2020 AMC 10A Problem 12

Call $P$ the point where the two medians cross. Perpendicular medians make a right angle at $P$, so aim for the right triangle $MPC$.

1. Lengths to $P$. The point where medians cross is two thirds of the way from each corner to the opposite midpoint. So
$$MP = CP = \tfrac23 \times 12 = 8.$$
2. Area of $\triangle MPC$. It has a right angle at $P$ and legs $8$ and $8$:
$$\text{area} = \tfrac12 \times 8 \times 8 = 32.$$
3. How big is $\triangle MPC$ compared with the whole? Triangles $MPC$ and $AMC$ share the base $MC$. Their heights above $MC$ compare like the distances from $P$ and from $A$ to that side. The third median, from $A$ to the midpoint of $MC$, also passes through $P$, and $P$ is two thirds of the way down it. So $P$ is one third as far from $MC$ as $A$ is, and $\triangle MPC$ has one third the area of $\triangle AMC$.
4. So the area of $\triangle AMC$ is $3 \times 32 = 96$.

The answer is $\boxed{\textbf{(C)}\ 96}$.

Two facts about medians do all the work. The crossing point splits each median into a $2$ to $1$ ratio, which turns the median lengths into the legs of a right triangle. And that crossing point sits one third of the way up from each side, which lets you scale the small triangle up to the whole triangle.

Look at the four-sided shape $MUVC$. Its diagonals $MV$ and $CU$ are perpendicular, and for such a shape the area is half the product of the diagonals: $\frac12 \times 12 \times 12 = 72$.

$U$ and $V$ are midpoints of $AM$ and $AC$, so triangle $AUV$ is a half-size copy of $AMC$, with one quarter of its area. That means $MUVC$ is the other three quarters. So the whole area is $72 \div \frac34 = 96$.