Three Tangent Circles

Solution 1: The area of the largest circle is $100 \pi$ , so the combined areas of the two smaller circles is $68$ % of $100 \pi$ , which is $68 \pi$ . Let the radii of the two smaller circles be denoted by $r$ and $R$ . Then we have $\pi r^2 + \pi R^2 = 68 \pi$ . Since all three circles are tangent to one another and the two smaller circles are inside the largest circle, we get $2r + 2R = 2 \cdot 10 = 20$ , or that $r + R = 10$ .

Squaring this equation, we get $r^2 + 2rR + R^2 = 100$ . Hence $2rR = 100 - 68 = 32$ , or $rR = 16$ .

Solution 2: Start as in the first paragraph of the previous solution.

Substituting $10 - r$ for $R$ gives us $\pi r^2 + \pi (10 - r)^2 = 68 \pi$ . Simplifying this equation gives us $2r^2 - 20r + 100 = 68$ , which further simplifies to $r^2 - 10r + 16 = 0$ . Factoring this equation gives us $(r-8)(r-2) = 0$ , so the radii of the two smaller circles must be $8$ and $2$ , and the product must be 16.