SOLUTION: Given regular hexagon ABCDEF,, with Centre O and sides of length 12. Let G be the midpoint of BC. Let H be the midpoint of DE. AH intersects EB at J and FG intersects EB at K..F

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Question 1120495: Given regular hexagon ABCDEF,, with Centre O and sides of length 12. Let G be the midpoint of BC. Let H be the midpoint of DE. AH intersects EB at J and FG intersects EB at K..Find JK.
Hint given in question ::: draw auxiliary lines HG and DA.
Thanks for any help .

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


I didn't see any use in the suggested auxiliary line segments HG and DA....

Perhaps there is a clever way to solve the problem using them; my solution is not very difficult.

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








Then the two midpoints in the problem have coordinates




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).)

So we can determine that the coordinates of J are .

The exact same similarity exists between triangles BKG and EKF, leading us to the coordinates of K as .

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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