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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:

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.

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:

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.

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

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 $3$.

(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:

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.

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:

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.

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 on 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 $3$.

(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:

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.

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 $3$.

(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:

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.