15. An equiangular hexagon has side lengths $6,7,8,9,10,11$ (not necessarily in this order). If the area of the hexagon is $k \sqrt{3}$, find the sum of all possible values of $k$.
(2 marks)
某等角六邊形各邊的長度為 6、7、8、9、10、11（不一定按此順序）。若這
六邊形的面積為 $k \sqrt{3}$, 求 $k$ 的所有可能值之和。
(2 分)
15. 213
15. We first observe that by extending three mutually nonadjacent sides of an equiangular hexagon, we get an equilateral triangle. From this we see that every equiangular hexagon is formed by removing three small equilateral triangles from a large equilateral triangle. Conversely, any such removal gives an equiangular hexagon.

It is thus easy to see that if the lengths of the consecutive sides of an equiangular hexagon are $a, b, c, d, e, f$, then we must have $a-d=-b+e=c-f$ (note that opposite sides form a pair and the lengths alternate in sign in the above equality; this also says that the sum of the lengths of three consecutive sides, which is also the side length of the equilateral triangle formed, may only take two different possible values, namely, either $a+b+c=c+d+e=e+f+a$ or $b+c+d=d+e+f=f+a+b)$. Furthermore, since an equilateral triangle of unit length has area $\frac{\sqrt{3}}{4}$, we see that the equiangular hexagon formed from removing equilateral triangles of sides lengths $a, c, e$ from a large equilateral triangle of side length $n$ has area $\frac{\sqrt{3}}{4}\left(n^{2}-a^{2}-c^{2}-e^{2}\right)$, where $n=a+b+c$.

Now we must divide the lengths $6,7,8,9,10,11$ into three pairs with equal difference. Since the perimeter is 51 , which is odd, the possible equal differences include 1 (the pairing is $\{6,7\}$, $\{8,9\},\{10,11\}$ ) and 3 (the pairing is $\{6,9\},\{7,10\},\{8,11\}$ ).

In the first case, the two sides adjacent to 6 must be 9 and 11 (i.e. the two larger elements of the other two pairs, since 6 is the smaller in its pair), and the side lengths in order must be (6, $9,10,7,8,11)$. In this case the area of the hexagon is $\frac{\sqrt{3}}{4}\left(25^{2}-6^{2}-10^{2}-8^{2}\right)=\frac{425}{4} \sqrt{3}$. (Another expression would be $\frac{\sqrt{3}}{4}\left(26^{2}-9^{2}-7^{2}-11^{2}\right)$ which is the same.)

Similarly, in the second case, the two sides adjacent to 6 must be 10 and 11, and the side lengths in order must be $(6,10,8,9,7,11)$, and the area of the hexagon in this case is $\frac{\sqrt{3}}{4}\left(24^{2}-6^{2}-8^{2}-7^{2}\right)=\frac{427}{4} \sqrt{3}$.

Combining these two cases, we see that the answer is $\frac{425}{4}+\frac{427}{4}=213$.