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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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    @JyrkiLahtonen, you are right, thanks for pointing this out, please, help me to derive the right proof.2012-06-22

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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...2012-06-24