The problem states:
Let $A = (0,0)$ and $B = (b,2)$ be points on the coordinate plane. Let $ABCDEF$ be a convex equilateral hexagon such that $\angle FAB = 120^\circ, \overline{AB}\parallel \overline{DE}, \overline{BC}\parallel \overline{EF,} \overline{CD}\parallel \overline{FA}$, and the y-coordinates of its vertices are distinct elements of the set $\{0,2,4,6,8,10\}$. The area of the hexagon can be written in the form m\sqrt {n}, where $m$ and $n$ are positive integers and n is not divisible by the square of any prime. Find $m + n.$
Clearly the y-coordinate of $F$ is $4$. With complex numbers it is easy to see that $F =\left(-\frac{8}{\sqrt{3}}, 4\right)$ $B= \left(\dfrac{10\sqrt{3}}{3},2 \right)$.
Thus the length of the sides of the hexagon is $4 \sqrt{7/3}$, then $FB = 4\sqrt{7}$. We must also have the height from A to FB is simply $2\sqrt{7/3}$. Clearly DEC and FAB are congruent and EFBC is a parallelogram therefore the final area is just:
$2\sqrt{7/3}(4\sqrt{7}) + 4\sqrt{7} \cdot 4\sqrt{7/3} =56\sqrt{3}$.
But the correct answer is 51, what is wrong with this?
Thanks
Best Answer
Area ABF = Area DEC, and for each of these triangles, the area = $\frac{1}{2}x^2sin 120^{\circ}$, or $\frac{1}{2}(4\sqrt{\frac{7}{3}})^2 \times \frac{\sqrt{3}}{2}$. The area of both of these triangles, then, is $56\frac{\sqrt{3}}{3}$.
Remaining is the parallelogram BCEF. The base is the length of FB, or $4\sqrt{7}$. The height is the perpendicular distance between the lines, given by $\frac{\vert{b_2 - b_1}\vert}{\sqrt{m^2 + 1}}$, or $\frac{22\sqrt{3}}{3\sqrt{7}}$. The parallelogram's area, then, is $88\frac{\sqrt{3}}{3}$. Adding it all together yields an area of $48\sqrt{3}$.
Note, I think it's much simpler to compute the area of ABDE and the other two triangles. The x-coordinate of B is $\sqrt{x^2 - 4}$, or $10\frac{\sqrt{3}}{3}$. Then the area of ABDE is 8 $\times$ $10\frac{\sqrt{3}}{3}$. The total area of the other two triangles BCD and EFA is 8 $ \times$ $8\frac{\sqrt{3}}{3}$, yielding a total area of $48\sqrt{3}$.