4. Points $M$ and $N$ are the midpoints of sides $AB$ and $AC$, respectively, of triangle $ABC$. Point $P$ lies on side $BC$ and satisfies $\overline{PC}=\frac{1}{3} \overline{BC}$. Lines $CM$ and $NP$ intersect at point $O$. Express the area of $\triangle OPC$ in terms of the area of $\triangle ABC$.
Solution. Let $P_{\triangle A B C}=P$, $P_{\triangle O P C}=P_{1}$, and $P_{\triangle N O C}=P_{2}$. From $\overline{C P}=\frac{1}{3} \overline{C B}$ it follows that

$$
P_{\triangle C N P}=\frac{1}{3} P_{\triangle B N C}
$$

However, $B N$ is the median of $\triangle A B C$, so

![](https://cdn.mathpix.com/cropped/2024_06_05_5d3747084f372969d7bdg-4.jpg?height=447&width=705&top_left_y=1778&top_left_x=777)

$$
P_{\triangle B N C}=\frac{1}{2} P_{\triangle A B C}=\frac{1}{2} P
$$

From this, we get

$$
P_{1}+P_{2}=P_{\Delta C N P}=\frac{1}{6} P
$$

Furthermore, $M$ is the midpoint of $A B$, so

$$
P_{\triangle A M O}=P_{\triangle M O B} \text { and } P_{\triangle A M C}=P_{\triangle M B C} \text {. }
$$

Since $O$ lies on the median $C M$, it follows that

$$
P_{\triangle A O C}=P_{\triangle A M C}-P_{\triangle A M O}=P_{\triangle M B C}-P_{\triangle M B O}=P_{\triangle O B C}
$$

Since $N$ is the midpoint of $A C$, it follows that $P_{\triangle A O C}=2 P_{\triangle N O C}=2 P_{2}$. Similarly, $P_{\triangle O B C}=3 P_{1}$. Therefore, $3 P_{1}=2 P_{2}$. From this and $P_{1}+P_{2}=\frac{1}{6} P$, we get $P_{1}+\frac{3}{2} P_{1}=\frac{1}{6} P$, i.e., $P_{1}=\frac{1}{15} P$.