Square to octagon dissection - how to cut the square?

Question

How to cut the square which tessellates to octagon using straightedge and compass? What are the exact measures of colored sides? What is the angle marked with red color?

Edit (I added vertices):

Edit. Acknowledgement after answers.

Thanks to Jean Marie, Blue and Daniel Mathias for tackling the problem. I accepted Blue's answer for its breakthrough, but I also weighed Daniel Mathias' solution for its geometric simplicity. Initially, I didn't expect the problem to be so laborious. I invite you to my puzzle in a similar flavor of dissections, which after many years still has no answer nor proof that the solution does not exist. Variation of Haberdasher problem of Henry Dudeney - dissection of equilateral triangle into square with flipping pieces

Answer 1

$\newcommand{\cis}{\operatorname{cis}}$

To reduce fraction-clutter, I'll define $\pi_n := \pi/n$; also, I'll abuse notation to define $\cis\theta:=(\cos\theta,\sin\theta)$.

Importantly, in the octagon figure, the central square is rotated everso-slightly —just-under $2^\circ$— from being "axis-aligned". This complicates things a bit. (Or not: Skip to the Note for an uncomplicated discussion.)

Suppose the octagon has circumradius $r$, so that its sides have length $s:=2r\sin\pi_8= \sqrt{2-\sqrt{2}}$. Define points $P_k := r\,\cis k\pi_8$, for $k=0, 1, 2, \ldots, 7$; assigning OP's labels, we have $C:=P_0$ and $D:=P_1$; also, rotational symmetry in the square arrangement guarantees $|DE|=|BC|$, so that $B$ and $E$ are necessarily midpoints of segments $P_7P_0$ and $P_1P_2$, respectively.

The square arrangement also indicates that $|AF|=|CD|$, so that $|OF|=s/\sqrt{2}$. We can thus write $$F = \frac{s}{\sqrt2}\cis\theta \qquad A = \frac{s}{\sqrt2}\cis(\theta-\pi_2)$$ for some $\theta$ that makes $A$, $E$, $F$ collinear. With a calculational assist from Mathematica, I find that $\theta\approx46.9698^\circ$ satisfies:

$$\cos\theta = \frac12 \left(\sqrt2-1\right) \left(\sqrt{2 (4 + 3 \sqrt2)} - \sqrt{2-\sqrt2}\right)$$ (Refining this value is left as an exercise to the reader.) OP's "red angle" is given by $$\angle DEF = \frac12\cdot 135^\circ - (\theta-45^\circ) \approx 65.5302^\circ$$ Finally, we can write $$\frac{|AF|}{|AE|} = \frac{8 \sin\pi_8}{2 + \sqrt2 + 4\sqrt2 \sin\pi_8 \cos\theta} \approx 0.62588$$

Since all the expressions here involve sums, differences, products, and ratios of nested square roots, the dissection is straightedge-and-compass constructible. See Notes for constructions.

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Note. Daniel Mathias helpfully demonstrates that, starting from the octagon itself, the construction of pentagon $ABCDE$ is dead-simple:

Let diagonals $P_1P_4$ and $P_2P_7$ meet at $I$, and let $\bigcirc O$ pass through $I$. Let the midpoint of $B$ and $E$ be $Q$, and let $\bigcirc Q$ pass through those points (and $O$). Then $A$ is the "lower" point where these circles meet (thanks to Thales' Theorem), and $F$ is the "other" point where line $AE$ meets $\bigcirc O$. (Verification is straightforward.)

Given that, we don't even need to know any lengths or angle-measures. Nevertheless, I'll give an easier derivation of the final ratio from my previous pass at this problem.

Let the sides of the octagon have lengths $2p$. Then we have $|AF|=|BC|=2p$ and $|BE|=2p+p\sqrt{2}=|AF|(1+\sqrt{2}/2)$. Recalling that $|EF|=|AB|$, we can invoke Pythagoras on $\triangle ABE$ thusly: $$ \begin{align} |AE|^2+(|AE|-|AF|)^2 = |BE|^2 &\quad\to\quad |AE| = |AF|\cdot\frac12 \left(1 + \sqrt{2 (1 + \sqrt2)}\right) \\[4pt] &\quad\to\quad \frac{|AF|}{|AE|}\approx 0.6255 \end{align}$$

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Note 2. For a carpentry-friendly construction that starts with the square ...

• Locate $E$ along the side $ZA$ (and corresponding points $B$, $E'$, $B'$ along the other sides), such that (leveraging previous calculations) $$\frac{|ZE|}{|ZA|}=\frac{|AE|-|AF|}{|AE|+(|AE|-|AF|)}\approx 0.2725$$ Double-check your measurements by verifying that lines $BB'$ and $EE'$ pass through the center of the square.
• Reflect $Z$ in $E$ to get $F$
• Locate $M$ as the midpoint of segment $FA$. Define $p := |MA|$.
• Locate $C$ and $D$ along at distance $p$ from $B$ and $E$ along segments $BB'$ and $EE'$; likewise $C'$ and $D'$.

Done!

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Note 3. Here's a straightedge-and-compass construction of the point $B$ in the above figure.

Let $O$ be square's center, $M$ the midpoint of its right side, and $P$ its upper-right vertex.

For follow-along calculations, take $|MP|=1$ to be half the square's side-length. Fiddling with the radical ratios above, we find that we want to construct $B$ such that $$|MB|=\frac{1}{\sqrt{2(1+\sqrt{2})}}$$ (This comports with @DanielMathias' comment to the question.)

