Mixing
Proving 2 chords are the same length in a circle divided into $n$ equal arcs
$begingroup$
Let us have a circle that is divided in to $n$ equal arcs by $n$ points on the circumference. There are $dfrac{n}{2}$ chords joining pairs of points. For what values of $n$ would it be necessary for there to be at least 2 chords with the same length, and how would you prove this?
geometry circle
asked Jan 6 at 2:08
user491842user491842
62
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add a comment |
$begingroup$
Let us have a circle that is divided in to $n$ equal arcs by $n$ points on the circumference. There are $dfrac{n}{2}$ chords joining pairs of points. For what values of $n$ would it be necessary for there to be at least 2 chords with the same length, and how would you prove this?
geometry circle
asked Jan 6 at 2:08
user491842user491842
62
$endgroup$
1
$begingroup$
$n$ is an even number then?
$endgroup$
– Christopher Marley
Jan 6 at 3:14
$begingroup$
For any n greater or equal to 3?
$endgroup$
– Moti
Jan 6 at 7:16
add a comment |
$begingroup$
Let us have a circle that is divided in to $n$ equal arcs by $n$ points on the circumference. There are $dfrac{n}{2}$ chords joining pairs of points. For what values of $n$ would it be necessary for there to be at least 2 chords with the same length, and how would you prove this?
geometry circle
asked Jan 6 at 2:08
user491842user491842
62
$endgroup$
Let us have a circle that is divided in to $n$ equal arcs by $n$ points on the circumference. There are $dfrac{n}{2}$ chords joining pairs of points. For what values of $n$ would it be necessary for there to be at least 2 chords with the same length, and how would you prove this?
geometry circle
geometry circle
asked Jan 6 at 2:08
user491842user491842
62
asked Jan 6 at 2:08
user491842user491842
62
asked Jan 6 at 2:08
user491842user491842
62
asked Jan 6 at 2:08
user491842user491842
62
asked Jan 6 at 2:08
user491842user491842
62
62
1
$begingroup$
$n$ is an even number then?
$endgroup$
– Christopher Marley
Jan 6 at 3:14
$begingroup$
For any n greater or equal to 3?
$endgroup$
– Moti
Jan 6 at 7:16
add a comment |
1
$begingroup$
$n$ is an even number then?
$endgroup$
– Christopher Marley
Jan 6 at 3:14
$begingroup$
For any n greater or equal to 3?
$endgroup$
– Moti
Jan 6 at 7:16
1
1
$begingroup$
$n$ is an even number then?
$endgroup$
– Christopher Marley
Jan 6 at 3:14
$begingroup$
$n$ is an even number then?
$endgroup$
– Christopher Marley
Jan 6 at 3:14
$begingroup$
For any n greater or equal to 3?
$endgroup$
– Moti
Jan 6 at 7:16
$begingroup$
For any n greater or equal to 3?
$endgroup$
– Moti
Jan 6 at 7:16
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The number of chords rather is ${nchoose 2}$ instead.
Your question then comes down to $${nchoose 2}=frac{n!}{(n-2)! 2!}=frac{n (n-1)}2>1$$ This solves to $n>2$, which then provides the number of points where at least 2 chords are available and therefore comparable.
But then, even so from $n>3$ on there are chords $(i, j)$ of different sizes, you always have a sequence of $n$ chords joining neighbouring points $(i, (i+1)mod n)$. And as those points have been equally spaced by assumption, the according $n>2$ chords will have the same length too; i.e. for any such $n$ individually.
--- rk
answered Jan 6 at 10:48
Dr. Richard KlitzingDr. Richard Klitzing
1,57016
$endgroup$
$begingroup$
There are $binom{n}{2}$ possible chords. The question states that (only) $frac{n}{2}$ of these chords are to be considered. For $n=6$ we can choose 3 chords with 3 distinct lengths, so it is not necessary for 2 chords to have the same length in this case.
$endgroup$
– Daniel Mathias
Jan 6 at 12:47
add a comment |
$begingroup$
For what values of $n$ would it be necessary for there to be at least 2 chords with the same length?
This is necessary for all odd $n$.
