calculating the angles of a figure that is an a convex curve. Take a convex curve. Select points on the curve and label them $1,2,3,...,n$. Do this the following way : Once you put a point down continue labeling the points while going clockwise and while keeping the points in order in that way. Now connect every other one with a line. So for example if I went all the way to $5$, I would connect $1$ to $3$, $3$ to $5$, $5$ to $2$, and $2$ to $4$. i.e. you are essentially making a star. When constructing a star in this way, what would $\angle 1 + ... + \angle(2n-1)$ be?
 A: I assume you want $n$ to be odd so that this process actually hits every vertex.
The sum of those angles equals $(n-4)\pi$. The way you can see this is to show that total sum of the exterior angles is $4\pi$; that amount then gets subtracted from $n\pi$ which is $n$ straight angles, leaving $(n-4)\pi$ as the sum of the interior angles.
The proof is basically the same as the proof that the total sum of the exterior angles of a convex polygon is $2\pi$, from which it follows that the sum of the interior angles is $(n-2)\pi$. 
For the star as you describe, keep track of how a tangent vector rotates as you go around the star. The tangent vector does not rotate at all as you go along an edge, and then it rotates by the exterior angle at a vertex, then it does not rotate at all as it goes along the next edge, then it rotates by the exterior angle at a vertex, etc. So, the total rotation of the tangent vector equals the sum of the exterior angles. On the other hand, by your construction it is evident that the tangent vector goes through 2 complete rotations, for a total angle of $4\pi$.
A: 


The angle sum is $\pi (n - 2k)$ 
where $k$ is the number of cycles. In the example above, $k=2$,
so the angles sum is $\pi (7-4) = 3 \pi$.

(Added: This is equivalent to Lee Mosher's proof.)
Call the points $p_i$. If you walk from $p_1$ to $p_3$ in the above example,
you are heading NorthEast.
At $p_3$, you must turn about $42^\circ$ clockwise to reorient your
direction to walk toward $p_5$. Call that $42^\circ$ the turn angle $\tau_3$ at
$p_3$. Call its supplement the internal angle $\theta_3 \approx 138^\circ$; $\tau_i + \theta_i = 180^\circ = \pi$.

If the full walk cycles $k$ times, your total turn angle must be $k (2 \pi)$ because
you end up facing the same as at the start. So
$$
2 k \pi = \sum \tau_i = \sum (\pi - \theta_i) = n \pi - \sum \theta_i \;,
$$
leading to $\sum \theta_i = \pi(n-2k)$.
As Lee noted, a special case is $k=1$, a polygon, when the familiar sum $(n-2)\pi$ is obtained.
