$15$ sheets of paper of various sizes and shapes lie on a desktop covering it completely. The sheets may overlap and may even hang over the edge. 
$15$ sheets of paper of various sizes and shapes lie on a desktop covering it completely. The sheets may overlap and may even hang over the edge. Show that $5$ of the sheets may be removed such that the remaining $10$ sheets cover at least $\frac{2}{3}$rd of the desktop.

What I Tried :- If some other other user (who never saw it before) reads this problem, they will be quite shocked and surprised like me . The question is about Extremal Principle, but how can the problem be true in reality ?
I also made a Counter-Example of this claim (in paint) , showing that the area is not always $\frac{2}{3}$rd of the desktop or a table for convenience .
Suppose this is the table (the picture is not that good) , and you remove the parts labelled $(11,12,13,14,15)$ , then you will only be left with $(1,2,$...$,10)$, and by looking I can say that the area covered by those parts is definitely far far lesser than $\frac{2}{3}$rd of the table .
As usual if you ask me, I don't know a solution myself . Whenever there is some strange problem like this it's usual for me not to get up with a solution .
So does anyone have any thoughts about this problem?
Note :- Maybe it can happen I didn't understand the problem quite properly . In that case can someone guide me what this problem actually means?
 A: As I mentioned in my comment, the question says : there is a choice of five sheets such that when you remove these sheets, what's left covers more than $\frac 23$ of the area. It does not say that any choice of five sheets removed, will lead to more than $\frac 23$ area covered.
Of course the extremal principle will be involved : once you sort out the sheets by cutting out overlaps and hanging parts, you just need to look at the $5$ smallest parts by area (smallest is where I use the extremal principle) and remove them.
The rest of the ten sheets put together , must cover at least twice as much area of the desktop as these five sheets of smallest area (why?) Now it is obvious that the five sheets of smallest area cannot have total area more than $\frac 13$ of the desktop, which finishes it.

While the extremal principle may be used ingeniously on certain occasions(like in geometry), you can see where it gets used "obviously" : if you are maximizing area, you would of course remove the sheets of smallest area first.
