Geometry puzzle

I misread that as a finite number of circles. Please disregard that portion.
It is still 2/10, because you are wrong.

draw 99 infinitely large circles anywhere.

No you don't understand.
You can still have patterns with infinitely many circles, at arbitrary distances.
You just cant say "oh yea this circle is an infinite distance away from this other one"
If you specify any two circles their distance is finite, otherwise it woldnt be specified at all which would be totally retarded.
That doesnt mean that you cant find a sequence of circles where the distance goes to infinity (although with the constraints of the problem I proved (veeeery shaky tho) that you wont find such a sequence)

If you make a infinitely large circle formed of circles intersecting at a single point, then any line anywhere can only pass through at most 4ish circles.

>infinitely large
not well defined
Unless I misunderstood but then I need a sketch or something to get it

...

*shrink or expand*

wait why cant I just draw a line right next to that blue circle

If my proof works then this would be a very nice analysis 1 problem

I don't want to read every detail of it but I see no mention of the '100' parameter.

Which n does your proof apply to if the assumption is "Assume the plane is filled with circles such that every line intersects at least one of them but no more than n of them"?

Then we expand our the radius of the major circle, a process which would go on to infinity. This was exactly my point in except I didn't look further into it. I love the limiting construction came up with.

any finite number n
come on you have to fix the arrangement of circles first and thwn show that any line fulfills the conditions... I disagree that having an ever expanding circle is well defined

to expand, I'm really curious if this proof works becaus it treats the problem for any number of circles and the circles can just be arbitrary bounded sets of points as well.

ITT you're all idiots.

So let me break down why this isn't possible to solve.

>Is it possible
Yes, this was completely unnecessary and even kind of misleading for a mathematical problem.

>To draw circles on the plane so that every line intersects at least one of them but no more than 100 of them?
'every line'?? You mean every line that exists on the plane?

What the fuck is this even trying to say. This is like one of those Veeky Forums 'be good look more like' moments where I cannot for the life of me grasp what you are trying to say.

this minus the reddit spacing

shut up it's a completely legitimate problem.
Can you find a set of circles such that for any line on the plane the number of circles it intersects is 0 < 1 < 101.
You put circles on a plane and then you arent supposed to be able to draw any line which contradicts the conditions.
How is this misleading or hard to get.

>0 < 1 < 101

Then it's a stupider problem than I ever imagined. You might as well have asked if it were possible to create one circle such that any line drawn in a plane would intersect it. It would be the same as wanting the same thing for any number between one and one hundred.

This is retarded.

This guy proves

if it is that trivial then show me a proof that is easier than mine!
Remember you can have infinitely many circles.

the answer is "no"
And you can't solve this with infinitely many circles because you can't have infinity > infinity - 100

Draw x circles such that x is a finite number

Draw a circle such that it touches two circles one time each and encloses all other circles.

Draw a tangent to this circle.

Done.

Now if more circles were added it could be proven that the circle would just get larger and the tangent not intersecting the circles farther away from the initial line satisfying the problem.

Thus, the answer is no, it is not possible.

>Draw x circles such that x is a finite number
finite number of circles is not required.

There are finitely or infinitely many circles on the plane but each line only intersects at most n circles where n is finite.

I actually did exactly what you did but it requires more work to show that you can find such a large circle when there might be infinitely many on the plane.

I'm not sure that x really has to be finite anyway in my proof. The final line of my proof involves me stating that you can increase the size of the overall circles again and again. So we start at a finite number and then end the proof with an increasing amount infinitely. I don't see anywhere you mentioned in your entire proof that the circle enclosing all other circles should touch two of the circles. That's important because it describes how to draw said circle.

The point was, you wanted a proof which was easier and shorter, and mine is valid, easier, and shorter. You can have all the fancy notation you want, mine is fundamentally correct and you have not provided any satisfactory objection to it yet.

Your's is invalid; it assumes the union of all the circles is bounded. This isn't necessarily true with infinite circles.

It is impossible to draw or conceive of an infinite shape.

The point is invalid, again.

To make things easier for you to understand, and infinite circle would really just resolve itself into being an equiangular hyperbola, if you tried to manifest it.

show me your big, all encompassing circle here (the tiling continues indefinetly)
>b..but there are obviously lines which intersect infinitely many lines
What if I found a complex infinite tiling which manages to solve the problem?

Did you not read the proof. If we start from the supposition of infinite circles, it cannot be completed. You are making a mistake of presupposing I start at this notion.

