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A curiosity from the Book of Numbers.
If you place n dots, irregularly, on the perimeter of a circle, how many separate regions are formed within the circle when all dots are joined in all possible ways? The dots should be placed such that only two lines intersect at any one point.
The sequence starts:
1 dot, 1 region.
2 dots, 2 regions.
3 dots, 4 regions.
...
Can you extend the sequence to 6 dots?
A curiosity from the Book of Numbers.
If you place n dots, irregularly, on the perimeter of a circle, how many separate regions are formed within the circle when all dots are joined in
all possible ways? The dots should be placed such that only two lines intersect at any one point.
The sequence starts:
1 dot, 1 region.
2 dots, 2 regions.
3 dots, 4 regions.
...
Can you extend the sequence to 6 dots?
A curiosity from the Book of Numbers.
If you place n dots, irregularly, on the perimeter of a circle, how many >separate regions are formed within the circle when all dots are joined in >all possible ways? The dots should be placed such that only two lines >intersect at any one point.
The sequence starts:
1 dot, 1 region.
2 dots, 2 regions.
3 dots, 4 regions.
...
Can you extend the sequence to 6 dots?
On Sun, 27 Jul 2025 11:42:35 -0000 (UTC), David Entwistle ><qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:
A curiosity from the Book of Numbers.
If you place n dots, irregularly, on the perimeter of a circle, how many >>separate regions are formed within the circle when all dots are joined in >>all possible ways? The dots should be placed such that only two lines >>intersect at any one point.
The sequence starts:
1 dot, 1 region.
2 dots, 2 regions.
3 dots, 4 regions.
...
Can you extend the sequence to 6 dots?
Even if you say irregular, the problem is rather complex as you can
more than two diagonals intersect at a point. (In the case of a
regular hexagon inscribed in a cirlce, three diagonals intersect at
the centre thereby reducing the number of regions by one --- oops,
I gave away the answer!) The condition that only two diagonals
can intersect at a point, leads to the maximum number of regions
into which the circle can be partioned.
I suggest a follow up question below .....
SPOILERS
SPOILERS
SPOILERS
SPOILERS
The problem is called Moser's Circle Problem and you can find the
answer at
https://en.wikipedia.org/wiki/Dividing_a_circle_into_areas
A beautiful video of one proof is provided by 3Bule1Brown
(Gareth Sanderson) in
https://www.youtube.com/watch?v=YtkIWDE36qU
.... which brings us to the more general, complex problem.
If n points are placed in a circle and all of them are connected,
what is the number of intersection points made by the diagonals?
I am not sure if this is solved, but the complexity can be seen
even if we condsider the following:
What is the number of intersection points made by the diagonals
of a regular polygon?
The answer to this is too long for me to type! But, I can
provide a reference. (Alas, I have a PDF of the paper, but as
we discussed a few days ago, this has been a text only group
and I do not want to violate that.)
I should thank Mathloger for bringing this to my attention. The
expression was in one of his slides, even though the topic was
something else.
Can you extend the sequence to 6 dots?
On Sun, 27 Jul 2025 11:42:35 -0000 (UTC), David Entwistle wrote:
Can you extend the sequence to 6 dots?
SPOILER.
POILER.
OILER.
ILER.
LER.
ER.
R.
.
Having spent a couple of hours refreshing my algebra skills, I think the candidate polynomial would be:
f(n) = (n^4)/24 - (n^3)/4 + 23*(n^2)/24 - 3*n/4 + 1
If n points are placed in a circle and all of them are connected,
what is the number of intersection points made by the diagonals?
You can check yourself against Wikipedia...
On Sun, 27 Jul 2025 14:40:40 -0400, Charlie Roberts wrote:
Hi Charlie.
If n points are placed in a circle and all of them are connected,
what is the number of intersection points made by the diagonals?
