Are these simple statements about polygons true?

Question

Let $P$ be an $n$-gon with $n \gt 3$. I'm looking for proofs or counterexamples for the following statements:

1. there exist consecutive vertices $A,B,C$ of $P$ such that $\triangle ABC \cap \partial P = \overline{AB} \cup \overline{BC}$
2. there exist consecutive vertices $A,B,C$ of $P$ such that $\triangle ABC \cap \partial P = \overline{AB} \cup \overline{BC}$ and the interior angle of $P$ at $B$ is convex
3. there exist different vertices $X,Y$ of $P$ such that $\overline{XY} \cap \partial P = \{X,Y\}$

For 1. and 2., if there is an offending vertex (must be if any other point is) in the triangle, maybe it could be put in some relation with the considered vertex, forming a finite dag which must have a maximal element, suitable as $B$.

For 3. it's actually easy (see my answer which I'm going to add in a moment) without an additional assumption, so let me also require $\overline{XY}\subset P$. Then, maybe decomposition into polygons (one a subset, one with disjoint interior) with fewer vertices by the external diagonal and applying induction would help?

Answer 1

I believe I have all the solutions now (though there may well be better ones, which of course are still welcome).

Let's start with proving the version of 3. from my previous answer: in any $n$-gon $P$ with $n>3$ there exist different vertices $X,Y$ of $P$ such that $\overline{XY} = \overline{XY} \cap P^\mathrm{o} \cup \{X,Y\}$.

Let $T$ be a vertex of $P$ at which the inner angle of $P$ is convex and $R,S$ be its neighbours. If $\overline{RS}$ satisfies the conclusion, we're done. Otherwise, take $T$ as $X$ and let's look for a suitable $Y$. If there is at least one vertex of $P$ on $\overline{RS}$ other than $R,S$, but none in $\triangle RST ^\mathrm{o}$, take any such vertex as $Y$. Otherwise, let $C$ be the convex hull of $(\triangle RST ^\mathrm{o} \cap \partial P)^⎯$. $C$ is a convex polygon.

Let $N$ be the point of $C$ nearest to $T$. If $N$ is a vertex of $P$, take $N$ as $Y$. Otherwise, $N$ is a boundary point of $C$ but not its vertex or one of its two vertices lying on $\overline{RS}$. Then take as $Y$ any end (there are 1 or 2) of a side of $C$ containing $N$ which belongs to $\triangle RST ^\mathrm{o}$ (let's call the straight line containing this side $d$). It is surely a vertex of $P$ and, given that $C$ is convex and contains a point (e.g. a vertex of $C$ lying on $RS$ but not on $d$) on the other side of $d$ than $T$, $C$ is contained in a half-plane to which none of the points of $\overline{TY}$ belongs except $Y$.

Since 2. is stronger than 1., let's turn to it next. I'll prove a strengthened statement: if $P$ is a polygon and $U,V$ are some adjacent vertices of $P$, there exist consecutive vertices $A,B,C$ of $P$ such that $B \notin \{U,V\}$, $\triangle ABC \cap \partial P = \overline{AB} \cup \overline{BC}$ and the interior angle of $P$ at $B$ is convex.

If $P$ is a triangle, take as $B$ the vertex which is neither $U$, nor $V$. Otherwise, $P$ must have more than 3 vertices, so the previous theorem applies. Let $(X,Y)$ be a pair of points whose existence it states. $\overline{XY}$ divides $P$ into two polygons with fewer vertices, one of which (call it $R$) hasn't got $\overline{UV}$ – the forbidden segment – as a side. The result follows by mathematical induction from considering $R$ with $\overline{XY}$ as the forbidden segment.

Additionally, observe that a quadrilateral has at least two such sequences $A,B,C$ of adjacent segments, disregarding permutations of $\{A,C\}$. The above proof can probably be slightly restructured to improve the conclusion and also find two in any $P$.

Edit

Indeed it can, albeit not slightly. Forget about forbidden segments and let's find $\{A,B,C\}$ with desired properties in both polygons resulting from the division. For a polygon thus obtained, if it's a triangle, the case is trivial, otherwise, start by determining whether there exists some $Z$ that the points $X,Y,Z$ or $Z,X,Y$ satisfy the conclusion. If so, cut $\triangle XYZ$ off. Continue cutting off with the new side (previously a diagonal) in the role of $\overline{XY}$ until it becomes impossible or only a triangle remains. Then invoke the inductive hypothesis. The cutting process ensures that $\triangle ABC$ obtained in the inductive step does share two sides with the original $P$.

Answer 2

For 3. without the additional requirement, if $P$ is convex, any 2 different non-adjacent vertices will do, otherwise suitable $X$ and $Y$ can be chosen as 2 adjacent vertices of the convex hull of $P$ (with degenerate vertices included, lying on the segment between its neighbours) which aren't adjacent in $P$.