Five Angles in a Star

• In a regular pentagram (5-pointed star), the angle in each point is 36 degrees, so the angles in all five points sum to 180 degrees:

What about an irregular pentagram, such as the following?

Now the angles might be all different from each other; the situation is much more complicated. Can you prove that the angles in all five points still sum to 180 degrees?

Restrictions (to make clear that this is neither a maths problem [as opposed to a maths puzzle] nor an exercise in computation or advanced Euclidean geometry):

• no arithmetic operations allowed (addition, multiplication, ...)
• you may draw one new line-segment on the star, but no more than that

Sorry rand, but I think this is just another math problem...("prove", "angles", "sum", "180 degrees")

@MarkN According to the canonical meta post on this, the sign of a maths *puzzle* as opposed to *problem* is to have a **clever or elegant solution, often an "aha" moment**, an **unexpected problem statement**, or an **unexpected or counterintuitive result**. The solution I have in mind definitely has the first of these features, and IMO the last two too.

This isn't a math puzzle - it's a logic puzzle. You just usually learn this logic from someone who also teaches math.

• Milo Brandt Correct answer

5 years ago

\$\hskip 1.5in\$

This is an image of an arrow sweeping each of the successive angles in the star. Notice that, after it traces all \$5\$ angles, its orientation is reversed - meaning it has rotated \$180^{\circ}\$ and that this must be the sum of the angles. We can do the same thing to the star in your figure, ergo, its angles too sum to \$180^{\circ}\$.

A Generalization:

We can do the same thing to a figure like this, whose angles sum to \$180^{\circ}\$: \$\hskip 1.5in\$
We can also do this to a triangle. The important property is this:

There must be no vertices of the star interior to the cone swept out by a ray traversing a given angle.

Satisfying this condition - which basically says that we never have to "ignore" vertices, but instead just rotate the arrow and see what it hits - we find that we can order the vertices in a "clockwise" manner, so that, at each angle, either the head or the tail of the arrow steps to the next vertex in the order (and they alternate which). Obviously, both head and tail will make a full revolution when twice as many angles as vertices have been traced, yielding the desired result.

(One might also express my condition as "assigning the vertices the numbers \$1\$ through \$2n+1\$ in clockwise order as seen from a central point, it must be that \$1\$ connects to \$n\$ and \$n+1\$, and all other points are connected analogously")

(Also, for what it's worth, I really liked this puzzle, even if my answer is not the intended one - I had a good, "Well that's *obvious*" moment, followed by a few hours of intense head-scratching, trying to figure why it was obvious, followed by "Aha! It *was* obvious!")

I take it your comment is a reference to this joke? =)

Accepted because it's even nicer than the answer I was also looking for, and also covers a generalisation.