Find the area of a quadrilateral $MBNP$ if the area of $\Delta ABC$ is $30\mathrm{cm}^2$

Question

$\Delta ABC$ is given with points $M\in\overline{AB}$ s. t. $\overrightarrow{AM}=2\overrightarrow{MB}$ and $N\in\overline{BC}$ s. t. $\overrightarrow{BN}=\overrightarrow{NC}$. $P$ is the intersection point of $\overline{AN}$ and $\overline{CM}$. Find the area of a quadrilateral $MBNP$ if the area of $\Delta ABC$ is $30\mathrm{cm}^2$.

My attempt:

Let $A_1\in\overline{BC}$ s. t. $\overline{AA_1}\perp\overline{BC}$ and let $C_1\in\overline{AB}$ s. t. $\overline{CC_1 }\perp\overline{AB}.$

$\overline{CC_1}$ is the common altitude of $\Delta AMC$ and $\Delta MBC$, so $$\overrightarrow{AM}=2\overrightarrow{MB}\implies\operatorname{Area}(\Delta AMC)=20\mathrm{cm}^2\;\&\;\operatorname{Area}(\Delta MBC)=10\mathrm{cm}^2$$ Similarly,

$\overline{AA_1}$ is the common altitude of $\Delta ABN$ and $\Delta ANC$, so $$\overrightarrow{BN}=\overrightarrow{BN}\implies\operatorname{Area}(\Delta ABN)=\operatorname{Area}(\Delta ANC)=15\mathrm{cm}^2$$ Then, $$\operatorname{Area}(\Delta AMP)+\operatorname{Area}(MBPN)=\operatorname{Area}(\Delta APC)+\operatorname{Area}(\Delta PNC)=15\mathrm{cm}^2$$ and $$\operatorname{Area}(\Delta AMP)+\operatorname{Area}(\Delta APC)=20\mathrm{cm}^2\;\&\;\operatorname{Area}(MBNP)+\operatorname{Area}(\Delta PNC)=10\mathrm{cm}^2$$ $$\implies\operatorname{Area}(\Delta APC)-\operatorname{Area}(MBPN)=5\mathrm{cm}^2\\\operatorname{Area}(\Delta AMP)-\operatorname{Area}(\Delta PNC)=5\mathrm{cm}^2$$

But this got circular.

Picture:

May I ask how to proceed? Thank you in advance!

Answer 1

The idea is based on the theorem "For triangles with the same altitude, the ratio of their areas is equal to the ratio of the corresponding bases."

According to the given ratios, the areas of different regions are as shown.

From the theorem, we can setup the following equations.

(1) z = 2y + y = 3y

(2) z + 2y = 2(x + x + y)

(3) 2y + y + x = 15

Eliminating z to get a relation between x and y. Then, find x + y from (3).

Answer 2

There are infinity of triangles of area $30\space cm^2$ in each of which the points $M, N, P$ can be constructed following the guidelines of the problem and then giving respective quadrilaterals of distinct shape. The requested area would be undetermined if these areas were unequal, which is not the case.

So we solve the problem by giving us a particular triangle comfortable to handle.

We put $A=(0,0),B=(9,0),C=(6,\frac{20}{3})$ and a simple calculation gives $N=(7.5,\frac{10}{3}),P=(6,\frac83),M=(6,0)$ which determines the corresponding quadrilateral whose area is equal to $7$.

Thus since this particular case is valid for all the other possible ones, the requested area is equal to $7cm^2$.