20041206, 15:58  #1 
Bronze Medalist
Jan 2004
Mumbai,India
804_{16} Posts 
Circle Puzzles 1
When spots are placed on the circumference of a circle and then joined, regions will be formed as follows. 1 spot 1 region 2 spots 2 regions 3 spots 4 regions 4 spots 8 regions 5 spots 16 regions 6 spots ? How many regions ? Is the answer 32 regions ? What exactly is it? The puzzle doesnt end there. Pl. await Circle puzzle 2 for more on this peculiar property of a circle. I promise it to be more mathematical than merely drawing circles and dividing them up into regions. Mally 
20041206, 17:02  #2  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
3^{3}·389 Posts 
Quote:
An old one but a good one. Thanks. Paul 

20041207, 00:23  #3 
Dec 2003
Hopefully Near M48
3336_{8} Posts 
If I remember correctly, the formula for the number of regions in a circle with n spots is actually a polynomial in n (I'm not sure whether its quartic or quintic.

20041207, 00:34  #4 
Jun 2003
The Texas Hill Country
1089_{10} Posts 
I think that there is something missing here.
If the spots are constrained to be evenly spaced, the number of regions is well defined. However, if the spots are allowed to occupy arbitrary locations, in some configurations, areas may degenerate into a point rather than an area with three distinct sides. 
20041207, 08:30  #5  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
10503_{10} Posts 
Quote:
However, your definition of "evenly spaced" is not quite the same as mine. Consider six points at the vertices of a hexagon. Three of the lines meet at the center of the circle. Paul 

20041208, 10:15  #6  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts 
Quote:
My reply was delayed due to a computer delay. Thank you Paul for your prompt reply and correct answer I am greately indebted to you for pointing out that the regions of of zero space are also a possibilty in the case of three intersecting lines in which case they are to be counted also. The first possible case occurs when there are 6 spots on the circle. To conform to the total of 31 regions when 3 lines intersect the point of zero space also has to be counted i.e. 30 regions and one zero space to give 31 regions which wil be the case. I have taken this problem from the book 'Power puzzes' by Philip Carter and Ken Russell in their original U.K. publication. Out of the 481 problems given I chose this as the best and the most mathematical. It is also an excelent book for word puzzles in which I think you said you were interested in. I submitted this problem to 'Mind Sport' of the Times of India which has a vast readership and got various solutions and responses. But no one pointed out the inclusion of zero space regions. Therefore your answer reveals great depth and insight which is truly commendable. Please note that the words 'spots' and 'regions' are very carefully chosen to avoid ambiguity with points and areas. In doodling around with this problem I observed that by joining all the points avalable in the diagram of a of a 5 sided polygon an infinite set of 5 pointed stars were formed one nesting in the the other approaching zero space resuting in a fractal with a simiarlity (or Hausdorf ) dimension. Coud this be simuated on a computer? Jinydu: The polynomial is a quartic function but please dont jump the gun and await Circle Puzzle 2. Mally 

20041208, 11:52  #7  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
3^{3}×389 Posts 
Quote:
As you did not specify that the spots must be at distinct points on the circumference of the circle, it could be that two or more could lie at the same point. Thus, for instance, with four spots all could lie at the same point (and all lines indeed meet at that same point) but it could be the case that the spots form two pairs, with two of the lines being of zero length and the other four lines being coincident. The intermediate case, with three coincident spots has three coincident lines and three lines of zero length. The final possibility, two coincident spots and two distinct ones, has one zerolength line, two pairs of coincident lines and one single line which joins the noncoincident spots. Still a good puzzle and one which shows the necessity for clear thinking rather than jumping to conclusions  which I did! Paul 

20041208, 16:40  #8 
Bronze Medalist
Jan 2004
Mumbai,India
100000000100_{2} Posts 
Circle puzzles 1
Off hand I think we should confine the problem to geometry and not analysis or algebra. Yes there are hypothetical cases where these cases may exist but by drawing diagrams we can eliminate most but I agree not all. For instance I did not think of the easier regular hexagon you explained and tackled the non regular 5 sided polygon and actually counted the regions and found the discrepancy that the no.of regions are 30 and not 31 as per theory. Thats when the zero region has to be counted to conform to that given in the problem. I follow Gauss' dictum 'Pauca sed matura' (few but ripe) and give much thought before I put it down in writing. However I will examine the case for 4 intersecting lines and hence at least one zero region. This requires at least 8 spots ( two end points) I think but untill I can draw it out I will not commit myself. I have worked out a rule for these and its not that easy to say this without a diagram. As the spots become larger in no. the diagrams become more difficult to make and be correct without making any omissions in counting the regions. Thats when the quartic takes over and one can predict the nth term without a diagram 'easily' but then the algebraic computation becomes more complicated if not difficult. Still I agree that its worth going into the zero lines and zero spaces theory, where literally the spots are ' gobbled' up. But we must sincerely follow it up further, which I hope too soon and enter my contribution. I hope you will do the same. Mally and a smoke! . 
20050706, 21:10  #9 
Jun 2005
Near Beetlegeuse
2^{2}·97 Posts 
I read the whole of this thread twice before posting this reply, and your βsolutionβ seems to me to be completely illogical.
You say that where three or more lines meet is a region with zero area, and yet you say that four spots gives 8 regions. If I put four spots on the circumference of a circle and join them up I get 8 regions with definite area and 4 regions with zero area making 12 regions all told. If I put 5 spots on the circumference of a circle I get 16 regions with definite area and 5 regions with zero area, making 21 regions all told. If I put 6 spots on the circumference of a circle I get 30 regions with definite area and 7 regions with zero area making 37 regions all told. Or is there some totally plausible explanation I have overlooked for a region with zero area not being able to be sited on the circumference? 
20050707, 13:12  #10 
Aug 2003
Upstate NY, USA
2·163 Posts 
are you trying to connect each of the exterior points to themselves to produce those 0 areas?
the zero areas above are referred to intersections of 3 or more lines as they create an interior triangle with area 0 
20050707, 19:54  #11 
Jun 2005
Near Beetlegeuse
2^{2}·97 Posts 
Hi Tom11784,
No, I am not trying to do anything clever like join spots to themselves. All I have done is read their own definition of a zero area region as being where three or more lines meet. By that definition, and as far as I can tell not excluded by anything else in the thread, when there are four spots on the circumference of the circle each of those spots has three lines meeting at the spot. Therefore, the spot itself is a region with zero area. It may be that they intended or understood their definition of a zero area region to be more specific than this, but they haven't actually said as much. So as far as I can tell all their totals with 4 or more spots are wrong. There may be a case for 3 spots also being wrong, but that depends upon their definition of a region, and as I am not trying to be deliberately contentious, just to understand what they are talking about I will leave that alone. 
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