# 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?

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 $$k \le N < N + 1$$.