# 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.

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

• $\frac{MNP}{ABC}=\frac{Area_{\triangle MNP}}{ Area_{\triangle ABC}}$ – Word Shallow Dec 17 '18 at 13:59
• $\triangle AMP$ 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. – Quang Hoang Dec 17 '18 at 15:42
• Got it. I edited something good in my solution, see now. – Word Shallow Dec 17 '18 at 15:55