For any even $n$ there are exactly $frac{n}{2}$ distinct chord lengths, as shown in the image for $n=8$. For odd $n$ there are $lfloorfrac{n}{2}rfloor$ distinct chord lengths. If the selection $frac{n}{2}$ chords for odd $n$ is rounded up, then there must be at least two chords with the same length.
answered Jan 6 at 13:17
Daniel MathiasDaniel Mathias
73617
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The number of chords rather is ${nchoose 2}$ instead.
Your question then comes down to $${nchoose 2}=frac{n!}{(n-2)! 2!}=frac{n (n-1)}2>1$$ This solves to $n>2$, which then provides the number of points where at least 2 chords are available and therefore comparable.
But then, even so from $n>3$ on there are chords $(i, j)$ of different sizes, you always have a sequence of $n$ chords joining neighbouring points $(i, (i+1)mod n)$. And as those points have been equally spaced by assumption, the according $n>2$ chords will have the same length too; i.e. for any such $n$ individually.
--- rk
answered Jan 6 at 10:48
Dr. Richard KlitzingDr. Richard Klitzing
1,57016
$endgroup$
$begingroup$
There are $binom{n}{2}$ possible chords. The question states that (only) $frac{n}{2}$ of these chords are to be considered. For $n=6$ we can choose 3 chords with 3 distinct lengths, so it is not necessary for 2 chords to have the same length in this case.
$endgroup$
– Daniel Mathias
Jan 6 at 12:47
add a comment |
$begingroup$
The number of chords rather is ${nchoose 2}$ instead.
Your question then comes down to $${nchoose 2}=frac{n!}{(n-2)! 2!}=frac{n (n-1)}2>1$$ This solves to $n>2$, which then provides the number of points where at least 2 chords are available and therefore comparable.
But then, even so from $n>3$ on there are chords $(i, j)$ of different sizes, you always have a sequence of $n$ chords joining neighbouring points $(i, (i+1)mod n)$. And as those points have been equally spaced by assumption, the according $n>2$ chords will have the same length too; i.e. for any such $n$ individually.
--- rk
answered Jan 6 at 10:48
Dr. Richard KlitzingDr. Richard Klitzing
1,57016
$endgroup$
$begingroup$
There are $binom{n}{2}$ possible chords. The question states that (only) $frac{n}{2}$ of these chords are to be considered. For $n=6$ we can choose 3 chords with 3 distinct lengths, so it is not necessary for 2 chords to have the same length in this case.
$endgroup$
– Daniel Mathias
Jan 6 at 12:47
add a comment |
$begingroup$
The number of chords rather is ${nchoose 2}$ instead.
Your question then comes down to $${nchoose 2}=frac{n!}{(n-2)! 2!}=frac{n (n-1)}2>1$$ This solves to $n>2$, which then provides the number of points where at least 2 chords are available and therefore comparable.
But then, even so from $n>3$ on there are chords $(i, j)$ of different sizes, you always have a sequence of $n$ chords joining neighbouring points $(i, (i+1)mod n)$. And as those points have been equally spaced by assumption, the according $n>2$ chords will have the same length too; i.e. for any such $n$ individually.
--- rk
answered Jan 6 at 10:48
Dr. Richard KlitzingDr. Richard Klitzing
1,57016
$endgroup$
The number of chords rather is ${nchoose 2}$ instead.
Your question then comes down to $${nchoose 2}=frac{n!}{(n-2)! 2!}=frac{n (n-1)}2>1$$ This solves to $n>2$, which then provides the number of points where at least 2 chords are available and therefore comparable.
But then, even so from $n>3$ on there are chords $(i, j)$ of different sizes, you always have a sequence of $n$ chords joining neighbouring points $(i, (i+1)mod n)$. And as those points have been equally spaced by assumption, the according $n>2$ chords will have the same length too; i.e. for any such $n$ individually.
--- rk
answered Jan 6 at 10:48
Dr. Richard KlitzingDr. Richard Klitzing
1,57016
answered Jan 6 at 10:48
Dr. Richard KlitzingDr. Richard Klitzing
1,57016
answered Jan 6 at 10:48
Dr. Richard KlitzingDr. Richard Klitzing
1,57016
answered Jan 6 at 10:48
Dr. Richard KlitzingDr. Richard Klitzing
1,57016
1,57016
$begingroup$
There are $binom{n}{2}$ possible chords. The question states that (only) $frac{n}{2}$ of these chords are to be considered. For $n=6$ we can choose 3 chords with 3 distinct lengths, so it is not necessary for 2 chords to have the same length in this case.
