AHSME 1999 Problem 21

A circle is circumscribed about a triangle with sides $20,21,$ and $29,$ thus dividing the interior of the circle into four regions. Let $A,B,$ and $C$ be the areas of the non-triangular regions, with $C$ be the largest. Then $\mathrm{(A) \ }A+B=C \mathrm{(B) \ }A+B+210=C \mathrm{(C) \ }A^2+B^2=C^2 \mathrm{(D) \ }20A+21B=29C \mathrm{(E) \ } \frac 1{A^2}+\frac 1{B^2}= \frac 1{C^2}$