sliding a shape with area= 5 on a grid so it covers at least 6 of it's points - riddle place a shape on an integer coordinates grid, which is continuous without holes, and that its area is 5. Explain why you can slide it (without twisting or warping it), so that it will cover at least six points of the grid?
 A: Let $A$ be any compact set with area 5. Fix a point $P$ in $A$, and define the function $f:[0,1]\times[0,1] \to \mathbb Z_{\ge0}$ by
$(x,y) \mapsto$ the number of lattice points covered by $A$ when $P$ is translated to $(x,y)$
Then the integral of $f$ over $[0,1]\times[0,1]$ is 5. So unless $f = 5$ almost everywhere, $f(x,y)$ must be $\ge$ 6 for some $(x_0,y_0)$.
Now, suppose that $f(x,y) < 5$ for some $(x,y)$. $A$ is compact, so its complement is open. Thus, when $P$ is at $(x,y)$, all lattice points that lie outside $A$ are bounded away from $A$. Hence there is a neighbourhood of $(x,y)$, with positive area, in which $f$ remains less than 5. (This uses the fact that $A$ is bounded, so only a finite number of lattice points can be close to $A$.) Therefore $f$ is not equal to 5 almost everywhere.
So now all we have to show is that $f$ can't be equal to 5 everywhere. Suppose otherwise, and imagine sliding $P$ from $(0,0)$ to $(1,0)$. At every point there are exactly 5 lattice points covered by $A$. But the set of lattice points covered when $P=(0,0)$ is not the same as the set of lattice points covered when $P=(1,0)$. So at some point $P=(a,0)$, a lattice point $L_1$ must get left behind by $A$ as it moves to the right. To compensate, a new lattice $L_2$ point must be consumed by $A$. But then at $P=(a,0)$, both $L_1$ and $L_2$ are in $A$, and we have $f(a,0) \ge 6$. QED
Note: It may be that more than one lattice point leaves $A$ at $P=(a,0)$. But then the same number of lattice points must be consumed by $A$, and the same argument applies.
A: Start with a shape that is 100 x 1/20 rectangle, and stick it on the x coordinate line. That covers 100 points 
