Can you tell me the solution to this little problem

Every angle of a hexagon is 120o, and the lengths of its sides are √(3-√3) and √(9-3√3), alternating. Prove that the area of the hexagon is an integer.

Note: √=square root

The trick on this one is to join every other vertex with an interior edge.

This splits the hexagon into four triangles:
(i) Three congruent exterior triangles (by "SAS") whose adjacent outer edges have lengths
a = √(3-√3) and b = √(9-3√3) = a√3, and included angle 120°.

Any of these triangles has area
(1/2)ab sin 120° = (1/2) a * a√3 * √3/2 = (3/4)a^2 = (3/4)(3 - √3).
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(ii) An interior equilateral triangle with side length "c" (which is the third side of the triangles mentioned in (i) above).

By the Law of Cosines, we have
c^2 = a^2 + b^2 - 2ab cos 120°
......= a^2 + (a√3)^2 - 2a * a√3 * (-1/2)
......= (4 + √3) a^2
......= (4 + √3) (3 - √3).
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So, the area of the equilateal triangle equals
(√3/4) c^2 = (√3/4)(4 + √3) (3 - √3).
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By (i) and (ii), the area of the hexagon equals
3 * (3/4)(3 - √3) + (√3/4)(4 + √3) (3 - √3)
= (1/4)(3 - √3) (9 + √3 (4 + √3)]
= (1/4)(3 - √3) (12 + 4√3)
= (3 - √3)(3 + √3)
= 6, which is an integer.

I hope this helps!

It's my pleasure!

Since the sides of the hexagon are not congruent, we can't simply split the vertices of the hexagon to having 60 degrees on both sides. So, it seems like at least some trigonometry is needed.

Nonetheless it is a neat problem.

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