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How many points are needed to define a circumference?
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$begingroup$
This doubt comes from a combinatorics problem in a textbook, which states:
Consider two strictly parallel lines and seven dots, four of which are over one of them, and three over the other. Three dots are chosen at random: what is the probability that they define a circumference?
It turns out, the solution is $1-{{4choose3}+1over{7choose3}}$. That is, all combinations in which the three chosen dots aren't collinear, divided by all combinations of three dots.
My question is: why mustn't they be collinear?
combinatorics geometry
asked Mar 20 at 11:42
Daniel OscarDaniel Oscar
36211
$endgroup$
add a comment |
$begingroup$
This doubt comes from a combinatorics problem in a textbook, which states:
Consider two strictly parallel lines and seven dots, four of which are over one of them, and three over the other. Three dots are chosen at random: what is the probability that they define a circumference?
It turns out, the solution is $1-{{4choose3}+1over{7choose3}}$. That is, all combinations in which the three chosen dots aren't collinear, divided by all combinations of three dots.
My question is: why mustn't they be collinear?
combinatorics geometry
asked Mar 20 at 11:42
Daniel OscarDaniel Oscar
36211
$endgroup$
2
$begingroup$
If three dots are collinear, that means they are on the same straight line. That straight line can't have more than two points in common with any given circle.
$endgroup$
– Gerry Myerson
Mar 20 at 11:45
$begingroup$
isn't a line like a circle with infinite radius
$endgroup$
– Sik Feng Cheong
Mar 20 at 11:46
$begingroup$
@SikFengCheong There are contexts where you would consider lines to be infinite-radius circles (like when using the flat plane as a model for hyperbolic or projective geometry), but this is not one of them.
$endgroup$
– Arthur
Mar 20 at 11:49
$begingroup$
The numerator in the "solution," $1-{4choose3}+1$, is negative. I think you meant $$1-{{4choose3}+1over{7choose3}}$$
$endgroup$
– Barry Cipra
Mar 20 at 12:42
$begingroup$
Have you already seen collinear points on a circle ?
$endgroup$
– Yves Daoust
Mar 20 at 14:38
add a comment |
$begingroup$
This doubt comes from a combinatorics problem in a textbook, which states:
Consider two strictly parallel lines and seven dots, four of which are over one of them, and three over the other. Three dots are chosen at random: what is the probability that they define a circumference?
It turns out, the solution is $1-{{4choose3}+1over{7choose3}}$. That is, all combinations in which the three chosen dots aren't collinear, divided by all combinations of three dots.
My question is: why mustn't they be collinear?
combinatorics geometry
asked Mar 20 at 11:42
Daniel OscarDaniel Oscar
36211
$endgroup$
This doubt comes from a combinatorics problem in a textbook, which states:
Consider two strictly parallel lines and seven dots, four of which are over one of them, and three over the other. Three dots are chosen at random: what is the probability that they define a circumference?
It turns out, the solution is $1-{{4choose3}+1over{7choose3}}$. That is, all combinations in which the three chosen dots aren't collinear, divided by all combinations of three dots.
My question is: why mustn't they be collinear?
combinatorics geometry
combinatorics geometry
asked Mar 20 at 11:42
Daniel OscarDaniel Oscar
36211
asked Mar 20 at 11:42
Daniel OscarDaniel Oscar
36211
edited Mar 20 at 16:33
Daniel Oscar
asked Mar 20 at 11:42
Daniel OscarDaniel Oscar
36211
asked Mar 20 at 11:42
Daniel OscarDaniel Oscar
36211
asked Mar 20 at 11:42
Daniel OscarDaniel Oscar
36211
36211
2
$begingroup$
If three dots are collinear, that means they are on the same straight line. That straight line can't have more than two points in common with any given circle.
$endgroup$
– Gerry Myerson
Mar 20 at 11:45
$begingroup$
isn't a line like a circle with infinite radius
$endgroup$
– Sik Feng Cheong
Mar 20 at 11:46
$begingroup$
@SikFengCheong There are contexts where you would consider lines to be infinite-radius circles (like when using the flat plane as a model for hyperbolic or projective geometry), but this is not one of them.
$endgroup$
– Arthur
Mar 20 at 11:49
$begingroup$
The numerator in the "solution," $1-{4choose3}+1$, is negative. I think you meant $$1-{{4choose3}+1over{7choose3}}$$
$endgroup$
– Barry Cipra
Mar 20 at 12:42
$begingroup$
Have you already seen collinear points on a circle ?
$endgroup$
– Yves Daoust
Mar 20 at 14:38
add a comment |
2
$begingroup$
If three dots are collinear, that means they are on the same straight line. That straight line can't have more than two points in common with any given circle.
$endgroup$
– Gerry Myerson
Mar 20 at 11:45
$begingroup$
isn't a line like a circle with infinite radius
$endgroup$
– Sik Feng Cheong
Mar 20 at 11:46
$begingroup$
@SikFengCheong There are contexts where you would consider lines to be infinite-radius circles (like when using the flat plane as a model for hyperbolic or projective geometry), but this is not one of them.
$endgroup$
– Arthur
Mar 20 at 11:49
$begingroup$
The numerator in the "solution," $1-{4choose3}+1$, is negative. I think you meant $$1-{{4choose3}+1over{7choose3}}$$
$endgroup$
– Barry Cipra
Mar 20 at 12:42
$begingroup$
Have you already seen collinear points on a circle ?
$endgroup$
– Yves Daoust
Mar 20 at 14:38
2
2
$begingroup$
If three dots are collinear, that means they are on the same straight line. That straight line can't have more than two points in common with any given circle.
