Question 1183719
<br>
I haven't come up with a standard geometric solution; but a solution using coordinate geometry works nicely.<br>
Let the origin (0,0) be the center of the regular hexagon; and for simplicity choose a "nice" number for the side length -- I chose 12.  My hexagon then has vertices
A(-12,0)
B(-6,6sqrt(3))
C(6,6sqrt(3))
D(12,0)
E(6,-6sqrt(3))
F(-6,6sqrt(3))<br>
From those, we get
G(0,6sqrt(3)) [midpoint of BC]
H(9,-3sqrt(3)) [midpoint of DE]<br>
{{{drawing(600,480,-15,15,-12,12
,line(-12,0,-6,6*sqrt(3)),line(-6,6*sqrt(3),6,6*sqrt(3)),line(6,6*sqrt(3),12,0)
,line(12,0,6,-6*sqrt(3)),line(6,-6*sqrt(3),-6,-6*sqrt(3)),line(-6,-6*sqrt(3),-12,0)
,locate(-12.5,0,A),locate(-6,6*sqrt(3)+1,B),locate(6,6*sqrt(3)+1,C),locate(12.5,0,D),locate(6.5,-6*sqrt(3),E),locate(-6.5,-6*sqrt(3),F)
,locate(0,6*sqrt(3)+1,G),locate(9.5,-3*sqrt(3),H)
,red(line(-6,6*sqrt(3),6,-6*sqrt(3)))
,green(line(-12,0,9,-3*sqrt(3)))
,blue(line(-6,-6*sqrt(3),0,6*sqrt(3)))
,locate(0,0,O),locate(1,-2*sqrt(3),J),locate(-1,2*sqrt(3),K)
)}}}<br>
Find the equations of lines AH, BE, and FG<br>
BE: y=(-sqrt(3))x
FG: y=(2sqrt(3))x+6sqrt(3)
AH: slope = (-3sqrt(3)/21) = -sqrt(3)/7
 equation: y=(-sqrt(3)/7)(x+12) = (-sqrt(3)/7)x-12sqrt(3)/7<br>
Find the coordinates of K, the intersection of FG and EB<br>
(-sqrt(3)x)=2sqrt(3)x+6sqrt(3)
-3sqrt(3)x=6sqrt(3)
x=-2
y=-sqrt(3)x = 2sqrt(3)<br>
Find the coordinates of J, the intersection of AH and BE<br>
(-sqrt(3)x)=(-sqrt(3)/7)x-12sqrt(3)/7
(-7sqrt(3)x=-sqrt(3)x-12sqrt(3)
-6sqrt(3)x=-12sqrt(3)
x=2
y=-sqrt(3)x=-2sqrt(3)<br>
So J is (2,-2sqrt(3)) and K is (-2,2sqrt(3))<br>
The length of JK is then given by the distance formula
(JK)=sqrt(4^2+(4sqrt(3))^2) = sqrt(16+48) = sqrt(64) = 8<br>
ANSWER: JK = 8<br>
Or, since I used a=12,<br>
ANSWER: JK = (2/3)a<br>
Note the symmetry of the two points J and K with respect to the center of the hexagon suggests that a standard geometric solution to the problem should be possible; but I haven't yet seen it.<br>