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Thread: Imo 1989/3

  1. #1
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    Imo 1989/3

    https://books.google.com.tr/books?id...0chord&f=false


    The question is in the link. There is also the solution.
    What I don't understand is the solution.It says : "An edge is counted more than once when it is a common chord of at least 2 circles. Since two circles can have at most one common cord and there are n such circles, the number of common chords, counted with repetition, is at most C(n,2)".

    How does he get C(n,2)?
    2 circles intersect at 2 points and if we join the 2 points with a line the line is a chord.How does he count the chords?
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    Re: Imo 1989/3

    There are $n$ circles. You get a common chord at the intersection of any two circles. If every circle intersects (all $n$ circles have two points of intersection with all $n-1$ other circles), then the number of common chords is equal to the number of ways that you can choose 2 circles from $n$ circles. Think Venn diagrams.
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    Re: Imo 1989/3

    Quote Originally Posted by kastamonu View Post
    https://books.google.com.tr/books?id...0chord&f=false


    The question is in the link. There is also the solution.
    What I don't understand is the solution.It says : "An edge is counted more than once when it is a common chord of at least 2 circles. Since two circles can have at most one common cord and there are n such circles, the number of common chords, counted with repetition, is at most C(n,2)".

    How does he get C(n,2)?
    2 circles intersect at 2 points and if we join the 2 points with a line the line is a chord.How does he count the chords?
    You must tell us the exact page and the exact problem to which you refer.
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    Re: Imo 1989/3

    Quote Originally Posted by SlipEternal View Post
    There are $n$ circles. You get a common chord at the intersection of any two circles. If every circle intersects (all $n$ circles have two points of intersection with all $n-1$ other circles), then the number of common chords is equal to the number of ways that you can choose 2 circles from $n$ circles. Think Venn diagrams.
    How did you see the question? After translation I see only the first 23 pages.
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    Re: Imo 1989/3

    Quote Originally Posted by Plato View Post
    How did you see the question? After translation I see only the first 23 pages.
    I didn't. I just interpreted the original question and thought about questions I might have had when I first took graph theory.
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    Re: Imo 1989/3

    Example 1.4.5.
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    Re: Imo 1989/3

    When you click the link the first question you see.
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    Re: Imo 1989/3

    Quote Originally Posted by kastamonu View Post
    When you click the link the first question you see.
    No that is not true. I had to have the text translated and scroll to near the end.

    Imo 1989/3-untitled13.jpg
    I know that I missed some edges.
    But you can see that the red edges would be counted more than once where as the blue would be counted only once.
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    Re: Imo 1989/3

    Many Thanks.
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    Re: Imo 1989/3

    There are 3 common chords counted with repetition in the figure:C(3,2)-WE CHOOSE 2 FROM 3 circles to find common chords counted with repetition.No problem.

    But in the book says :"Clearly,the points of S on C(P) determine at least C(k,2) edges. As there are n points P in S,the total number of these edges, counted with repetiton, is at least nC(k,2)."
    What is k and n in the figure above?
    How do we find the total number counted with repetition?
    How do we understand that
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    Re: Imo 1989/3

    Quote Originally Posted by kastamonu View Post
    There are 3 common chords counted with repetition in the figure:C(3,2)-WE CHOOSE 2 FROM 3 circles to find common chords counted with repetition.No problem.

    But in the book says :"Clearly,the points of S on C(P) determine at least C(k,2) edges. As there are n points P in S,the total number of these edges, counted with repetiton, is at least nC(k,2)."
    What is k and n in the figure above?
    How do we find the total number counted with repetition?
    How do we understand that
    I found the answer. I understood better after drawing 4 circles. Many Thanks.
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