If I start at a finite number of circles, then the solution to the proof can be expanded to an infinite number of circles.

that's not how proofs work.
You are proving the problem for any finite number of circles.
But any finitee number is not equivalent to infinite number.
I will show you the equivalent of your logic with series:
for any sum of natural numbers up to some n we can find an upper bound, i.e. 1+2+3 < 10
1+2+3+4+5+6+7 < 100 etc.
Therefore, the infinite sum of all natural numbers is bounded.
This is what you are doing.

Technically not so. With the concept of infinity you can make the argument that infinity itself is not smaller than anything. But that's not what I'm doing.

I'm not saying that the infinite number of circles you are proposing is smaller than anything, but I am proposing that there will never be a collection of circles that you cannot treat the same as the circles in the finite example.

To your first point: this is exactly how proofs work. Mathematicians will see if a line intersection a parabola, for instance, is parallel to the asymptote to determine if it will ever meet the parabola. They need not know the entire length of the line, to know that the lines are parallel.

A better example: Archimedes used a finite number of triangles to approximate an infinite number. He used 92 triangles and Euclid's I.47 to define pi between 3 (1/7) and 3 (10/71).

This is a better example because it is more like what we are trying to do here. If I want to define the property of an infinite number of circles, I must start from a known finite property of circles, much like how Archimedes started with a finite property of triangles.

The proof holds. No doubt, you misunderstand what bounded means as well, literally does not mean enclosing with a circle.

ok I thought about it some more and I fucked up, the last (*) part is not correct that way.
So to be more clear let A = [0, 2pi) be the set of lines through Pand r: A -> [0, inf) be a the function which assigns every line its radius (distance from P where it intersects no more circles).

First, I'll explaim what is wrong about my proof.
I assumed the opposite (R not bounded) and found a sequence of angles a_n where r(a_n) diverges to infinity.
I then concluded that there exists a subsequence of a_n which I will call b_n which converges. r(b_n) also diverges to infinity.
Now I assumed that r is continuous without realizing and this is wrong.
Ill try to prove it from this point on but ffs it's suddenly pretty complicated.
Im not even sure if that answer is "no" at all anymore.

Your proof literally proofs that given any set of circles there is a line which does not intersect a single one of them.
You dont even use the constraints given by the problem.
I will give you a set of circles where there is no line which intersects all none of them (although your proof would say there is)
C := {circle of radius 2 at a point P where P = (n, m) for two natural numbers}
Which is just a grid of circles.
Every line intersects some of them (infinitely many in fact).
If you show that any finite grid can be encompassed and just say "yea increase the circle if there are more circles im the grid" then you'd prove that there is a line not intersecting any circles on the plane which is wrong.

>then you'd prove that there is a line not intersecting any circles on the plane which is wrong.
That's literally the point, considering what OP is asking for is a line intersecting none to prove him wrong.

Do you even understand why I proved it that way?

Can't you just draw a ring of circles on the plane?

ignore what is said in OPs post. Does your proof not claim that even if I make my tiling infintely big there will always be a line "outside" which does not intersect any circle?

This is the case, yes.

But only determinable because of what is observed with a finite number of circles.

So what we are talking about cannot possibly be represented by any figures, axes, pictures, etc.

Except one thing. It's important you understand that the circle constructed always touches two circles. This is important because that's not exactly what your diagram shows.

fml I've changed my mind, R is not bounded and the answer to the problem is yes.
I can't find a line here which intersects too many or too few circles..

>So what we are talking about cannot possibly be represented by any figures, axes, pictures, etc.
let N^2 be the set of the circles centers.
let 2 be the radius of each circle.

You can represent it with variables and words. Your picture is not an accurate representation of infinity, nor will it ever be.

Besides you also got the original proof wrong. You have failed to raise a reasonable objection to the proof besides stating that for whatever reason(none from what I can see) you want me to START with infinity, instead of end with it.

Good night.

This looks like a probable result, really changed my mind. Could it be drawn with four arms instead of three too?

>It's easy to place infinitely many circles in the plane such that every line only intersects at most 100 circles.
oh really? tell me about it
because your pattern sure can't be periodic, otherwise every line that intersects one circle would intersect an infinity of them

I don't know lad, what shape do you think the "arms" should have
because if they have an asymptote, then a line that is the asymptote itself will intersect an infinite number of circles
so they should be parabolas, maybe, and the radii of the circles should indeed decrease the further away you get from the center

this is actually a very interesting related problem.
I am fairly certain that it is possible to place strictly onfinitely many circles in a way that every line only intersects finitely many (for example place them on a parabola).
But it might bery well be possible that for any infinite circles you place, you can find a line with arbitrarily many intersections.
If this is the case then we have also disproven OPs problem for the infine case (finite is trivial).

