Octagon inscribed in a rectangle.

Let the pink sides of the four right-angled triangles be $a>b$ and the side of the octagon (given) be $s$ , so that $2a+s=w$ and $2b+s=w-2$ . Also, by Pythagoras, $a^2+b^2=s^2$ .

Since we want to find $w$ , we'll eliminate $a$ and $b$ . For convenience below, write $x=w-s$ : $\begin{aligned} 2a &=x \\ 2b &= x-2 \\ 4a^2+4b^2 &=x^2+(x-2)^2=4s^2 \end{aligned}$

Tidying up and solving, $x=1 \pm \sqrt{2s^2-1}$ . Clearly we can ignore the negative root; so we find $w=s+1+\sqrt{2s^2-1}=\boxed6$

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The simplest way to find $s+1+\sqrt{2s^2-1}$ is by considering the quadratic that $s$ solves; we have $(s+5)^2=51$ , which can be rewritten as $s^2=26-10s$ .

Now, $2s^2-1=51-20s=(s+5)^2-20s=(s-5)^2$ , and, noting that $s<5$ , the rest is easy.

Let the side lengths of the legs of the right triangle cut off be $a$ and $b$ as seen. Then,

$\begin{cases} w = 2a + \sqrt{51} - 5 & ...(1) \\ w-2 = 2b + \sqrt{51} - 5 & ...(2) \end{cases} \implies (1) - (2): \quad 2a - 2b = 2 \implies b = a-1$

Note that:

$\begin{aligned} a^2 + b^2 = (\sqrt{51}-5)^2 \\ 2a^2 - 2a + 1 & = 76 - 10\sqrt{51} \\ 2a^2 - 2a - 75+10\sqrt{51} & = 0 \end{aligned}$

The roots to the quadratic equation are $a$ and $b$ . By Vieta's formula, $a+b = 1$ . Let $a = \frac 12 + u$ and $b=\frac 12 - u$ , then

$\begin{aligned} \left(\frac 12 + u \right)\left(\frac 12 - u \right) & = \frac {10\sqrt{51}-75}2 \\ \left(u + \frac 12\right)\left(u-\frac 12 \right) & = \frac {75-10\sqrt{51}}2 \\ u^2 - \frac 14 & = \frac {75-10\sqrt{51}}2 \\ u^2 & = \frac {151-20\sqrt{51}}4 \\ \implies u & = \frac {10 - \sqrt{51}}2 \\ \implies a & = u + \frac 12 = \frac {11-\sqrt{51}}2 \\ \implies w & = 2a + \sqrt{51} - 5 = \boxed 6 \end{aligned}$

$w - 2x = w - 2 - 2y \implies x - y = 1 \implies y = x - 1$

and

$x^2 + y^2 = (w - 2x)^2 \implies x^2 + (x - 1)^2 = w^2 - 4wx + 4x^2 \implies$

$2x^2 - 2(2w - 1)x + w^2 - 1 = 0 \implies x = \dfrac{2w - 1 \pm \sqrt{2w^2 - 4w + 3}}{2}$

Using $x = \dfrac{2w - 1 + \sqrt{2w^2 - 4w + 3}}{2} \implies w - 2x < 0:$

$w - 2x = 1 - w - \sqrt{2w^2 - 4w + 3}$ and $(1 - w)^2 - (2w^2 - 4w + 3) =$

$-(w^2 - 2w + 2) < 0$ for all real $w$

$\implies |1 - w| < \sqrt{2w^2 - 4w + 3} \therefore$ we drop $x = \dfrac{2w - 1 + \sqrt{2w^2 - 4w + 3}}{2}$

and choose $x = \dfrac{2w - 1 - \sqrt{2w^2 - 4w + 3}}{2} \implies$

$w - 2x = 1 - w + \sqrt{2w^2 - 4w + 3} = \sqrt{51} - 5 \implies$

$2w^2 - 4w + 3 = ((\sqrt{51} - 6) + w)^2 = 87 - 12\sqrt{51} + 2(\sqrt{51} - 6)w + w^2 \implies$

$w^2 + (8 - 2\sqrt{51})w + 12\sqrt{51} - 84 = 0 \implies$

$w = \dfrac{-8 + 2\sqrt{51} \pm 2\sqrt{151 - 20\sqrt{51}}}{2} =$ $\dfrac{-8 + 2\sqrt{51} \pm 2\sqrt{(10 - \sqrt{51})^2}}{2} =$

$\dfrac{-8 + 2\sqrt{51} \pm (10 - \sqrt{51})}{2} = 6, -14 + 2\sqrt{51}$

$w = 2\sqrt{51} - 14 \implies w - 2 < 0 \therefore$ choose $\boxed{w = 6}$

$\begin{aligned} \bigg(\frac{w-(\sqrt{51}-5)}{2}\bigg)^2 + \bigg(\frac{(w-2) - (\sqrt{51} - 5)}{2}\bigg)^2 &= (\sqrt{51}-5)^2 \\ \implies w &= \boxed{6} \end{aligned}$