[amelius]

How do we know there is only 1 stable position?

(A perfect circle would have many, but you could consider that a degenerate configuration.)

[eternauta3k]

There isn't necessarily just one. A rectangle-ish form could have 4.

[amelius]

Yes, rectangle would also be degenerate. Basically any configuration where adding an epsilon weight somewhere generates a non-epsilon shift so to speak.

[eternauta3k]

To me, "degenerate" means that it has a symmetry so that the different states are somehow identical (e.g. equal energy in a quantum system). If it's rectangle-ish (say, an imperfect hand-drawn rectangle) then it has 4 distinct equilibrium positions. Not sure if we're thinking of the same with "degenerate".

[s1artibartfast]

A plus shape "+" has four stable equilibriums which all stable optimums, resistant to weight and self correcting.

[marcosdumay]

You mean unstable?