MHB Center of mass of a two particle system

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The center of mass (CM) of a two-particle system is defined as the weighted average of their positions based on their masses. To prove that the CM lies on the line joining the two particles, one can assume, for contradiction, that it does not. By expressing the positions of the particles and the CM mathematically, it becomes evident that the CM must lie on the line segment connecting the two particles. The parameterization of the line segment shows that as the weightings change, the CM remains confined to this line. Thus, the proof confirms that the center of mass indeed lies on the line joining the two particles.
Dustinsfl
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How does one prove the center of mass of a two particle system lies on the line joining them?

Would we do this by contradiction?
Suppose on the contrary that the CM doesn't lie on the line joining the two particles. Where do I go from here though?
 
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Definition of CM of two particles of mass $m_{1}$ and $m_{2}$:
$$ \mathbf{r}_{ \text{cm}}= \frac{m_{1} \mathbf{r}_{1}+m_{2} \mathbf{r}_{2}}{m_{1}+m_{2}}.$$
We can view the line segment from $\mathbf{r}_{1}$ to $\mathbf{r}_{2}$ as follows:
$$\{\mathbf{r}| \exists\,t\in[0,1] \; \text{s.t.} \; \mathbf{r}=t \mathbf{r}_{1}+(1-t)\mathbf{r}_{2} \}.$$
You can see that $t=0$ means $\mathbf{r}=\mathbf{r}_{2}$ and $t=1$ corresponds to $\mathbf{r}=\mathbf{r}_{1}$. As $t$ varies in the interval $[0,1]$, the vector $\mathbf{r}$ sweeps out the line segment from $\mathbf{r}_{2}$ to $\mathbf{r}_{1}$. Now compare this expression to the expression for the center of mass.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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