A crack in the rectangle

Let the width of the rectangle be $w$ , and the length be $w + 2$ , and place the rectangle on a coordinate system so that the bottom left corner is on the origin.

Then the line on the diagonal of the rectangle is $y = -\frac{w}{w + 2} + w$ , and the line on the diagonal of the dark green square is $y = x$ , and these lines intersect at $(\frac{w(w + 2)}{2(w + 1)}, \frac{w(w + 2)}{2(w + 1)})$ , so the side of the dark green square is $s = \frac{w(w + 2)}{2(w + 1)}$ .

Now look at just three green squares in a row, and let their widths be $a$ , $b$ , and $c$ .

Then the two triangles are similar, which means $\frac{a - b}{a} = \frac{b - c}{b}$ , which can be rearranged to $\frac{b}{a} = \frac{c}{b}$ , which means $a$ , $b$ , and $c$ follow a geometric progression.

This means that for some ratio $r_l$ , the length of the rectangle is $w + 2 = s + r_l(s + r_l(s ...$ , or $w + 2 = s + r_l(w + 2)$ , which solves to $r_l = 1 - \frac{s}{w + 2}$ , and the sum of the areas of the light green squares along the length is $A_l = r_l^2s^2 + r_l^2(r_l^2s^2 + r_l^2(r_l^2s^2 ...$ or $A_l = r_l^2s^2 + r_l^2A_l$ , which solves to $A_l = \frac{r_l^2s^2}{1 - r_l^2}$ . For $s = \frac{w(w + 2)}{2(w + 1)}$ , $r_l$ solves to $r_l = \frac{w + 2}{2(w + 1)}$ and $A_l$ solves to $A_l = \frac{w(w + 2)^4}{4(w + 1)^2(3w + 4)}$ .

Likewise, for some ratio $r_w$ , the width of the rectangle is $w = s + r_w(s + r_w(s ...$ , or $w = s + r_ww$ , which solves to $r_w = 1 - \frac{s}{w}$ , and the sum of the areas of the light green squares along the width is $A_w = r_w^2s^2 + r_w^2(r_w^2s^2 + r_w^2(r_w^2s^2 ...$ or $A_w = r_w^2s^2 + r_w^2A_w$ , which solves to $A_w = \frac{r_w^2s^2}{1 - r_w^2}$ . For $s = \frac{w(w + 2)}{2(w + 1)}$ , $r_w$ solves to $r_w = \frac{w}{2(w + 1)}$ and $A_w$ solves to $A_w = \frac{w^4(w + 2)}{4(w + 1)^2(3w + 2)}$ .

The area of all the green squares is $A_G = s^2 + A_l + A_2 = \frac{w^2(w + 2)^2}{4(w + 1)^2} + \frac{w(w + 2)^4}{4(w + 1)^2(3w + 4)} + \frac{w^4(w + 2)}{4(w + 1)^2(3w + 2)} = \frac{w(w+2)(3w^2 + 6w + 4)(5w^2 + 10w + 4)}{4(w + 1)^2(3w + 2)(3w + 4)}$ .

If the area of the rectangle is $A_R = w(w + 2)$ , then by symmetry the area of the yellow crack is $A_R - 2A_G$ , and the fraction of the yellow crack to the rectangle is $\frac{A_R - 2A_G}{A_R} = 1 - \frac{2A_G}{A_R} = 1 - \frac{(3w^2 + 6w + 4)(5w^2 + 10w + 4)}{2(w + 1)^2(3w + 2)(3w + 4)} = \frac{w(w + 2)(3w^2 + 6w + 2)}{2(w + 1)^2(3w + 2)(3w + 4)}$ .

Setting $\frac{w(w + 2)(3w^2 + 6w + 2)}{2(w + 1)^2(3w + 2)(3w + 4)} = \frac{705}{4576}$ and simplifying gives $(w - 3)(w + 5)(519 w^2 + 1038 w + 376) = 0$ , which solves to $w = 3$ for $w > 0$ .

Therefore, the width is $w = 3$ , the length is $w + 2 = 5$ , and the perimeter is $3 + 5 + 3 + 5 = \boxed{16}$ .