= Solution
Use the bead <angles> $\theta_1,\theta_2,\theta_3$ as <generalized coordinates>. The straight-line spring length between two beads follows from the <law of cosines>:
$$
\ell_{ij}=\sqrt{2-2\cos(\theta_i-\theta_j)}.
$$
The hoop is horizontal, so the <gravitational potential energy> is constant and can be omitted. The <kinetic energy> is $(m/2)\sum_i\dot\theta_i^2$, giving the <Lagrangian>
$$
\boxed{L=\frac m2\sum_{i=1}^3\dot\theta_i^2-\frac k2\sum_{i<j}\left[\sqrt{2-2\cos(\theta_i-\theta_j)}-\sqrt3\right]^2.}
$$
At $\theta_i^0=\Theta+2\pi(i-1)/3$, every spring has its natural length, so the <potential energy> is zero and its first <derivatives> vanish. This is an <equilibrium> for every constant $\Theta$. It is stable against relative displacement because each spring energy is nonnegative; the overall rotation is a neutral direction.
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