Six circles (including their circumferences and interiors) are arranged in the plane so that no one of them contains the center of another. Prove that they [the six circles] cannot have a point in common.
This should be solved by application of pigeonhole principle. Don't have the right idea.
Let $B_i$ denote the $i$-th ball. Assume $P$ to be one particular point satisfying
$$P \in \bigcap_{i=1}^6 B_i$$
Then since $P\in B_i \quad\forall\ i$, we must have that the centers, call them $m_i$ lie within $B_P$, the ball of the same radius around $P$.
Now we take the balls of half radius $B'_i$ and apply the pigeonhole principle to six circles of radius $\frac r2$ (the $B'_i$) inside a circle of radius $r$ ($B_P$) and the constraint that no $B'_i$ may touch another.
