= Solution
For the <normal modes of an equilateral spring triangle>, the complement of the three rigid motions has dimension three. One convenient mutually <orthogonal> basis is
$$
b=(-\sqrt3,\sqrt3,0;-1,-1,2),\qquad d=(-\sqrt3,\sqrt3,0;1,1,-2),\qquad s=(-1,-1,2;-\sqrt3,\sqrt3,0).
$$
These vectors are also orthogonal to $t_x,t_y,r$, and each has squared length $12$. Their geometry comes from applying symmetric linear deformations to the equilibrium triangle: $b$ is uniform expansion, $d$ stretches one axis while compressing the other, and $s$ is the second independent shear pattern. Their extensions are
$$
\begin{array}{c|ccc}&e_{12}&e_{13}&e_{23}\\\hline
b&2\sqrt3&2\sqrt3&2\sqrt3\\
d&2\sqrt3&-\sqrt3&-\sqrt3\\
s&0&3&-3
\end{array}.
$$
Insert each row into the force equations. For example, $b$ gives $m\ddot x_1=3\sqrt3 k=-3k b_{x_1}$ and $m\ddot y_3=-6k=-3k b_{y_3}$, with the same proportionality in every component. The other two rows similarly give force $-3k/2$ times their displacement vector. Thus
$$
Kb=3kb,\qquad Kd=\tfrac32kd,\qquad Ks=\tfrac32ks,
$$
and the remaining <normal modes> and <normal frequencies> are
$$
\boxed{\begin{array}{c|c|c}\text{pattern}&\omega^2&\omega\\\hline
b&3k/m&\sqrt{3k/m}\\
d&3k/(2m)&\sqrt{3k/(2m)}\\
s&3k/(2m)&\sqrt{3k/(2m)}
\end{array}.}
$$
Any orthogonal pair spanning the same two-dimensional shear subspace is equally valid, because its <eigenvalue> is repeated. The six displayed vectors are nonzero and mutually orthogonal, so they supply all six independent <normal modes> without solving a characteristic polynomial.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2002/paper-2-triangle-normal-modes.png]
{title=Rotation, breathing and two shear patterns of the three-mass triangle}
{description=Arrows show infinitesimal displacements using the equilibrium coordinates in the paper. Translations supply two further zero modes. The breathing frequency squared is 3k/m; both shear frequencies squared are 3k/(2m).}
For explicit <normal coordinates>, normalize $t_x,t_y$ by $\sqrt3$ and $r,b,d,s$ by $\sqrt{12}$, and write $q=\sum_jQ_j\widehat v_j$. Since all masses are equal,
$$
T=\frac m2\sum_j\dot Q_j^2,\qquad V=\frac m2\sum_j\omega_j^2Q_j^2,\qquad\ddot Q_j+\omega_j^2Q_j=0.
$$
Positive-frequency coordinates have $Q_j=A_j\cos(\omega_jt)+B_j\sin(\omega_jt)$; the three rigid coordinates have the linear form found above.
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