A point is taken on each median of the triangle, dividing the median into the ratio 3:1. How many times the area of triangle... 
A point is taken on each median of the triangle, dividing the median into the ratio 3:1, counting from the top. How many times the area of triangle with vertices  at these three points are less than the area of the original triangle.

My attempt at answering : I understand from the figure I drew that there will be a smaller triangle inside of the bigger one. But, I don't understand how to find the  relationship between the area of the bigger triangle and the smaller one. Answer given is "64 times smaller". Some pointers to find the answer would be much appreciated. Thanks.
 A: Perform a linear transformation of the original triangle onto the triangle with vertices $(0,0),$ $(8,0),$ and $(0,8)$ in a Cartesian plane.
The transformation does not generally preserve area, of course, but it does preserve the ratios of areas.
The smaller triangle then has vertices at
$(3,3),$ $(2,3),$ and $(3,2).$
The ratio of areas is
$$ \frac{\frac12(8^2)}{\frac12{1^2}} = 64.$$
A: 
\begin{align}
A_m&=\tfrac12\,(B+C),\quad
B_m=\tfrac12\,(A+C),\quad
C_m=\tfrac12\,(B+A)
,\\
D&=\tfrac14\,(A+3\,A_m),\quad
E =\tfrac14\,(B+3\,B_m),\quad
F =\tfrac14\,(C+3\,C_m)
,\\
D-E&=\tfrac18\,(B-A),\quad
E-F =\tfrac18\,(C-B),\quad
F-D =\tfrac18\,(A-C)
,\\
\frac{S_{ABC}}{S_{DEF}}
&=
\frac{1^2}{(\tfrac18)^2}=64
.
\end{align} 
A: The small triangle is obtained from the large triangle through a linear stretching from the centroid $M$ by the factor $-{1\over8}$, so that we have an area ratio of $64$.
For a proof consider, e.g., the median $AA_m$ of length $|AA_m|=:\ell$. We then have $|AM|={2\over3}\ell$ and $|DM|=\bigl({3\over4}-{2\over3}\bigr)\ell$, so that
$${|DM|\over|AM|}={1\over8}\ .$$
