The Plank Problem 2 Dimensions We were trying to solve this wonderful problem, but have not succeeded to solve.
It goes like this:
Let $R=[0,1]^2$, and $D\subseteq R$ be a convex set which intersects each side of $R$. Define a plank to be a set of the kind $[a,b]\times[0,1]$ or $[0,1]\times[a,b]$. The planks are parallel to the axes. Show that for each set of planks that covers $D$, it holds that the sum of the areas of the planks is at least 1.
Thanks,
LLAP.
A hint from the oracle: The Marriage theorem. (We still have yet to solve)
 A: As promised, here is a geometric solution, without "Marriage Theorem" (I also would like to see a solution which uses it).  
A remark concerning probabilistic methods.: I do not think that the Lebesgue measure on $R$ is the right object here, since $D$ can have Lebesgue measure zero (if $D$ equals a diagonal of $R$). I do suspect however that there is a proof using a non-Lebesgue measure on $R$ which detects "widths" of subsets rather than their areas (see my comment on George Lowther's proof). However, I do not know how to construct one (at least, without first proving Theorem 1). 
Below is my solution almost completely rewritten (I originally missed two cases). 
I will label the vertices of $R$ as $a=(0,0), b=(0,1), c=(1,1), d=(1,0)$.  By the "width" of a plank I mean the length of its shorter side. I let $\pi_1, \pi_2$ denote the projections of $R$ to its horizontal side $ad$ and vertical side $ab$ respectively. 
Theorem 1. For every finite collection $C$  of planks covering $R$, the total width $W$ of $C$ is $\ge 1$. 
Proof. The proof is by induction on the total number $N$ of planks. If $N=1$, it is clear that $W\ge 1$. 
Now, consider the case $N=2$, when $C$ consist of one vertical plank $P$ and one horizontal plank $Q$ of the widths $v$ and $w$ respectively. (The case $N=2$ with just two horizontal or two vertical planks reduces immediately to $N=1$.) 
Lemma 1. If $N=2$ then $W=v+w\ge 1$. 
Proof. Since $D$ intersects all four sides of $R$, consider the quadrilateral $D'=x_1y_1x_2 y_2\subset D$ where $x_1\in ad\cap D, x_2\in bc\cap D', y_1\in ab\cap D', y_2\in cd \cap D$. The planks $P$ and $Q$ break edges of $R$ in subsegments of the lengths $g_1, g_2$ and  $h_1, h_2$, see figure below.  I will consider only the critical case depicted in the figure, where the segments $y_1x_2, x_2y_2, y_2x_1, x_1y_1$ pass through the four intersection points $z_1, z_2, z_3, z_4$ of the boundaries of planks (if this is not the case, one can modify $D'$, $P$ and $Q$ so that the total width of $P$ and $Q$ strictly decreases and they still cover $D'$).   
Consider the shaded rectanges as in the figure: Rectangles marked with the same color have the same area. The total area of the four shaded rectanges adjacent to the corners of $R$ equals 
$$
(h_1+h_2)(g_1+g_2)= (1-v)(1-w). 
$$
On the other hand, the area of the rectange $z_1 z_2 z_3 z_4$ equals $vw$ and, clearly, the total area of the shaded rectangles inside  $z_1 z_2 z_3 z_4$ equal $(1-v)(1-w)$. From this, we get
$$
(1-v)(1-w)\le vw
$$
$$
1\le v+w=W,
$$
as claimed. qed.  

Now, let $N\ge 3$ and assume that for each collection of $<N$ planks, $W\ge 1$. I will now consider only collections of planks where each plank is different from the entire $R$ (otherwise, the claim is clear). 
Consider a collection $C$ of planks which consists of $m$ vertical planks $P_i$ and $n$ horizontal planks $Q_j$, where $m\le n$ (otherwise we swap the vertical and horizontal coordinates) and $N=m+n\ge 3$. I number the vertical planks $P_i$ from left to right. Without loss of generality, we can assume that vertical planks are pairwise disjoint and horizontal planks are also pairwise disjoint (otherwise, we amalgamate the non-disjoint vertical/horizontal planks, without increasing the total width $W$ and reducing the number of planks). 
For the same reason, we can assume that any proper subset of $C$ does not cover $D$. 
The complement
$$
D\setminus \bigcup_{i=1}^m P_i
$$
is a union of convex sets $D_j$. I also number $D_j$'s from left to right. Each $D_j$ is covered by exactly one horizontal plank (by the disjointness property). Hence, the number $k$ of the convex sets $D_j$ is $\le n$. Furthermore, $k\in \{m-1, m, m+1\}$ since there exists exactly one $D_j$ between each consecutive pair $P_i, P_{i+1}$. Since $m\le n$, it follows that at most one of the planks $Q_i$ covers two different components $D_j$, which means that
$$
k-1\le n\le k. 
$$
By combining these inequalities, we see that
$$
m\le k\le m+1,
$$
and, hence, either $P_1$ or $P_m$ does not contain one of the two vertical sides of $R$. (I assume that it is $P_1$ which does not contain $ab$, otherwise, we change the orientation.) Lastly, if $k=m$ (i.e, $P_m$ contains $cd$) then $n=k=m$ and, hence, each plank $Q_i$ covers exactly one of the components $D_j$. In any case, I let $Q_1$ denote the horizontal plank covering $D_1$.  
