Problem:
A hexagon that is inscribed in a circle has side lengths 22,22,20,22,22, and 20 in that order. The radius of the circle can be written as p+q, where p and q are positive integers. Find p+q.
Solution:
Let r be the radius of the circle, let α be the central angle of each arc cut off by chords of length 22, and let β be the central angle of each arc cut off by chords of length 20. Then 4α+2β=2π. Thus α and β/2 are complementary angles, so that cosα=sin(β/2).
By the Law of Cosines, 222=r2+r2−2r2cosα=2r2(1−cosα), so cosα=1−2r2222. Note that sin(β/2)=r10.
Together these observations give the equation
1−2r2222=r10
which implies
r2−10r−242=0
The solutions are r=5±267. Only the positive sign gives a positive value of r. The requested sum is 5+267=272.
OR
Note that any diagonal of the hexagon that connects opposite vertices is a diameter of the circle. Let ABCD be the isosceles trapezoid where AD=BC=22,AB=20, and CD is a diameter of the circle. Let the center of the circle be O, and let H be a point on CD such that AH⊥CD. Apply the Pythagorean Theorem to △AOH to get r2=AH2+OH2, and to △ADH to get (r−OH)2+AH2=AD2. Given that OH=21AB=10 and AD=22, conclude that r2=222−(r−10)2+102 and r2−10r−242=0, which is the same equation as in the previous solution.
