15. Find the ratio of the area of $\triangle A B C$ to the area of the triangle formed by its medians.

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15. Let point $O$ be the intersection of the medians of $\triangle ABC$ (Figure 17). On the extension of median $\mathrm{BE}$, construct $\mathrm{ED}=0 \mathrm{E}$. From the properties of medians, the sides of $\triangle \mathrm{CDO}$ are $\frac{2}{3}$ of the sides of the triangle formed by the medians. If $S_{1}$ represents the area of the triangle formed by the medians, then
$$
S_{i}=\frac{9}{4} S_{C D O .}
$$

On the other hand, $\triangle C D O$ consists of two triangles equal in area to $\triangle C E O$, and $\triangle \mathrm{ABC}$ is composed of six such triangles, so $S_{C D O}=\frac{1}{3} S_{\triangle B}$. Therefore,
$$
\frac{S_{1}}{S_{\triangle B C}}=\frac{3}{4}
$$