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This is the same question as here in SE.

I have a conjecture, it is like this:

Suppose there is a non-self-intersecting polygon lies inside a closed square of length $1$. The polygon has every side length of $1$. Suppose the polygon is not the boundary of the unit square (i.e. a square frame.) Prove or disprove that the number of sides of the polygon is an odd number.

I have pondered this problem for quite a while. What I have tried is this: Suppose the unit square is dissected into four $1/2\times 1/2$ quadrants, like $\begin{array}{|c|c|}\hline1 & 2\\\hline3&4\\\hline\end{array}$. Therefore, we want to prove that the line segments that go through the quadrants diagonally ($1$ to $4$, or $2$ to $3$) is odd number. To prove this, I want to prove the intermediate steps:

(1) We don't have both kinds of diagonals. That is, no edges from $1$ to $4$ and $2$ to $3$.

Update (from comments:) we may need to use the fact that it is a polygon. Generally If we place the segments in a good way, there can be two segments such that one segment has end points in regions $1$, $4$, And the other can have the segment has endpoint in regions $2$, $3$. For example, the segment connects $(0,0)$ and $(\sqrt 3/2−\epsilon_1,1/2+\epsilon_2)$; and the segment connects $(1−\sqrt 3/2,1)$ to $(1−\epsilon_3,1/2−\epsilon_4)$ for some small $\epsilon$'s (There does not need to be the same $\epsilon$.)

(2) All the diagonal-crossing segments are together.

(3) The diagonal crossing segment has odd number of them.

However, I stuck at all of them... Is there a way to solve this problem (even if not using the ideas from above)? Any help are welcome!

P.S. we can construct the polygon as this for $2k+1$ sides, as the zigzag shape:

enter image description here

The square of size $1$ is the smallest possible, because we can have this shape for any square $1+\varepsilon$: we have the edges as the original square's edge and have a spike inside.

enter image description here

P.S. Someone the MSE suggested that we can divide the square into nine regions, instead of four in the problem...

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    $\begingroup$ I don't understand the first diagram. It looks like all the points are on a side of the square, so most of the lengths exceed one. $\endgroup$
    – Gerry Myerson
    Commented Oct 27 at 8:43
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    $\begingroup$ @GerryMyerson You can try it yourself here... In the diagram all the lengrhs are $1$, And if you consider the extreme case that $F, H$, etc Is on the lower side, The height for each isosceles triangle does not exceed $1$. Now move $F$ and $H$ little up, then you can get the polygon. Sorry for the picture because it's not really easy to be distinguishable by human eyes because the distance to the edges is $O(1/k^2)$ So it looks like $E,F,G,H,I$ are on the edge, but it's not... I am sorry for the confusion. I will try to update the picture with smaller points and thinner lines. $\endgroup$
    – JetfiRex
    Commented Oct 27 at 12:49
  • $\begingroup$ @JetfiRex I think you may be able to prove that any edge from 1 to 4 must cross any edge from 2 to 3 by considering where the endpoints of a diagonal edge must be. There is a forbidden region in the center of the original unit square (any point in this forbidden region is less than 1 unit away from any other point in the square). $\endgroup$
    – Timothy Chow
    Commented Oct 28 at 11:46
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    $\begingroup$ @TimothyChow Thank you, but I have thought about this before. But generally If we place the segments in a good way, there can be two segments such that one segment has end points in regions 1, 4, And the other can have the segment has endpoint in regions 2, 3. For example, the segment connects $(0,0)$ and $(\sqrt 3/2-\epsilon_1,1/2+\epsilon_2)$; and the segment connects $(1-\sqrt 3/2,1)$ to $(1-\epsilon_3, 1/2-\epsilon_4)$ for some small $\epsilon$'s (There does not need to be the same $\epsilon$...) $\endgroup$
    – JetfiRex
    Commented Oct 28 at 12:06
  • $\begingroup$ If you follow the edges of the polygon, you always make an acute turn left or right. One idea is to distinguish cases based on the number of consecutive turns in the same direction (e.g. the example has 4, and perhaps this is the only possibility). The simplest case would be alternating left and right turns. Considering alternating points on the polygon, we obtain two new polygons: an outer polygon P and an inner polygon Q. Consider the area of the triangles based on the inside of the edges of polygon P. Fixing the edge lengths of P, its area is maximized when its vertices are concyclic. $\endgroup$
    – 1001
    Commented Nov 11 at 3:51

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