The cyclic octagon ABCDEFGH has sides a, a, a, a, b, b, b, b respectively. Find the radius of the circle that circumscribes ABCDEFGH in terms of a and b.

(In reply to A Start? by Steve Herman)

While I have a couple of ideas, I'm not really sure how to express them.

Draw the spokes of the octagon from its vertices to the centre. Then there will be 3 sets of pairs of triangles whose angles at the centre  add up to 90 degrees. The remaining two angles at the edge of the octagon, where a side of length a meets one of length b, are fixed for all {a,b} and are 135 degrees each.

 So the octagon can be simplified:

Construct a chord AB,  so that a line BC at a 135 degree angle to AB intersects the edge of the circle. The origin of the circle is O, and angle AOC is a right angle. Now AO,BO,CO, are all radii of length r, so that:

a^2=2r^2-2r^2cos(x), b^2=2r^2-2r^2cos(90-x) where x is the angle at O of the triangle whose base is a.

Now (2r^2-2r^2cos(x))/(2r^2-2r^2cos(90-x)) = (cos(x)-1)/(cos(90-x)-1) = a^2/b^2, so if any two of {a,b,x} are known then r can be determined.

Edited on July 17, 2013, 11:42 am

Posted by broll on 2013-07-17 11:23:46