Haselbauer-Dickheiser Test no. 3: Circle divided by lines between a blue dotsWhat is the next number in the series: 21, 21, 23, 20, 5, 25, 31, 24,?Polygonal PizzaHikers Meeting in the MiddleProfessor Halfbrain and the dissection of a rectangleA moderate visual number puzzleHalf the pentagon!Find the maximum number of right claimsInner Triangles in the circle
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Haselbauer-Dickheiser Test no. 3: Circle divided by lines between a blue dots
What is the next number in the series: 21, 21, 23, 20, 5, 25, 31, 24,?Polygonal PizzaHikers Meeting in the MiddleProfessor Halfbrain and the dissection of a rectangleA moderate visual number puzzleHalf the pentagon!Find the maximum number of right claimsInner Triangles in the circle
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;
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$begingroup$
This is the test no. 3 from Haselbauer-Dickheiser Test.
3.
These three circles below all have blue dots on their circumference which are connected by straight lines. These lines divide the circles into smaller regions. The first circle, with two blue dots, is separated into two regions. The second circle, with three blue dots, is separated into four regions. The third circle, with four blue dots, is separated into 8 regions.
Given a circle where seven blue dots are placed anywhere on its circumference, what would be the maximum number of regions into which it could be divided?
(Original image)
mathematics visual geometry
edited Oct 5 at 7:46
Andrew T.
1096 bronze badges
asked Sep 18 at 14:26
CuriousSuperheroCuriousSuperhero
2031 silver badge7 bronze badges
$endgroup$
add a comment
|
$begingroup$
This is the test no. 3 from Haselbauer-Dickheiser Test.
3.
These three circles below all have blue dots on their circumference which are connected by straight lines. These lines divide the circles into smaller regions. The first circle, with two blue dots, is separated into two regions. The second circle, with three blue dots, is separated into four regions. The third circle, with four blue dots, is separated into 8 regions.
Given a circle where seven blue dots are placed anywhere on its circumference, what would be the maximum number of regions into which it could be divided?
(Original image)
mathematics visual geometry
edited Oct 5 at 7:46
Andrew T.
1096 bronze badges
asked Sep 18 at 14:26
CuriousSuperheroCuriousSuperhero
2031 silver badge7 bronze badges
$endgroup$
add a comment
|
$begingroup$
This is the test no. 3 from Haselbauer-Dickheiser Test.
3.
These three circles below all have blue dots on their circumference which are connected by straight lines. These lines divide the circles into smaller regions. The first circle, with two blue dots, is separated into two regions. The second circle, with three blue dots, is separated into four regions. The third circle, with four blue dots, is separated into 8 regions.
Given a circle where seven blue dots are placed anywhere on its circumference, what would be the maximum number of regions into which it could be divided?
(Original image)
mathematics visual geometry
edited Oct 5 at 7:46
Andrew T.
1096 bronze badges
asked Sep 18 at 14:26
CuriousSuperheroCuriousSuperhero
2031 silver badge7 bronze badges
$endgroup$
This is the test no. 3 from Haselbauer-Dickheiser Test.
3.
These three circles below all have blue dots on their circumference which are connected by straight lines. These lines divide the circles into smaller regions. The first circle, with two blue dots, is separated into two regions. The second circle, with three blue dots, is separated into four regions. The third circle, with four blue dots, is separated into 8 regions.
Given a circle where seven blue dots are placed anywhere on its circumference, what would be the maximum number of regions into which it could be divided?
(Original image)
mathematics visual geometry
mathematics visual geometry
edited Oct 5 at 7:46
Andrew T.
1096 bronze badges
asked Sep 18 at 14:26
CuriousSuperheroCuriousSuperhero
2031 silver badge7 bronze badges
edited Oct 5 at 7:46
Andrew T.
1096 bronze badges
asked Sep 18 at 14:26
CuriousSuperheroCuriousSuperhero
2031 silver badge7 bronze badges
edited Oct 5 at 7:46
Andrew T.
1096 bronze badges
edited Oct 5 at 7:46
Andrew T.
1096 bronze badges
edited Oct 5 at 7:46
Andrew T.
