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.


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 $k=\frac{3}{2}$

  • $\begingroup$ $\frac{MNP}{ABC}=\frac{Area_{\triangle MNP}}{ Area_{\triangle ABC}}$ $\endgroup$ – Word Shallow Dec 17 '18 at 13:59
  • $\begingroup$ $\triangle AMP$ and $\triangle ABC$ are not, in general, similar, because as you pointed out $AM:AB = k:(k+1)$ while $AP:AC = 1:(k+1)$. However, your others arguments still apply. $\endgroup$ – Quang Hoang Dec 17 '18 at 15:42
  • $\begingroup$ Got it. I edited something good in my solution, see now. $\endgroup$ – Word Shallow Dec 17 '18 at 15:55

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