2
$\begingroup$

A diagonal $\overline{uv}$ in simple polygon $P$ is called short if the distance from $u$ to $v$ is 2 in $P$ (two segments between them). Prove or disprove that every simple polygon has a short diagonal.

I always have a problem with the proofs like that, it's hard to prove obvious things, and very often it's very misleading, what seems obvious for me actually cannot be true any more in some extreme case.

If you have any idea about the proof, please, share it with us. Thanks!

  • 3
    Does a side of a triangle count as a diagonal?2012-06-22
  • 0
    As for your proof: How do you know there is only *one* "lowest vertex"?2012-06-22
  • 0
    May be I misunderstood, but I don't think that your method works. Consider the 4-gon with vertices at (in clockwise order) $(0,0)$, $(-1,2)$, $(0,1)$ and $(1,2)$ - an arrowhead pointing down if you wish. The origin has the smallest $y$-coordinate, but the two vertices adjacent to it don't form a short diagonal. There is, of course, a short diagonal in this 4-gon though.2012-06-22
  • 0
    Or IOW, having a convex angle is not enough, because the resulting diagonal may not be in the interior of the polygon.2012-06-22
  • 0
    What is a "diagonal", e.g., is it assumed that a diagonal does not intersect any sides? In this case the diagonal at your "convex vertex" might be illegal, in so far as there might be another spike of $P$ within the spike you are cutting off.2012-06-22
  • 0
    @JyrkiLahtonen, you are right, thanks for pointing this out, please, help me to derive the right proof.2012-06-22

1 Answers 1