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? .

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

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)#