2010 AMC 10A Problem 19

Every interior angle of an equiangular hexagon is $120^\circ$. Drawing $AC$, $CE$, $EA$ splits the hexagon into $\triangle ACE$ and three corner triangles $ABC$, $CDE$, $EFA$, each having sides $1$ and $r$ with a $120^\circ$ angle between them. The corner triangles are congruent, so the picture has $120^\circ$ rotational symmetry and $\triangle ACE$ is equilateral.

Each corner triangle has area $\tfrac12 \cdot 1 \cdot r \cdot \sin 120^\circ = \dfrac{\sqrt3}{4} r$. By the Law of Cosines,
$$AC^2 = 1 + r^2 - 2r\cos 120^\circ = r^2 + r + 1,$$
so the equilateral $\triangle ACE$ has area $\dfrac{\sqrt3}{4}(r^2 + r + 1)$.

The hexagon's area is the sum: $\dfrac{\sqrt3}{4}(r^2 + r + 1 + 3r) = \dfrac{\sqrt3}{4}(r^2 + 4r + 1)$. The condition says
$$r^2 + r + 1 = \frac{7}{10}(r^2 + 4r + 1) \;\Longrightarrow\; 10r^2 + 10r + 10 = 7r^2 + 28r + 7 \;\Longrightarrow\; 3r^2 - 18r + 3 = 0,$$
that is, $r^2 - 6r + 1 = 0$. Its discriminant $36 - 4 = 32$ is positive and both roots are positive (sum $6$, product $1$), so both are valid, and by Vieta their sum is $6$.

The answer is $\boxed{\textbf{(E)}\ 6}$.

Diagonals from alternate vertices cut a hexagon into a central triangle plus three corner triangles; with all angles $120^\circ$ every piece has a known area formula. "Sum of all possible values" is the cue to stop at the quadratic and read off Vieta rather than solving. Note the two roots are reciprocals, matching the symmetry of swapping the roles of $1$ and $r$ after scaling.

• Solving the quadratic and reporting one root, or dropping the r term in AC^2 = r^2 + r + 1 by using cos 120 with the wrong sign.
• Setting $\triangle ACE$ equal to $70\%$ of the three corner triangles instead of $70\%$ of the whole hexagon.