20041215, 17:54  #1 
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts 
Circles part 2
We have seen that if to a circle spots are added then the resutling regions formed are as follows. No.of spots are 1, 2 3 4 5 6 No. of regionsl 1, 2 4 8 16 31 The 1st 5 terms are in G.P but from the 6th T6 =31 and not 32 as expected. We must note that any point where 3 or more lines intersect has to be counted as a zero area region [Thanks Paul] Concurrent lines like the spots themselves are not to be counted. Quest: What are the next two terms? This is sufficient data to to work out the nth term Tn. Derive formula for Tn. Hint: There are at least three different ways in which Tn can be expressed. Mally 
20041218, 06:00  #2 
Mar 2004
ARIZONA, USA
27_{8} Posts 
a novice's guess
(n>5)=(2^n1)1
7=(2^6)1=63 8=(2^7)1=127 I am most likely wrong, but learning is failing, correct? Last fiddled with by THILLIAR on 20041218 at 06:04 
20041218, 08:52  #3  
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}×3×599 Posts 
Quote:
That assumption would probably not survive a thoughtful comparison of diagrams with n points and the connecting diagonals, for n= 2, 3, 4, 5, ... One would soon note, for example, that each time a point is added, one more region bordering the circle is created. There are [b]n[/b] such regions between the circle and the inscribed [b]n[/b]sided polygon. When one subtracts [b]n[/b] from T[b]n[/b], one is left with the number of regions inside the polygon. But soon that doesn't jibe with a 2^([b]n[/b]1) assumption ... This would lead to one of your three ways of representing Tn, I suppose.  P.S. Why does the spoiler tag interfere with wordwrapping when I don't explicitly include line breaks? Or is that just on my browser? Last fiddled with by cheesehead on 20041218 at 09:04 

20041218, 11:40  #4 
Sep 2002
Vienna, Austria
DB_{16} Posts 
(120 + 94 n + 5 n^2 + 25 n^3  5 n^4 + n^5)/120

20041218, 16:25  #5  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts 
Circles part 2
Quote:
Thanks to your response to this problem in two parts. I think they should be upgraded to the math forum as they contain deep math concepts. Kindly note Xyzzy if you read this. Thrillar:. A good spirited attempt but your formula does not agree with the 6th term onwards. Yes every time you fall Get UP and don’t remain fallen! Cheesehead: A good explanation but not rigorously correct. The spots on the circle formed by concurrent lines are not to be confused with points formed by intersecting lines which tho’ of zero space are to be counted. The spots are not included in the summation. Wpolly: You are on the right track. You are out by a factor of appx. 5n, which if you divide by it, your answer will be nearly correct but not exactly. I’m sure if you rework your calculations you should have the right answer. If you recall in ‘Circle 1’ I had mentioned that the function is a quartic and not the quintic you have given. Now to give a further hint I give the next three terms after 6th term T6 = 31. These are T7 =57; T8 =99 ; T9 ==163. I give two methods: 1) Use the method of finite differences and the first 8 terms are sufficient to evaluate the function. 2) You may also use Pascals triangle and make an appropriate diagonal cut from the 6th line downwards. The summation of each line gives a term. From the 6th line omit the no. 1 on the R.H.S.. On the 7th. line omit two no.s. viz 6 and ! on the R.H.S and form the diagonal downwards omitting the terms on the R.H.S. of the diagonal. [R.H.S. =Right hand side.] Try it out as it’s quite easy. If you have understood any of these 2 methods (the second is easier and straight forward) you should be able to predict the tenth term. T10 = ? I await your answers after which I will work out the quartic in full and give a concise formula how to remember it in only 3 terms given by an American math’ cian (Name concealed ) who worked it out. . Mally 

20041218, 20:52  #6  
"Richard B. Woods"
Aug 2002
Wisconsin USA
7188_{10} Posts 
Quote:


20041221, 09:24  #7 
Sep 2002
Vienna, Austria
11011011_{2} Posts 
(24  18 n + 23 n^2  6 n^3 + n^4)/24
Last fiddled with by wpolly on 20041221 at 09:29 
20041221, 09:26  #8 
Sep 2002
Vienna, Austria
3×73 Posts 
T10=256
Last fiddled with by wpolly on 20041221 at 09:30 
20041221, 09:31  #9 
Sep 2002
Vienna, Austria
3×73 Posts 
http://mathworld.wolfram.com/CircleD...nbyChords.html
http://www.research.att.com/cgibin/...i?Anum=A000127 Warning: spoilers inside.... 
20041221, 12:13  #10  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts 
Circles part 2
Quote:
Thank you wpoly. You are 'spot' on in both T10 and the formula. Have you connected this with the one given in Combination form? They are both the same. And what about Pascals triangle? Your websites were excellent but there are some mistakes in both which I will comment upon later Mally l 

20041227, 15:17  #11  
Bronze Medalist
Jan 2004
Mumbai,India
4004_{8} Posts 
Circles part 2
Quote:
Since I have not received any more replies or answers I reveal here the precise and concise formula to the nth. term first derived by Math'cian Leo Moser whose name it bears. Leo Mosers formula is Tn = N + (N)C4 +(N1)C2 C stands for combination Written in full it is (n^4  6n^3 + 23n^2  18n + 24 )/ 24 Mally 

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