The Maximum number of points of intersection of 4 distinct circles and 8 distinct straight lines is

Question

1)66

2)64

3)104

4)40

Can anyone please help me to solve this problem?

My attempt : I have seen this link--Maximum number of points of intersection. But I can not understand how they are so sure that all intersection points are distinct.

Clearly ,it was solved by @Archaick in this link--Maximum number of points of intersection. — Abdul Halim, Apr 22 '20 at 12:45
You can think draw a first circle,second cycle can touch maximum two intersection points,third cycle can intersect maximum 4 intersection points and the last one can touch maximum 6 intersection points and similar thinking for straight line then you can get similar way of thinking nicely..i.e.permutation and combination. — Abdul Halim, Apr 22 '20 at 12:49
I know what you mean OP; it would be great if someone could algebraically using matrices or something prove this - wouldn't that cross with linear algebra? I think it would take a long time to prove it with this way though. — , Apr 22 '20 at 12:50
Try this first for a smaller number of circles and lines. First draw the lines so that no three are concurrent and no two are parallel (go line-by-line), then try to fit the circles in (go circle-by-circle). Most likely, you will not need to retrace steps if you do this right. Try this for $4$ lines and $2$ circles first, then extrapolate the logic. The idea essentially is this : if you have any issues during placement, they are usually resolved by a very tiny perturbation of the last line or circle you are trying to place. — Sarvesh Ravichandran Iyer, Apr 22 '20 at 12:56
You are right to be unsure of the fact that the combinatorial maximum is achievable : however, I think in these problems involving intersection of lines/circles in your book, the combinatorial maximum is achievable (they are not testing your geometry, they are testing your combinatorial ability). Therefore you can safely assume this in your computations : nobody will ask you for the diagram of eight lines and four circles. — Sarvesh Ravichandran Iyer, Apr 22 '20 at 12:58

Parcly Taxel · Accepted Answer · 2020-04-22T14:09:31.343

4

I will construct an explicit example. Take the following two configurations:

The second configuration is the eight-line solution to my generous lazy caterer problem. Now scale down the eight-line configuration so that all its intersections are inside the intersection of all the four circles (the small squarish region). If it happens that there is a multiple intersection, we can just tweak to remove it. Then we are guaranteed that every line intersects every circle twice, obtaining the maximum of $104$ intersections.

This is the result:

To generalise to any number of circles and lines:

Set the sizes of the circles to be equal and make them encircle a point. This way every circle intersects every other circle twice, and there is a region inside all circles.
Take a solution to the lazy caterer problem for the requisite number of lines – lines that all intersect each other – and scale the configuration down so that all intersections are inside the intersection of all circles. Generally, there is no multiple intersection, meaning that every line intersects every circle twice and we are finished. If there are multiple intersections, shifting the lines by a small amount (not enough to create multiple intersections elsewhere) suffices to remove the multiple intersection.

edited Apr 22 '20 at 14:09

answered Apr 22 '20 at 13:09

Parcly Taxel

103,344

1

There are so many intersection points(Circle - Circle) outside the squarish intersection region. How are you being so sure that after placing the configuration in this region , no straight line will pass through any of this intersection points.@Parcly Taxel – cmi Apr 22 '20 at 13:49
1

@cmi We can tweak. I'm making one big picture for you. – Parcly Taxel Apr 22 '20 at 13:50
Okay I am also waiting for your further explaination.@ParclyTaxel – anonymous Apr 22 '20 at 13:51
@sani done, added the picture. – Parcly Taxel Apr 22 '20 at 13:55
Thank You so much...But Can we not prove this by logic?@ParclyTaxel – anonymous Apr 22 '20 at 13:56
As to logic: see my answer – oliver Apr 22 '20 at 13:56
@sani I have proven it by logic, I have constructed an example! – Parcly Taxel Apr 22 '20 at 13:57
@ParclyTaxel: can you extend your example to higher numbers by induction? – oliver Apr 22 '20 at 13:58
@oliver We don't need induction, but it is trivial to generalise my solution to higher numbers of circles and lines. – Parcly Taxel Apr 22 '20 at 13:59
Can we prove the second part after placing the second configuration in the squarish intersection , the straightlines which coming out from this region will not pass through any of those intersection points..@ParclyTaxel – anonymous Apr 22 '20 at 14:00
@sani Yes, we tweak! It is sound and simple. – Parcly Taxel Apr 22 '20 at 14:01
@ParclyTaxel: If you talk about logic, then you always talk about the ability to generalize. The ability to generalize means understanding. If you just provide an example, there need not be understanding. – oliver Apr 22 '20 at 14:02
@ParclyTaxel I am trying to imagine. – anonymous Apr 22 '20 at 14:02
@oliver It is generalisable by merely adding more circles tight enough around a centre and then adding more lines to the lazy caterer's solution. – Parcly Taxel Apr 22 '20 at 14:03
@oliver Done. I added the generalisation. – Parcly Taxel Apr 22 '20 at 14:09

oliver · Answer 2 · 2020-04-22T13:47:50.250

The trick is that you can place the individual objects (circles + lines) an arbitrarily tiny (but nonzero) distance/angle apart. Then maximum intersection is guaranteed as long as you make sure that there are no multiple intersections. For example, proceed as follows:

Draw an orientation point and make sure that it is inside of all the circles. Think of a thin pole where you drop your 4 rings over. Then every ring (circle) must necessarily have at least two intersection points with every other ring.
Now you have to avoid multiple intersections. Since every added circle just has to avoid a finite set of previous intersection points, that is easy to achieve: just avoid them when fine-tuning the ring positions. You have continuous area between the other points. ;-)
Now there is a small area around the center pole that is part of all circles (the intersection area of all the circles)
Take infinitely long spaghettis (lines) and drop them randomly. As long as neither two of the spaghettis are in parallel, they will intersect maximally (one point for each pair). If they are in parallel, adjust them. If multiple intersections occur, adjust as well. As in (2) you have plenty of space for the adjustment due to continuity between the previous intersection points.
Finally you have to make sure that all spaghetties go through the interior of all the circles. This is the case, if at least one point of every line is inside the intersection area of all the circles. It is enough to make sure that all the spaghetti-to-spaghetti intersections are inside the intersection area (3) of all the circles. Since the former is a finite set, you can scale and move your spaghetti configuration accordingly until they are all inside.

Et voila: you have reached the number of intersections that are combinatorially possible under the given requirements.

Once I am done with the previous answer . I will read yours and try to understand. Thank You..@Oliver — anonymous, Apr 22 '20 at 13:58

score 1 · Answer 3 · answered Mar 25 '21 at 20:22

An interesting way to understand the 8-line situation is with inversion. As we know a circle that passes through the center of inversion is inverted into a line. Choose the original as the center of inversion, and choose 8 evenly spaced points on the unit circle with angles ranging from 0 to $\frac{7\pi}{8}$ as centers and draw 8 circles passing through the origin, we easily get 8 circles that intersect with each other at distinctive points. Invert the 8 circles centered at the origin, we get 8 lines intersecting each other at distinct points. 8 circles inverted into 8 lines

The Maximum number of points of intersection of 4 distinct circles and 8 distinct straight lines is

3 Answers3