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At the south pole $\mathbf{r}=-R\hat{z}$. Choose plate C as the instantaneous reference frame. Since $R=4\times 10^9$ mm, the two components of relative angular speed give 20 and 48 mm/year. The plate velocities in the fixed tangent-plane axes areThe plate velocity triangle therefore has sidesThe physical boundary rays point along $+x$ for AB, $+y$ for AC and $(12,-5)$ for BC. Plate A occupies the quadrant between the first two rays, B the narrow wedge between AB and BC, and C the remainder. The stated overriding relations make AB and AC subduction trenches, while BC remains a transform fault: this is a trench-trench-transform triple junction. Put the subduction teeth on B's side of AB and A's side of AC, identifying B above A and A above C.

For kinematic stability there must be a single junction velocity $\mathbf{U}$ compatible with all three boundary-normal motions. Use the standard idealization that each trench's normal velocity is that of its overriding plate, with no independent rollback. AB is parallel to $x$, so $U_y=v_{B,y}=20$. AC is parallel to $y$, so $U_x=v_{A,x}=-48$. The transform requires $\mathbf{U}-{\mathbf{v}}_C$ to be parallel to $(12,-5)$, which is true for this intersection. HenceThe southern triple junction is instantaneously kinematically stable. It is stationary relative to B, migrates along AB relative to A, and migrates along BC relative to C. This compatibility result does not promise long-term stability if plate rotations or trench rollback change. Notice that the three relative velocities close even though the overriding order is not a statement that all three plates move toward one common stationary point.

At the north pole $\mathbf{r}=+R\hat{z}$, so the velocities reverse: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: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, $({\mathbf{v}}_B-{\mathbf{v}}_A)\cdot ({\mathbf{v}}_C-{\mathbf{v}}_A)=0$. The two ridge directions are then perpendicular, as they are here. 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.

For pure normal symmetric spreading, the two ridge constraints are perpendicular bisectors of two edges of the plate velocity triangle. The transform constraint lies along the third edge. For a nondegenerate triangle their common intersection exists only when the angle opposite that third edge is a right angle; the junction velocity is then the midpoint of the transform edge.