2010 AMC 10A Problem 19

"Equiangular" means all six angles are equal. A hexagon's angles add to $720^\circ$, so each angle is $120^\circ$.

Step 1: cut the hexagon into pieces. Draw $AC$, $CE$, $EA$. The hexagon splits into the middle triangle $ACE$ and three corner triangles $ABC$, $CDE$, $EFA$. Each corner triangle has sides $1$ and $r$ with a $120^\circ$ angle between them, so all three are identical. That makes $AC = CE = EA$, so triangle $ACE$ is equilateral.

Step 2: area of one corner triangle. Take triangle $ABC$ with base $BC = r$. Extend $BC$ past $B$ and drop a line from $A$ straight down to it, meeting it at $P$. The angle at $B$ outside the triangle is $180^\circ - 120^\circ = 60^\circ$, so triangle $ABP$ is a $30$-$60$-$90$ triangle with longest side $AB = 1$. Its sides are
$$BP = \tfrac12, \qquad AP = \tfrac{\sqrt3}{2}.$$
The height of triangle $ABC$ over base $r$ is $AP$, so its area is $\tfrac12 \cdot r \cdot \tfrac{\sqrt3}{2} = \tfrac{\sqrt3}{4}\, r$.

Step 3: the length $AC$. In right triangle $APC$, the legs are $AP = \tfrac{\sqrt3}{2}$ and $PC = PB + BC = \tfrac12 + r$. So
$$AC^2 = \left(r + \tfrac12\right)^2 + \tfrac34 = r^2 + r + \tfrac14 + \tfrac34 = r^2 + r + 1.$$

Step 4: the areas. An equilateral triangle with side $s$ has area $\tfrac{\sqrt3}{4} s^2$. So
- triangle $ACE$: $\tfrac{\sqrt3}{4}(r^2 + r + 1)$;
- hexagon: triangle $ACE$ plus three corners, $\tfrac{\sqrt3}{4}(r^2 + r + 1 + 3r) = \tfrac{\sqrt3}{4}(r^2 + 4r + 1)$.

Step 5: use the 70%. The $\tfrac{\sqrt3}{4}$ cancels on both sides:
$$r^2 + r + 1 = \frac{7}{10}(r^2 + 4r + 1).$$
Multiply by $10$: $10r^2 + 10r + 10 = 7r^2 + 28r + 7$. Move everything to the left: $3r^2 - 18r + 3 = 0$. Divide by $3$:
$$r^2 - 6r + 1 = 0.$$

Step 6: add the possible values. The quadratic formula gives $r = 3 + 2\sqrt2$ and $r = 3 - 2\sqrt2$. Both are positive (the second is about $0.17$), so both are real side lengths. Their sum is $6$.

(Shortcut: for $r^2 - 6r + 1 = 0$, the two solutions always add to $6$, the number in front of $r$ with its sign flipped.)

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

Joining every other corner of a hexagon cuts it into a middle triangle and three corner triangles, and with all angles $120^\circ$ every piece is easy to measure using $30$-$60$-$90$ triangles. When a question asks for the "sum of all possible values," it usually leads to a quadratic, and the sum of its two solutions can be read straight from the equation.

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 $\boxed{\textbf{(E)}\ 6}$.