geometry puzzle

[Anonymous No.9487130]

I saw this interesting geometry puzzle but I'm stuck:

"Is it possible to draw circles on the plane so that every line intersects at least one of them but no more than 100 of them?"

Can someone give me some pointers?

[Anonymous No.9487404]

How many circles are you supposed to draw?

[Anonymous No.9487411]

>>9487130
Yes, draw a circle with an infinite radius. Otherwise it's not possible, since for an arbitrary amount of circles (let's say, from some distance to the origin) you can always find a line sufficiently far away from the origin and at such and angle that is intersects no lines.

[Anonymous No.9487412]

>>9487411
*intersects no circles

[Anonymous No.9487426]

>>9487411
>Yes, draw a circle with an infinite radius.
A line is not a circle. A circle must have a positive, real-valued radius.
>Otherwise it's not possible, since for an arbitrary amount of circles (let's say, from some distance to the origin) you can always find a line sufficiently far away from the origin and at such and angle that is intersects no lines.
You could just use infinitely many circles, you know? It's easy to place infinitely many circles in the plane such that every line only intersects at most 100 circles. The difficulty stems from the additional requirement, that every line should intersect at least one.

[Anonymous No.9487607]

>>9487411
>for an arbitrary amount of circles (let's say, from some distance to the origin) you can always find a line sufficiently far away from the origin and at such and angle that is intersects no lines.
[citation needed]

[Anonymous No.9487636]

yes it's possible
first think about this problem with points instead of circles
then use small circles instead of points
you should be able to fill in the details

[Anonymous No.9487637]

>>9487130
Put the circles in a circle ya dingus.

[Anonymous No.9487696]

No
I will prove by contradiction.
Assume the plane is filled with circles such that every line intersects at least one of them but no more than 100 of them.

I will use an Axiom: (A) "The distance between any two circles is finite"

Let P be a point on the plane.
For every line i through P, there exists a distance r_i such that the line does not intersect any more circles if you go down the line a distance of r_i in any direction starting from P. This follows from (A) and that the line only intersects finitely many circles.
Basically any line has to stop intersecting circles at some point, otherwise it would intersect infinitely many.
I will call the r_i for any line i the "radius" of the line.
The set R = {r: r is the radius of a line through P} must be bounded. This is fiddly to prove and I will do it below at (*).
Because R is bounded we can find a line k with a maximum radius.
Draw a circle around P with a radius greater than r_k.
No line intersects any circle at a point further away from P than r_k.
This means that there are no points belonging to circles furhter away from P than r_k.
Draw a line somewhere outside our big circle.
The line does not intersect any circles.

(*) proof that R is bounded:
Assume R is not bounded, then, there exists a sequence L_n of lines where the radius goes to infinity.
Each line shall be identified and characterized by it's angle.
The angle of the lines is bounded. (on a circle) This means that the lines c_n (their angles) lie on a bounded set.
By the bolzano weierstrass theorem there must exist a subsequence of c_n where the angles (i.e. just the lines) converge.
Because the set of lines is continuous, there exists a line M such that the radius of this line is the limit of the radii of the subsequence.
But the limit of the radii of the subsequence is infinite, just as the limit of the parent sequence c_n.
Therefore the radius of M is infinite, which violates (A).
Therefore R must be bounded.

[Anonymous No.9487706]

>>9487696
>The distance between any two circles is finite
Well sure, if you guarantee your answer beforehand, ofc your answer will be right.
2/10

[Anonymous No.9487707]

>>9487706
>>9487696
I misread that as a finite number of circles. Please disregard that portion.
It is still 2/10, because you are wrong.

[Anonymous No.9487709]

>>9487130
draw 99 infinitely large circles anywhere.

[Anonymous No.9487716]

>>9487706
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)

[Anonymous No.9487740]

>>9487716
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.

[Anonymous No.9487745]

>>9487740
>infinitely large
not well defined
Unless I misunderstood but then I need a sketch or something to get it

[Anonymous No.9487759]

>>9487745

[Anonymous No.9487761]

>>9487759
*shrink or expand*

[Anonymous No.9487792]

>>9487759
wait why cant I just draw a line right next to that blue circle

[Anonymous No.9487838]

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

[Anonymous No.9488393]

>>9487696
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"?

[Anonymous No.9488526]

>>9487792
Then we expand our the radius of the major circle, a process which would go on to infinity. This was exactly my point in >>9487411 except I didn't look further into it. I love the limiting construction >>9487759 came up with.

[Anonymous No.9489350]

>>9488393
any finite number n
>>9488526
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

[Anonymous No.9489384]

>>9489350
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.

[Anonymous No.9489385]

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 4chan 'be good look more like' moments where I cannot for the life of me grasp what you are trying to say.

[Anonymous No.9489392]

>>9489385
this minus the reddit spacing

[Anonymous No.9489402]

>>9489392
>>9489385
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.

