Area in a triangle divided by a segment that goes through the centroid.

Question

In the image, $G$ is the centroid of $\triangle ABC$. If $BN=4NC$, and the area of $ACNM$ is $19$, find the area of $\triangle MNB$.

My try: I draw a segment parallel to $AC$ from $M$ to $BC$. I did the same from $N$ to $AC$. Then I tried to apply some similarity of triangles, drawing the heights of the similar triangles and trapezoids that are created by those segments, but I couldn't proceed further. Any help to do this problem?

P.S: I'm searching a solution that doesn't involve trigonometry, but all solutions are appreciated.

Answer 1

Let [.] denote areas and $\frac{AM}{AB} = x$. Given that $G$ is the centroid, we have $\frac{AD}{DC}=1$ and $\frac{BG}{GD} = 2$.

Then, evaluate the areas below in terms of $I=[ABC]$,

$$[CND] = \frac15 [BDC]=\frac15\cdot \frac12I$$ $$[MBN] = (1-x)[ABN] = (1-x)\cdot \frac45 I$$ $$[MDN] =\frac12 [MBN]= (1-x)\cdot \frac25 I$$ $$[AMD] = x [ABD] =\frac x2 I$$

The sum of above four areas is equal to $I$, which leads to

$$\frac 12 x + \frac65 (1-x)+ \frac1{10}= 1$$

Solve to obtain $x = \frac37$. Then, $[MBN] = (1-x)\frac45 I=\frac{16}{35}I$ and

$$\frac{[MBN]}{[ACNM]}= \frac{[MBN]}{I-[MBN]} = \frac{\frac{16}{35}I}{I-\frac{16}{35}I}=\frac{16}{19}$$

Thus, the area of $MBN$ is

$$[MBN] = \frac{16}{19}[ACNM]= 16$$

Answer 2

Let $\vec{BA}=\vec{a},$ $\vec{BC}=\vec{c},$ $\vec{BM}=x\vec{a}$ and $\vec{MG}=y\vec{MN}.$

Thus, $$-\vec{BM}+\vec{BG}=\vec{MG}$$ or $$-x\vec{a}+\frac{1}{3}\vec{a}+\frac{1}{3}\vec{c}=y\left(-x\vec{a}+\frac{4}{5}\vec{c}\right),$$ which gives the following system: $$-x+\frac{1}{3}=-xy$$ and $$\frac{1}{3}=\frac{4}{5}y,$$ which gives $$y=\frac{5}{12},$$ $$x=\frac{4}{7}$$ and $$\frac{S_{\Delta BMN}}{S_{\Delta ABC}}=\frac{4}{7}\cdot\frac{4}{5}=\frac{16}{35}.$$ Thus, $$\frac{S_{ACMN}}{S_{\Delta ABC}}=\frac{19}{35},$$ $$S_{\Delta ABC}=35$$ and $$S_{\Delta BMN}=16.$$ Another way:

Let $BK$ be a median of $\Delta ABC$, $L\in AM$ such that $LK||MN,$ $LK\cap BC=\{P\}$, $Q\in BC$ such that $AQ||MN$ and $BN=4x$.

Thus, $NC=x$ and by the Thales's theorem $$\frac{NP}{4x}=\frac{NP}{BN}=\frac{GK}{BQ}=\frac{1}{2},$$ which gives $$NP=2x$$ and $$PC=x.$$ Now, since $KP||AQ$ and $AK=KC$, we obtain: $$PQ=PC=x,$$ which gives $$BM:ML:LA=BN:NP:PQ=4:2:1$$ and $$BM=\frac{4}{7}BA.$$ The rest is the same.