= Solution
Use the equilibrium coordinates specified in the paper; their distances from the origin are one and their mutual separation is $\sqrt3$. Let $u_i=(x_i,y_i)$ be the small displacements. Each initially unstretched <spring> has, to first order, extension equal to the relative displacement projected along its original bond. Choose the three oriented extensions
$$
e_{12}=x_2-x_1,\qquad e_{13}=\tfrac12(x_3-x_1)+\tfrac{\sqrt3}{2}(y_3-y_1),\qquad e_{23}=-\tfrac12(x_3-x_2)+\tfrac{\sqrt3}{2}(y_3-y_2).
$$
The <kinetic energy> is exact in these displacement coordinates, while the <potential energy> needed for <small oscillations> is its quadratic part:
$$
\boxed{T=\frac m2\sum_{i=1}^3(\dot x_i^2+\dot y_i^2),\qquad V=\frac k2(e_{12}^2+e_{13}^2+e_{23}^2).}
$$
The factor $\sqrt3$ in the equilibrium bond length cancels when forming the unit bond directions. Applying the <Euler-Lagrange equation>, or differentiating $V$ for the force on each mass, gives
$$
\boxed{\begin{aligned}
m\ddot x_1&=k(e_{12}+\tfrac12e_{13}),&m\ddot y_1&=\tfrac{\sqrt3 k}{2}e_{13},\\
m\ddot x_2&=-k(e_{12}+\tfrac12e_{23}),&m\ddot y_2&=\tfrac{\sqrt3 k}{2}e_{23},\\
m\ddot x_3&=-\tfrac k2(e_{13}-e_{23}),&m\ddot y_3&=-\tfrac{\sqrt3 k}{2}(e_{13}+e_{23}).
\end{aligned}}
$$
The total force vanishes componentwise, as expected for three masses coupled only by internal <spring> forces.
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