Feedback on proof of "clipping a convex N-polygon creates a convex polygon with at most N+1 vertices" I want to state that "when a convex polygon with $N$ vertices is clipped against a half-plane, the resulting will be a convex polygon of at most $N+1$ vertices".
I know now how to argue about convexity. This is my argument about the $N+1$ part:
An intersection of any convex polygon with a half-plane consists of at most one line segment.
For there to be an intersection at all, at least one point of the polygon must be on a different side of the plane than all others.
Therefore, at least one point of the polygon is "cut off".
The intersection line creates 0 to 2 new vertices: a vertex in each intersection with a polygon edge and no vertex if it passes through an existing polygon vertex.
Because there are at most two intersections of the line and the convex polygon - at most 2 additional polygon vertices can be created.
Thus, the result of clipping a convex polygon of $N$ vertices with a plane will remove at least one vertex and create at most two vertices. New polygon will not have more than $N+1$ vertices.
It sounds clumsy and amateur to me myself. How can I improve it? Make it more sound?
 A: How 'bout this (assuming strict convexity):
There are two cases: Either the halfplane boundary meets some vertex or vertices of the polygon, or it meets no vertex of the polygon. We'll address the second case.
At most two edges of the polygon cross the halfplane boundary. Let's renumber so that these edges are $(p_1, p_2)$ and $(p_k, p_{k+1})$, with $p_1$ inside the halfplane. [note that it's possible that $k = N$, and $p_{k+1}$ therefore denotes $p_1$. In particular, however, $k \le N$.] 
We then know that $p_2, p_3, \ldots, p_k$ are in the halfplane as well, and $p_{k+1}, \ldots, p_n$ are outside the halfplane. The intersection of the  halfplane edge with $(p_1, p_2)$ is a single point we'll denote $q_1$; the intersection with the other edge is a single point $q_{k+1}$. The clipped polygon now has vertices $q_1, p_2, \ldots, p_k, q_{k+1}$. Because $k \le N$, we see that the clipped polygon has at most $N+1$ vertices. 
The first case --- one or more vertices lie on the clip-edge --- is similar. If one lies on the clip edge, let it be called $p_1$ as before, and let $q_1 = p_1$; then the prior argument  works fine. If two lie on the clip edge, and they're adjacent, then the clipped polygon is the original (or is a single edge, depending on the orientation of the half-plane). If two non-adjacent vertices lie on the clip edge, label the first $p_1$, and the adjacent out-of-halfplane vertex $p_2$; continue numbering until you reach the next vertex, $p_k$, on the halfplane edge; the clipped polygon now has vertices $p_1, \ldots, p_k$, and $k \le N < N + 1$. 
It's not really much better, but involves a little less informality. 