• Transfer $P$ to $P'$ on the horizontal midline via rotation about $M$. $$|MP'| = 1$$

• Transfer $P$ to $P''$ on the midline via rotation about $P'$. $$|MP''| = 1+\sqrt{2}$$

• Transfer $M$ to $M'$ on the midline via rotation about $P''$. $$|MM'| = 2(1+\sqrt{2})$$

• Transfer $M'$ to $M''$ on the side-line $MP$ via the semicircle with diameter $OM'$.
  $$|MM''| = \sqrt{|MM'|} = \sqrt{2(1+\sqrt{2})}$$

• Draw line $M''O$; along the perpendicular to this line at $O$, transfer $O$ to $B$ on the side-line. $$|MB| = \frac{1}{|MM''|} = \frac{1}{\sqrt{2(1+\sqrt{2})}}$$

Done!

Answer 2

I cannot beat the nice presentation made by @Blue.

It is why I have erased a previous version that did not bring much.

What I would like to do here is to place this octagon dissection in the "big picture", i.e. in the way the left and the right figure can be found and used in a certain very nice plane tessellation.

Have a look at the figure from this site :

Fig. 1.

or this one, found here :

Fig. 2.

that I have programmed :

Fig. 3.

If you are interested, see at the bottom of this answer this Matlab program based on the inclination angle $\alpha=\approx 20.53°$ (depicted in red) of the "green grid" over the "reference grid" (with the blue octagons and the red squares).

This is precisely this angle $\alpha$ that I would like to analyse.

Let us make a zoom on the bottom left part of fig. 3 (Geogebra figure this time) :

Fig. 4.

Let us work on this Fig.4 . I don't give details for evident choices of coordinates. I have taken the names $ABCDEF$ given by the author of the question to the vertices irregular polygon.

Let $\theta=\frac{\pi}{8}$.

As the sides of the octagon have a length equal to $2$, we have $OD=OH=\frac{1}{sin\theta}$. Therefore, here are the coordinates of some points :

$$\begin{cases}D&=&(OD \cos\theta, OD \sin\theta) &=& (\cot\theta,1)&=&(1+\sqrt{2},1)\\H&=&(1,1+\sqrt{2})&&&&\\ E&=&\frac12(D+H)&=&(1+\frac12 \sqrt{2}) \times &&(1,1)\end{cases}\tag{1}$$

We can now express that points $A,F,E$ have to be aligned. This condition is equivalent to the condition that rotation with angle $\alpha$ transforms the line defined by $A,F,E$ into a vertical line, i.e. that the image of $E$ by this rotation is a point with abscissa $1$.

Working with complex numbers, we can express this condition by :

$$\Re \left(e^{i \alpha}(1+\frac12 \sqrt{2})\underbrace{(1+i)}_{\sqrt{2}e^{i \pi/4}}\right)=1$$

Otherwise said,

$$\Re \left(e^{i (\alpha+\pi/4)}(\sqrt{2}+1)\right)=1$$

$$\cos(\alpha+\pi/4)=\frac{1}{1+\sqrt{2}}=\sqrt{2}-1$$

giving finally:

$$\alpha=\operatorname{acos}(\sqrt{2}-1)-\pi/4 \approx 0.358319 rd = 20.5302°$$

Remark : Of course, knowing all the coordinates of the points, it is easy to retrieve the lengths you are looking for.

Matlab program for the generation of fig. 3:

ns=2; % figure size 
r=1/sin(pi/8); % radius of circumcircle to octagon
a=acos(sqrt(2)-1)-pi/4, % rot. angle 0.3583
b=2*r/sqrt(sqrt(2))-1;
sh=r*cos(pi/8)+1; % shift
oct=r*exp(i*pi/8)*exp(i*(0:8)*pi/4); % octagon
sq=[1-i,1+i,-1+i,-1-i,1-i]; % square
plot(sq,'-k');
for p=0:ns
   for q=0:ns
      plot(2*sh*(p+i*q)+oct,'-b');
      plot(2*sh*(p+i*q+1/2)+sq,'-r');
      plot(2*sh*(p+i*q+i/2)+sq,'-r');
   end;
end;
% green square + legs :
for p=0:2*ns
   for q=0:2*ns
      if mod(p+q,2)==0
         for k=0:3
            plot(sh*(p+i*q)+exp(-i*a)*i^k*[1-i,1+b*i],'-g');
         end;
      end;
   end;
end;

Answer 3

Construction of point $B$, derived from my dissection of the octagon, which has been incorporated into Blue's answer and is shown here for reference. Critically, this creates the required angle $\angle OBP_1$ equal to $\angle DEF$.

$O$ is the center of the square.
$P_1$ is the midpoint of a side.
Circle with center $A$ and passing through $P_1$ intersects $OA$ at $P_2$.
Circle with center $O$ and passing through $P_2$ intersects $OP_1$ at $P_3$.
Circle with center $P_1$ passes through $P_3$.
Circle with center $P_4$ (midpoint of $OA$) passes through $O$, $A$, and $P_1$.
These two circles, respective of those in the octagon dissection, intersect outside the square at $P_5$.
Line through $O$ and $P_5$ intersects the square at points $B$ and $B'$.
Circle with center $O$ and passing through $B$ intersects the square at points $E$ and $E'$.
Circle with center $B$ and passing through $P_1$ intersects $BB'$ at $C$.
Circle with center $O$ and passing through $C$ intersects $BB'$ at $C$ and $EE'$ at $D$ and $D'$.
Points $F\;(EF=EZ)$ and $M\;(FM=MA)$ are marked. Though not a necessary part of the construction, $EM$ is congruent to $AP_1$ with $EF=AB$ and $FM=BP_1$ confirming the construction of point $C$.

Complete dissection of the square: