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This is quite a tricky question and I can't answer it.

Aspiring Mathlete
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    chasing the angles and using triangle angle sum yields nothing? – evil999man Apr 13 '14 at 11:23
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    Make believe the vertices lie on a circle, and use the theorem that says that the angle subtended by a chord at a point on the circle is half the angle subtended by that chord at the center. Or type $$\rm sum\ of\ the\ angles\ of\ a\ star$$ into Google and see what comes up. – Gerry Myerson Apr 13 '14 at 11:24
  • Nice question! Playing with GeoGebra confirms that the answer is 180 degrees, however. – Sawarnik Apr 13 '14 at 12:15
  • Any thoughts on the two answers that have been posted, user108104? – Gerry Myerson Apr 15 '14 at 05:47

3 Answers3

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Ok, I finally solved it and can prove the sum is 180 degrees.

Let's mark the vertices clockwise, from A to G. I will refer to A, B .. G as angles of the 'star'.

We form and take as example AED triangle; we know the sum of all angles is 180. Thus A + ADE + DEA = A + E + BED + EDG + D = 180 ( 1 ).

Let be M the point of intersection between BE and GD. In EMD triangle, we have angle EMD = 180 - BED - EDG ( 2 ).

From ( 1 ) and ( 2 ) => A + E + D = EMD ( 3 ). (similar procedure here for others)

Let be N the point of intersection between AD and FC. Let be P the point of intersection between AE and FC. In ANP triangle we have sum of all angles: A + APN + ANP = 180 ( 4 ).

But actually, ANP = DNC. DNC = G + D + C as shown above in similar procedure as ( 3 ). The same: APN = FPE = F + B + E.

If we put all things together, formula ( 4 ) becomes A + B + C + D + E + F = 180 (Q.E.D)

Cristi
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You can assume that the external vertices of your star are the vertices of a regular heptagon (changing "not too much" the position of one vertex does not affect the sum of the angles), hence any angle equals $\frac{\pi}{7}$ by considering the circumscribed circle. This gives that the original angles sum up to $\pi$.

Jack D'Aurizio
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Here's a variation on the proof by @Cristi.

We start the same way; label the vertices clockwise, A to G, and we look at triangle ADE, whose angles sum to $\pi$ (measuring angles in radians, of course).

There are 7 triangles like ADE (the others are BEF, CFG, DGA, EAB, FBC, and GCD), so adding up all their angles gives $7\pi$.

But adding up all their angles also adds up all the angles of the heptagon ABCDEFG, but counting the angles of the star 3 times each. So, $$7\pi={\rm sum\ of\ angles\ of\ heptagon\ plus\ twice\ sum\ of\ angles\ of\ star}$$ Now the angles of the heptagon sum to $5\pi$ (the angles of any convex $n$-gon sum to $(n-2)\pi$), so $$7\pi=5\pi{\rm \ plus\ twice\ sum\ of\ angles\ of\ star}$$ so the angles of the star add up to $\pi$.

Gerry Myerson
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