# Which polygons can be turned inside out by a smooth deformation?

Take a non-degenerate polygon with side lengths $$\{a_1,\dots,a_n\}$$ in a convex configuration. What is the condition on the $$a_i$$'s so that the polygon can be turned inside out by a continuous motion that preserves the side lengths.

For example any triangle cannot be, but the unit square and unit hexagon can be, since you can collapse both to form a degenerate shape with zero area. A four sided figure can sometimes be inverted, sometimes not if the smallest side is too small.

Just working out the four sided figure could be interesting I think.

I suspect that it is possible iff you can find a configuration of zero area, where the area is defined as $$\frac{1}{2}(x_1y_2-x_2y_1+x_2y_3-x_3y_2+\dots+x_ny_1-x_1y_n)$$ where the $$(x_i,y_i)$$ are vertices of the polygon with i increasing from 1 to n as you progress around the polygon in order. It is trivial that if a deformation exists the area must change sign and hence be zero but the reverse implication may be more challenging. (Of course this doesn't solve the problem as this is not a condition on the side lengths but might be a useful observation).

Update: My conjecture that you can invert a polygon iff you can deform it into a configuration of zero area is incorrect due to the following counterexample based on the figure included in Joseph O'Rourke's answer:

The polygon on the left has 3 side lengths $$\frac{1}{\sqrt 3}$$ and $$1+\frac{2}{\sqrt 3}$$ and hence does not satisfy the criteria in the paper because $$2(1+\frac{2}{\sqrt 3})>(1+\frac{2}{\sqrt 3})+\frac{3}{\sqrt 3}.$$ However you can deform the left configuration into the one on the right which has zero area.

In addition you can slightly increase the short sides AB, CD, EF to create a negative area on the right configuration but still not satisfy the criteria for inversion. Therefore we cannot modify the conjecture to the obvious one that there must be mutually deformable configurations of both strictly positive and strictly negative area.

Therefore the two equivalence classes of configurations for a polygon that doesn't satisfy the criteria must sometimes contain a mixture of positive and negative area configurations.

• What do you mean by `smooth motion'? Should squashing a square to a line really count? – wnx Aug 28 '19 at 10:38
• @WalterNeff I mean a continuous motion that preserves the side lengths. In the case of a square squashing to line I meant that you take opposite vertices of the square and move them together keeping all sides equal so that the squares becomes a rhombus of smaller and smaller area until the area vanishes. I'll add this as a clarification to the question thanks. – Ivan Meir Aug 28 '19 at 10:54
• Yes it should be, thanks! – Ivan Meir Aug 29 '19 at 8:05
• You shouldn't update your question to include an answer. – CJ Dennis Aug 30 '19 at 4:02
• @CJDennis It actually wasn't an answer to my question rather a refutation of a conjecture that I posed as part of my own observations related to the problem. – Ivan Meir Aug 30 '19 at 7:35

This question was explored here:

Lenhart, William J., and Sue H. Whitesides. "Reconfiguring closed polygonal chains in Euclidean $$d$$-space." Discrete & Computational Geometry 13, no. 1 (1995): 123-140; DOI: 10.1007/BF02574031, eudml.

From the Abstract: "It is shown that in three or more dimensions, reconfiguration is always possible, but that in dimension two this is not the case. Reconfiguration is shown to be always possible in two dimensions if and only if the sum of the lengths of the second and third longest links add to at most the sum of the lengths of the remaining links."

• See also e.g. the work of Kapovich and Millson math.ucdavis.edu/~kapovich/EPR/plane.pdf or Farber and Schuetz arxiv.org/abs/math/0609140 for topological studies of moduli spaces of polygons. – Mark Grant Aug 28 '19 at 12:38
• Thank you Joseph, that's a very elegant result that should be better know given it is such a natural question to ask. I am still thinking about the question of whether having a configuration with 0 area is also equivalent to a shape being reconfigurable. I think this follows from the paper but I need to think a bit more about this. – Ivan Meir Aug 28 '19 at 13:01
• @MarkGrant Thank you - interesting to see such sophisticated results arising from an elementary problem. – Ivan Meir Aug 28 '19 at 13:08
• @IvanMeir: I added their Fig.1, which supports your zero-area criterion. – Joseph O'Rourke Aug 28 '19 at 13:12
• @JosephO'Rourke Actually I believe I found a counterexample to the zero area conjecture based on the figure in your answer - see my update. – Ivan Meir Aug 28 '19 at 18:09