<PRE><B>Question 1120495</B>
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I didn&#39;t see any use in the suggested auxiliary line segments HG and DA....&lt;br&gt;
Perhaps there is a clever way to solve the problem using them; my solution is not very difficult.&lt;br&gt;
Consider the regular hexagon with the center O at (0,0), with sides AB and DE horizontal; A in quadrant III, B in quadrant IV, D in quadrant I, and E in quadrant II.  Then the coordinates of the vertices of the hexagon are&lt;br&gt;
{{{A(-6,-6*sqrt(3))}}}
{{{B(6,6*sqrt(3))}}}
{{{C(12,0)}}}
{{{D(6,6*sqrt(3))}}}
{{{E(-6,6*sqrt(3))}}}
{{{F(-12,0)}}}&lt;br&gt;
Then the two midpoints in the problem have coordinates&lt;br&gt;
{{{G(9,-3*sqrt(3))}}}
{{{H(0,6*sqrt(3))}}}&lt;br&gt;
Now triangles HJE and AJB are similar, with the ratio of similarity 1:2 because of the lengths of the bases EH (length 6) and AB (length 12).  Since the height of the hexagon is 12*sqrt(3), we can determine that the y coordinate of point J is 2*sqrt(3).  (It is 1/3 of the way from side DE -- y value 6*sqrt(3) -- and side AB -- y value -6*sqrt(3).)&lt;br&gt;
So we can determine that the coordinates of J are {{{J(-2,2*sqrt(3))}}}.&lt;br&gt;
The exact same similarity exists between triangles BKG and EKF, leading us to the coordinates of K as {{{K(2,-2*sqrt(3))}}}.&lt;br&gt;
Then, making JK the hypotenuse of a right triangle, the Pythagorean Theorem (or the observation that we have a 30-60-90 right triangle, with legs 4 and 4*sqrt(3)) gives us the length of JK as 8.
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