# 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.