3. In triangle $ABC$, a point $M$ is taken on side $AC$, and a point $N$ is taken on side $BC$. Segments $AN$ and $BM$ intersect at point $O$. The areas of triangles $AMO, ABO, BNO$ are $P_{1}, P_{2}, P_{3}$ respectively. Calculate the area of triangle $CMN$.
Solution. The area of triangle $CMN$ will be denoted by $P$ and the area of triangle $MON$ will be denoted by $Q$.

Triangles $AMO$ and $MON$ have equal heights to $AO$ and $ON$, respectively, and triangles $AOB$ and $NOB$ have equal heights to $AO$ and $ON$, respectively, so $\frac{P_{1}}{Q}=\frac{\overline{AO}}{\overline{ON}}=\frac{P_{2}}{P_{3}}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_134071f890a711206f81g-6.jpg?height=312&width=358&top_left_y=1632&top_left_x=1104)

From the last equality, we get $Q=\frac{P_{1} P_{3}}{P_{2}}$. From the pairs of triangles $CMN$ and $BMN$, and the pairs of triangles $ANC$ and $ANB$ (each pair has equal heights dropped to $CN$ and $NB$, respectively), we get: $\frac{P}{Q+P_{3}}=\frac{\overline{CN}}{\overline{NB}}=\frac{P_{1}+Q+P}{P_{2}+P_{3}}$.

According to this, $P\left(P_{2}-Q\right)=Q^{2}+Q P_{1}+Q P_{2}+P_{1} P_{3}$, i.e., $P=\frac{Q^{2}+Q P_{1}+Q P_{2}+P_{1} P_{3}}{P_{2}-Q}$, and if we substitute $Q=\frac{P_{1} P_{3}}{P_{2}}$ into the last equality, we get

$$
P=\frac{\left(\frac{P_{3} P_{3}}{P_{2}}\right)^{2}+\frac{P_{1} P_{3}}{P_{2}} P_{1}+\frac{P_{1} P_{3}}{P_{3}} P_{3}+P_{1} P_{3}}{P_{2}-\frac{P_{1} P_{3}}{P_{2}}}=\frac{P_{1} P_{3}\left(P_{1}+P_{2}\right)\left(P_{2}+P_{3}\right)}{P_{2}\left(P_{2}^{2}-P_{1} P_{3}\right)}.
$$