, 2021

Solution:

Assume $AC=1$. Note that from symmetry, it can be seen that all angles in $ACEG$ must be equal. Further, by similar logic all sides must be equal which means that $ACEG$ is a square. Additionally, as $AB=BC$, $ABC$ is an isosceles triangle, which means the octagon consists of a unit square with four isosceles triangles of area $1/2$ attached.

Now, if the side length of the octagon is $s$, and $\angle B=2\theta$, then we obtain that
$$\frac{1}{2} s^{2} \sin (2 \theta)=\frac{1}{2} \Longrightarrow 2 s^{2} \cos (\theta) \sin (\theta)=1$$
Further, since the length $AC$ is equal to $1$, this means that $s \sin (\theta)=\frac{1}{2}$. From this, we compute
$$2 s \cos (\theta)=\frac{2 s^{2} \sin (\theta) \cos (\theta)}{s \sin (\theta)}=\frac{1}{\frac{1}{2}}=2$$
So $\tan (\theta)=\frac{s \sin (\theta)}{s \cos (\theta)}=\frac{1}{2}$. From this, $\sin (\theta)=\frac{1}{\sqrt{5}}$ and $\cos (\theta)=\frac{2}{\sqrt{5}}$, which means $\sin (B)=\sin (2 \theta)=2 \cdot \frac{1}{\sqrt{5}} \cdot \frac{2}{\sqrt{5}}=\frac{4}{5}$.