A geometric problem on number of points and lines
$begingroup$
There are P points and L lines such that
- Each line contains 8 points
- Each point lies on 8 lines
- Any two distinct lines intersect in a unique point
- Any two distinct points lie on a unique line.
Lines may be straight or curved. What is $Ptimes L$ ?
My answer, is that, using last two conditions, if $P$ is no of points then L = $binom{P}{2}$. Also if no of lines is $L$ then $P = binom{L}{2}$, from here we need $binom{binom{L}{2}}{2} = L$ which gives 3 as L. This is certainly incorrect.
How to approach this geometric problem?
combinatorics geometry
$endgroup$
add a comment |
$begingroup$
There are P points and L lines such that
- Each line contains 8 points
- Each point lies on 8 lines
- Any two distinct lines intersect in a unique point
- Any two distinct points lie on a unique line.
Lines may be straight or curved. What is $Ptimes L$ ?
My answer, is that, using last two conditions, if $P$ is no of points then L = $binom{P}{2}$. Also if no of lines is $L$ then $P = binom{L}{2}$, from here we need $binom{binom{L}{2}}{2} = L$ which gives 3 as L. This is certainly incorrect.
How to approach this geometric problem?
combinatorics geometry
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1
$begingroup$
What happened to the eight points on a line/eight lines through a point when you were counting?
$endgroup$
– Mark Bennet
Mar 10 at 7:35
$begingroup$
@MarkBennet yes that went unused!
$endgroup$
– Max Payne
Mar 10 at 7:45
add a comment |
$begingroup$
There are P points and L lines such that
- Each line contains 8 points
- Each point lies on 8 lines
- Any two distinct lines intersect in a unique point
- Any two distinct points lie on a unique line.
Lines may be straight or curved. What is $Ptimes L$ ?
My answer, is that, using last two conditions, if $P$ is no of points then L = $binom{P}{2}$. Also if no of lines is $L$ then $P = binom{L}{2}$, from here we need $binom{binom{L}{2}}{2} = L$ which gives 3 as L. This is certainly incorrect.
How to approach this geometric problem?
combinatorics geometry
$endgroup$
There are P points and L lines such that
- Each line contains 8 points
- Each point lies on 8 lines
- Any two distinct lines intersect in a unique point
- Any two distinct points lie on a unique line.
Lines may be straight or curved. What is $Ptimes L$ ?
My answer, is that, using last two conditions, if $P$ is no of points then L = $binom{P}{2}$. Also if no of lines is $L$ then $P = binom{L}{2}$, from here we need $binom{binom{L}{2}}{2} = L$ which gives 3 as L. This is certainly incorrect.
How to approach this geometric problem?
combinatorics geometry
combinatorics geometry
edited Mar 10 at 8:43
Maria Mazur
49.9k1361125
49.9k1361125
asked Mar 10 at 7:25
Max PayneMax Payne
2,552825
2,552825
1
$begingroup$
What happened to the eight points on a line/eight lines through a point when you were counting?
$endgroup$
– Mark Bennet
Mar 10 at 7:35
$begingroup$
@MarkBennet yes that went unused!
$endgroup$
– Max Payne
Mar 10 at 7:45
add a comment |
1
$begingroup$
What happened to the eight points on a line/eight lines through a point when you were counting?
$endgroup$
– Mark Bennet
Mar 10 at 7:35
$begingroup$
@MarkBennet yes that went unused!
$endgroup$
– Max Payne
Mar 10 at 7:45
1
1
$begingroup$
What happened to the eight points on a line/eight lines through a point when you were counting?
$endgroup$
– Mark Bennet
Mar 10 at 7:35
$begingroup$
What happened to the eight points on a line/eight lines through a point when you were counting?
$endgroup$
– Mark Bennet
Mar 10 at 7:35
$begingroup$
@MarkBennet yes that went unused!
$endgroup$
– Max Payne
Mar 10 at 7:45
$begingroup$
@MarkBennet yes that went unused!
$endgroup$
– Max Payne
Mar 10 at 7:45
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
There are $binom P2$ pairs of points, and each pair gives a line. However, each line has $8$ points on it, meaning that the pairs of points are divided into groups of $binom 82=28$, where any pair in a group gives the same line. So the number of distinct lines is $L= frac{binom P2}{28}$.
Similarly, the number of distinct points is $P= frac{binom L2}{28}$.
This gives the following set of equations:
$$
cases{P=frac{L(L-1)}{56}\L=frac{P(P-1)}{56}}
$$
$endgroup$
$begingroup$
makes perfect sense, thanks a lot!
$endgroup$
– Max Payne
Mar 10 at 7:46
add a comment |
$begingroup$
Because of first two condititons we have $$Ltimes 8 = Ptimes 8implies L=P$$
Then from the last condition we have $${8choose 2}L = {Pchoose 2}implies P= 57$$
Such a configuration really exists, it is a projective plane of order $7$ it has $7^2+7+1$ points and the same number of lines.