$endgroup$
– Daniel Mathias
Jan 6 at 12:47
add a comment |
$begingroup$
There are $binom{n}{2}$ possible chords. The question states that (only) $frac{n}{2}$ of these chords are to be considered. For $n=6$ we can choose 3 chords with 3 distinct lengths, so it is not necessary for 2 chords to have the same length in this case.
$endgroup$
– Daniel Mathias
Jan 6 at 12:47
$begingroup$
There are $binom{n}{2}$ possible chords. The question states that (only) $frac{n}{2}$ of these chords are to be considered. For $n=6$ we can choose 3 chords with 3 distinct lengths, so it is not necessary for 2 chords to have the same length in this case.
$endgroup$
– Daniel Mathias
Jan 6 at 12:47
$begingroup$
There are $binom{n}{2}$ possible chords. The question states that (only) $frac{n}{2}$ of these chords are to be considered. For $n=6$ we can choose 3 chords with 3 distinct lengths, so it is not necessary for 2 chords to have the same length in this case.
$endgroup$
– Daniel Mathias
Jan 6 at 12:47
add a comment |
$begingroup$
For what values of $n$ would it be necessary for there to be at least 2 chords with the same length?
This is necessary for all odd $n$.
For any even $n$ there are exactly $frac{n}{2}$ distinct chord lengths, as shown in the image for $n=8$. For odd $n$ there are $lfloorfrac{n}{2}rfloor$ distinct chord lengths. If the selection $frac{n}{2}$ chords for odd $n$ is rounded up, then there must be at least two chords with the same length.
answered Jan 6 at 13:17
Daniel MathiasDaniel Mathias
73617
$endgroup$
add a comment |
$begingroup$
For what values of $n$ would it be necessary for there to be at least 2 chords with the same length?
This is necessary for all odd $n$.
For any even $n$ there are exactly $frac{n}{2}$ distinct chord lengths, as shown in the image for $n=8$. For odd $n$ there are $lfloorfrac{n}{2}rfloor$ distinct chord lengths. If the selection $frac{n}{2}$ chords for odd $n$ is rounded up, then there must be at least two chords with the same length.
answered Jan 6 at 13:17
Daniel MathiasDaniel Mathias
73617
$endgroup$
add a comment |
$begingroup$
For what values of $n$ would it be necessary for there to be at least 2 chords with the same length?
This is necessary for all odd $n$.
For any even $n$ there are exactly $frac{n}{2}$ distinct chord lengths, as shown in the image for $n=8$. For odd $n$ there are $lfloorfrac{n}{2}rfloor$ distinct chord lengths. If the selection $frac{n}{2}$ chords for odd $n$ is rounded up, then there must be at least two chords with the same length.
answered Jan 6 at 13:17
Daniel MathiasDaniel Mathias
73617
$endgroup$
For what values of $n$ would it be necessary for there to be at least 2 chords with the same length?
This is necessary for all odd $n$.
For any even $n$ there are exactly $frac{n}{2}$ distinct chord lengths, as shown in the image for $n=8$. For odd $n$ there are $lfloorfrac{n}{2}rfloor$ distinct chord lengths. If the selection $frac{n}{2}$ chords for odd $n$ is rounded up, then there must be at least two chords with the same length.
answered Jan 6 at 13:17
Daniel MathiasDaniel Mathias
73617
answered Jan 6 at 13:17
Daniel MathiasDaniel Mathias
73617
answered Jan 6 at 13:17
Daniel MathiasDaniel Mathias
73617
answered Jan 6 at 13:17
Daniel MathiasDaniel Mathias
73617
73617
add a comment |
add a comment |
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1
$begingroup$
$n$ is an even number then?
$endgroup$
– Christopher Marley
Jan 6 at 3:14
$begingroup$
For any n greater or equal to 3?
$endgroup$
– Moti
Jan 6 at 7:16