9. (20 points) Inside an acute triangle $A B C$, a point $M$ is marked. The lines $A M, B M$, $C M$ intersect the sides of the triangle at points $A_{1}, B_{1}$ and $C_{1}$ respectively. It is known that $M A_{1}=M B_{1}=M C_{1}=3$ and $A M+B M+C M=43$. Find $A M \cdot B M \cdot C M$.
Answer: 441.

Solution. Let $A M=x, B M=y, C M=z$. Note that $\frac{M C_{1}}{C C_{1}}=\frac{S_{A M B}}{S_{A B C}}$ and similarly for the other two segments. From the equality $\frac{S_{A M B}}{S_{A B C}}+\frac{S_{B M C}}{S_{A B C}}+\frac{S_{A M C}}{S_{A B C}}=1$ it follows that $\frac{3}{x+3}+\frac{3}{y+3}+\frac{3}{z+3}=1$, which transforms into $x y z=2 \cdot 3^{3}+3^{2}(x+y+z)=2 \cdot 3^{3}+3^{2} \cdot 43$