1096 bronze badges
1096 bronze badges
asked Sep 18 at 14:26
CuriousSuperheroCuriousSuperhero
2031 silver badge7 bronze badges
asked Sep 18 at 14:26
CuriousSuperheroCuriousSuperhero
2031 silver badge7 bronze badges
asked Sep 18 at 14:26
CuriousSuperheroCuriousSuperhero
2031 silver badge7 bronze badges
2031 silver badge7 bronze badges
add a comment
|
add a comment
|
4 Answers
4
active
oldest
votes
$begingroup$
The answer is
57.
This is a well-known problem called
Moser's circle problem. The sequence given by "maximal number of regions with $n$ blue dots" for increasing values of $n$ is $1,2,4,8,16,31,57,dots$ (OEIS A000127). It's famously deceptive because the first few terms make it look like it's going to be simply the powers of 2, as another answer guessed, but it isn't.
answered Sep 18 at 14:42
Rand al'ThorRand al'Thor
89.7k18 gold badges253 silver badges512 bronze badges
$endgroup$
$begingroup$
Well done, you got me again!
$endgroup$
– Weather Vane
Sep 18 at 14:49
$begingroup$
What are other deceptive sequences? (non-trivial ones that have real applications)?
$endgroup$
– smci
Sep 18 at 23:07
$begingroup$
@smci Dunno. This is my go-to example for a sequence that seems to go one way and actually goes another.
$endgroup$
– Rand al'Thor
Sep 19 at 5:12
add a comment
|
$begingroup$
An answer from @Randal'Thor was posted while I prepared this.
My (independent) answer is
57
Which I obtained by counting successive diagrams.
This is confirmed by the sequence
2,4,8,16,31,57
which is shown by OEIS to be A000127
Maximal number of regions obtained by joining n points around a circle by straight lines.
answered Sep 18 at 14:47
Weather VaneWeather Vane
6,8091 gold badge4 silver badges27 bronze badges
$endgroup$
$begingroup$
this is what I thought of by seeing the picture but the thing is if each point is connected by the line then there is? 42 lines right? the region formula I got is wrong?
$endgroup$
– Sayed Mohd Ali
Sep 18 at 14:55
$begingroup$
@SayedMohdAli that linked page gives the forrmula $(n^4 - 6n^3 + 23n^2 - 18n + 24)/24$
$endgroup$
– Weather Vane
Sep 18 at 15:00
1
$begingroup$
@SayedMohdAli the numbers of lines is half that because each each is shared by two points. So $n(n-1)/2$
$endgroup$
– Weather Vane
Sep 18 at 15:08
$begingroup$
I saw later :P previously I calculated number of lines wrong it should be 7*6/2 and I did 7*6... but later I corrected it :P I created sets. but my ideas was exactly same as yours but with little more research I got another way.. :D +1
$endgroup$
– Sayed Mohd Ali
Sep 18 at 19:06
add a comment
|
$begingroup$
My answer is reference
Regions of a Circle Cut by Chords to n Points
---------------------------------------------- n points are distributed round the circumference of a circle and each point is
joined to every other point by a chord of the circle. Assuming that
no three chords intersect at a point inside the circle we require the
number of regions into which the circle is divided.
With no lines the circle has just one region. Now consider any
collection of lines. If you draw a new line across the circle which
does not cross any existing lines, then the effect is to increase the
number of regions by 1. In addition, every time a new line crosses an
existing line inside the circle the number of regions is increased by
1 again.
So in any such arrangement
number of regions = 1 + number of lines + number of interior
intersections
= 1 + C(n,2) + C(n,4)
Note that the number of lines is the number of ways 2 points can be
chosen from n points. Also, the number of interior intersections is
the number of quadrilaterals that can be formed from n points, since
each quadrilateral produces just 1 intersection where the diagonals
of the quadrilateral intersect.
Examples:
n=4 Number of regions = 1 + C(4,2) + C(4,4) = 8
n=5 Number of regions = 1 + C(5,2) + C(5,4) = 16
n=6 " " = 1 + C(6,2) + C(6,4) = 31
n=7 " " = 1 + C(7,2) + C(7,4) = 57
answered Sep 18 at 14:38
Sayed Mohd AliSayed Mohd Ali
1,2261 silver badge20 bronze badges
$endgroup$
$begingroup$
I will update the answer counting :P the total lines wait.
$endgroup$
– Sayed Mohd Ali
Sep 18 at 14:39
$begingroup$
It is asking for the number of regions, not the number of lines.
