Question 21A 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 $C$ being the largest. Then$A + B = C$$A + B + 210 = C$$A^2 + B^2 = C^2$$20A + 21B = 29C$$\dfrac{1}{A^2} + \dfrac{1}{B^2} = \dfrac{1}{C^2}$