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(A) For a mass $m$ at the centroid of the hexagon (at the origin), the net force is zero when $\Sigma F_x = 0$ and $\Sigma F_y = 0$.The gravitational

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$0$

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(A) For a mass $m$ at the centroid of the hexagon (at the origin), the net force is zero when $\Sigma F_x = 0$ and $\Sigma F_y = 0$.The gravitational force exerted by a mass $m_k$ at distance $r$ from the origin on mass $m$ is $F_k = \frac{G m m_k}{r^2}$.The $x$-component of the force is $F_{kx} = -\frac{G m m_k}{r^2} \cos 	heta_k$ (directed towards the mass).Summing the $x$-components: $\Sigma F_x = -\frac{G m M}{r^2} \sum_{k=1}^{6} k^i |\cos 	heta_k| \cos 	heta_k = 0$.For the vertices of a regular hexagon, the angles $	heta_k$ are $0^{\circ}, 60^{\circ}, 120^{\circ}, 180^{\circ}, 240^{\circ}, 300^{\circ}$.Substituting these values:$1^i |\cos 0^{\circ}| \cos 0^{\circ} + 2^i |\cos 60^{\circ}| \cos 60^{\circ} + 3^i |\cos 120^{\circ}| \cos 120^{\circ} + 4^i |\cos 180^{\circ}| \cos 180^{\circ} + 5^i |\cos 240^{\circ}| \cos 240^{\circ} + 6^i |\cos 300^{\circ}| \cos 300^{\circ} = 0$$1^i(1)(1) + 2^i(1/2)(1/2) + 3^i(1/2)(-1/2) + 4^i(1)(-1) + 5^i(1/2)(-1/2) + 6^i(1/2)(1/2) = 0$$1^i + \frac{2^i}{4} - \frac{3^i}{4} - 4^i - \frac{5^i}{4} + \frac{6^i}{4} = 0$Testing $i=0$:$1 + \frac{1}{4} - \frac{1}{4} - 1 - \frac{1}{4} + \frac{1}{4} = 0$. This holds true.Thus, the value of $i$ is $0$.

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