Circles inside a square 
There are several circles inside a square of side length $1$. The sum of the circumferences of the circles is $10$. Prove that there exists a line that intersects at least $4$ of the circles.

We need to solve this using expected value, I tried making some diagrams, but gained nothing out of it. I really need some help.
 A: Since we need to solve this using expected value, it is more "useful" to think of probability. And we are lucky that the square is a unit square:

Therefore, the probability of a line intersecting a circle, is simply the diameter of that circle (think of vertical movement of the green line!). Let's move on: We know from the question $$\sum_i \pi d_i=10\implies \sum_i d_i=\dfrac{10}{\pi},$$ where $d_i$ are diameter of each circle $i$. Then $$\mathbb{E}[\text{line intersects with circle}]=\mathbb{E}[C_1 \text{ intersect}]+\mathbb{E}[C_2 \text{ intersect}]+\cdots+\mathbb{E}[C_n \text{ intersect}]=\sum_i \mathbb{E}[C_i \text{ intersect}]=\sum_i d_i=\dfrac{10}{\pi}\approx \dfrac{10}{3.14}>3$$
This reads: it is expected that there are more than 3 circles that intersect with the line. We can then conclude that there exists a line that intersects at least 4 of the circles.
A: Let's say the radii of the circles are $r_1, \dots, r_n$.  We know that $2 \pi \sum r_i = 10$  Consider a random vertical line uniformly distributed on support of the square.  The probability of hitting circle $i$, say, is exactly $2r_i/ 1$.
Therefore, the expectation of the sum of indicators for all the circles (which by linearity is the expected number of intersects), is
$ 2 \sum r_i = 10/ \pi > 3$.  Since the number of intersections must be an integer, there exists a vertical line that intersects at least 4 circles.
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. $\diamond$
