= Solution
At the north pole $\mathbf r=+R\hat z$, so the <velocities> reverse:
$$
\mathbf v_C=(0,0),\qquad\mathbf v_A=(48,0),\qquad\mathbf v_B=(48,-20)\quad\mathrm{mm\,yr^{-1}}.
$$
The triangle still has side lengths 20, 48 and 52 mm/year. Across AB, A moves toward positive $y$ relative to B; across AC, A moves toward positive $x$ relative to C. Both boundaries therefore open and are <spreading ridges>, with full opening <speeds> 20 and 48 mm/year. Together with BC they form a <ridge-ridge-transform triple junction>.
In the symmetric normal-spreading model, a ridge's normal <velocity> is the mean normal <velocity> of its two plates. AB gives $U_y=(0-20)/2=-10$, and AC gives $U_x=(48+0)/2=24$. Their intersection lies on the BC transform <velocity> line:
$$
\boxed{\mathbf U_N=(24,-10)\,\mathrm{mm\,yr^{-1}}=\frac{\mathbf v_B+\mathbf v_C}{2}.}
$$
\b[This northern junction is also instantaneously kinematically stable.] The diagrams use the same fixed geocentric tangent axes at both poles; reversing <velocities> is not the same operation as reflecting a map drawn from an exterior observer's viewpoint.
Not every ridge-ridge-transform geometry works. For pure normal spreading, each ridge <velocity> line is the <perpendicular bisector> of the corresponding side of the <plate velocity triangle>. Their intersection is the triangle's <circumcenter>. A transform junction must also lie on the third side's line. A nondegenerate triangle's <circumcenter> lies there only at that side's midpoint, so the opposite angle is a right angle. Equivalently, $\big(\mathbf v_B-\mathbf v_A\big)\cdot\big(\mathbf v_C-\mathbf v_A\big)=0$. The two ridge directions are then perpendicular, as they are here. \b[RRF junctions are conditionally stable, with the right-angle ridge geometry in this symmetric pure-normal model; they are not all stable.] Oblique or asymmetric accretion changes the velocity-line assumptions and must be specified separately.
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