[ Current line of the triangle ] $[\quad$ Center of mass $\quad]$ $A A_{1}$ is the median of triangle $A B C$. Point $C_{1}$ lies on side $A B$, and $A C_{1}: C_{1} B=1: 2$. Segments $A A_{1}$ and $C C_{1}$ intersect at point $M$. Find the ratios $A M: M A_{1}$ and $C M: M C_{1}$.
First method. Through point $A_{1}$, draw a line parallel to $C C_{1}$ until it intersects $A B$ at point $K$. $A_{1} K$ is the midline of triangle $C B C_{1}$, so $C C_{1}=A_{1} K, B K=K C_{1}=A C_{1}$.

Therefore, $M C_{1}$ is the midline of triangle $A_{1} A K_{1}$, which means $M C_{1}=1 / 2 A_{1} K=1 / 4 C C_{1}$, and $A M=M A_{1}$.

Second method. Place unit masses at points $B$ and $C$, and a mass of 2 at point $A$. Then $M$ is the center of mass of the resulting system of material points. Replacing the masses at $B$ and $C$ with a mass of 2 placed at point $A_{1}$, we get $A M$ : $M A_{1}=2: 2$. Replacing the masses at $A$ and $B$ with a mass of 3 placed at point $C_{1}$, we get $C M: M C_{1}=3: 1$.