Polygon can be covered by circle. 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.
 A: 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?
A: 
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:
$$d_1<a_1+a_2$$
$$d_2<d_1+a_3 \rightarrow d_2<a_1+a_2+a_3$$
$$D<d_2+a_4 \rightarrow D<a_1+a_2+a_3+a_4$$
We can make the same idea for the downside and conclude that:
$$2D<a_1+a_2+...+a_n \rightarrow D<\frac{a_1+a_2+...+a_n}{2} \rightarrow D< 1002$$
Then take a circle such that the diameter contain $AE$ and once the diameter $1002$, no other distance can goes outside of that circle. 
