Question

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? .

Answer 1

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

`(0+sqrt(2)+sqrt(2)+0)/4` and `(1+1+(-1-sqrt(2))+(-1-sqrt(2)))/4`

So the center point is `((sqrt(2))/2,-sqrt(2)/2)`

Translate the center point to zero (change all x values by `-(sqrt(2))/2` and increase all y values by `+sqrt(2)/2`)

Then, rotate each point by 90 degrees. (`x = y and y = -x`)

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

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

Answer 2

`C(1+sqrt2, 0), D(1+sqrt2, -sqrt2), G(-1, -sqrt2) and H(-1, 0)`

Explanation:

As AB is in y-direction, side of the octagon L = `y_B-y_A=sqrt2`

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)`xx(1/2+cos45^o)`

`=sqrt2(1/2+1/sqrt2)=1+1/sqrt2`

`d_y=1/2L=1/sqrt2`

From symmetry, the remaining vertices are

`C(x_M+d_x, y_M+d_y),` giving `C(1+sqrt2, 0)`

`D(x_M+d_x, y_M-d_y),` giving `D(1+sqrt2, -sqrt2)`

`G(x_M-d_x, y_M+-d_y),` giving G(-1, -sqrt2)#

`H(x_M+d_x, y_M+d_y),` giving H(-1, 0)#