# ABCDEFGH is a regular convex octagon, with A( 0, 1 ), B( sqrt2, 1 ), E( sqrt 2, -1-sqrt 2 ) and F( 0, -1-sqrt2 ). How do you find the coordinates of the remaining vertices? .

Feb 8, 2017

Not so sure, but here is what I think one would do if they had solve this problem (or a similar one with different points).

Average all x values and y values

$\frac{0 + \sqrt{2} + \sqrt{2} + 0}{4}$ and $\frac{1 + 1 + \left(- 1 - \sqrt{2}\right) + \left(- 1 - \sqrt{2}\right)}{4}$

So the center point is $\left(\frac{\sqrt{2}}{2} , - \frac{\sqrt{2}}{2}\right)$

Translate the center point to zero (change all x values by $- \frac{\sqrt{2}}{2}$ and increase all y values by $+ \frac{\sqrt{2}}{2}$)

Then, rotate each point by 90 degrees. ($x = y \mathmr{and} y = - x$)

Finally, translate each of these new points back to their original positions. (increase all x values by $+ \frac{\sqrt{2}}{2}$ and change all y values by $- \frac{\sqrt{2}}{2}$)

Depending on how the octagon was set up,
A' = G, B'=H, F'=D, E'=C

Feb 9, 2017

$C \left(1 + \sqrt{2} , 0\right) , D \left(1 + \sqrt{2} , - \sqrt{2}\right) , G \left(- 1 , - \sqrt{2}\right) \mathmr{and} H \left(- 1 , 0\right)$

#### Explanation:

As AB is in y-direction, side of the octagon L = ${y}_{B} - {y}_{A} = \sqrt{2}$

From the averages of the coordinates of A, B, E and F, the center M

has coordinates

(x_M, y_M)=(1/sqrt2, -1/sqrt2).

C, D, G and H are equidistant from M. in the directions of the x and y

axes. These have the common distances

d_x = (side of octagon)$\times \left(\frac{1}{2} + \cos {45}^{o}\right)$

$= \sqrt{2} \left(\frac{1}{2} + \frac{1}{\sqrt{2}}\right) = 1 + \frac{1}{\sqrt{2}}$

${d}_{y} = \frac{1}{2} L = \frac{1}{\sqrt{2}}$

From symmetry, the remaining vertices are

$C \left({x}_{M} + {d}_{x} , {y}_{M} + {d}_{y}\right) ,$ giving $C \left(1 + \sqrt{2} , 0\right)$

$D \left({x}_{M} + {d}_{x} , {y}_{M} - {d}_{y}\right) ,$ giving $D \left(1 + \sqrt{2} , - \sqrt{2}\right)$

$G \left({x}_{M} - {d}_{x} , {y}_{M} \pm {d}_{y}\right) ,$ giving G(-1, -sqrt2)

$H \left({x}_{M} + {d}_{x} , {y}_{M} + {d}_{y}\right) ,$ giving H(-1, 0)#