We have a regular octagon $ABCDEFGH$ with sides of equal length $\sqrt{2-\sqrt{2}}$, taking vertices $A, C$, and $E$ as centers, arcs of radii $AG, CA$, and $EG$ are drawn respectively, which intersect at points $P$ and $Q$ ($P$ and $Q$ being interior to the polygon). Calculate $PQ$.(Answer:($\sqrt3-1)$)
I made the drawing but I need to demonstrate that the $\angle PEO = 30^o$ or that the $\angle GPE = 150^o$. So I can finish because as the remarkable triangle we discover the radius $1$.
$\triangle PKO_{isosceles-rectangle)} \implies PK = OK$
$PQ = \frac{PK}{2}$
$ \triangle EPK: tg 30^o = \frac{PK}{1-PK}$ and it will be resolved
Observe that the triangles $\Delta ACP$ and $\Delta AGQ$ are equilateral by construction, all sides are congruent.
The constructions of $P,Q$ are symmetric / reflected (step by step) w.r.t. the line $AE$, which is a symmetry line of the given octogon. So $PQ\|GC$. (Since $G,C$ correspond by the same reflection.) The angle in $P$ between $PQ$ and $PC$ is thus $$ \widehat{QPC} = \widehat{PCG} = \widehat{PCA} - \widehat{GCA} = 60^\circ-45^\circ =15^\circ\ . $$ We also know $ \widehat{ECA} =90^\circ$ in the octogon. (Or in the square $ECAG$.) Changing the view, looking to the circle centered in $C$ with radius $CA=CE$ we have thus the arc $\overset\frown{EA} = 90^\circ$. The angle $ \widehat{EPA}$ is thus in measure $\frac 12\cdot 270^\circ=135^\circ$. This shows that $$ \widehat{EPQ} = 135^\circ-15^\circ-60^\circ=60^\circ\ . $$ By the reflected argument, or by reflection, $\widehat{EQP} = 60^\circ$.
So $\Delta EPQ$ is equilateral, its angle in $E$ is $60^\circ$, and after drawing its bisector $EOA$ we obtain $\widehat{PEO}=\frac 12\cdot 60^\circ=\bbox[yellow]{\ 30^\circ\ }$.
We have one objection, the above picture is the same as in the OP picture, but not exactly what the problem, taken mot-a-mot, wants:
To switch from one picture to the other one use a rotation of $90^\circ$ around the symmetry center of the octogon. (So instead of $\Delta EPQ$ from the solution and the first picture, we have now an equilateral triangle $\Delta GPQ$ in the second picture. And all the positions of the points $P,Q$ that can be obtained by similar constructions using the centers $A,C,E,G$ for the $90^\circ$ arcs are building a square.)