3
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ $\frac{MNP}{ABC}=\frac{Area_{\triangle MNP}}{ Area_{\triangle ABC}}$ $\endgroup$ – 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.