old-fashioned star

This is a very tough question. I don't know whether I would do at the time of GMAT because it took me a while to do it.

First of all you need to close the extremities of your star in triangles, so that now you have a pentagon inside the star.

The formula (n-2)*180, where n is the number of sides, gives you the sum of the intern angles.

Thus, inside the pentagon you have (5-2)*180 = 3*180. Keep this number.

Now if you pick one of the angles inside the pentagon you will notice that together with a angle inside the triangle, the sum is 180.

Let's call the angles of the pentagon 1º, 2º, 3º, 4º, 5º
And the angles of the triangles, v, w, x, y, z. (Notice that there are two angles v, two angles w, ...)

Summing each one of the angles inside the pentagon with one angle inside the triangle we have.

v + 1º + w + 2º + x + 3º + y + 4º + z + 5º = 5*180
As we know that 1º + 2º + 3º + 4º + 5º = 3*180 (the sum of the angles of the pentagon.),
we have v + w + x + y + z = 2*180

Now if you sum all the angles of all triangles you will have
2v + 2w + 2x + 2y + 2z + a + b + c + d + e = 5*180
2*(v + w + x + y + z) + a + b + c + d + e = 5*180

As we know that v + w + x + y + z = 2*180

2*(2*180) + a + b + c + d + e = 5*180
4*180 + a + b + c + d + e = 5*180
a + b + c + d + e = 180

Ufa!! Ok now I really don't want a question like this on real GMAT...

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