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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?

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  • $\begingroup$ Maybe a usefull observation is that for a given $P$ the best ratio is obtained by a quadrilatere $Q$ whose vertices are also vertices of $P$. $\endgroup$ – Gilles Bonnet May 5 '14 at 9:16
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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:

Quadrilateral ABCD and B'

$B'$ 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 $ABCD$:

Quadrilateral ABCD and parallelogram Q'

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.

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