$endgroup$
add a comment |
$begingroup$
- Any two distinct lines intersect in a unique point
- Any two distinct points lie on a unique line.
These two rules are, essentially, the incidence axioms for two-dimensional projective geometry. As such, we have a standard model for the system described here: the projective plane over the field $mathbb{F}_7=mathbb{Z}/7$ with seven elements.
We can use that model to count the points and lines - or we can just do things directly. Suppose we have a projective plane such that each line has $n+1$ points and each point is on $n+1$ lines.
Then, consider the system of all $n+1$ lines through a given point $P$. These lines each contain $n$ points other than $P$. Also, each point other than $P$ is on exactly one of these lines. That's $n(n+1)$ points other than $P$; add $P$ back in, and we have a total of exactly $n^2+n+1$ points.
By symmetry between points and lines, we also have exactly $n^2+n+1$ lines; we could run a similar count using some particular line $ell$, the $n+1$ points on it, and the $n$ other lines through each of those points.
For our example with $n+1=8$, that's $7^2+7+1=57$ each points and lines, for a product of $3249$.
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are $binom P2$ pairs of points, and each pair gives a line. However, each line has $8$ points on it, meaning that the pairs of points are divided into groups of $binom 82=28$, where any pair in a group gives the same line. So the number of distinct lines is $L= frac{binom P2}{28}$.
Similarly, the number of distinct points is $P= frac{binom L2}{28}$.
This gives the following set of equations:
$$
cases{P=frac{L(L-1)}{56}\L=frac{P(P-1)}{56}}
$$
$endgroup$
$begingroup$
makes perfect sense, thanks a lot!
$endgroup$
– Max Payne
Mar 10 at 7:46
add a comment |
$begingroup$
There are $binom P2$ pairs of points, and each pair gives a line. However, each line has $8$ points on it, meaning that the pairs of points are divided into groups of $binom 82=28$, where any pair in a group gives the same line. So the number of distinct lines is $L= frac{binom P2}{28}$.
Similarly, the number of distinct points is $P= frac{binom L2}{28}$.
This gives the following set of equations:
$$
cases{P=frac{L(L-1)}{56}\L=frac{P(P-1)}{56}}
$$
$endgroup$
$begingroup$
makes perfect sense, thanks a lot!
$endgroup$
– Max Payne
Mar 10 at 7:46
add a comment |
$begingroup$
There are $binom P2$ pairs of points, and each pair gives a line. However, each line has $8$ points on it, meaning that the pairs of points are divided into groups of $binom 82=28$, where any pair in a group gives the same line. So the number of distinct lines is $L= frac{binom P2}{28}$.
Similarly, the number of distinct points is $P= frac{binom L2}{28}$.
This gives the following set of equations:
$$
cases{P=frac{L(L-1)}{56}\L=frac{P(P-1)}{56}}
$$
$endgroup$
There are $binom P2$ pairs of points, and each pair gives a line. However, each line has $8$ points on it, meaning that the pairs of points are divided into groups of $binom 82=28$, where any pair in a group gives the same line. So the number of distinct lines is $L= frac{binom P2}{28}$.
Similarly, the number of distinct points is $P= frac{binom L2}{28}$.
This gives the following set of equations:
$$
cases{P=frac{L(L-1)}{56}\L=frac{P(P-1)}{56}}
$$
answered Mar 10 at 7:39
ArthurArthur
123k7122211
123k7122211
$begingroup$
makes perfect sense, thanks a lot!
$endgroup$
– Max Payne
Mar 10 at 7:46
add a comment |
$begingroup$
makes perfect sense, thanks a lot!
$endgroup$
– Max Payne
Mar 10 at 7:46
$begingroup$
makes perfect sense, thanks a lot!
$endgroup$
– Max Payne
Mar 10 at 7:46
$begingroup$
makes perfect sense, thanks a lot!
$endgroup$
– Max Payne
Mar 10 at 7:46
add a comment |
$begingroup$
Because of first two condititons we have $$Ltimes 8 = Ptimes 8implies L=P$$
Then from the last condition we have $${8choose 2}L = {Pchoose 2}implies P= 57$$
Such a configuration really exists, it is a projective plane of order $7$ it has $7^2+7+1$ points and the same number of lines.
$endgroup$
add a comment |
$begingroup$
Because of first two condititons we have $$Ltimes 8 = Ptimes 8implies L=P$$
Then from the last condition we have $${8choose 2}L = {Pchoose 2}implies P= 57$$
Such a configuration really exists, it is a projective plane of order $7$ it has $7^2+7+1$ points and the same number of lines.
$endgroup$
add a comment |
$begingroup$
Because of first two condititons we have $$Ltimes 8 = Ptimes 8implies L=P$$
Then from the last condition we have $${8choose 2}L = {Pchoose 2}implies P= 57$$
Such a configuration really exists, it is a projective plane of order $7$ it has $7^2+7+1$ points and the same number of lines.
