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^22r^2cos(x), b^2=2r^22r^2cos(90x) where x is the angle at O of the triangle whose base is a.
Now (2r^22r^2cos(x))/(2r^22r^2cos(90x)) = (cos(x)1)/(cos(90x)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 20130717 11:23:46 