MHB A center of mass determined triangle: Find the angles of the triangle P_1P_2P_3

[lfdahl]

Masses $m, 2m$ and $\sqrt{3}m$ are located at points $P_1, P_2$ and $P_3$ on a circle $C$
so that their center of mass coincides with the center of $C$.
Find the angles of the triangle $P_1P_2P_3$.

 

[lfdahl]

Here´s the suggested solution:

Spoiler

Let $C$ be the unit circle of the $xy$-plane, with mass $m$ at the point $(1; 0)$.

If \[\angle P_1OP_2 = \alpha, \: \: \: \angle P_1OP_3 = \beta\] – then\[m\cdot 0+2m\sin \alpha +\sqrt{3}m\sin \beta = 0\: \: \: \: (1). \\ m\cdot 1 +2m\cos \alpha +\sqrt{3}m\cos \beta = 0\: \: \: \: (2).\]From $(1)$: \[\frac{\sin \alpha }{\sin \beta }= -\frac{1}{2}\sqrt{3}\]substitute in $(2)$ and obtain\[1 + 2 \cos \alpha +\sqrt{3}\sqrt{1-\frac{4}{3}\sin^2\alpha } = 0,\]whence\[\cos \alpha = -\frac{1}{2}, \: \: \alpha = 120^{\circ}.\]so that\[\sin \beta = -1, \: \: \beta = 270^{\circ}.\]Thus, the angles of $\bigtriangleup P_1P_2P_3$ are: $45^{\circ}, 60^{\circ}, 75^{\circ}$.