[Ratio of areas of similar triangles]

Find the area of a triangle whose medians are 12, 15, and 21.

#
Let the area of triangle $ABC$ be $S$. We will prove that the area of the triangle whose sides are equal to the medians of triangle $ABC$ is $\frac{3}{4}$. Let $M$ be the point of intersection of the medians of triangle $ABC$, and $B_1$ be the midpoint of side $AC$. Extend the median $BB_1$ beyond point $B_1$ and mark a segment $B_1K$ equal to $B_1M$. Since $AMCK$ is a parallelogram, then $KC = AM$. Therefore, the sides of triangle $MCK$ are $\frac{2}{3}$ of the sides of the triangle formed by the medians of triangle $ABC$. Consequently, the desired triangle is similar to triangle $MCK$ with a similarity ratio of $\frac{3}{2}$, and its area is $\frac{9}{4}$ of the area of triangle $MCK$, i.e.,

$$
S_1 = \frac{9}{4} \cdot 2 \cdot \frac{1}{6} S = \frac{3}{4} S
$$

The area $S_1$ of the triangle with sides 12, 15, and 21 can be found using Heron's formula:

$$
S_1 = \sqrt{24 \cdot 12 \cdot 9 \cdot 3} = 36 \sqrt{6}
$$

Therefore,

$$
S_{\triangle ABC} = S = \frac{4}{3} S_1 = \frac{4}{3} \cdot 36 \sqrt{6} = 48 \sqrt{6}.
$$

## Answer

$48 \sqrt{6}$.