imc

First, realize that each triangle is congruent, a right triangle and that the two legs are equal. Also, each side of the octagon is equal, because of the definition of regular shapes. Let $s$ be the length of a leg of the isosceles right triangle. In terms of $s$, the hypotenuse of the isosceles right triangle, which is also the length of a side of the regular octagon, is $s\sqrt{2}$. Since the length of each side of the square is 2000, the length of each side of the regular octagon is equal to the length of a side of the square ($2000$) subtracted by $2$ times the length of a leg of the isosceles right triangle ( the total length of the side is $2s+o$, $o$ being the length of a side of the regular octagon), which is the same as $2s$. As an expression, this is $2000-2s$, which we can equate to $s\sqrt{2}$, ( since the octagon is regular, meaning all of the side's lengths are congruent) giving us the following equation:$2000-2s=s\sqrt{2}$. By isolating the variable and simplifying the right side, we get the following: $2000=s(2+\sqrt{2})$. Dividing both sides by $(2+\sqrt{2})$, we arrive with $\frac{2000}{2+\sqrt{2}}=s$, now, to find the length of the side of the octagon, we can plug in $s$ and use the equation $2000-2s=o$, $o$ being the length of a side of the octagon, to derive the value of a side of the octagon. After plugging in the values, we derive $2000-2(\frac{2000}{2+\sqrt{2}})$, which is the same as $2000-(\frac{4000}{2+\sqrt{2}})$, factoring out a $2000$, we derive the following: $2000(1-(\frac{2}{2+\sqrt{2}}))$, by rationalizing the denominator of $\frac{2}{2+\sqrt{2}}$, we get $2000(1-(2-\sqrt{2}))$, after expanding, finally, we get $\boxed{2000(\sqrt{2}-1)}$ !(not a factorial symbol, just an exclamation point)