The longest side must have length at least $ \frac {2001} {1415}$, thus the 1414 remaining sides has total length at most $2001 - \frac {2001} {1415} = \frac {2001 \times 1414} {1415}$. Now divide this set of 1414 sides into 707 pairs of consecutive sides. There is two consecutive sides, say $AB$ and $BC$, such that $AB+BC \leq \frac {2001 \times 1414} {1415 \times 707} < 2 \sqrt {2}$. Now, let $S$ be the area of triangle $ABC$. Using AM/GM, we have : ${ \sqrt {S} = \frac {1} { \sqrt {2} } \sqrt { \sin {B}} \sqrt {AB \times BC} \leq \frac {1} { \sqrt {2} } \sqrt {AB \times BC} \leq \frac {1} {2 \sqrt {2} } (AB + BC)} < \frac {2 \sqrt {2}} {2 \sqrt {2}} = 1.$ Pierre.

My solution: Let we can't find 3 vertices of $N$ which form a triangle of area smaller than 1. Then $S(A_1A_2A_3),S(A_2A_3A_4),...,S(A_{1415}A_1A_2)\geq 1,.$ We know $2\le 2S(A_1A_2A_3)\le A_1A_2\cdot A_2A_3,$ Also we know from $AM-GM,$ $A_1A_2+A_2A_3\ge 2\sqrt{A_1A_2\cdot A_2A_3}\ge 2\sqrt{2}.$ Then (we find similarly way other one ) $2001\cdot 2=2(A_1A_2+A_2A_3+...+A_{1415}A_1)\ge 1415\cdot 2\sqrt{2},$ Then we find $4004001\ge 4004450,$ Which is a contradiction.

Ferid.---. wrote: We know $2\le 2S(A_1A_2A_3)\le A_1A_2\cdot A_2A_3,$ How did you get this?

Duarti wrote: Ferid.---. wrote: We know $2\le 2S(A_1A_2A_3)\le A_1A_2\cdot A_2A_3,$ How did you get this? We know $S(A_1A_2A_3)=\frac{A_1A_2\cdot A_2A_3\cdot Sin(A_1A_2A_3)}{2}\le \frac{A_1A_2\cdot A_2A_3}{2}.$

i agreed with you