Calculating the minimum size of some intersection (Olympiad problem)... any hints or solution? Please PROBLEM: 
A person has a coat of area $1$ composed of five possibly overlapping patches. The area of each patch is at least $\frac{1}{2}$. Prove that there are two patches whose overlap has area at least $\frac{1}{5}$.
IDEAS THAT HAVE COME TO MY MIND: 
-Using the inclusion-exclusion principle to have an equality involving the intersections of the patches
-Trying to argue by contradiction (assuming the intersection of any two is less than $\frac{1}{5}$). I found out  that the sum of the areas of the patches must be between $\frac{5}{2}$ and $\frac{6}{2}$.
 A: (Note: this argument is reconstructed from writing the problem as a linear program in $2^5$ variables, solving the dual, and then interpreting the dual solution.)

Define $P_i$ for $1 \le i \le 5$ to be the size of patch $i$, and $P_{ij}$ for $1 \le i < j \le 5$ to be the size of the overlap between patches $i$ and $j$. We measure in units of coats, so that $\frac12$ is half the coat's area. 
Let $z = \max_{i,j} P_{ij}$, the size of the largest overlap. Since $P_i \ge \frac12$ for all $i$, and $P_{ij} \le z$ for all $i$ and $j$, we have $$\frac15 \sum_i P_i - \frac1{10} \sum_{i<j} P_{ij} \ge \frac12 - z.\tag{$\star$}$$ 
Now we look at how much each bit of coat contributes to the left-hand side of $(\star)$. (By "bit of coat" we can mean a part of the coat specified by which patches it's in.)
If the bit of coat is in $k$ patches, it's contained in $\binom{k}{2}$ overlaps, so it contributes $\frac15 k - \frac1{10}\binom{k}{2} = \frac{k(5-k)}{20}$ of its area, which is maximized at $k=2$ or $k=3$ by $\frac{3}{10}$. If the left-hand side of $(\star)$ counts each bit of coat with a multiplier of at most $\frac{3}{10}$, then its value can be at most $\frac{3}{10}$ of the coat's total area.
Therefore $(\star)$ tells us that $\frac{3}{10} \ge \frac12 - z$, or $z \ge \frac15$, and we're done. (In fact, the average size of an overlap, not just the maximum, must be at least $\frac15$.)

The primal solution to the same linear program shows that this bound is tight. Divide the coat into $10$ regions $\{R_1, R_2, \dots, R_{10}\}$ of equal size. Let 
\begin{align}
 A &= R_1 \cup R_2 \cup R_6 \cup R_7 \cup R_8 \\
 B &= R_3 \cup R_4 \cup R_6 \cup R_7 \cup R_9 \\
 C &= R_1 \cup R_3 \cup R_8 \cup R_9 \cup R_{10} \\
 D &= R_4 \cup R_5 \cup R_6 \cup R_8 \cup R_{10} \\
 E &= R_2 \cup R_5 \cup R_7 \cup R_9 \cup R_{10}
\end{align}
be the five patches, and you can check that each patch has area $\frac12$ but intersects with every other patch in at two regions, so each overlap has area $\frac15$.
