Sawtooth-like polygonal chains in $\mathbb R^2$ must self-intersect.

Question

We consider closed polygonal chains in the 2-dimensional plane with an even number of sides, say $2n$, numbered as $A_1B_1A_2B_2\dots A_nB_nE$, where $E = A_1$.

We require additionally, that each $B_i$ lies right below $A_i$, like with sawtooth.

Edit: Since right below seems ambiguous, let's state this line explicitly as that the $y$-coordinate of each $B_i$ be strictly less than that of $A_i$.

Is it true that any such polygonal chain must self-intersect?

In the 3-dimensional case, there is the obvious counterexample of traversing the vertices of any prism in the obvious order. I have however yet to find a proof or counterexample in the 2-dimensional case.

Answer 1

For convenience, extend the indices cyclically to all of $\Bbb Z$, so that e.g. $A_0=A_n$, $A_{n+1}=A_1$.

Proof by induction on $n$.

The case $n=1$ is trivial and the case $n=2$ is clear: $B_1A_2$ and $B_2A_1$ are the diagonals of the trapezoid $A_1B_1B_2A_2$.

With $n>2$ consider a sawtooth polygonal chain $A_1B_1A_2\ldots B_nA_1$ and assume it is a simple polygon. Introduce coordinates $A_k=(x_k,y_k)$ and $B_k=(x_k,z_k)$ (where $z_k<y_k$). If for some $k$, we have $x_{k+1}=x_k$ (i.e., $A_{k+1}$ is on the line $A_kB_k$), then either $y_{k+1}>z_k$ and we have an immediate self-intersection, or $y_{k+1}\le z_k$ and we can drop $B_k$, $A_{k+1}$ from the vertices and describe the same sawtooth polygonal chain with only $2\cdot(n-1)$ vertices - which is self-intersecting by induction hypothesis, contradiction! Hence we may assume that $x_{k+1}\ne x_k$ for all $k$.

As $x_{n+1}=x_1$, the sequence $x_1,x_2,\ldots$ cannot be monotonic. Hence there must exist $k$ with $x_{k-1}<x_k>x_{k+1}$; call such $k$ locally rightmost. Among all locally rightmost $k$, pick one with minimal $x_k$. Depending on the situation, we consider the following trapezoid $A_kB_kPQ$:

• If $x_{k-1}=x_{k+1}$, let $P=A_{k+1}$, $Q=B_{k-1}$
• If $x_{k-1}>x_{k+1}$, let $Q=B_{k-1}$ and $P$ the point on $B_{k}A_{k+1}$ on the same vertical line as $Q$
• If $x_{k-1}<x_{k+1}$, let $P=A_{k+1}$ and $Q$ the point on $B_{k-1}A_k$ on the same vertical line as $P$

As $B_kA_{k+1}$ does not intersect $B_{k-1}A_k$, we have $P$ below $Q$ in each case, i.e., $A_kB_kPQ$ is a convex trapezoid. Let $U$ be the interior of $A_kB_kPQ$ together with the interior of line segment $PQ$ (in other words, the closed filled trapezoid $A_kB_kPQ$ but with the three edges $QA_kB_kP$ removed). We make two observations:

• Not all $A_i$ are $\in U$. In fact, at least $A_{k-1},A_k,A_{k+1}\notin U$.
• $A_i\in U\iff B_i\in U$. This follows because the line segment $A_iB_i$ can enter/leave $U$ only across the also vertical line segment $PQ$.

Assume some $A_i$ is in $U$. Extend this to a maximal set $I=\{r,r+1,\ldots, s\}$ of (cyclically) consecutive indices with $A_i\in U$ for all $i\in I$, but $A_{r-1}, A_{s+1}\notin U$. As observed, then also $B_{r-1}\notin U$ and $B_s\in U$. As the only way "allowed" to enter/leave $U$ is across $PQ$, we conclude $x_{r-1}<x_r$ and $x_s>x_{s+1}$. It follows that some $j\in I$ is locally rightmost. By minimality of $x_k$, we have $x_j\ge x_k$, but that contradicts $A_j\in U$! We conclude that no $A_i$ is $\in U$.

By our observation, also no $B_i$ is $\in U$. Finally, if some arbitrary point $R$ on the polygonal chain is $\in U$, then at least one of the endpoints of the line segment containing $R$ must be $\in U$, but we just saw that no $A_i,B_i$ is $\in U$.

We conclude that the interior of line segment $PQ$ has no point in common with our polygonal chain. This allows us to take $QP$ as a shortcut, i.e., the sawtooth polygonal chain $A_1B_1\ldots A_{k-1}B_{k-1}QPA_{k+1}B_{k+1}\ldots B_nA_1$, or rather

• $A_1B_1\ldots A_{k-1}B_{k+1}\ldots B_nA_1$ if $x_{k-1}=x_{k+1}$
• $A_1B_1\ldots A_{k-1}PA_{k+1}B_{k+1}\ldots B_nA_1$ if $x_{k-1}=x_{k+1}$ if $x_{k-1}>x_{k+1}$
• $A_1B_1\ldots A_{k-1}B_{k-1}QB_{k+1}\ldots B_nA_1$ if $x_{k-1}=x_{k+1}$ if $x_{k-1}<x_{k+1}$

is not self-intersecting - whereas by induction hypothesis it is!

Answer 2

Yes. For simplicity let us assume that "polygon" means that consecutive edges cannot be parallel (no 180-degree angles at vertices). In particular this implies that the outgoing edges from $B_i$ are not vertical.

If the polygon does not intersect itself, it is oriented either clockwise or counterclockwise.

If clockwise, consider any leftmost vertex $v$, and the vertical line $L$ passing through it. No vertex of the polygon lies to the left of $L$. Because we do not allow 180-degree angles, and because of the sawtooth condition, exactly one of the edges incident at $v$ will be vertical, and it must be oriented up because the polygon is oriented clockwise. This is impossible, because all vertical edges must be oriented down by the sawtooth condition.

If the polygon is oriented counterclockwise, repeat the same reasoning at any rightmost vertex; because the polygon is oriented counterclockwise, the incident vertical edge must be oriented up, which is impossible.