TITLE: The Plank Problem 2 Dimensions
QUESTION [25 upvotes]: 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)

REPLY [8 votes]: Here's an alternative proof, using a probabilistic 'coupling' method, via the following Lemma. Throughout, I let $D$ be as in the question: a convex subset of $R=[0,1]^2$ intersecting each of the four sides of $R$.

Lemma: There exists random variables $U_1,U_2$ (defined on some probability space) which are both uniform on $[0,1]$, and such that $(U_1,U_2)$ lies in $D$ with probability 1.

[Note: Although $U_1$ and $U_2$ here have the uniform distribution, there is no requirement for the joint distribution of $(U_1,U_2)$ to be uniform, or even have an absolutely continuous distribution. For example, if $D$ Is one of the diagonals of the unit square, which has zero Lebesgue measure but satisfies the requirements, then we would take $U_1=U_2$ (or $U_1=1-U_2$). This ensures that $(U_1,U_2)\in D$ and, if $U_2$ is chosen with the uniform distribution, then $U_1$ is also uniform.]
If $A\subseteq[0,1]$ is the set of $x$-coordinates of the vertical planks, then it is a finite union of closed intervals, the vertical planks cover the region $A\times[0,1]$ and have total area $\lambda(A)$ ($\lambda$ is Lebesgue measure). Similarly, if $B\subseteq[0,1]$ is the set of $y$-coordinates of the horizontal planks then they cover the region $[0,1]\times B$ and have total area $\lambda(B)$. The following corollary of the above lemma is then a positive answer to the original question.

Corollary: If $A,B$ are measurable subsets of $[0,1]$ such that $D\subseteq(A\times[0,1])\cup([0,1]\times B)$, then $\lambda(A)+\lambda(B)\ge1$.

Proof: Letting $U_1,U_2$ be random variables uniformly distributed on $[0,1]$ with $(U_1,U_2)\in D$ (almost surely),
\begin{align}
\lambda(A)+\lambda(B)&=\mathbb{P}(U_1\in A)+\mathbb{P}(U_2\in B)\\
&\ge\mathbb{P}(U_1\in A{\rm\ or\ }U_2\in B)\\
&=\mathbb{P}\left((U_1,U_2)\in(A\times[0,1])\cup([0,1]\times B)\right)\\
&\ge\mathbb{P}\left((U_1,U_2)\in D\right)\\
&=1.
\end{align}

I'll give a proof of the lemma now, which will require going through several cases. Letting $\pi_i\colon\mathbb{R}^2\to\mathbb{R}$ ($i=1,2$) denote projection onto the $x$ and $y$ coordinates respectively, $\pi_i(x_1,x_2)=x_i$, we need to find a random variable $X$ taking values in $D$ such that $\pi_1(X)$ and $\pi_2(X)$ are both uniform on $[0,1]$. The lemma will then be answered by taking $U_i=\pi_i(X)$.
I'll make repeated use of the fact that if $X$ is uniformly distributed on one of the diagonals of a rectangle $[a_1,b_1]\times[a_2,b_2]$ then $\pi_i(X)$ is uniformly distributed on $[a_i,b_i]$. The idea is to find a finite set of rectangles $R^{(k)}=[a^{(k)}_1,b^{(k)}_1]\times[a^{(k)}_2,b^{(k)}_2]$ and numbers $p^{(k)}\ge0$ summing to $1$ such that each $R^{(k)}$ has a pair of opposite corners in $D$ (and, by convexity, a diagonal in $D$), then we can choose $X$ to be uniformly distributed on the diagonal of $R^{(k)}$ with probability $p^{(k)}$. Then, $U_i=\pi_i(X)$ is uniformly distributed on the interval $[a^{(k)}_i,b^{(k)}_i]$ with probability $p^{(k)}$, so has probability density
$$
p_i(x)=\sum_kp^{(k)}(b^{(k)}_i-a^{(k)}_i)^{-1}1_{\{a^{(k)}_i < x\le b^{(k)}\}}.
$$
I'll do this so that the probability density is $1$ on $(0,1]$.
First, let $P_1=(x_1,0)$, $Q_1=(0,y_1)$, $P_2=(x_2,1)$ and $Q_2=(1,y_2)$ be points of intersection of $D$ with the edges of $R$. Then, the quadrilateral $P_1Q_1P_2Q_2$ is contained in $D$.
By symmetry, we suppose that $x_1\le x_2$, $x_1\le y_1$. There are a few cases to consider (I include diagrams, with the shaded area being the convex hull of the points $P_1,Q_1,P_2,Q_2$, which is contained in $D$, and the red lines are the rectangle diagonals supporting the distribution of $X$).

