One circle in a plane can produce one closed region at most(It produces one closed region at least).Two circles in a plane can produce at most three regions(They produce at least two regions).Three circles can produce seven closed regions at most(They produce three closed regions at least). How many distinct closed regions could be produced by 4, 5 and n circles at most? If n circles can produce f(n) closed regions at most(Of course they produce n closed regions at least), can they produce any number of closed regions between n and f(n)?


Question and Answers Forum

All Questions Topic List
Geometry Questions
Previous in All Question Next in All Question
Previous in Geometry Next in Geometry
Question Number 4033 by Rasheed Soomro last updated on 27/Dec/15

$One circle in a plane can produce one closed$ $region at most (It produces one closed region$ $at least) . Two circles in a plane can produce$ $at most three regions (They produce at least$ $two regions) . Three circles can produce seven$ $closed regions at most (They produce three$ $closed regions at least) .$ $How many distinct closed regions could be produced$ $by 4, 5 and n circles at most ?$ $If n circles can produce f (n) closed regions$ $at most (Of course they produce n closed$ $regions at least), can they produce any$ $number of closed regions between n and f (n) ?$
Commented by Rasheed Soomro last updated on 27/Dec/15

$For three circles 3, 4, 5, 6 and 7 closed regions$ $can be made .$
Commented by Rasheed Soomro last updated on 27/Dec/15

$Which area of mathematics does this problem$ $belong to ?$
Commented by Yozzii last updated on 27/Dec/15

Commented by Yozzii last updated on 27/Dec/15

$I o n l y f i r s t m e t t h i s i d e a i n s e t t h e o r y$ $b u t f (n) = 2^{n} - 1 i s v a l i d o n l y i f 0 ⩽ n ⩽ 3 .$ $4 c i r c l e s a p p e a r t o g i v e 13 c l o s e d$ $r e g i o n s a t m o s t .$
Commented by Yozzii last updated on 27/Dec/15

Commented by Yozzii last updated on 27/Dec/15

$5 c i r c l e s a p p e a r t o p r o d u c e a t m o s t$ $21 c l o s e d r e g i o n s .$ $n = 0 : f (0) = 0$ $n = 1 : f (1) = 1$ $n = 2 : f (2) = 3$ $n = 3 : f (3) = 7$ $n = 4 : f (4) = 13$ $n = 5 : f (5) = 21$ $n = 6 : f (6) = 31$ $f (1) - f (0) = 1 = 1 \times 2^{0}$ $f (2) - f (1) = 2 = 1 \times 2$ $f (3) - f (2) = 4 = 2 \times 2$ $f (4) - f (3) = 6 = 3 \times 2$ $f (5) - f (4) = 8 = 4 \times 2$ $f (6) - f (5) = 10 = 5 \times 2$ $n ⩾ 1, f (n + 1) - f (n) = 2 n, f (1) = 1$ $f (n + 1) = f (n) + 2 n$ $f (n + 1) - f (n) = 2 n$ $f (n) - f (n - 1) = 2 (n - 1)$ $f (n - 1) - f (n - 2) = 2 (n - 2)$ $⋮$ $f (4) - f (3) = 2 \times 3$ $f (3) - f (2) = 2 \times 2$ $f (2) - f (1) = 2 \times 1$ $\Rightarrow f (n + 1) - f (1) = 2 \underset{r = 1}{\sum^{n}} r$ $f (n + 1) = f (1) + 2 \times 0.5 n (n + 1)$ $f (n + 1) = 1 + n (n + 1)$ $\Rightarrow f (n) = 1 + n (n - 1)$ $f (n) = n^{2} - n + 1 n ⩾ 1$
Commented by prakash jain last updated on 27/Dec/15

$n - c i r c l e w i l l g i v e maximum 2^{n} - 1 closed region .$ $^{n} C_{1} +^{n} C_{2} + . . +^{n} C_{n} = 2^{n} - 1$ $Each term on LHS^{n} C_{r} is created by number$ $of ways r circles can intersect . Each of those$ $intersection form a closed area .$ $Since all circles can be independently drawn$ $you can create any number of closed regions$ $from n to 2^{n} - 1 .$
Commented by Rasheed Soomro last updated on 27/Dec/15

$But sir {Yozzii}^{'} s experiment shows that 4 circles$ $can produce 13 closed regions (See image of$ $four circles above) while your formula suggests 2^{4} - 1 = 15$ $Is there any mistake in experiment ?$
Commented by Yozzii last updated on 27/Dec/15

$D o w n l o a d t h e a p p l i c a t i o n c a l l e d G e o G e b r a .$ $I u s e d i t t o e x p e r i m e n t . I c o u l d c e r t a i n l y$ $b e w r o n g, s o u s e t h e a p p t o s e e f o r$ $y o u r s e l f h o w y o u c a n o b t a i n 15 . I$ ${w a s n}^{'} t s u c c e s s f u l i n o b t a i n i n g 15$ $o n m y o w n .$
Commented by Rasheed Soomro last updated on 27/Dec/15

$The process should be as follows .$ $The second circle must cut the first$ $circle, the third circle must cut first$ $and second circle at distinct points .$ $In this way each next circle must cut$ $all the previous circles at distinct points,$ $in order to get maximum number of$ $closed regions . I think you have done well .$
Commented by prakash jain last updated on 27/Dec/15

$I will try experimenting myself .$ $In a way you are saying a venn diagram$ $is not possible with 4 sets A, B, C, D to show$ $all of the following .$ $individual 4$ $2 pairs 6$ $3 paris 4$ $all intersection 1$ $I will do some more study to recall .$ $You may be right . I may have forgotten$ $a few things .$
Commented by Yozzii last updated on 27/Dec/15

$T h e d i a g r a m o u t l i n i n g a l l p o s s i b l e$ $e v e n t s f o r 4 s e t s d o e s n o t u t i l i s e$ $a c i r c l e, a s I s a w i n a b o o k I l o o k e d$ $a t r e c e n t l y . S o m e t h i n g l o o k i n g l i k e$ $a w r e n c h w a s u s e d, n o t a c i r c l e . {T h a t}^{'} s w h y I$ $d o u b t e d e m p l o y i n g f (n) = 2^{n} - 1 .$ $I t h i n k t h a t i f a c i r c l e w e r e p o s s i b l e$ $t h e a u t h o r {w o u l d n}^{'} t h a v e u s e d t h a t$ $w r e n c h s h a p e t o y i e l d a l l 16 r e g i o n s .$
Commented by prakash jain last updated on 27/Dec/15

$You are correct n circle can at most$ $creat = n^{2} - n + 2 regions .$
Commented by Yozzii last updated on 27/Dec/15

$I n c l u d i n g e x t e r n a l r e g i o n r i g h t ?$
Commented by prakash jain last updated on 27/Dec/15

$Yes .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum