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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 $\frac{1}{n}(x_1y_2-x_2y_1+x_2y_3-x_3y_2+\dots+x_ny_1-x_1y_n)$ 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.

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.

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 $\frac{1}{n}(x_1y_2-x_2y_1+x_2y_3-x_3y_2+\dots+x_ny_1-x_1y_n)$ 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).

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 $\frac{1}{n}(x_1y_2-x_2y_1+x_2y_3-x_3y_2+\dots+x_ny_1-x_1y_n)$ 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.

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 smooth motion.

For example any triangle cannot be, but the unit square and unit hexagon can be, since you can squash both to form a line. 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 $\frac{1}{n}(x_1y_2-x_2y_1+x_2y_3-x_3y_2+\dots+x_ny_1-x_1y_n)$ 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).

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 $\frac{1}{n}(x_1y_2-x_2y_1+x_2y_3-x_3y_2+\dots+x_ny_1-x_1y_n)$ 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).