Could you clarify the above, or provide a pointer to the question? Is
that, n points are placed on the perimeter of a circle, or n points are >placed somewhere within a circle? If the second, are we extending straight >lines through the points to the point they intersect with the circle,
rather than having the line end at the points? When you say diagonals, I'm >thinking chords?
Thanks.
Sorry, I should have been clearer. What I meant was a generalsation of
the problem you posed. The n points are all on the circumference of the circle just as in your case. But, no condition is made on their
(relative) positions.
2) the vertices of a regular polygon -- which is treated in the paper I mentioned, or
On Tue, 29 Jul 2025 09:20:19 -0400, Charlie Roberts wrote:
2) the vertices of a regular polygon -- which is treated in the paper I
mentioned, or
The number of intersections of chords from the vertices of the regular >polygons, with an odd number of vertices, looks to be more my level of >problem. That'll limit the number of intersecting chords to a maximum of >two. The polygons with even vertices 'looks' hard.
Yes, at least to me. I was amazed when I saw the expression for the
number of intersections and regions for the case of regular polygons.
A curiosity from the Book of Numbers.
If you place n dots, irregularly, on the perimeter of a circle, how many separate regions are formed within the circle when all dots are joined in all possible ways? The dots should be placed such that only two lines intersect at any one point.
Yes, at least to me. I was amazed when I saw the expression for the
number of intersections and regions for the case of regular polygons.
Are you interested in a reference to the paper?
I'm happy that the number of intersections for the regular polynomial,
I'm not sure but isn't one way of specifying the restriction to say that
the points are "in general position"?
This problem clicks my nostalgia button. My excellent high school math teacher, seeing I needed a challenge, assigned me this problem; I
developed difference equations to get a quartic solution; and went back
to day-dreaming.
I'm five times older now, but, in some ways, thrice as dumb.
On Wed, 30 Jul 2025 09:23:08 -0400, Charlie Roberts wrote:
Yes, at least to me. I was amazed when I saw the expression for the
number of intersections and regions for the case of regular polygons.
Are you interested in a reference to the paper?
Yes, that would be interesting.
I've looked at the sequence provided by the OEIS (https://oeis.org/
A006561), wrote a small program to do the leg-work of calculating the >various finite differences: 1st, 2nd, 3rd etc. It hasn't arrived at a >constant for the differences up the 40th. That suggests, either I've made
a mistake, or the sequence isn't represented by a single polynomial, or it >is of a very high order.
I'm happy that the number of intersections for the regular polynomial,
with an odd number of vertices, is polynomial (see above). It may be that >the regular polygon with an even number of vertices has the same total >number of intersections, but it's just that more than two diagonals >intersect at the same point - reducing the overall count. Perhaps I should
look at the difference between the "odd-vertices prediction" and the--
actual. My mathematics doesn't extend to telling me that would definitely
be a sensible and useful thing to do. I'm happy to be guided by curiosity.
David, Here it is.
THE NUMBER OF INTERSECTION POINTS MADE BY THE
DIAGONALS OF A REGULAR POLYGON
BJORN POONEN AND MICHAEL RUBINSTEIN
AT&T Bell Laboratories, Murray Hill, NJ 07974, USA
Current address: University of California at Berkeley, Berkeley, CA 94720-3840, USA
Email address:poonen@@math.berkeley.edu
AT&T Bell Laboratories, Murray Hill, NJ 07974, USA
Current address: Princeton University, Princeton, NJ 08544-1000, USA
Email address:miker@@math.princeton.edu
You will find a link to the PDF file if you go to
https://math.mit.edu/~poonen/
and search for the title. The link is the fifth from the bottom of
the list of "Research articles".
So, the special cases, which deviate from the easier odd vertices case,
are polygons where the number of vertices is divisible by: 2, 6 and 30.
That sequence suggests products of primes. 2, 2 x 3,-a and 2 x 3 x 5. If
it isn't a daft question, is there any easy-to-understand reason why the special cases don't continue 2, 6, 30, 210...?