2020 AMC 10A Problem 12

Let the centroid be the origin. Since a centroid divides each median in a $2:1$ ratio, we may place median $MV$ horizontally with $M=(8,0)$ and $V=(-4,0)$, and median $CU$ vertically with $C=(0,8)$ and $U=(0,-4)$. Because $U$ is the midpoint of $AM$, we get $A=2U-M=(-8,-8)$. The area of $\triangle AMC$ is $\dfrac12 |(16,8)\times(8,16)|$ $=\dfrac12(256-64)=96$. Thus, C is the correct answer.