A very big hexagon

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:

Big hexagon

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.


One thought on “A very big hexagon

  1. Bryce Herdt November 12, 2016 / 8:03 pm

    Eh? That’s a little more work than strictly necessary.
    The sides of length 2000 and 2009 are parallel. Because we’re dealing with an equiangular hexagon, the sum of the unknown sides must be 4014 on account of equaling 2006+2008. We can now calculate the other sides, but can also skip it. The perimeter is 4014+4014+2000+2009 =12037.


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