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+qβ, where p and q are positive integers. Find p+q.
Solution:
Let r be the radius of the circle, let Ξ± be the central angle of each arc cut off by chords of length 22, and let Ξ² be the central angle of each arc cut off by chords of length 20. Then 4Ξ±+2Ξ²=2Ο. Thus Ξ± and Ξ²/2 are complementary angles, so that cosΞ±=sin(Ξ²/2).
By the Law of Cosines, 222=r2+r2β2r2cosΞ±=2r2(1βcosΞ±), so cosΞ±=1β2r2222β. Note that sin(Ξ²/2)=r10β.
Together these observations give the equation
1β2r2222β=r10β
which implies
r2β10rβ242=0
The solutions are r=5Β±267β. Only the positive sign gives a positive value of r. The requested sum is 5+267=272β.
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 AHβ₯CD. Apply the Pythagorean Theorem to β³AOH to get r2=AH2+OH2, and to β³ADH to get (rβOH)2+AH2=AD2. Given that OH=21βAB=10 and AD=22, conclude that r2=222β(rβ10)2+102 and r2β10rβ242=0, which is the same equation as in the previous solution.