$endgroup$
– Jaap Scherphuis
Sep 18 at 14:41
1
$begingroup$
In case you are still wondering, the region formula you previously used does not apply to this case. It assumes that every pair of lines intersect in a unique point, and counts all the regions. In this case we have points where more than 2 lines intersect (the blue points). We also have lines intersecting outside the circle (e.g. non-adjacent edges) leading to extra regions outside the circle that we are not interested in counting here.
$endgroup$
– Jaap Scherphuis
Sep 18 at 15:29
add a comment
|
$begingroup$
64 - the number appears to be doubling with each additional point.
answered Sep 18 at 14:36
AndyJ97AndyJ97
1
$endgroup$
8
$begingroup$
Nope. This is a famously deceptive sequence.
$endgroup$
– Rand al'Thor
Sep 18 at 14:43
add a comment
|
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer is
57.
This is a well-known problem called
Moser's circle problem. The sequence given by "maximal number of regions with $n$ blue dots" for increasing values of $n$ is $1,2,4,8,16,31,57,dots$ (OEIS A000127). It's famously deceptive because the first few terms make it look like it's going to be simply the powers of 2, as another answer guessed, but it isn't.
answered Sep 18 at 14:42
Rand al'ThorRand al'Thor
89.7k18 gold badges253 silver badges512 bronze badges
$endgroup$
$begingroup$
Well done, you got me again!
$endgroup$
– Weather Vane
Sep 18 at 14:49
$begingroup$
What are other deceptive sequences? (non-trivial ones that have real applications)?
$endgroup$
– smci
Sep 18 at 23:07
$begingroup$
@smci Dunno. This is my go-to example for a sequence that seems to go one way and actually goes another.
$endgroup$
– Rand al'Thor
Sep 19 at 5:12
add a comment
|
$begingroup$
The answer is
57.
This is a well-known problem called
Moser's circle problem. The sequence given by "maximal number of regions with $n$ blue dots" for increasing values of $n$ is $1,2,4,8,16,31,57,dots$ (OEIS A000127). It's famously deceptive because the first few terms make it look like it's going to be simply the powers of 2, as another answer guessed, but it isn't.
answered Sep 18 at 14:42
Rand al'ThorRand al'Thor
89.7k18 gold badges253 silver badges512 bronze badges
$endgroup$
$begingroup$
Well done, you got me again!
$endgroup$
– Weather Vane
Sep 18 at 14:49
$begingroup$
What are other deceptive sequences? (non-trivial ones that have real applications)?
$endgroup$
– smci
Sep 18 at 23:07
$begingroup$
@smci Dunno. This is my go-to example for a sequence that seems to go one way and actually goes another.
$endgroup$
– Rand al'Thor
Sep 19 at 5:12
add a comment
|
$begingroup$
The answer is
57.
This is a well-known problem called
Moser's circle problem. The sequence given by "maximal number of regions with $n$ blue dots" for increasing values of $n$ is $1,2,4,8,16,31,57,dots$ (OEIS A000127). It's famously deceptive because the first few terms make it look like it's going to be simply the powers of 2, as another answer guessed, but it isn't.
answered Sep 18 at 14:42
Rand al'ThorRand al'Thor
89.7k18 gold badges253 silver badges512 bronze badges
$endgroup$
The answer is
57.
This is a well-known problem called
Moser's circle problem. The sequence given by "maximal number of regions with $n$ blue dots" for increasing values of $n$ is $1,2,4,8,16,31,57,dots$ (OEIS A000127). It's famously deceptive because the first few terms make it look like it's going to be simply the powers of 2, as another answer guessed, but it isn't.
answered Sep 18 at 14:42
Rand al'ThorRand al'Thor
89.7k18 gold badges253 silver badges512 bronze badges
edited Sep 19 at 5:12
answered Sep 18 at 14:42
Rand al'ThorRand al'Thor
89.7k18 gold badges253 silver badges512 bronze badges
answered Sep 18 at 14:42
Rand al'ThorRand al'Thor
89.7k18 gold badges253 silver badges512 bronze badges
answered Sep 18 at 14:42
Rand al'ThorRand al'Thor
89.7k18 gold badges253 silver badges512 bronze badges
89.7k18 gold badges253 silver badges512 bronze badges
$begingroup$
Well done, you got me again!
