Another geometrical problem

[gugo82]

The title says it all... I think also this one is kinda classical.

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Problem:

Let [tex]$D$[/tex] be the unit disc and [tex]$N\in \mathbb{N}$[/tex].
It is easy to see that exactly [tex]$N$[/tex] rectangles with height [tex]$2$[/tex] and width [tex]$\frac{2}{N}$[/tex] can be used to cover the whole disc; for example, they can be placed as in the following picture (where [tex]$N=5$[/tex]):
[asvg]xmin=-1;xmax=1;ymin=-1;ymax=1;
noaxes();
circle([0,0],1);
stroke="red";
rect([-1,-1],[1,1]);
line([-0.6,-1],[-0.6,1]); line([-0.2,-1],[-0.2,1]); line([0.2,-1],[0.2,1]); line([0.6,-1],[0.6,1]);[/asvg]
Trivially, any number [tex]$k>N$[/tex] of such rectangles can be used to cover the whole of [tex]$D$[/tex].

Prove that there's no chance to cover the whole disc [tex]$D$[/tex] with any number [tex]$k < N$[/tex] of rectangles with height [tex]$2$[/tex] and width [tex]$\frac{2}{N}$[/tex].

[gugo82]

The idea, or the trick, is to lift the problem in dimension [tex]$3$[/tex].

The circle [tex]$D$[/tex] may be interpreted as the section of a solid tridimensional sphere [tex]$S$[/tex] with any of its equatorial planes [tex]$\Pi$[/tex]; consequently, the rectangles may be thought as the traces on [tex]$\Pi$[/tex] of "slices" of [tex]$S$[/tex], each "slice" being made cutting [tex]$S$[/tex] with a couple of parallel planes at distance [tex]$h=\frac{2}{N}$[/tex] both orthogonal to [tex]$\Pi$[/tex].

Archimedes' hat-box theorem states that the surface area of each "slice" equals [tex]$2\pi\ h=\frac{4\pi}{N}$[/tex].

And now...

Proof: Assume by contradiction that it is actually possible to cover the whole of [tex]$D$[/tex] with [tex]$k < N$[/tex] rectangles; then the sum of the surface areas of the correspondig "slices" equals [tex]$4\pi\ \frac{k}{N}$[/tex]. But [tex]$4\pi\ \frac{k}{N}$[/tex] is strictly less than the total surface area of [tex]$S$[/tex] (which is [tex]$4\pi$[/tex] of course) and this yields a contradiction.
In fact, if [tex]$\xi \in \partial S$[/tex] then [tex]$x:=\text{proj}_\Pi \xi \in D$[/tex] and therefore exists a [tex]$2\times \frac{2}{N}$[/tex] rectangle [tex]$R$[/tex] s.t. [tex]$x\in R$[/tex]; this fact implies that [tex]$\xi$[/tex] lies in the curved part of the boundary of the "slice" corresponding to [tex]$R$[/tex]. Hence the class of the curved parts of the "slices" boundary covers [tex]$\partial S$[/tex] and (by standard measure theoretic arguments) the sum of their surface areas has to be greater than or equal to [tex]$4\pi$[/tex].


This problem was proposed by prof. Hajłasz during SMI 2010 Summer Course in Cortona.
The solution was found by Cezar Lupu.

[gugo82]

Uh-oh, I fogot to give you a...

Hint: Use Archimedes' hat-box theorem in a very creative way.