Can a convex polytope with $f$ facets have more than $f$ facets when projected into $\mathbb{R}^2$? Let $P$ be a convex polytope in $\mathbb{R}^d$ with $n$ vertices and $f$ facets.
Let $\text{Proj}(P)$ denote the projection of $P$ into $\mathbb{R}^2$.
Can $\text{Proj}(P)$ have more than $f$ facets?
In the general case, each successive projection can increase the number of facets from $f$ to $\left\lfloor \frac{f^2}{2} \right\rfloor$, but I'm wondering if $\mathbb{R}^2$ is a special case.
 A: Consider the polytope in $\mathbb{R}^3$ with $8$ vertices at coordinates $(\pm 1, \pm 2, 1), (\pm 2, \pm 1, -1)$.  Geometrically this looks like a cube where the top face is stretched in the direction of the $y$-axis and the bottom face is stretched in the direction of the $x$-axis, but it still has the face structure of a cube and in particular has $6$ facets.
It's projection onto the first two coordinates is clearly an octagon, and hence has two more facets than the original polytope.
A: Your question is essentially about extension complexity.  In general, the extension complexity of a polytope $P$ is the minimum number of facets over all polytopes $Q$ which project to $P$.  You are interested in the extension complexity of polygons.  Fiorini, Rothvoß, and Tiwary proved that regular $n$-gons have extension complexity $O(\log n)$.  For lower bounds, they give examples of $n$-gons which have extension complexity $\sqrt{2n}$.
It is an open question if there exists for infinitely many $n$, an $n$-gon with extension complexity $\Omega(n)$.
Edit. Shitov has recently shown that every convex $n$-gon has extension complexity at most $147 n^{2/3}$, which resolves the open question in the negative.
As a bonus, here is a picture of the polytope in Nate's answer (courtesy of Samuel Fiorini).

A: Another, combinatorially minimal, counterexample of such a polytope $P$ (with only five facets) is the convex hull of the six vertices $(\pm2, 0, 0)$, $(\pm1, \pm1, 1)$. Its projection to the $xy$-plane is a hexagon. The minimality of the number of vertices and the number of facets is easy to prove.