$endgroup$
– Weather Vane
Sep 18 at 14:49
$begingroup$
What are other deceptive sequences? (non-trivial ones that have real applications)?
$endgroup$
– smci
Sep 18 at 23:07
$begingroup$
@smci Dunno. This is my go-to example for a sequence that seems to go one way and actually goes another.
$endgroup$
– Rand al'Thor
Sep 19 at 5:12
add a comment
|
$begingroup$
Well done, you got me again!
$endgroup$
– Weather Vane
Sep 18 at 14:49
$begingroup$
What are other deceptive sequences? (non-trivial ones that have real applications)?
$endgroup$
– smci
Sep 18 at 23:07
$begingroup$
@smci Dunno. This is my go-to example for a sequence that seems to go one way and actually goes another.
$endgroup$
– Rand al'Thor
Sep 19 at 5:12
$begingroup$
Well done, you got me again!
$endgroup$
– Weather Vane
Sep 18 at 14:49
$begingroup$
Well done, you got me again!
$endgroup$
– Weather Vane
Sep 18 at 14:49
$begingroup$
What are other deceptive sequences? (non-trivial ones that have real applications)?
$endgroup$
– smci
Sep 18 at 23:07
$begingroup$
What are other deceptive sequences? (non-trivial ones that have real applications)?
$endgroup$
– smci
Sep 18 at 23:07
$begingroup$
@smci Dunno. This is my go-to example for a sequence that seems to go one way and actually goes another.
$endgroup$
– Rand al'Thor
Sep 19 at 5:12
$begingroup$
@smci Dunno. This is my go-to example for a sequence that seems to go one way and actually goes another.
$endgroup$
– Rand al'Thor
Sep 19 at 5:12
add a comment
|
$begingroup$
An answer from @Randal'Thor was posted while I prepared this.
My (independent) answer is
57
Which I obtained by counting successive diagrams.
This is confirmed by the sequence
2,4,8,16,31,57
which is shown by OEIS to be A000127
Maximal number of regions obtained by joining n points around a circle by straight lines.
answered Sep 18 at 14:47
Weather VaneWeather Vane
6,8091 gold badge4 silver badges27 bronze badges
$endgroup$
$begingroup$
this is what I thought of by seeing the picture but the thing is if each point is connected by the line then there is? 42 lines right? the region formula I got is wrong?
$endgroup$
– Sayed Mohd Ali
Sep 18 at 14:55
$begingroup$
@SayedMohdAli that linked page gives the forrmula $(n^4 - 6n^3 + 23n^2 - 18n + 24)/24$
$endgroup$
– Weather Vane
Sep 18 at 15:00
1
$begingroup$
@SayedMohdAli the numbers of lines is half that because each each is shared by two points. So $n(n-1)/2$
$endgroup$
– Weather Vane
Sep 18 at 15:08
$begingroup$
I saw later :P previously I calculated number of lines wrong it should be 7*6/2 and I did 7*6... but later I corrected it :P I created sets. but my ideas was exactly same as yours but with little more research I got another way.. :D +1
$endgroup$
– Sayed Mohd Ali
Sep 18 at 19:06
add a comment
|
$begingroup$
An answer from @Randal'Thor was posted while I prepared this.
My (independent) answer is
57
Which I obtained by counting successive diagrams.
This is confirmed by the sequence
2,4,8,16,31,57
which is shown by OEIS to be A000127
Maximal number of regions obtained by joining n points around a circle by straight lines.
answered Sep 18 at 14:47
Weather VaneWeather Vane
6,8091 gold badge4 silver badges27 bronze badges
$endgroup$
$begingroup$
this is what I thought of by seeing the picture but the thing is if each point is connected by the line then there is? 42 lines right? the region formula I got is wrong?
$endgroup$
– Sayed Mohd Ali
Sep 18 at 14:55
$begingroup$
@SayedMohdAli that linked page gives the forrmula $(n^4 - 6n^3 + 23n^2 - 18n + 24)/24$
$endgroup$
– Weather Vane
Sep 18 at 15:00
1
$begingroup$
@SayedMohdAli the numbers of lines is half that because each each is shared by two points. So $n(n-1)/2$
$endgroup$
– Weather Vane
Sep 18 at 15:08
$begingroup$
I saw later :P previously I calculated number of lines wrong it should be 7*6/2 and I did 7*6... but later I corrected it :P I created sets. but my ideas was exactly same as yours but with little more research I got another way.. :D +1
$endgroup$
– Sayed Mohd Ali
Sep 18 at 19:06
add a comment
|
$begingroup$
An answer from @Randal'Thor was posted while I prepared this.
