prove that all line can be intersected with a unit circle

Question

I need to solve this problem.. from geometry. suppose there are n lines in the plane such that every three of them can be intersected with a unit circle. Prove that all of them can be intersected with a unit circle.

Answer 1

For the $i$th line, draw a closed, infinite strip $C_i$ of width $2$ with the line down the middle. In other words, $C_i$ is the set of all points at a distance $\leq 1$ from the $i$th line. The problem is to show that if the $C_i$'s have empty intersection, then three of them already have empty intersection.

[EDIT: At this point, we can stop the proof, as the result follows directly from Helly's theorem: http://en.wikipedia.org/wiki/Helly%27s_theorem .]

The problem is easy if all the lines are parallel (the two outermost ones are at a distance $\leq 2$ from each other), so we can assume that there are two intersecting lines, which we can assume are the first two. Thus $C_1 \cap C_2$ is a parallelogram.

Now let $p$ be the last integer for which $K = \cap_{i = 1}^{p} C_i \ne \emptyset$. $K$ is a compact, convex set with a polygonal boundary, and it is at a strictly positive distance from $C_{p+1}$. Let $a$ be a point with minimal distance to $C_{p+1}$, which exists since $K$ is compact. We may assume that $a$ is one of the vertices of $K$.

Then the sides of $K$ that meet at $a$ belong to two different, non-parallel strips $C_i$ and $C_j$. Neither of these sides can point from $a$ towards $C_{p+1}$, because $a$ is minimally distant from $C_{p+1}$. Thus the intersection of $C_i$ and $C_j$ is a parallelogram containing $K$, with a vertex at $a$, and sides extending the sides of $K$ that meet at $a$.

It follows from these facts that $C_i \cap C_j$ already has empty intersection with $C_{p+1}$.