Sum of circles in a square 
Within a square with side length $1$ there are a finite number of circles that are allowed to overlap.
The sum of all the circumferences of the circles is equal to $10$.
Conjecture: There is a straight line which intersects or touches at least four of these circles.

So I tried to approach this problem by showing, that $3$ circles cannot provide the circumfence sum $S_C=10.$ I want to prove this problem with more general terms, so let the side of the square be $a$ so that $S_C=10a.$ I guess this should not affect the solution. Moreover the largest circle with radius $r=\frac{a}{2}$ has a circumfence of $C_{1}=2\pi\cdot \frac{a}{2}=\pi a$ and as $3\pi a < 10a,$ we need $4$ circles. Now I tried different things like putting other large circles in the square but I do not know how to go on with the proof and how to show the conjecture with the line.
Maybe someone of you has got a nice idea. Thanks in advance. P.S.: This is a german math olympiad problem from 2007/2008.
 A: $\textit{Solution}$
Let $A$ and $B$ be two adjacent vertices of the square. Tightening up the problem, show that there is a straight line perpendicular to $A B$ with the required property.
If the circles are projected perpendicular to the line $\overline{A B}$, one obtains (possibly overlapping) partial lines of the line $\overline{A B}$ (see Figure).

The length of such a line coincides with the diameter of the respective circle. A straight line perpendicular to $A B$ intersects such a section exactly when this straight line meets the corresponding circle.
The existence of a straight line perpendicular to $\overline{A B}$ that meets at least four circles is thus equivalent to the existence of a point $C$ on the line $\overline{A B}$ that belongs to at least four of these partial lines.
We now assume that there is no such point $C$. Then every point of $\overline{A B}$ belonged to at most three partial distances. The sum of the lengths of all partial distances is therefore at most $3 \cdot|A B|=3$. Consequently, the sum of the circumferences of the circles is not greater than $3 \pi$ and thus less than $10$. Since this contradicts the premise, the assertion is proved. $\blacksquare$
$\textit{Remark 1}$
The assertion of the problem can be generalised as follows: Let the width of a convex figure be the width of the narrowest strip that can accommodate this figure. If there are finitely many convex figures within a figure of width $h$ and the sum of their widths is greater than $k \cdot h$, then there is a straight line that meets at least $k+1$ of these figures.
$\textit{Remark 2}$
To show that the sum of the lengths of the above partial lines is at most $3 \cdot|A B|$, one could argue as follows:
Only finitely many points of $\overline{A B}$ are endpoints of the given partial distances. These points decompose $\overline{A B}$ into finitely many new sublines, each of which contains no further endpoint. Each of these new partial lines is contained in at most three of the original partial lines, and their total length is $|A B|$. The sum of the lengths of the original partial lines is thus not greater than three times the sum of the new partial lines. $\diamond$