My (independent) answer is
57
Which I obtained by counting successive diagrams.
This is confirmed by the sequence
2,4,8,16,31,57
which is shown by OEIS to be A000127
Maximal number of regions obtained by joining n points around a circle by straight lines.
answered Sep 18 at 14:47
Weather VaneWeather Vane
6,8091 gold badge4 silver badges27 bronze badges
$endgroup$
An answer from @Randal'Thor was posted while I prepared this.
My (independent) answer is
57
Which I obtained by counting successive diagrams.
This is confirmed by the sequence
2,4,8,16,31,57
which is shown by OEIS to be A000127
Maximal number of regions obtained by joining n points around a circle by straight lines.
answered Sep 18 at 14:47
Weather VaneWeather Vane
6,8091 gold badge4 silver badges27 bronze badges
answered Sep 18 at 14:47
Weather VaneWeather Vane
6,8091 gold badge4 silver badges27 bronze badges
answered Sep 18 at 14:47
Weather VaneWeather Vane
6,8091 gold badge4 silver badges27 bronze badges
answered Sep 18 at 14:47
Weather VaneWeather Vane
6,8091 gold badge4 silver badges27 bronze badges
6,8091 gold badge4 silver badges27 bronze badges
$begingroup$
this is what I thought of by seeing the picture but the thing is if each point is connected by the line then there is? 42 lines right? the region formula I got is wrong?
$endgroup$
– Sayed Mohd Ali
Sep 18 at 14:55
$begingroup$
@SayedMohdAli that linked page gives the forrmula $(n^4 - 6n^3 + 23n^2 - 18n + 24)/24$
$endgroup$
– Weather Vane
Sep 18 at 15:00
1
$begingroup$
@SayedMohdAli the numbers of lines is half that because each each is shared by two points. So $n(n-1)/2$
$endgroup$
– Weather Vane
Sep 18 at 15:08
$begingroup$
I saw later :P previously I calculated number of lines wrong it should be 7*6/2 and I did 7*6... but later I corrected it :P I created sets. but my ideas was exactly same as yours but with little more research I got another way.. :D +1
$endgroup$
– Sayed Mohd Ali
Sep 18 at 19:06
add a comment
|
$begingroup$
this is what I thought of by seeing the picture but the thing is if each point is connected by the line then there is? 42 lines right? the region formula I got is wrong?
$endgroup$
– Sayed Mohd Ali
Sep 18 at 14:55
$begingroup$
@SayedMohdAli that linked page gives the forrmula $(n^4 - 6n^3 + 23n^2 - 18n + 24)/24$
$endgroup$
– Weather Vane
Sep 18 at 15:00
1
$begingroup$
@SayedMohdAli the numbers of lines is half that because each each is shared by two points. So $n(n-1)/2$
$endgroup$
– Weather Vane
Sep 18 at 15:08
$begingroup$
I saw later :P previously I calculated number of lines wrong it should be 7*6/2 and I did 7*6... but later I corrected it :P I created sets. but my ideas was exactly same as yours but with little more research I got another way.. :D +1
$endgroup$
– Sayed Mohd Ali
Sep 18 at 19:06
$begingroup$
this is what I thought of by seeing the picture but the thing is if each point is connected by the line then there is? 42 lines right? the region formula I got is wrong?
$endgroup$
– Sayed Mohd Ali
Sep 18 at 14:55
$begingroup$
this is what I thought of by seeing the picture but the thing is if each point is connected by the line then there is? 42 lines right? the region formula I got is wrong?
