15. Let $A=(0,0), B=(b, 2)$ and $A B C D E F$ be a convex equilateral hexagon such that $\angle F A B$ $=120^{\circ}$. Suppose $A B / / E D, B C / / F E, C D / / A F$, and the $y$-coordinates of $C, D, E, F$ are some permutation of $4,6,8,10$. If 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 $m+n$.
(2 marks)
設 $A=(0,0) 、 B=(b, 2)$, 而 $A B C D E F$ 是一個等邊的凸六邊形, 其中 $\angle F A B=$ $120^{\circ}$ 。假設 $A B / / E D 、 B C / / F E 、 C D / / A F$, 而且 $C 、 D 、 E 、 F$ 的 $y$ 坐標是 4、6、 8、10 的某個排列。已知六邊形的面積可寫成 $m \sqrt{n}$ 的形式, 其中 $m 、 n$ 是正整數, 而 $n$ 不可被任何質數的平方整除。求 $m+n$ 。
(2 分)
15. 51
15. Since $A B C D E F$ is convex, it must be either $C$ or $F$ which has $y$-coordinate 4 . Suppose $C$ has $y$-coordinate 4 . Then, since $A B=B C$, either $C$ must be either $(2 b, 4)$ or $(0,4)$, both of which are not allowed (as the former implies that $A, B$, $C$ are collinear while the latter implies that $A B C D E F$ is concave). Hence $F$ has $y$ coordinate 4. Furthermore, since $C D / / A F$, the $y$-coordinate of $D$ exceeds that of $C$ by 4 . It follows that the $y$-coordinates of $D$ and $C$ are 10
and 6 respectively, and thus the $y$-coordinate of $E$ is 8 .
Let $C=(c, 6), D=(d, 10), E=(e, 8)$ and $F=(f, 4)$. From the fact that $B C=C D$, we know that $b=d$, i.e. $B D$ is vertical. As $A B$ and $E D$ are equal and parallel, $A B D E$ is a parallelogram, and hence $A E$ is also vertical, so $e=0$. Furthermore, $\triangle A E F$ and $\triangle B C D$ are congruent by the SSS condition. Note also that $f<0$ by the convexity condition.
Now the area of $A B C D E F$ is equal to $2[A F E]+[A B D E]=2 \times \frac{1}{2} \times 8 \times(-f)+8 b=8(b-f)$, so we must find the values of $b$ and $f$. Let $x$ be the side length of the hexagon. Computing the lengths of $A B$ and $A F$ respectively, we have
$$
x^{2}=b^{2}+4=f^{2}+16 \text {. }
$$

On the other hand, since $\angle F A B=120^{\circ}$, applying cosine law in $\triangle F A B$ gives
$$
(b-f)^{2}+(2-4)^{2}=x^{2}+x^{2}-2(x)(x) \cos 120^{\circ}=3 x^{2} \text {. }
$$

Solving, we get $b=\frac{10}{\sqrt{3}}$ and $f=-\frac{8}{\sqrt{3}}$. Hence the area of $A B C D E F$ is $8(b-f)=48 \sqrt{3}$, and hence the answer is $48+3=51$.