2010 AMC 12A Problem 17

Note that $\triangle ACE$ is equilateral. Using the Law of Cosines in $\triangle ABC,$ we get $AC^2=r^2+1^2-2r\cos120^\circ$ $=r^2+r+1.$ The area of $\triangle ACE$ is then $$\dfrac{\sqrt3}{4} (r^2 + r + 1).$$ The three corner triangles $\triangle ABC,$ $\triangle CDE,$ and $\triangle EFA$ each have area $\frac12\cdot1\cdot r\cdot\sin120^\circ=\frac{r\sqrt3}{4}.$ Thus the hexagon has area $\dfrac{\sqrt3}{4}(r^2+r+1)$ $+3\cdot\dfrac{r\sqrt3}{4}$ $=\dfrac{\sqrt3}{4}(r^2+4r+1).$ The condition $[ACE]=70\%\cdot[ABCDEF]$ gives $$r^2+r+1=\dfrac{7}{10}(r^2+4r+1),$$ so $r^2-6r+1=0.$ By Vieta’s formulas, the sum of the possible values of $r$ is $6.$ Thus, E is the correct answer.