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I saw this problem from a math forum but it has been left unanswered for months.

The problem states:

"Prove that every polygon with perimeter $2004$ can be covered by a circle with diameter $1002$ "

I have tried the following methods but i keep failing, Any hints for a possible method are appreciated:

$1)$ I tried proving that all triangles with perimeter $2004$ can be covered by circle with diameter $1002$ and then use strong induction to say that all such $n$-gons can be covered and then try to prove it for alla the $n+1$-gons

$2)$ I tried to use contradiction but also failed.

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    *Hint*:: The maximum possible length of the longest side of polygon will be less than $1002$. – 2017-01-06
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    I had already noticed that but i guess i haven't used it properly, anyways i will try it again.Thanks for the answer. – 2017-01-06
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    your polygon is regular ? – 2017-01-06
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    not necessarily – 2017-01-06

2 Answers 2

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Let's say that $D$ is the biggest distance between two vertex.

Just for better undestanding let's consider that upside of $AE$ has only four sides (it doesn't change anything). By triangle inequality we have:

$$d_1

We can make the same idea for the downside and conclude that:

$$2D

Then take a circle such that the diameter contain $AE$ and once the diameter $1002$, no other distance can goes outside of that circle.

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Hints: Say we have two points on the perimeter of the polygon that are separated by 1002+$\epsilon$ units ($\epsilon>0$). Say these points are $\mathbf{x}_1$ and $\mathbf{x}_2$. Draw a line $L$ through these two points. Assuming the segment between these two points isn't a face of the polygon (the other case follows similarly), we have that there exist line segments, which aren't parallel to $L$ that cross $L$ and go through points $\mathbf{x}_1$ and $\mathbf{x}_2$ defining distinct faces of the polygon. What is the minimum perimeter of a polygon that closes the four end points of these segments? Can it possibly have a perimeter of 2004?