$endgroup$
– Sayed Mohd Ali
Sep 18 at 14:55
$begingroup$
@SayedMohdAli that linked page gives the forrmula $(n^4 - 6n^3 + 23n^2 - 18n + 24)/24$
$endgroup$
– Weather Vane
Sep 18 at 15:00
$begingroup$
@SayedMohdAli that linked page gives the forrmula $(n^4 - 6n^3 + 23n^2 - 18n + 24)/24$
$endgroup$
– Weather Vane
Sep 18 at 15:00
1
1
$begingroup$
@SayedMohdAli the numbers of lines is half that because each each is shared by two points. So $n(n-1)/2$
$endgroup$
– Weather Vane
Sep 18 at 15:08
$begingroup$
@SayedMohdAli the numbers of lines is half that because each each is shared by two points. So $n(n-1)/2$
$endgroup$
– Weather Vane
Sep 18 at 15:08
$begingroup$
I saw later :P previously I calculated number of lines wrong it should be 7*6/2 and I did 7*6... but later I corrected it :P I created sets. but my ideas was exactly same as yours but with little more research I got another way.. :D +1
$endgroup$
– Sayed Mohd Ali
Sep 18 at 19:06
$begingroup$
I saw later :P previously I calculated number of lines wrong it should be 7*6/2 and I did 7*6... but later I corrected it :P I created sets. but my ideas was exactly same as yours but with little more research I got another way.. :D +1
$endgroup$
– Sayed Mohd Ali
Sep 18 at 19:06
add a comment
|
$begingroup$
My answer is reference
Regions of a Circle Cut by Chords to n Points
---------------------------------------------- n points are distributed round the circumference of a circle and each point is
joined to every other point by a chord of the circle. Assuming that
no three chords intersect at a point inside the circle we require the
number of regions into which the circle is divided.
With no lines the circle has just one region. Now consider any
collection of lines. If you draw a new line across the circle which
does not cross any existing lines, then the effect is to increase the
number of regions by 1. In addition, every time a new line crosses an
existing line inside the circle the number of regions is increased by
1 again.
So in any such arrangement
number of regions = 1 + number of lines + number of interior
intersections
= 1 + C(n,2) + C(n,4)
Note that the number of lines is the number of ways 2 points can be
chosen from n points. Also, the number of interior intersections is
the number of quadrilaterals that can be formed from n points, since
each quadrilateral produces just 1 intersection where the diagonals
of the quadrilateral intersect.
Examples:
n=4 Number of regions = 1 + C(4,2) + C(4,4) = 8
n=5 Number of regions = 1 + C(5,2) + C(5,4) = 16
n=6 " " = 1 + C(6,2) + C(6,4) = 31
n=7 " " = 1 + C(7,2) + C(7,4) = 57
answered Sep 18 at 14:38
Sayed Mohd AliSayed Mohd Ali
1,2261 silver badge20 bronze badges
$endgroup$
$begingroup$
I will update the answer counting :P the total lines wait.
$endgroup$
– Sayed Mohd Ali
Sep 18 at 14:39
$begingroup$
It is asking for the number of regions, not the number of lines.
$endgroup$
– Jaap Scherphuis
Sep 18 at 14:41
1
$begingroup$
In case you are still wondering, the region formula you previously used does not apply to this case. It assumes that every pair of lines intersect in a unique point, and counts all the regions. In this case we have points where more than 2 lines intersect (the blue points). We also have lines intersecting outside the circle (e.g. non-adjacent edges) leading to extra regions outside the circle that we are not interested in counting here.
$endgroup$
– Jaap Scherphuis
Sep 18 at 15:29
add a comment
|
$begingroup$
My answer is reference
Regions of a Circle Cut by Chords to n Points
---------------------------------------------- n points are distributed round the circumference of a circle and each point is
joined to every other point by a chord of the circle. Assuming that
no three chords intersect at a point inside the circle we require the
number of regions into which the circle is divided.
With no lines the circle has just one region. Now consider any
collection of lines. If you draw a new line across the circle which
does not cross any existing lines, then the effect is to increase the
number of regions by 1. In addition, every time a new line crosses an
existing line inside the circle the number of regions is increased by
1 again.
So in any such arrangement
number of regions = 1 + number of lines + number of interior
intersections
= 1 + C(n,2) + C(n,4)
Note that the number of lines is the number of ways 2 points can be
chosen from n points. Also, the number of interior intersections is
the number of quadrilaterals that can be formed from n points, since
each quadrilateral produces just 1 intersection where the diagonals
of the quadrilateral intersect.
Examples:
n=4 Number of regions = 1 + C(4,2) + C(4,4) = 8
n=5 Number of regions = 1 + C(5,2) + C(5,4) = 16
n=6 " " = 1 + C(6,2) + C(6,4) = 31
n=7 " " = 1 + C(7,2) + C(7,4) = 57
answered Sep 18 at 14:38
Sayed Mohd AliSayed Mohd Ali
1,2261 silver badge20 bronze badges
$endgroup$
$begingroup$
I will update the answer counting :P the total lines wait.
