The famous result by Bang is that if a convex compact set $K\subset \mathbb{R}^n$ is covered by a finite number of open planks, then the sum of their widths is greater than a width of $K$. [The closed, correspondingly open, plank with normal $\theta$ of width $h\geqslant 0$ is the set of points lying between, correspondingly strictly between, two planes at distance $h$, both orthogonal to a unit vector $\theta$. The width $w(K)$ of $K$ is defined as the minimum of widths of closed planks containing $K$.]

If $n=2$ and $K$ is a unit disc, there is a short proof using the lifting to the third dimension, also mentioned in the answer by Noam Elkies : considering $K$ as a section of the unit ball in $\mathbb{R}^3$, for any plank $S$ of width $h$ its lifting $S\times \mathbb{R}=\{(s,x)\in \mathbb{R}^3: s\in S, x\in \mathbb{R}\}$ intersects the unit sphere by a set of area (at most) $2\pi h$ (this fact belongs to Archimedes himself). Since the whole unit sphere, which has area $4\pi$, must be covered by the liftings of our planks, we immediately get the desired lower bound 2 for the sum of their widths, it is strict for open planks.

Now about the general case, we again use the lifting but differently.

We use the following

**Lemma.** If $K\subset \mathbb{R}^n$ is a convex compact set and $f\in \mathbb{R}^n$, $\|f\|\leqslant w(K)=:h$, then

a) $K\cap (K+f)\ne \emptyset$;

b) $w(K\cap (K+f)) \geqslant h-\|f\|$.

**Proof.** a) Assume the contrary. Then by Hahn -- Banach $K$ and $K+f$ may be separated by a plane $\langle x,\theta\rangle=c$. That is, $\langle x,\theta\rangle< c<\langle x+f,\theta\rangle$ for any $x\in K$. Thus $K$ may be covered by an open plank of width $\langle f,\theta\rangle \leqslant \|f\|\leqslant h$, a contradiction.

b) Denote $g=f\cdot \frac{h}{\|f\|}$ (if $f\ne 0$, the case $f=0$ is trivial). Then $\|g\|=h$ and by a) there exists a point $a\in K\cap (K+g)$. We have by convexity $$\frac{h-\|f\|}h(K-a)\subset K-a,\\ \frac{h-\|f\|}h(K+g-a)\subset K+g-a,$$
that is equivalent to $a+\frac{h-\|f\|}h(K-a)\subset K\cap (K+f)$.
Therefore $w(K\cap (K+f))\geqslant w(a+\frac{h-\|f\|}h(K-a))=h-\|f\|$.

Now assume that $\sum h_i\leqslant h=w(K)$ and the open planks $S_i=\{x:|\langle x-x_0,\theta_i\rangle|< \frac{h_i}2 \}$, $i=1,\ldots,N$, cover $K$. In other words, we assume that there exists a point, called $x_0$, which belongs to all the middle planes of the planks ($x_0$ may belong to $K$ or not).

The $2^N$ sets
$K\pm \frac{h_1}2 \theta_1 \pm \frac{h_2}2 \theta_2\pm \ldots \pm \frac{h_N}2 \theta_N$ have a non-empty intersection: this follows from applying Lemma $N$ times (we start with $w((K-\frac{h_1}2\theta_1)\cap (K+ \frac{h_1}2\theta_1))=w(K\cap (K+h_1\theta_1))\geqslant h- h_1$ and proceed naturally, using the obvious inclusions like $(A\cap B)+x\subset (A+x)\cap (B+x)$.)

So, for certain $p\in \mathbb{R}^n$, the set $\Omega=\{p\pm \frac{h_1}2 \theta_1 \pm \frac{h_2}2 \theta_2\pm \ldots \pm \frac{h_N}2 \theta_N\}$ is contained in $K$. Choose the point $q\in \Omega$ on the maximal distance from $x_0$. We should have $|\langle q-x_0,\theta_i\rangle| <h_i/2$ for some $i$, and this implies (easily seen from the picture) that both points $q+h_i\theta_i$, $q-h_i\theta_i$ are further from $x_0$ than $q$. But one of these two points belongs to $\Omega$, a contradiction.

Now a general case. Assume that $K$ is covered by $N$ planks. If the normals of our planks are linearly independent, there middle planes have a common point and we are done. If $N\leqslant n$, we may move our planks a bit so that their normals become linearly independent and they still cover $K$. Finally, if $N>n$, we lift $K$ to a cylinder $C:=K\cdot [0,M]^{N-n}\subset \mathbb{R}^N$ (where $M$ is so large that $w(C)=w(K)$, $M=h$ is enough) and lift the planks $S_i$ to $S_i\times \mathbb{R}^{N-n}$. The problem is reduced to the case which is already done.

20more comments