Task B-1.4. Points $M$ and $N$ belong to sides $\overline{A C}$ and $\overline{B C}$ of triangle $A B C$, respectively. Point $O$ is the intersection of segments $\overline{A N}$ and $\overline{B M}$. Determine the area of triangle $C M N$ if the areas of triangles $O M A, O A B$, and $O B N$ are 2, 3, and 4, respectively.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_38fd7d226823545bc90fg-03.jpg?height=596&width=658&top_left_y=1705&top_left_x=690)

Let

$$
P_{\triangle O M A}=P_{1}=2, \quad P_{\triangle O A B}=P_{2}=3, \quad P_{\triangle O B N}=P_{3}=4, \quad \text { and } P_{\triangle O N M}=P_{4}
$$

The area of triangle $M N C$, which we are looking for, is denoted by $P$.

Triangles $O A B$ and $O B N$ share the same height $x$. Similarly, triangles $O M A$ and $O N M$ share the same height $y$. From the ratios of their areas, we get

$$
\begin{aligned}
& \frac{P_{3}}{P_{2}}=\frac{4}{3} \\
& \frac{\frac{|O N| \cdot x}{2}}{\frac{|O A| \cdot x}{2}}=\frac{4}{3} \\
& \frac{|O N|}{|O A|}=\frac{4}{3}
\end{aligned}
$$

Similarly,

$$
\begin{aligned}
& \frac{P_{4}}{P_{1}}=\frac{\frac{|O N| \cdot y}{2}}{\frac{|O A| \cdot y}{2}} \\
& \frac{P_{4}}{2}=\frac{|O N|}{|O A|}
\end{aligned}
$$

From (1) and (2), it follows that $P_{4}=\frac{8}{3}$.

Next, consider triangles $M N C$ and $A N C$ with the common height $z$. As in the previous case, from the ratios of their areas, we get the ratio of the sides

$$
\frac{\frac{|C M| \cdot z}{2}}{\frac{|A C| \cdot z}{2}}=\frac{P}{2+\frac{8}{3}+P}
$$

Thus,

$$
\frac{|C M|}{|A C|}=\frac{3 P}{14+3 P}
$$

Similarly, from triangles $M B C$ and $A B C$ with the common height $v$, we get

$$
\frac{\frac{|C M| \cdot v}{2}}{\frac{|A C| \cdot v}{2}}=\frac{4+\frac{8}{3}+P}{9+\frac{8}{3}+P}
$$

Thus,

$$
\frac{|C M|}{|A C|}=\frac{20+3 P}{35+3 P}
$$

From (3) and (4), we get

$$
\frac{20+3 P}{35+3 P}=\frac{3 P}{14+3 P}
$$

The area of the desired triangle is

$$
P=\frac{280}{3}
$$