<s>[INST] Calculate the ratio of the sides of a given triangle given the ratio of areas. Given a triangle $\triangle ABC$, points $M$, $N$, $P$ are drawn on the sides of the triangle in a way that $\frac{|AM|}{|MB|} = \frac{|BN|}{|NC|}= \frac{|PC|}{|PA|}=k$, where $k>0$.

Calculate $k$, given that the area of the triangle $\triangle MNP$ and the area of the triangle $ \triangle ABC$ are in the following ratio: $Area_{\triangle MNP} = \frac{7}{25} \times Area_{\triangle ABC}$.

I have tried Heron formula, bot the calculations seem to be too complicated, I have also tried to simplify the problem and assume that k is equal to 1 and then calculate the ratio of the triangles area but it also doesn't help.
I was also looking for similar triangles.
I would appreciate some hint.
 [/INST] Draw $BB1$ is perpendicular to $AC$; $MM1//AC$ and perpendicular to $AC$, then we have $\frac{AM}{AB}=\frac{MM1}{BB1}$
So $$\frac{AMP}{ABC}=\frac{1/2 \cdot AP\cdot MM1}{1/2\cdot AC \cdot BB1} =\frac{AP\cdot AM}{AC\cdot AB}$$
$$\Rightarrow \frac{AMP}{ABC} = \frac{CPN}{ABC} = \frac{BMN}{ABC} = \frac{AM}{AB}. \frac{AP}{AC} = \frac{k}{(k+1)^2}$$
Or $$\frac{MNP}{ABC} = 1 - \frac{AMP}{ABC} - \frac{CPN}{ABC} - \frac{BMN}{ABC} = 1 - \frac{3k}{(k+1)^2}=\frac{7}{25}$$
It is not difficult to find $k=\frac{2}{3}$ or $k=\frac{3}{2}$
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