Well, the lad over there proved that the "infinitely many circles on a line" solutions can't work, see the first answer:
math.stackexchange.com/questions/480450/circles-on-the-plane-such-that-every-line-intersects-at-least-one-of-them-but-no

I am not sure...
>because if they have an asymptote, then a line that is the asymptote itself will intersect an infinite number of circles

this is not that easy, we enter some really fucking complicated territory here.
you have to consider the dircles distances and radii and maybe they dont need to touch eachother. All those values could converge or diverge...

But if it is an asymptote or a parabola I believe that in both cases you can find a line which intersects arbitrarily many (not infinite though.. even in the hyperbola case infinte might be preventable.

Well without having read that yet, I guess it would be nice to plug that in the proof earlier and say that therefore R is bounded to conclude the proof.

Can't you just use any old function from [0, ∞) that goes off to infinity, as long as it's curvy enough (i.e. not a straight line)? We wouldn't need the unit circle anymore either, and the radii of the circles sitting on this curve would depend on both its curvature and on the maximum allowed number of intersections. The only problem is you still get limiting behaviour when considering tangents (see ).

when I said "infinite circles" I meant "an infinite number of circles", not "a circle of infinite radius". This was obvious to everyone but you.

Holy fuck can we start testing people before letting them post here, to keep morons like this out?

Well I think that in your example and most parabolas/hyperbolas) for any n you can find a line with morem than n intersections (if you go and follow the trajectory of the parabola it becomes more and more aligned with your line so you cut more circles unles the radii start doing some weird shit.)

this question doesn't even make any sense
that's like asking if it's possible to draw 100 circles in R^2 that are not in R^2

...

1. Start with three circles touching (in a triangular pattern)

2. Black out all lines in the plane that go through more than one circle.

3. Add another circle as close as you can to the starting three while not inside any black part of the plane.

4. Go back to step 2.

This is repeated indefinitely so all parts of the plane are blacked out, and every line only goes through two circles

QED

How the fuck does this overcome this objection???

This basically /thread right there.

It's not possible.

Just one (hopefully not too irresponsible) guess - instead of looking at the problem directly, one might look at the space of all lines. You can describe them as [math] \mathbb{S}^1 \times \mathbb{R}_0^+ [/math], with the points [math] (n,0) [/math] and [math] (-n,0) [\math] identified (the point [math] (n,a) [/math] represents the line given by [math] n \cdot x = a [/math]).

Each circle now occupies some subspace of this space - i.e. the subspace of lines going through that circle.
Essentially, what you want to know is if there is a covering by such sets such that every point is contained in just finitely many such sets.
Probably a topological interpretation might help (or it won't, it's just playing around with ideas).

* [math] (n,0) [/math] and [math] (-n,0) [/math] identified (the point [math] (n,a) [/math] represents...

>such that x is a finite number
already wrong. a solution might involve infinite cirlces

what about something like:

the centers of the circles are on the line f(x) = e^x (and f(x < 0) = e^-x) with decreasing radius?

Nice one! It seems that this is the solution of this problem.

There is just one small circle at the origin missing. You need to care for the line y = 0.

>It seems that this is the solution of this problem.
do you have the answer key or something?

What is the definition of 'every line'?

any possible line you can draw on the plane

If I had one I wouldn't have said "seems", would I? Just a feeling that it could work out.

But, rethinking the idea, there might still be the possibility that any line passes through arbitrarily many, though finitely many, circles... which is not what we want.

Any circle drawn on a plane has infinity many lines intersecting it have you all go retarded?

>might still be the possibility that any line passes through arbitrarily many, though finitely many, circles... which is not what we want.
yeah. you'd want the radius of the circles to change at a rate that limits that. I'm not sure how to do it.

I think you misunderstood the problem. Just read again.

No.
It isn't specified that the line has to be on the plane.

And it still does.

What is so hard for you retards to understand this logic.

By definition, in order to solve for a situation in which infinity exists, you must first create the finite geometrical situation.

Here is a better board for people like you

using the axiom of choice it's fairly easy to construct such a set of open subsets of [math]\mathbb{R}^2[/math]

>what is transfinite induction
You can't just prove something is true of countably infinite circles by proving it for finitely many.

Am I a brainlet because I don't understand this problem? What lines are we talking about?

I assume you had another problem in mind - the one of constructing a set of points such that each line contains exactly two of these points.
I tried something similar but it doesn't really work - taking a maximal set (by Zorn's lemma) of circles such that each line intersects at most, say, 100 of these circles, we may ask if there still could be a line which intersects none of these circles. The problem is that now putting a circle anywhere on this line may indeed could make one line intersect 101 circles...
What I want to say is - there might be no contradiction in assuming that there is a line intersecting no circle in a maximal set of circles.

You might just give a sketch of your idea.