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[Virex]

The answer depends on your definition of a line. If it's a half-infinite line, going from (0,0) in a certain direction but never stopping, then the minimum number is 6:


You can completely encase the center circle with 3 circles, however, this leaves 3 gaps, namely those for which the line goes straight through the points where those 3 circles touch. These lines can be caught by adding 3 more circles so that those 3 lines cross them.


If you're talking about finite length lines with l > r and l freely chosen, then the number of circles required is infinite, this because it is impossible to perfectly tile a surface with a finite amount of circles, there will always be gaps between the circles surrounding the center circle.

[Another]

I guess that I would better illustrate my idea.

Spoiler (click to show/hide)

That red dot in the middle is actually a circle of radius 0.1. Green circles with radii 20 have about 0.1 gaps between them through which the blue lines were drawn. The room to move the lines is not enough that any of them could be touching the red circle. I can provide precise numbers used to plot this if it is of any interest.

If the additional circles can be made arbitrary close to each other than the their minimum radius R (assuming all 3 are equal) can be found by making the target circle tangential to 2 of the lines and one of the bigger circles. Alternatively at some minimal separation distance between the big circles all points inside become "visible" from outside.

[Christes]

The issue that Virex is bringing up is essentially the same one that I was.  If by circle, ed boy meant the interiors (what I would call open disks), then Virex is saying that a line can slip through between the two circles.

[G-Flex]

The answer depends on your definition of a line. If it's a half-infinite line, going from (0,0) in a certain direction but never stopping, then the minimum number is 6:


You can completely encase the center circle with 3 circles, however, this leaves 3 gaps, namely those for which the line goes straight through the points where those 3 circles touch. These lines can be caught by adding 3 more circles so that those 3 lines cross them.


If you're talking about finite length lines with l > r and l freely chosen, then the number of circles required is infinite, this because it is impossible to perfectly tile a surface with a finite amount of circles, there will always be gaps between the circles surrounding the center circle.

There's a problem with terminology here.

A line extends infinitely in both directions. A ray extends infinitely but only in one direction (from a starting vertex). A line segment is finite in length.

[Christes]

I've heard the term "line" colloquially used to refer to all three.  Also, ed boy referred to the line as having an "end point", so I assume he meant ray.

******

Incidentally, if tangent circles are bad (either because ed boy meant just the interiors of the circles or being tangent counts as "overlapping"), then it should be possible with four.  Here's a picture:

Spoiler (click to show/hide)

Edit: scratch that - I think it's possible with 3: small, big and VERY big.  Here's a pic:

Spoiler (click to show/hide)

The blue one is a circle of radius 10000.  The arrangement will work for a very small target circle (such as radius .001).  It can be scaled up.

So I think the optimal answer is still going to be three - assuming that I understand the problem correctly.  It clearly requires more than 2.

[Another]

I think that the original problem how I understand it can has a somewhat practical interpretation:
"Given N large cylindrical barrels of radius R (or different R_i) - could you hide a smaller cylindrical barrel of radius r inside them? The barrels are not allowed to perfectly touch and there is always a visible gap between them."

For the solution with 3 large enough equal circles geometry tells me that it is possible when r<(1-sqrt(8 )/3)R~0.0572R.

There are also arrangements of 3 squares or 3 triangles without touching that leave some area inside totally obscured. Who knows, maybe you will need it one day to hide yourself .

[MagmaMcFry]

Next question: How many open spheres do you need to hide a point P in n-dimensional space?

Spoiler: Obscure Terminology (click to show/hide)

"Hide" as in "block the point's view of infinity in every direction" as in "every ray starting at P intersects at least one sphere".
An open sphere is a sphere interior not including the surface, so tangent lines aren't blocked.

Spoiler: Answer (click to show/hide)

n+1.
You can split the directions to infinity into n+1 pieces with n-simplex symmetry. Every one of these pieces of infinity fits into one hemisphere of all the directions, even leaving some space thereby. So you can block the point's view of a single piece by simply putting a sphere close enough to the point, but not touching P. Now you can shrink the sphere about P, until the new sphere fits between the old sphere and P. Repeat this until you have n+1 spheres of greatly varying size. Now distribute these n+1 spheres equally about the n+1 pieces of infinity. Then every direction is in at least one piece of infinity, therefore blocked by at least one sphere.

[Christes]

Heh, that's basically a generalization of the algorithm I used to make my last solution.  Sweet.

[MagmaMcFry]

Here's another nice problem I found somewhere a while ago. Have at it!

Let S be a finite set of at least two points in the plane, with no three points collinear. Let l be a line (no ends) that passes through exactly one point P of S. Now rotate l clockwise around P, until l first meets another point Q in S, and continue rotating l clockwise around Q. Repeat this process indefinitely. This infinite process is called a windmill about S, and a complete windmill is a windmill that uses every point in S as a pivot point infinitely often. Show that for every S there exists a complete windmill.


PS: I don't think that this still classifies as "Mathematics Help".

[ed boy]

I don't think that works. Imagine a pentagon with another point in the middle. The only possible time to rotate around the centre point is if you start with it, and that only happens once.

[MagmaMcFry]

Well, let's call the point in the middle C, and then the points outside 1, 2, 3, 4, 5 (clockwise). Then the rotation order is like this, it helps to draw this:

C, 1, 4, C, 2, 5, C, 3, 1, C, 4, 2, C, 5, 3, C, 1, 4, etc., and that's a complete windmill.

Remember: You don't rotate a ray, but a line, and a line has two non-ends you have to watch out for.

[Christes]

In other words, there exists a complete windmill, but not every starting line induces one.

[ed boy]

In other words, there exists a complete windmill, but not every starting line induces one.

My bad, I thought it meant that every point would do it.

[ZetaX]

IMO 2011, problem 2 

[MagmaMcFry]

how did you find it?