AMC10 2010 A

Answer (E): Triangles $ABC$, $CDE$ and $EFA$ are congruent, so $\triangle ACE$ is equilateral. Let $X$ be the intersection of the lines $AB$ and $EF$ and define $Y$ and $Z$ similarly as shown in the figure. Because $ABCDEF$ is equiangular, it follows that $\angle XAF=\angle AFX=60^\circ$. Thus $\triangle XAF$ is equilateral. Let $H$ be the midpoint of $XF$. By the Pythagorean Theorem, $AE^2=AH^2+HE^2=\left(\frac{\sqrt{3}}{2}r\right)^2+\left(\frac{r}{2}+1\right)^2=r^2+r+1$ Thus, the area of $\triangle ACE$ is $\frac{\sqrt{3}}{4}AE^2=\frac{\sqrt{3}}{4}(r^2+r+1).$ The area of hexagon $ABCDEF$ is equal to $[XYZ]-[XAF]-[YCB]-[ZED]=\frac{\sqrt{3}}{4}\left((2r+1)^2-3r^2\right)=\frac{\sqrt{3}}{4}(r^2+4r+1)$ Because $[ACE]=\frac{7}{10}[ABCDEF]$, it follows that $r^2+r+1=\frac{7}{10}(r^2+4r+1)$ from which $r^2-6r+1=0$ and $r=3\pm2\sqrt{2}$. The sum of all possible values of $r$ is $6$.