Approximating convex sets with disjoint rectangles in an optimal way Let $O \subset \mathbb{R}^2$ be a convex open set of finite Lebesgue measure $1=m(O)$. Let's call a collection $P$ of $n$ disjoint open rectangles contained in $O$ a "partial cover of $n$ pieces". (The rectangles may be tilted.) Let us denote $C_n$ the collection of all $n$ piece partial covers of $O$. Write $$A_n = \sup_{P \in C_n} m(P).$$
Obviously, $A_n \to 1$ from below. From the convexity condition, it's also intuitively clear that convergence will be "good" - the intuition being that a convex set cannot differ by "too much" from a rectangle. Is it possible to make this intuition precise by writing a lower bound of some sort for $A_n$, perhaps depending on the diameter of $O$?
 A: I can pull a lower bound for some subsequence of $A_n$. I start dividing the convex set in two parts and add one rectangle in each part. Then, considering the spaces not covered by the first two rectangle, these form $6$ disjoint convex subsets of the original sets, over which I apply one rectangle each : I am now left with $18$ convex subsets... and I keep tripling the number of convex subsets left, thus I am adding 
$
x_n \, \overset{def}{=} \, \, 2 + 2\cdot 3 + 2 \cdot 3^2 + ... + 2 \cdot 3^{n-1} = 2(3^n-1)/(3-1) = 3^n - 1
$
rectangles at each step of my algorithm (See drawings below). I claim that $A_{x_n} \ge a_n$ (and $a_n$ is also described below, and converges to $1$ at a reasonable speed).
Let $\mathrm{diam}(O)$ be the diameter of $O$ and let two points $x$ and $y$ on $\partial O$ such that $\| x - y \| = \mathrm{diam}(O)$. (The function which takes two points in $O$ and outputs their distance is continuous, and $\partial O$ is compact, so we indeed have such a couple $(x,y)$.) This means that at point $x$, the tangent parallel to $\partial O$ which goes through $x$ is orthogonal to the line that goes from $x$ to $y$. See drawing below.

Now since $x$ and $y$ are chosen so that their distance is maximal, the vectors tangent to $\partial O$ at $x$ and $y$, in both "up" and "down" directions (see drawings) are going to point inwards (because they have maximal distance), so that the "height" of points on $\partial O$ relatively to the $x-y$ axis will be a concave function (minus the function is convex), thus has a maximum. Choose the point $z_1$ which has maximum height for the top zone, and $z_2$ for the bottom zone. I have made another drawing which considers the bottom case, and the top case is symmetric.
In the drawing I am defining $f(t)$ to be the area of a certain rectangle. I define $f : (0,a) \to \mathbb R$ by letting $t$ be a distance from the intersection between the $x-y$ axis and the point dropped orthogonally from $z_2$ (which is the bottom point) and some point to the left of that point on the $x-y$ axis. To draw the rectangle, you drop down vertically, then turn orthogonally until you hit the boundary, and then turn orthogonally and stop at the $x-y$ axis again. I am defining the rectangle in this manner because the worst convex set that has $x$, $y$ and $z$ at those places is just two triangles, hence area will be at its minimum and I will derive a lower bound.

Using geometry, you can easily see that 
$$
f(t) = (t+\beta)(h-\alpha) = \left(t+\frac{t(d-a)}a \right)\left( h - \frac{ht}a \right) = \frac{dh}a \left( t \left( 1 - \frac ta \right) \right).
$$
which gives the maximum at $t = \frac a2$ (natural, isn't it) and a maximum of $\frac{3}{16} dh.
Now the area of both rectangles is $\frac{3}{16}dh_1 +\frac{3}{16}dh_2 = \frac{3}{16}(d(h_1+h_2)) \ge \frac{3}{16}$ and the inequality comes from the fact that you can enclose your convex set in a box of width $d$ and height $h_1+h_2$ where $h_1$ is the maximal height of one of the constructions and $h_2$ is the other one.
At step $1$, I have removed at least $r=3/16$ of the set $O$ with 2 rectangles. The space left (imagine a circle with a square inscribed in it, there are $4$ zones left... these are the new convex sets I am iterating over). Now on the $(1-r)$ area left, I can remove at least $3/16$'s of it, hence there is $r + (1-r)r$ area removed now. Iterating in this manner, I obtain
$$
a_1 = r, \qquad a_n = a_{n-1} + (1-a_{n-1})r
$$
and since this sequence is clearly increasing and bounded above, it converges, and letting the limit $a = a + (1-a)r$, you find $a = 1$.
I think geometric convergence is "good enough". Don't you think? 
Thanks to Theo Buehler for the hint on using screenshots.
