geometry – Arbitrary $n-text{masses}$ for $n-text{points}$ along the same cirunference create a $n-text{regular}$ polygon, for $n=3, 4, 5.$


Suppose three points $(q_1,q_2,q_3)$ of abritrary masses $(m_1,m_2,m_3)$ are on a circunference with center at the origin, creating a triangle (non necessarily equilateral).

Suppose the the barycenter and the center of mass are in the origin. Prove that all masses are equal $m_1=m_2=m_3$, and prove that the triangle is equilateral.

I began by using the definition of barycenter $G = frac{1}{3}(q_1+q_2+q_3)$

As the barycenter is in the origin, $G=0$, so I conclude that $q_1=-q_2-q_3$.

Then, as $m_1q_1+m_2q_2+m_3q_3=0$ because the center os mass is in the origin, I followed that
$$-m_1q_2-m_1q_3m_2q_2+m_3q_3=0$$

Thus, $m_1=m_2=m_3$.

For a lemma I proved at class I concluded that the triangle is equilateral.

Now, my doubt:

First, I don’t know if I forgot something, or I am using what I want to prove. Second, how do I prove the same thing por $n=4, 5$ making a square and a pentagon, respectively.