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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!

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Does a side of a triangle count as a diagonal? – Brian M. Scott Jun 22 '12 at 17:30
As for your proof: How do you know there is only one "lowest vertex"? – Blue Jun 22 '12 at 17:40
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. – Jyrki Lahtonen Jun 22 '12 at 17:55
Or IOW, having a convex angle is not enough, because the resulting diagonal may not be in the interior of the polygon. – Jyrki Lahtonen Jun 22 '12 at 17:57
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. – Christian Blatter Jun 22 '12 at 17:58

Any three conscutive vertices $ABC$ form a diagonal of length $2$. If a polygon has no short diagonal then $AC$ lies outside (if the polygon is a closed region), and the other side of the line from the outside must be the inside, therefore the interior angle at $B$ is $\geq 180^\circ$. If $ABC=180^\circ$ then $B$ is on an edge and is not a vertex. The same also applies to all other angles, therefore a polygon with no short sides must have all interior angles $>180^\circ$. But the sum of the interior angles in a polygon is $180(n-2)<180n$.

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Your solution doesn't really discuss the problem raised by Christian Blatter and myself. It is possible that the parts of the polygon that you did not draw "fold back in" in the sense that the polygon may have a vertex in the intersection $\Delta ABC\cap\Delta BCD$ in the r.h.s. figure. As such a vertex will rule out both $BD$ and $AC$ as a solution, a proof should show that this cannot happen for all the convex angles... – Jyrki Lahtonen Jun 24 '12 at 10:05

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