$endgroup$
– Sayed Mohd Ali
Sep 18 at 14:39
$begingroup$
It is asking for the number of regions, not the number of lines.
$endgroup$
– Jaap Scherphuis
Sep 18 at 14:41
1
$begingroup$
In case you are still wondering, the region formula you previously used does not apply to this case. It assumes that every pair of lines intersect in a unique point, and counts all the regions. In this case we have points where more than 2 lines intersect (the blue points). We also have lines intersecting outside the circle (e.g. non-adjacent edges) leading to extra regions outside the circle that we are not interested in counting here.
$endgroup$
– Jaap Scherphuis
Sep 18 at 15:29
add a comment
|
$begingroup$
My answer is reference
Regions of a Circle Cut by Chords to n Points
---------------------------------------------- n points are distributed round the circumference of a circle and each point is
joined to every other point by a chord of the circle. Assuming that
no three chords intersect at a point inside the circle we require the
number of regions into which the circle is divided.
With no lines the circle has just one region. Now consider any
collection of lines. If you draw a new line across the circle which
does not cross any existing lines, then the effect is to increase the
number of regions by 1. In addition, every time a new line crosses an
existing line inside the circle the number of regions is increased by
1 again.
So in any such arrangement
number of regions = 1 + number of lines + number of interior
intersections
= 1 + C(n,2) + C(n,4)
Note that the number of lines is the number of ways 2 points can be
chosen from n points. Also, the number of interior intersections is
the number of quadrilaterals that can be formed from n points, since
each quadrilateral produces just 1 intersection where the diagonals
of the quadrilateral intersect.
Examples:
n=4 Number of regions = 1 + C(4,2) + C(4,4) = 8
n=5 Number of regions = 1 + C(5,2) + C(5,4) = 16
n=6 " " = 1 + C(6,2) + C(6,4) = 31
n=7 " " = 1 + C(7,2) + C(7,4) = 57
answered Sep 18 at 14:38
Sayed Mohd AliSayed Mohd Ali
1,2261 silver badge20 bronze badges
$endgroup$
My answer is reference
Regions of a Circle Cut by Chords to n Points
---------------------------------------------- n points are distributed round the circumference of a circle and each point is
joined to every other point by a chord of the circle. Assuming that
no three chords intersect at a point inside the circle we require the
number of regions into which the circle is divided.
With no lines the circle has just one region. Now consider any
collection of lines. If you draw a new line across the circle which
does not cross any existing lines, then the effect is to increase the
number of regions by 1. In addition, every time a new line crosses an
existing line inside the circle the number of regions is increased by
1 again.
So in any such arrangement
number of regions = 1 + number of lines + number of interior
intersections
= 1 + C(n,2) + C(n,4)
Note that the number of lines is the number of ways 2 points can be
chosen from n points. Also, the number of interior intersections is
the number of quadrilaterals that can be formed from n points, since
each quadrilateral produces just 1 intersection where the diagonals
of the quadrilateral intersect.
Examples:
n=4 Number of regions = 1 + C(4,2) + C(4,4) = 8
n=5 Number of regions = 1 + C(5,2) + C(5,4) = 16
n=6 " " = 1 + C(6,2) + C(6,4) = 31
n=7 " " = 1 + C(7,2) + C(7,4) = 57
answered Sep 18 at 14:38
Sayed Mohd AliSayed Mohd Ali
1,2261 silver badge20 bronze badges
edited Sep 18 at 18:57
answered Sep 18 at 14:38
Sayed Mohd AliSayed Mohd Ali
1,2261 silver badge20 bronze badges
answered Sep 18 at 14:38
Sayed Mohd AliSayed Mohd Ali
1,2261 silver badge20 bronze badges
answered Sep 18 at 14:38
Sayed Mohd AliSayed Mohd Ali
1,2261 silver badge20 bronze badges
1,2261 silver badge20 bronze badges
$begingroup$
I will update the answer counting :P the total lines wait.
$endgroup$
– Sayed Mohd Ali
Sep 18 at 14:39
$begingroup$
It is asking for the number of regions, not the number of lines.
