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Suppose there are n lines in the plane such that for any three of them there is a unit circle intersecting each of them. Prove that there is a unit circle intersecting all n lines.

How I have imagined them pictorially is that there will be some straight lines passing through the circle and some forming tangents. At first, I thought its an example of Helly's theorem but I read about Jung's theorem. And I thought maybe its an example of it. Moreover, to apply Helly's theorem we need to have convex sets which are nowhere mentioned explicitly in the question. This is what I thought. And I found the answer of it here prove that all line can be intersected with a unit circle where someone has asked the similar question and someone pointed out for n=2 it is false for parallel lines.

And I am still not able to understand it. I will be thankful if someone tells me.

Logo
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  • The failure for $n = 2$ is just a technicality in the statement. The requirement that every trio of lines has a unit circle intersecting all three is "vacuously true", since there are no trios of lines.So two parallel lines that are separated by a distance of more than $2$ satisfies the hypothesis, but there is no unit circle that intersects both. This is not anything insightful. It just takes advantage of a tiny gap left open by sloppy wording. A little more care in how the statement is said closes the gap and disallows this counter-example. – Paul Sinclair Dec 08 '20 at 02:11
  • The two proofs (first by Helly's theorem, then a direct proof) given by Dave in the other thread both seem servicable to me. If you can say where it is you lose the argument, I can help explain it. I would suggest asking Dave directly, but they apparently disappeared from this forum just two days after that post. – Paul Sinclair Dec 08 '20 at 02:29
  • @PaulSinclair what I basically understood from the first half is this: For the ith line, we can draw a closed, infinite strip Ci of width 2 with the ith line in the middle of this strip. In other words, Ci is the set of all points at a distance ≤ 1 from the ith line. These sets are convex by construction & since there is a unit circle intersecting any three lines, thus, Ci ∩ Cj ∩ Ck is not empty for i, j, k ∈ {1, ..., n}. – Logo Dec 08 '20 at 17:07
  • @PaulSinclair Now since we are dealing with $R^2$ and n ≥ 3 thus using Helly's theorem the intersection of all Ci's is non-empty, which implies, there is a unit circle intersecting all the n lines. – Logo Dec 08 '20 at 17:11
  • @PaulSinclair But I am not able to follow the second part where is considering parallel lines – Logo Dec 08 '20 at 17:13

1 Answers1

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The second part is a separate proof from Helly's theorem. It looks like Dave wrote this proof out first, then later realized Helly's theorem was applicable at the indicated point and added the edit.

For his own idea, Dave needs a pair of lines that are not parallel to each other. But of course, this isn't guaranteed, so he breaks the proof into two cases: Either

  1. All $n$ lines are parallel to each other, or
  2. There are at least two lines that intersect.

Then he examines the first case: If all the lines are parallel, then they have to be in order, with two outside lines and all the others inbetween. If you take the two outside lines and pick any of the middle lines (this is where the proof fails for the $n = 2$ case, as mentioned in the comments), then you have a triple of lines, and per the hypothesis, there is some circle intersecting all three. But the only way the circle can intersect both outside lines is to pass through all the lines in the middle. So it intersects every line. The parallel case is proved.

Next he proves the second case: Two of the lines $\ell_1$ and $\ell_2$ intersect. The cylinders of radius $1$ about them, $C_1$ and $C_2$ therefore intersect to form a parallelogram (actually, a rhombus). This is a closed polygon. Number the cylinders about the remaining lines as $3$ and on in some arbitrary order.

Now we note a couple things about the Sets $P_k = \bigcap_{i=1}^k C_i$, for any $k \ge 2$. Because the intersection of convex sets is also convex, each $P_k$ is convex. Further, if $P_{k-1}$ is a polygon and $P_k \ne \varnothing$, then $P_k = C_k \cap P_{k-1}$ is also a polygon. This is because the intersection just cuts off from $P_{k-1}$ the portions that extend beyond $C_k$'s two boundary lines. And since $P_2$ is a polygon, by induction, so are all the $P_k$ for $k > 2$.

Now suppose that $\bigcap_{k=1}^n C_k = \varnothing$. Then there has to be some largest $p < n$ such that $P_p = \bigcap_{k=1}^p C_k \ne \varnothing$. Now since $p$ is the largest possible, $P_p\cap C_{p+1} = \varnothing$, and in particular $P_p$ does not intersect the line $\ell_{p+1}$. Dave uses an optimization argument to prove the intuitively obvious fact that the closest point of the solid polygon $P_p$ to $\ell_{p+1}$ will be some vertex $v$ of $P_p$. Label the two boundary segments meeting at $v$ by $b_i$ and $b_j$. Because of how the polygon $P_p$ is formed, these segments are parts of the boundaries of two cylinders $C_i, C_j$ for some $i, j \le p$. Now $b_i$ and $b_j$ are at best parallel to $\ell_{p+1}$, or else they slant away from $\ell_{p+1}$ as you move away from $v$. Otherwise $v$ would not be the closest point of $P_p$ to $\ell_{p+1}$.

But the intersection $C_i\cap C_j$ is a rhombus lying on the same side of $b_i$ and $b_j$ as its subset $P_p$. Which means that $v$ is also the closest point of $C_i\cap C_j$ to $\ell_{p+1}$. But $v \notin C_{p+1}$, so it is a distance of more than $1$ away from $\ell_{p+1}$. Thus the same must be true of every point of $C_i\cap C_j$ to $\ell_{p+1}$. In particular, $C_i\cap C_j \cap C_{p+1} = \varnothing$, which means there can be no circle intersecting all three of $\ell_i, \ell_j, \ell_{p+1}$, a contradiction.

Paul Sinclair
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