoint of the opposite side.

1. Pick the right median. Let $O$ be the center of the circle. Since $AB$ is a diameter, $O$ is the midpoint of $AB$. So the segment from $C$ to $O$ is the median from $C$.
2. Place the centroid. The centroid $G$ is two thirds of the way from $C$ to $O$. That is the same as one third of the way from $O$ to $C$.
3. Measure. $OC$ is a radius, $12$. So
$$OG = \tfrac13 \times 12 = 4.$$
4. The curve. $O$ never moves, and $G$ always stays $4$ away from it. So as $C$ goes around the circle, $G$ traces a circle of radius $4$ around $O$. (The two missing points come from $C = A$ and $C = B$, which are not allowed. They do not change the area.)
5. Area.
$$\pi \times 4^2 = 16\pi \approx 16 \times 3.14 = 50.2\ldots,$$
which rounds to $50$.

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

When two corners stay fixed and one corner moves, the centroid copies the moving corner's path at one third the size, shrunk toward the midpoint of the fixed side. Shrinking lengths to one third shrinks area to one ninth: $144\pi \div 9 = 16\pi$.

Use coordinates. Put $O$ at $(0,0)$, $A = (-12, 0)$, $B = (12, 0)$, and $C = (a, b)$ with $a^2 + b^2 = 144$. The centroid is the average of the three corners:
$$G = \left(\frac{-12 + 12 + a}{3}, \frac{0 + 0 + b}{3}\right) = \left(\frac{a}{3}, \frac{b}{3}\right).$$
Its distance from $O$ is $\frac13\sqrt{a^2 + b^2} = \frac{12}{3} = 4$, the same circle of radius $4$.