Maximum area of convex quadrilateral in convex polygon I’m wondering if we can say anything about the following situation: Given a convex polygon $P$, you want to draw a convex quadrilateral $Q$ which is contained in $P$. You want to maximize the ratio $\dfrac{\text{Area }Q}{\text{Area }P}$. What is the minimum ratio that you can always achieve, no matter what $P$ is? Or can you not guarantee anything better than zero?
 A: There's always a quadrilateral $Q$ in $P$ with $\text{Area}(Q)\ge\frac12\text{Area}(P)$:
Proof. By a standard compactness argument, there exists a convex quadrilateral $Q$ in $P$ with maximum area.  Let it be $ABCD$.  Draw the line through $B$ parallel to $AC$, and consider a point $B'$ which is on the other side of that line from the quadrilateral:

$B'$ cannot be in $P$, since otherwise by convexity the quadrilateral $AB'CD$ would be contained in $P$, contrary to our assumption that $ABCD$ has maximum area for such quadrilaterals.  Thus $P$ lies entirely on the same side of the line we drew through $B$ as the quadrilateral does.
Repeating this argument for all four vertices yields that $P$ is contained in a parallelogram $Q'$ whose sides are parallel to the diagonals of $ABCD$:

So $\text{Area}(P) \le \text{Area}(Q') = 2\text{Area}(Q)$.  End of proof.
This $\frac12$ doesn't seem to be the best possible constant.  For one thing, we cannot get equality in the above proof, since we'd need $P=Q'$, in which case we could have gotten a quadrilateral with larger area than $Q$.  For another thing, the same result actually holds even under the restriction that $Q$ must be a rectangle: for this and more, see Marek Lassak, "Approximation of convex bodies by rectangles", Geom. Dedicata 47 (1993), 111–117, doi:10.1007/BF01263495.