$endgroup$
– Jaap Scherphuis
Sep 18 at 14:41
1
$begingroup$
In case you are still wondering, the region formula you previously used does not apply to this case. It assumes that every pair of lines intersect in a unique point, and counts all the regions. In this case we have points where more than 2 lines intersect (the blue points). We also have lines intersecting outside the circle (e.g. non-adjacent edges) leading to extra regions outside the circle that we are not interested in counting here.
$endgroup$
– Jaap Scherphuis
Sep 18 at 15:29
add a comment
|
$begingroup$
I will update the answer counting :P the total lines wait.
$endgroup$
– Sayed Mohd Ali
Sep 18 at 14:39
$begingroup$
It is asking for the number of regions, not the number of lines.
$endgroup$
– Jaap Scherphuis
Sep 18 at 14:41
1
$begingroup$
In case you are still wondering, the region formula you previously used does not apply to this case. It assumes that every pair of lines intersect in a unique point, and counts all the regions. In this case we have points where more than 2 lines intersect (the blue points). We also have lines intersecting outside the circle (e.g. non-adjacent edges) leading to extra regions outside the circle that we are not interested in counting here.
$endgroup$
– Jaap Scherphuis
Sep 18 at 15:29
$begingroup$
I will update the answer counting :P the total lines wait.
$endgroup$
– Sayed Mohd Ali
Sep 18 at 14:39
$begingroup$
I will update the answer counting :P the total lines wait.
$endgroup$
– Sayed Mohd Ali
Sep 18 at 14:39
$begingroup$
It is asking for the number of regions, not the number of lines.
$endgroup$
– Jaap Scherphuis
Sep 18 at 14:41
$begingroup$
It is asking for the number of regions, not the number of lines.
$endgroup$
– Jaap Scherphuis
Sep 18 at 14:41
1
1
$begingroup$
In case you are still wondering, the region formula you previously used does not apply to this case. It assumes that every pair of lines intersect in a unique point, and counts all the regions. In this case we have points where more than 2 lines intersect (the blue points). We also have lines intersecting outside the circle (e.g. non-adjacent edges) leading to extra regions outside the circle that we are not interested in counting here.
$endgroup$
– Jaap Scherphuis
Sep 18 at 15:29
$begingroup$
In case you are still wondering, the region formula you previously used does not apply to this case. It assumes that every pair of lines intersect in a unique point, and counts all the regions. In this case we have points where more than 2 lines intersect (the blue points). We also have lines intersecting outside the circle (e.g. non-adjacent edges) leading to extra regions outside the circle that we are not interested in counting here.
$endgroup$
– Jaap Scherphuis
Sep 18 at 15:29
add a comment
|
$begingroup$
64 - the number appears to be doubling with each additional point.
answered Sep 18 at 14:36
AndyJ97AndyJ97
1
$endgroup$
8
$begingroup$
Nope. This is a famously deceptive sequence.
$endgroup$
– Rand al'Thor
Sep 18 at 14:43
add a comment
|
$begingroup$
64 - the number appears to be doubling with each additional point.
answered Sep 18 at 14:36
AndyJ97AndyJ97
1
$endgroup$
8
$begingroup$
Nope. This is a famously deceptive sequence.
$endgroup$
– Rand al'Thor
Sep 18 at 14:43
add a comment
|
$begingroup$
64 - the number appears to be doubling with each additional point.
answered Sep 18 at 14:36
AndyJ97AndyJ97
1
$endgroup$
64 - the number appears to be doubling with each additional point.
answered Sep 18 at 14:36
AndyJ97AndyJ97
1
answered Sep 18 at 14:36
AndyJ97AndyJ97
1
answered Sep 18 at 14:36
AndyJ97AndyJ97
1
answered Sep 18 at 14:36
AndyJ97AndyJ97
1
1
8
$begingroup$
Nope. This is a famously deceptive sequence.
$endgroup$
– Rand al'Thor
Sep 18 at 14:43
add a comment
|
8
$begingroup$
Nope. This is a famously deceptive sequence.
$endgroup$
– Rand al'Thor
Sep 18 at 14:43
8
8
$begingroup$
Nope. This is a famously deceptive sequence.
$endgroup$
– Rand al'Thor
Sep 18 at 14:43
$begingroup$
Nope. This is a famously deceptive sequence.
$endgroup$
– Rand al'Thor
Sep 18 at 14:43
add a comment
|
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