Case I: $y_1\ge y_2$
In this case,the four rectangles with diagonals $P_1Q_1$, $Q_1P_2$, $P_2Q_2$, $Q_2P_1$ do not intersect. Together with the center rectangle, $[x_1,x_2]\times[y_1,y_2]$, which is contained in $D$ they partition $R$. Explicitly, the rectangles $[0,x_1]\times[0,y_1]$, $[0,x_2]\times[y_1,1]$, $[x_2,1]\times[y_2,1]$, $[x_1,1]\times[0,y_2]$ and $[x_1,x_2]\times[y_2,y_1]$ partition $R$ and each have a diagonal in $D$. So, we can choose $X$ to be uniformly distributed on a diagonal of any these rectangles with probability equal to the area of the rectangle.
In the remaining cases, the rectangles with diagonals $Q_1P_2$ and $Q_2P_1$ overlap, so the construction in Case I does not work.
Case II: $x_1\le y_1\le x_2\le y_2$. In this case $P_1Q_1$ is the diagonal of a rectangle which is taller than in is wide. We add a rectangle to the right of this to make a square, and take diagonals of these. Similarly, we add a rectangle below the one with diagonal $P_2Q_2$ to make a square, and take diagonals of these. These two squares do not overlap, so we can add the line segment in $D$ joining the nearest corners of these squares (which is a diagonal of the square $[y_1,x_2]^2$). More explicitly, using 'BL-TR' to mean 'bottom-left to top-right diagonal', etc.

$R^{(1)}=[0,x_1]\times[0,y_1]$, $p^{(1)}=x_1$ (TL-BR = $P_1Q_1$ is in $D$).
$R^{(2)}=[x_1,y_1]\times[0,y_1]$, $p^{(2)}=y_1-x_1$ (BL-TR is in $D$).
$R^{(3)}=[y_1,x_2]\times[y_1,x_2]$, $p^{(3)}=x_2-y_1$ (BL-TR is in $D$).
$R^{(4)}=[x_2,1]\times[x_2,y_2]$, $p^{(4)}=y_2-x_2$ (BL-TR is in $D$).
$R^{(5)}=[x_2,1]\times[y_2,1]$, $p^{(5)}=1-y_2$ (BR-TL = $P_2Q_2$ is in $D$).

Case III: $x_1\le y_1\le y_2\le x_2$. This is a similar arrangement to Case II except that the rectangle with diagonal $P_P2Q_2$ is taller than it is wide, so we add a rectangle to the left of this rather than below to make a square.

$R^{(1)}=[0,x_1]\times[0,y_1]$, $p^{(1)}=x_1$ (TL-BR = $P_1Q_1$ is in $D$).
$R^{(2)}=[x_1,y_1]\times[0,y_1]$, $p^{(2)}=y_1-x_1$ (BL-TR is in $D$).
$R^{(3)}=[y_1,y_2]\times[y_1,y_2]$, $p^{(3)}=y_2-y_1$ (BL-TR is in $D$).
$R^{(4)}=[y_2,x_2]\times[y_2,1]$, $p^{(4)}=x_2-y_2$ (BL-TR is in $D$).
$R^{(5)}=[x_2,1]\times[y_2,1]$, $p^{(5)}=1-x_2$ (TL-BR = $P_2Q_2$ is in $D$).

Case IV: $x_1\le x_2 < y_1 < y_2$. In this case, any two squares (with standard orientation) containing diagonals $P_1Q_1$ and $P_2Q_2$ overlap, so the construction in cases I and II does not work.
The point $(x_2,y_1)$ is in $D$, and there exists unique $a_0\in[0,y_1]$, $b_0\in[x_2,1]$ such that $(x_2,a_0)$ and $(b_0,y_1)$ are on $Q_2P_1$. Explicitly, $a_0=y_2\frac{x_2-x_1}{1-x_1}$ and $b_0=y_2^{-1}(y_1+x_1(y_2-y_1))$.
Choosing $a\in[a_0,y_1]$ and $b\in[x_2,b_0]$, so that $(x_2,a)$ and $(b,y_1)$ are in $D$, take

$R^{(1)}=[0,x_1]\times[0,y_1]$, $p^{(1)}=x_1$ (TL-BR = $P_1Q_1$ is in $D$).
$R^{(2)}=[x_1,x_2]\times[0,a]$, $p^{(2)}=x_2-x_1$ (BL-TR is in $D$).
$R^{(3)}=[x_2,b]\times[a,y_1]$, $p^{(3)}=y_1-x_2$ (BL-TR is in $D$).
$R^{(4)}=[b,1]\times[y_1,y_2]$, $p^{(4)}=y_2-y_1$ (BL-TR is in $D$).
$R^{(5)}=[x_2,1]\times[y_2,1]$, $p^{(5)}=1-y_2$ (BR-TL = $P_2Q_2$ is in $D$).

It can be seen that using this distribution, $U_2=\pi_2(X)$ is uniform on $[y_1,1]$ with probability $1-y_1$ and is uniform on each of $[0,a]$ and $[a,y_1]$. The probability of $U_2$ being in $[0,a]$ is $x_1\frac{a}{y_1}+(x_2-x_1)$. For $U_2$ to be uniform on $[0,1]$ we need to solve $x_1\frac{a}{y_1}+(x_2-x_1)-a=0$. It can be checked that this is negative for $a=a_0$ and positive for $a=y_1$, so it can be solved for some $a$ in $[a_0,y_1]$. Explicitly, $a=y_1\frac{x_2-x_1}{y_1-x_1}$. Similarly, choosing $b=(y_2-x_2)^{-1}(y_1+x_2(y_2-y_1-1))$ gives the uniform distribution for $U_1=\pi_1(X)$.