# Find the center of the octagon, complex plane

The problem given read as follows,

Two consecutive vertices of a regular convex octagon are $(1,2)$ and $(3,-2)$. Find the center of the octagon.

I would like to know if my answer is correct. Here's what I have thought:

I've drawn the complex plane and set the affixes given accordingly (I've called them $z_{1} = (1,2)$ and $z_{2}=(3,-2)$).

Since $z_{1}, z_{2}, ... z_{8}$ compose a regular octagon, the angle between $z_{1}$ and $z_{2}$ must be $\frac{\pi}{4}$. So I have an angular relation at my disposal.

Say $z_{0} = x+iy$ is the center of the polygon, then I have two vectors, namely $\vec{v} = \vec{z_{0} z_{2}}$ and $\vec{u} = \vec{z_{0} z_{1}}$ and since I have an angle relation between them, I can write:

$\vec{u} = \vec{v}.e^\frac{i \pi}{4} \iff z_{1}-z_{0} = (z_{2} - z_{0}).cis \left(\frac{\pi}{4} \right) (*)$

Substituting with my known affixes, I get:

$1+2i-z_{0}=(3-2i-z_{0}).e^\frac{i \pi}{4}$

Working out some algebra, I've got:

$z_{0} = \frac{3e^{\frac{i \pi}{4}} - 1}{e^\frac{i \pi}{4} - 1} + i\frac{-2-2 e^\frac{i \pi}{4}}{e^\frac{i \pi}{4} - 1}$

Converting from polar form to trigonometric form and doing a lengthy algebra, I've got:

$\boxed{z_{0} = 2 + i \frac{\sqrt{2}}{2-\sqrt{2}}}$

I believe that my reasoning is correct, although substituting my result at $(*)$ I get:

$\left(1 - \sqrt{2}, \frac{4-3\sqrt{2}}{2-\sqrt{2}}\right) \neq \left(\frac{3 \sqrt{2} - 2}{2}, \frac{1-2 \sqrt{2}}{2-\sqrt{2}}\right)$

I would appreciate if someone could help me figure out my mistake.

• The reasoning is correct, except you'll need to consider $\operatorname{cis}\left(\color{red}{\pm}\frac{\pi}{4}\right)$ which will give two solutions in the end, symmetrical with respect to the segment $z_1z_2$.
– dxiv
Commented Sep 10, 2017 at 20:34
• No. The octagons having consecutive vertices at $(1,2)$ and $(3,-2)$ are two and their centres are $$\left(-2 \sqrt{2},-1-\sqrt{2}\right);\;\left(4+2 \sqrt{2},1+\sqrt{2}\right)$$ Commented Sep 10, 2017 at 20:50

$z_0=x+i y;z_1=1+2 i;z_2=3-2 i$

$z_1-z_0=e^{\frac{i \pi }{4}} (z_2-z_0)$

$-e^{\frac{i \pi }{4}} (-x-i y+(3-2 i))-x-i y+(1+2 i)=0$

$\left\{ \begin{array}{c} \frac{x}{\sqrt{2}}-x-\sqrt{2}-\frac{y}{\sqrt{2}}-\frac{3}{\sqrt{2}}+1=0 \\ \frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}-y+\sqrt{2}-\frac{3}{\sqrt{2}}+2=0 \\ \end{array} \right.$

$x=-2 \sqrt{2},y=-1-\sqrt{2}$

but there is also another possibility, as you can see in the picture

$z_2-z_0=e^{\frac{i \pi }{4}} (z_1-z_0)$

which leads to the other solution

$x=2 \left(2+\sqrt{2}\right),y=1+\sqrt{2}$

Hope this is useful

Given two points on a plane, we seek to create an octagon and determine its centroid. The vertices of the octagon can be found sequentially by creating a chain linkage. Here we utilize the fact that each line segment has the same length and angular offset. Thus, each new segment turns by $\pi/4$ from the current one to create an internal angle of $3\pi/4$ for the octagon. Thus, given $z_1$ and $z_2$ we have

$$s=|z_{1}-z_{2}|\\ z_k=z_{k-1}+se^{i(\arg(z_{k-1}-z_{k-2})+\pi/4)}\quad k=3...8$$

It's that simple. The centroid is then the average or mean value of $z$, i.e.,

$$z_c=\langle z\rangle$$

In the present case we find that

$$z_c=2(2+\sqrt{2})+i(1+\sqrt{2})$$