20061115, 01:08  #1 
Nov 2006
3 Posts 
How to divide a circle into 16 pieces with only 5 lines????
My son has to divide a circle into 16 pieces with only 5 lines and his whole class is stumped but the teachcher won't help! You'd have to know her to understand. Can any one help us??? Thanks!

20061115, 02:42  #2 
Aug 2003
Snicker, AL
2^{3}·3·5^{2} Posts 
I came up with fairly simple solutions that result in 10, 11, 12, 13, 14, 15, and 16 pieces.
The big question is whether or not we should be helping do someone's homework. The teacher may have a reason for asking the kids to solve this puzzle. I will tell you that the secret is the first two lines you draw which must have an intersection near the edge of the circle. Fusion Last fiddled with by Fusion_power on 20061115 at 02:51 
20061115, 03:47  #3 
"Mike"
Aug 2002
17033_{8} Posts 
http://www.research.att.com/~njas/sequences/A000127
We've included a solution, zipped, so it isn't easily viewed. We think it is more interesting to think it out. The fact we were able to figure it out puts this math problem at around the third grade level of difficulty. 
20061115, 03:59  #4 
Nov 2006
3 Posts 
Thanks a bunch
We had figured it out with his brothers help after his concert but wanted to check if we'd done it right! Thanks again! When I usually ask for help i will show what work we have so far but didn't know how to show our work for this one! Thanks again! M and Mom

20061115, 04:03  #5 
"Mike"
Aug 2002
1E1B_{16} Posts 
Stick around and let the kids read the forum. They'll learn all sorts of cool stuff. We're 36 years old physically but we have the mind of a 12 year old. We learn new stuff here every day.
(When we say "we" we mean just me. We don't know how we picked up this annoying habit.) 
20061115, 04:19  #6  
Jun 2005
2·191 Posts 
Quote:


20061115, 05:05  #7 
Jun 2005
2·191 Posts 
I think this problem can be generalized in a novel way. Given the following conditions:
1. a total of n lines 2. no two lines are parallel (each pair has an intersection) 3. no 3 lines share an intersection 4. a circle is drawn large enough to enclose every intersection I believe the circle will be divided into n^{2}/2+n/2+1 regions. Does anyone concur? Xyzzy, the site you referenced described something different. It's connecting a given number of points along the circle (3 points yields 4 regions...etc)...on that page, 16 refers to 5 points connected by 10 lines yielding 16 regions. Drew Last fiddled with by drew on 20061115 at 05:15 
20061115, 07:55  #8 
Dec 2005
3×5×13 Posts 
I agree with this formula.
Next question, what would be the answer if in stead of straight lines we allowed for arcs ? 
20061115, 08:24  #9 
Aug 2003
Snicker, AL
600_{10} Posts 
n^2/2 + n/2 + 1 = ?
4^/2 + 4/2 + 1 = ? 8 + 2 + 1 = 11 Drew, your formula predicts 11 pieces can be made from 4 lines. Can you find a solution with 4 lines that gives 11 pieces? How does your formula fare with only 3 lines? I came up with 3 different solutions that resulted in 16 pieces for 5 lines drawn. There are probably a few more valid solutions. Drew, I would point out that your solution is not the preferred solution, but I won't tell you why. You will have to solve the puzzle for a circle divided by 4 lines to figure this out. Said another way, there is a generalized pattern of drawing the lines that yields the maximum number of pieces whether 3, 4, 5, or more lines are drawn. I did not however carry this to a logical conclusion so it is possible more than 6 lines could prevent the pattern from working. The interesting part to me was coming up with solutions for less than 16 pieces. The minimum seems to be 10 pieces from 5 lines drawn. Given this information maybe you could come up with a formula that would express the relationship of minimum and maximum pieces vs number of lines. Fusion 
20061115, 09:12  #10 
Dec 2005
C3_{16} Posts 
well, with 5 lines it seems that 6 pieces should be possible as well

20061115, 11:47  #11  
"Mike"
Aug 2002
3×7×367 Posts 
Quote:
http://www.research.att.com/~njas/sequences/A000124 

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