$endgroup$
– Gerry Myerson
Mar 20 at 11:45
$begingroup$
If three dots are collinear, that means they are on the same straight line. That straight line can't have more than two points in common with any given circle.
$endgroup$
– Gerry Myerson
Mar 20 at 11:45
$begingroup$
isn't a line like a circle with infinite radius
$endgroup$
– Sik Feng Cheong
Mar 20 at 11:46
$begingroup$
isn't a line like a circle with infinite radius
$endgroup$
– Sik Feng Cheong
Mar 20 at 11:46
$begingroup$
@SikFengCheong There are contexts where you would consider lines to be infinite-radius circles (like when using the flat plane as a model for hyperbolic or projective geometry), but this is not one of them.
$endgroup$
– Arthur
Mar 20 at 11:49
$begingroup$
@SikFengCheong There are contexts where you would consider lines to be infinite-radius circles (like when using the flat plane as a model for hyperbolic or projective geometry), but this is not one of them.
$endgroup$
– Arthur
Mar 20 at 11:49
$begingroup$
The numerator in the "solution," $1-{4choose3}+1$, is negative. I think you meant $$1-{{4choose3}+1over{7choose3}}$$
$endgroup$
– Barry Cipra
Mar 20 at 12:42
$begingroup$
The numerator in the "solution," $1-{4choose3}+1$, is negative. I think you meant $$1-{{4choose3}+1over{7choose3}}$$
$endgroup$
– Barry Cipra
Mar 20 at 12:42
$begingroup$
Have you already seen collinear points on a circle ?
$endgroup$
– Yves Daoust
Mar 20 at 14:38
$begingroup$
Have you already seen collinear points on a circle ?
$endgroup$
– Yves Daoust
Mar 20 at 14:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If three collinear points $ABC$ belong to a circle, then there exists a point $O$ such that $$AO=BO=CO.$$
Let's show this is not possible. From the above equality it follows that $OAB$ and $OAC$ are two isosceles triangles, hence
$$
angle OAB=angle OBA =angle OCB.
$$
But then, in triangle $OBC$ we have an external angle $angle OBA$ equal to internal angle $angle OCB$, and that is impossible by Euclid's exterior angle theorem, QED.
answered Mar 20 at 14:35
AretinoAretino
25.8k31545
$endgroup$
add a comment |
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$begingroup$
If three collinear points $ABC$ belong to a circle, then there exists a point $O$ such that $$AO=BO=CO.$$
Let's show this is not possible. From the above equality it follows that $OAB$ and $OAC$ are two isosceles triangles, hence
$$
angle OAB=angle OBA =angle OCB.
$$
But then, in triangle $OBC$ we have an external angle $angle OBA$ equal to internal angle $angle OCB$, and that is impossible by Euclid's exterior angle theorem, QED.
answered Mar 20 at 14:35
AretinoAretino
25.8k31545
$endgroup$
add a comment |
$begingroup$
If three collinear points $ABC$ belong to a circle, then there exists a point $O$ such that $$AO=BO=CO.$$
Let's show this is not possible. From the above equality it follows that $OAB$ and $OAC$ are two isosceles triangles, hence
$$
angle OAB=angle OBA =angle OCB.
$$
But then, in triangle $OBC$ we have an external angle $angle OBA$ equal to internal angle $angle OCB$, and that is impossible by Euclid's exterior angle theorem, QED.
answered Mar 20 at 14:35
AretinoAretino
25.8k31545
$endgroup$
add a comment |
$begingroup$
If three collinear points $ABC$ belong to a circle, then there exists a point $O$ such that $$AO=BO=CO.$$
Let's show this is not possible. From the above equality it follows that $OAB$ and $OAC$ are two isosceles triangles, hence
$$
angle OAB=angle OBA =angle OCB.
$$
But then, in triangle $OBC$ we have an external angle $angle OBA$ equal to internal angle $angle OCB$, and that is impossible by Euclid's exterior angle theorem, QED.
answered Mar 20 at 14:35
AretinoAretino
25.8k31545
$endgroup$
If three collinear points $ABC$ belong to a circle, then there exists a point $O$ such that $$AO=BO=CO.$$
Let's show this is not possible. From the above equality it follows that $OAB$ and $OAC$ are two isosceles triangles, hence
$$
angle OAB=angle OBA =angle OCB.
$$
But then, in triangle $OBC$ we have an external angle $angle OBA$ equal to internal angle $angle OCB$, and that is impossible by Euclid's exterior angle theorem, QED.
answered Mar 20 at 14:35
AretinoAretino
25.8k31545
answered Mar 20 at 14:35
AretinoAretino
25.8k31545
answered Mar 20 at 14:35
AretinoAretino
25.8k31545
answered Mar 20 at 14:35
AretinoAretino
25.8k31545
25.8k31545
add a comment |
add a comment |
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$begingroup$
If three dots are collinear, that means they are on the same straight line. That straight line can't have more than two points in common with any given circle.
$endgroup$
– Gerry Myerson
Mar 20 at 11:45
$begingroup$
isn't a line like a circle with infinite radius
$endgroup$
– Sik Feng Cheong
Mar 20 at 11:46
$begingroup$
@SikFengCheong There are contexts where you would consider lines to be infinite-radius circles (like when using the flat plane as a model for hyperbolic or projective geometry), but this is not one of them.
$endgroup$
– Arthur
Mar 20 at 11:49
$begingroup$
The numerator in the "solution," $1-{4choose3}+1$, is negative. I think you meant $$1-{{4choose3}+1over{7choose3}}$$
$endgroup$
– Barry Cipra
Mar 20 at 12:42
$begingroup$
Have you already seen collinear points on a circle ?
$endgroup$
– Yves Daoust
Mar 20 at 14:38