Can all polygons outside of the largest inscribed rectangle in a convex polygon be concave Let $C$ be a convex polygon and $R$ the largest (by area) rectangle lying within $C$. Does there exist a convex polygon $C$ such that when $R$ is removed, all remaining polygons are concave? In other words: Does there exist a polygon $C$ such that all parts of $C \setminus R$ are concave?
Edit: I am quite sure the answer is yes, and I think this site gives some examples, but not a formal proof.
 A: I think this is an example of the existence of such a polygon. Let me know if it fails in some way if I am still misunderstanding your question. Here is $C$:

And in red is our largest rectangle by area, $R$, which I have not proven, but is rather clear I hope. And what is left over in blue is concave.

A: Answer is no (I believe).
This answer is incomplete

Suppose there is such a shape $C$.
Consider the largest rectangle $R$ in $C$. If 2 consecutive vertices of $R$ touch $C$, then the part that is cut out will be convex. Hence, in order to fulfill your conditions, at least 1 of any 2 adjacent vertices must not lie on the boundary of $C$. This implies that 2 opposite vertices of $R$ do not lie on the boundary of $C$.
If 4 vertices of $R$ do not lie on the boundary of $C$, it is easy to see that a small expansion will result in a larger $R'$, contradicting the maximality of $R$.
If 3 vertices of $R$ do not lie on the boundary of $C$, notice that due to convexity of $C$, the boundary of $C$ can only touch the perimeter of $R$ at that one vertex. Then, once again, a small expansion from that vertex will result in a larger $R'$, contradicting the maximality of $R$.
Hence, we must have 2 opposite vertices of $R$ (labelled $a, c$) which do not lie no the boundary of $C$, and the other 2 (labeled $b, d$) will lie on the boundary of $C$.
If the boundary of $C$ touches $ba$ or $da$ at a point that is not $b$ or $d$, then the convexity of $C$ implies that the boundary of $C$ touches $a$, which is a contradiction. As such, the boundary of $C$ only touches $R$ at the points $b$ and $d$.
Let $O$ be the center of $R$ (intersection of diagonals). Consider the circle with center $O$ and diameter $bd$. If $ac$ is not the perpendicular (diameter) to $bd$, then by rotating points $a,c$ with respect to $O$, we can find a rectangle $bedf$ inscribed in $O$ with a larger area.
This leaves us with $abcd$ being a square. WLOG, let's use a coordinate system with $O=(0,0), b = (-1, 0), a = (0,1), d = (1, 0), c = (0, -1)$.
Since $a$ and $c$ do not touch the boundary of $C$, let $ 2 \epsilon$ be the minimum of the distance between either of points $a$ or $c$ with the boundary of $C$. Furthermore, set $ \epsilon < 0.1$. Let $ a' = (0, 1+ \epsilon)$, $c' = (0, -1 - \epsilon)$, neither of which points touch the boundary of $C$. By the convexity of $C$, the quadrilateral $a'bc'd$ only touches $C$ at $b$ and $d$.
I feel that we should be able to find a rectangle of area > 2 here, but am unable to do so easily.
A: Try this: 
$C$ is the hexagon with vertices $(0,\pm 1)$, $\bigl(\pm(1+\epsilon),\>\pm\epsilon\bigr)$, where $\epsilon\ll1$. I conjecture that the largest rectangle inscribed in $C$ is the square with vertices $(\pm 1,0)$, $(0,\pm1)$.
