2020 AMC 10A Problem 12

Let $P$ be the intersection of the medians, the centroid. The centroid divides each median in a $2 : 1$ ratio measured from the vertex, so
$$MP = CP = \tfrac{2}{3} \cdot 12 = 8.$$
Since the medians are perpendicular, $\triangle MPC$ is a right triangle with legs $MP$ and $CP$:
$$[MPC] = \tfrac12 \cdot 8 \cdot 8 = 32.$$

The three medians split $\triangle AMC$ into six triangles of equal area, and $\triangle MPC$ is made of two of them (the third median through $A$ passes through $P$ and the midpoint of $MC$). So $[MPC] = \frac{1}{3}[AMC]$, giving
$$[AMC] = 3 \cdot 32 = 96.$$

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

Two facts about medians do all the work: the $2:1$ centroid ratio converts median lengths into the legs of a right triangle, and the "six equal areas" fact scales that small triangle up to the whole. The isosceles condition is only there to make the medians symmetric; the argument never needs it.

Quadrilateral $MUVC$ has perpendicular diagonals $MV$ and $CU$, so its area is $\frac12 \cdot 12 \cdot 12 = 72$. Since $UV$ is a midsegment, $\triangle AUV$ is one quarter of $\triangle AMC$, so $MUVC$ is three quarters of it: $[AMC] = \frac{4}{3} \cdot 72 = 96$.

• Using the full medians as the legs of the right triangle, (1/2)(12)(12) = 72, and stopping at choice (B).
• Taking $\triangle MPC$ to be one sixth or one half of the triangle rather than one third.