[Anonymous No.9489405]

>>9489402
>0 < 1 < 101

[Anonymous No.9489408]

>>9489402
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.

[Anonymous No.9489415]

>>9487696
This guy proves

[Anonymous No.9489417]

>>9489408
if it is that trivial then show me a proof that is easier than mine!
Remember you can have infinitely many circles.

[Anonymous No.9489421]

>>9489417
>>9487130
the answer is "no"
And you can't solve this with infinitely many circles because you can't have infinity > infinity - 100

[Anonymous No.9489446]

>>9489417
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.

[Anonymous No.9489453]

>>9489446
>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.

[Anonymous No.9489457]

>>9489453
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.

[Anonymous No.9489469]

>>9489457
Your's is invalid; it assumes the union of all the circles is bounded. This isn't necessarily true with infinite circles.

[Anonymous No.9489476]

>>9489469
It is impossible to draw or conceive of an infinite shape.

The point is invalid, again.

[Anonymous No.9489477]

>>9489476
>>9489469
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.

[Anonymous No.9489483]

>>9489457
>>9489446
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?

[Anonymous No.9489489]

>>9489483
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.

[Anonymous No.9489493]

>>9489489
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.

[Anonymous No.9489505]

>>9489493
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.

[Anonymous No.9489509]

>>9487696
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.

[Anonymous No.9489517]

>>9489505
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.

[Anonymous No.9489520]

>>9489517
>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?

[Anonymous No.9489521]

>>9487130
Can't you just draw a ring of circles on the plane?

[Anonymous No.9489528]

>>9489520
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?

[Anonymous No.9489532]

>>9489528
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.

[Anonymous No.9489536]

>>9489509
>>9487696
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..

[Anonymous No.9489544]

>>9489532
>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.

[Anonymous No.9489549]

>>9489544
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.

[Anonymous No.9489644]

>>9489536
This looks like a probable result, really changed my mind. Could it be drawn with four arms instead of three too?

[Anonymous No.9489662]

>>9487426
>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

[Anonymous No.9489678]

>>9489644
>>9489536
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

[Anonymous No.9489686]

>>9489662
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).

[Anonymous No.9489691]

>>9489536
Well, the lad over there proved that the "infinitely many circles on a line" solutions can't work, see the first answer:
https://math.stackexchange.com/questions/480450/circles-on-the-plane-such-that-every-line-intersects-at-least-one-of-them-but-no

[Anonymous No.9489693]

>>9489678
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.

[Anonymous No.9489694]

>>9489691
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.

[Anonymous No.9490007]

>>9489693
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 >>9487759).

[Anonymous No.9490051]

>>9489477
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.

[Anonymous No.9490057]

>>9489549
Holy fuck can we start testing people before letting them post here, to keep morons like this out?

[Anonymous No.9490261]

>>9490007
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.)

[Anonymous No.9490269]

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

[Anonymous No.9490289]

[Anonymous No.9490696]

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

[Anonymous No.9494013]

>>9490696
How the fuck does this overcome this objection???

>>9489446


This basically /thread right there.

It's not possible.

[Anonymous No.9494499]

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).

[Anonymous No.9494506]

>>9494499
* [math] (n,0) [/math] and [math] (-n,0) [/math] identified (the point [math] (n,a) [/math] represents...

[Anonymous No.9494554]

>>9489446
>such that x is a finite number
already wrong. a solution might involve infinite cirlces

[Anonymous No.9494669]

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?

[Anonymous No.9494727]

>>9494669
Nice one! It seems that this is the solution of this problem.

[Anonymous No.9494748]

>>9494727
There is just one small circle at the origin missing. You need to care for the line y = 0.

[Anonymous No.9494817]

>>9494727
>It seems that this is the solution of this problem.
do you have the answer key or something?

[Anonymous No.9494855]

>>9487130
What is the definition of 'every line'?

[Anonymous No.9494944]

>>9494855
any possible line you can draw on the plane

[Anonymous No.9494965]

>>9494817
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.

[Anonymous No.9494977]

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

[Anonymous No.9494996]

>>9494965
>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.

[Anonymous No.9495010]

>>9494977
I think you misunderstood the problem. Just read again.

[Anonymous No.9495017]

>>9487130
No.
It isn't specified that the line has to be on the plane.

[Anonymous No.9495124]

>>9494554
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
>>>/pol/

[Anonymous No.9496561]

>>9489662
using the axiom of choice it's fairly easy to construct such a set of open subsets of [math]\mathbb{R}^2[/math]

[Anonymous No.9497139]

>>9495124
>what is transfinite induction
You can't just prove something is true of countably infinite circles by proving it for finitely many.

[Anonymous No.9497155]

>>9487130
Am I a brainlet because I don't understand this problem? What lines are we talking about?

[Anonymous No.9497438]

>>9496561
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.

[Anonymous No.9497460]

>>9496561
You might just give a sketch of your idea.