$endgroup$
Because of first two condititons we have $$Ltimes 8 = Ptimes 8implies L=P$$
Then from the last condition we have $${8choose 2}L = {Pchoose 2}implies P= 57$$
Such a configuration really exists, it is a projective plane of order $7$ it has $7^2+7+1$ points and the same number of lines.
edited Mar 10 at 8:39
answered Mar 10 at 8:22
Maria MazurMaria Mazur
49.9k1361125
49.9k1361125
add a comment |
add a comment |
$begingroup$
- Any two distinct lines intersect in a unique point
- Any two distinct points lie on a unique line.
These two rules are, essentially, the incidence axioms for two-dimensional projective geometry. As such, we have a standard model for the system described here: the projective plane over the field $mathbb{F}_7=mathbb{Z}/7$ with seven elements.
We can use that model to count the points and lines - or we can just do things directly. Suppose we have a projective plane such that each line has $n+1$ points and each point is on $n+1$ lines.
Then, consider the system of all $n+1$ lines through a given point $P$. These lines each contain $n$ points other than $P$. Also, each point other than $P$ is on exactly one of these lines. That's $n(n+1)$ points other than $P$; add $P$ back in, and we have a total of exactly $n^2+n+1$ points.
By symmetry between points and lines, we also have exactly $n^2+n+1$ lines; we could run a similar count using some particular line $ell$, the $n+1$ points on it, and the $n$ other lines through each of those points.
For our example with $n+1=8$, that's $7^2+7+1=57$ each points and lines, for a product of $3249$.
$endgroup$
add a comment |
$begingroup$
- Any two distinct lines intersect in a unique point
- Any two distinct points lie on a unique line.
These two rules are, essentially, the incidence axioms for two-dimensional projective geometry. As such, we have a standard model for the system described here: the projective plane over the field $mathbb{F}_7=mathbb{Z}/7$ with seven elements.
We can use that model to count the points and lines - or we can just do things directly. Suppose we have a projective plane such that each line has $n+1$ points and each point is on $n+1$ lines.
Then, consider the system of all $n+1$ lines through a given point $P$. These lines each contain $n$ points other than $P$. Also, each point other than $P$ is on exactly one of these lines. That's $n(n+1)$ points other than $P$; add $P$ back in, and we have a total of exactly $n^2+n+1$ points.
By symmetry between points and lines, we also have exactly $n^2+n+1$ lines; we could run a similar count using some particular line $ell$, the $n+1$ points on it, and the $n$ other lines through each of those points.
For our example with $n+1=8$, that's $7^2+7+1=57$ each points and lines, for a product of $3249$.
$endgroup$
add a comment |
$begingroup$
- Any two distinct lines intersect in a unique point
- Any two distinct points lie on a unique line.
These two rules are, essentially, the incidence axioms for two-dimensional projective geometry. As such, we have a standard model for the system described here: the projective plane over the field $mathbb{F}_7=mathbb{Z}/7$ with seven elements.
We can use that model to count the points and lines - or we can just do things directly. Suppose we have a projective plane such that each line has $n+1$ points and each point is on $n+1$ lines.
Then, consider the system of all $n+1$ lines through a given point $P$. These lines each contain $n$ points other than $P$. Also, each point other than $P$ is on exactly one of these lines. That's $n(n+1)$ points other than $P$; add $P$ back in, and we have a total of exactly $n^2+n+1$ points.
By symmetry between points and lines, we also have exactly $n^2+n+1$ lines; we could run a similar count using some particular line $ell$, the $n+1$ points on it, and the $n$ other lines through each of those points.
For our example with $n+1=8$, that's $7^2+7+1=57$ each points and lines, for a product of $3249$.
$endgroup$
- Any two distinct lines intersect in a unique point
- Any two distinct points lie on a unique line.
These two rules are, essentially, the incidence axioms for two-dimensional projective geometry. As such, we have a standard model for the system described here: the projective plane over the field $mathbb{F}_7=mathbb{Z}/7$ with seven elements.
We can use that model to count the points and lines - or we can just do things directly. Suppose we have a projective plane such that each line has $n+1$ points and each point is on $n+1$ lines.
Then, consider the system of all $n+1$ lines through a given point $P$. These lines each contain $n$ points other than $P$. Also, each point other than $P$ is on exactly one of these lines. That's $n(n+1)$ points other than $P$; add $P$ back in, and we have a total of exactly $n^2+n+1$ points.
By symmetry between points and lines, we also have exactly $n^2+n+1$ lines; we could run a similar count using some particular line $ell$, the $n+1$ points on it, and the $n$ other lines through each of those points.
For our example with $n+1=8$, that's $7^2+7+1=57$ each points and lines, for a product of $3249$.
answered Mar 10 at 8:57
jmerryjmerry
17.1k11633
17.1k11633
add a comment |
add a comment |
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$begingroup$
What happened to the eight points on a line/eight lines through a point when you were counting?
$endgroup$
– Mark Bennet
Mar 10 at 7:35
$begingroup$
@MarkBennet yes that went unused!
$endgroup$
– Max Payne
Mar 10 at 7:45