From the 2008 Netherlands Junior Maths Olympiad:
A hexagon has six angles all equal to 120 degrees. The lengths of four consecutive sides are 2000, 2006, 2008 and 2009. Determine the perimeter of the hexagon.
The first time I attempted this, I drew a lot of lines inside the hexagon. I ended up with the correct answer, but it was all a bit untidy, so I looked for a simpler solution. Thankfully there is one:
Here, ABCDEF is the original hexagon, G is the point where lines AB and CD meet, and H is the point where AF and DE meet.
Now, since all the interior angles of the hexagon are 120 degrees, opposite sides are parallel. So in fact, AHDG is a parallelogram. Also, it can be seen that each of the angles in the triangles BGC and EHF are equal to 60 degrees, so they are equilateral.
This means that GC = GB = 2006, and EH = FH = x.
It is a property of parallelograms that their opposite sides are equal in length, and this is how we calculate x and y. Since DH = AG, x + 2009 = 2000 + 2006, so
x = 1997.
Further, we have AH = DG, so x + y = 2008+2006, so
y = 2017.
Thus the perimeter of the hexagon is 2000 + 2006 + 2008 + 2009 + 1997 + 2017 = 12037.
An interesting little problem, requiring a few sensibly chosen lines to be drawn in, followed by a bit of algebra and geometry.