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I have a convex polytope $P$ in $R^2$ of $\dim P = \dim\operatorname{aff} P = 2$, $\dim\operatorname{aff}P$ is the dimension of the affine hull of $P$. Let $L$ be the line orthogonal to the normal vector $v$. I look at the orthogonal projection of $P$ to $L$, and would like to show that for each inner point $p$ in the projection there are $2$ points on the boundary of $P$ which are projected into $p$.

I hope someone is able to help me give a rigorous proof.

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You can get proper formatting for operators like $\operatorname{aff}$ using operatorname{aff}. (Also note that variable names are usually italicized not just in formulas but also when they appear in the text by themselves.) –  joriki May 2 '12 at 22:10
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If we replace $P$ by its closure, it has the same boundary and the projection onto $L$ has the same inner points. Thus we may assume without loss of generality that $P$ is closed.

Let $L_p$ be the line projected onto a point $p$ of the projection. Since $P$ is convex and closed, the set of points on $L_p$ that belongs to $P$ is a closed interval. We want to show that this interval is not a single point if $p$ is an inner point of the projection.

If the interval were a single point for all points $p$ in the projection, then $P$ would have to be a line, which contradicts the assumption that $\dim\operatorname{aff}P=2$. Thus there is at least one point $q$ in the projection for which the interval contains two distinct points. For any inner point $p$, we can find a point $r$ in the projection on the other side of $p$ than $q$. Since $r$ is in the projection, at least one point $a$ in $P$ is projected onto $r$. Since $P$ is convex, connecting the two distinct points in the interval for $q$ with $a$ yields two distinct points in $P$ that are projected onto $p$. Thus the interval on $L_p$ belonging to $P$ cannot be a single point.

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