Circles on the plane such that every line intersects at least one of them but no line intersects more that 100 of them

Question

I have a serious problem with this problem: Is it possible to Draw circles on the plane such that every line intersects at least one of them but no line intersects more that 100 of them !?

Any help or suggestion would be helpful.

Answer 1

I have only a partial (negative) solution.

I can prove that there is no chain of circles of infinite length. That is, no sequence of circles without repetitions, where every two consecutive circles intersect or touch and the sum of the radii of the circles diverges. In particular, this rules out the suggestions with hyperbolas or parabolas, if I understand them correctly.

Proof by contradiction. Let $C$ be one of the circles of the chain and let $P$ be its center. We will prove that some ray starting in $P$ intersects infinitely many circles. Let's call $P$-rays the rays starting at $P$.

A circle of radius $r$ with center at distance $\ell$ from $P$ intersects at least $2r/(\ell\cdot 2\pi)$ fraction of the $P$-rays. (This is not true if $r>\ell\cdot \pi$, of which we take care at the end.)

Let $r_i$ be the sequence of radii of the circles along the chain starting with $C$. Then the distance $\ell_i$ between $P$ and the center of the $i$-th circle is at most $2(r_1+r_2+\cdots + r_i)$. So the $i$-th circle blocks at least $(1/(2\pi)) \cdot r_i/(r_1+\cdots +r_i)$ fraction of $P$-rays. Since the chain has infinite length, the sum $r_1 + r_2 + \cdots$ diverges. Then also the sum of $r_i$ divided by partial sums diverges. That is, the sum of fractions of blocked $P$-rays goes to infinity and some $P$-ray is blocked infinitely many times.

We still need to do something with the circles of the chain with $r>\ell\cdot \pi$. First of all, $C$ is one of them: we counted that it blocks $(1/(2\pi)) \cdot r_1/r_1$ fraction, and that's true. For others, we may have counted that they block many times more than 100% percent of the $P$-rays. But since each such circle contains $P$ and crosses every $P$-ray, there are at most $99$ of such circles. So the excess counted to the sum is finite and the corrected sum still diverges.