Case 1: $k=n$, i.e., each $Q_i$ covers exactly one $D_i$, $i=1,...,n$. As we noted above, 
the  component $D_1$ has nonempty intersection with the vertical segment $ab$. Let $h_1$ denote the length of the projection $\pi_2(D_1)\subset ab$  and let $v_1$ denote the length of the projection $\pi_1(D_1)$ of $D_1$. Since $D_1$ is covered by a $Q_1$, it is clear that width of $Q_1$ is $\ge h_1$. 
Lemma 2. $v_1\le h_1$. 
Proof. Consider a point $x\in D_1\cap ab$ and the segment $yz=D_1\cap P_1$. The triangle $\Delta=xyz$ is contained in $D_1$. Moreover, convexity of $D$ and the fact that it intersects all four sides of $R$ imply that the rays from $x$ extending the segments $xy$ and $xz$ both intersect the union of the horizontal segments $bc\cup da$. It is then an elementary geometric calculation to conclude that the height $v_1$ (from the vertex $x$) of $T$ is $\le $ the length of the base $yz$ of $T$. The latter is $\le h_1$. qed.  
Let $P_0\subset R$ denote the vertical plank which is the product $\pi_1(D_1)\times ab$ ($P_0$ is not a part of our collection $C$ of planks). By Lemma 1, the width of $P_0$ is $\le$ the width of $Q_1$.
 Then replace $Q_1$ with $P_0\cup P_1$. Since $Q_1$ covers only the component $D_1$, it follows that the new collection $C'$ of planks still covers $R$. By Lemma 1, $C'$ has overall width $W'\le W$ and, by construction, $C'$ consists of $N-1$ planks. Therefore, by the induction assumption, $1\le W'\le W$ and we are done. This concludes the proof of Theorem 1 in Case 1. 
Case 2: $n=k-1$. Then $k=m+1$, $n=m$. 
Subcase 2a: $Q_1$ covers $D_1$ and (after renumbering if necessary) a distinct horizontal plank $Q_n$ covers $D_{n+1}$. Then we repeat the argument from Case 1 and conclude that $W\ge 1$. 
Subcase 2a: $Q=Q_1$ covers both $D_1$ and $D_{n+1}$. We then switch roles of horizontal and vertical directions (this is OK since $n=m$).  Repeating the above case-by-case analysis we conclude that there exists a vertical plank  $P:=P_\ell$ which covers two components $E_1, E_{n+1}$ of
 $$
 R\setminus \bigcup_{i=1}^n Q_i. 
 $$
 The components $E_1, E_{n+1}$ have nonempty intersections with the horizontal sides of $R$. 
 Now, pick from each subset $D_1, D_{m+1}, E_1, E_{n+1}$ one point $y_1, y_2, x_1, x_2$ which belongs to a side of $R$. 
Lemma 3. The convex hull $D'$ of $x_1, x_2, y_1, y_2$ is covered by the planks $P$ and $Q$. 
Proof. It suffices to show that every edge of $D'$ is contained in $K=P\cup Q$. Suppose that there exists an edge of $D'$ (after relabeling, we can assume that this edge is $x_1y_1$) which is not contained in $P\cup Q$. 
Let $x_1'\in x_1y_1$ and $y_1'\in x_1y_1$ be the points of intersection of the boundary of $K$ with $x_1y_1$  (these intersection points are labelled so that $x_1'$ belongs to the left side $s_1$ of $P$ and $y_1'$ belongs the bottom side $t_1$ of $Q$. Let $z_1'=s_1\cap t_1$; then the solid right angled triangle $x_1'y_1'z_1'$ is not covered by $P\cup Q$. However, by the disjointness property of planks, if follows that a small corner of $x_1'y_1'z_1'$ at the vertex $z_1'$ is not covered by any plank in $C$. 

Contradiction. qed. 
In view of Lemma 3, we can now apply Lemma 1 (since we have a covering of $D'$ by two planks) and conclude that the total width of $P$ and $Q$ is $\ge 1$. 
This concludes the proof of Theorem 1. qed 
I gave a proof of Theorem 1 for finite collections of planks. An easy approximation argument implies that the same conclusion holds for countable collections of planks. (For uncountable collections, the statement is meaningless, as far as I can tell.) 
Theorem 1 is a special case of much harder theorem (theorem 2 below) proven by T.Bang in
A solution to the "Plank Problem", Proceedings of AMS, 2 (1951).
The "Plank Problem" itself was posed by A.Tarski in 1932. 
Theorem 2: Let $D\subset R^d$ be a bounded convex subset and $C$ is a countable collection of planks covering $D$. Then the total width of $C$ is $\ge$ the width of $D$. 
Here are the definitions: A plank in $R^d$ is the closed region bounded by two parallel hyperplanes. The width of a plank is the distance between the hyperplanes. The total width of $C$ is the sum of width of planks in $C$. The width of $D$ is the infimum of lengths of its orthogonal projections to straight lines in $R^d$. 
Bang's proof is clever and nonelementary. It has led to a number of problems some of which are still open. Read here to find more about it. 
