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  1. MHB Journeyman
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    #1
    Each of ten segments has integer length and each one's length is greater than 1cm and less than 55cm. Prove that you can select three sides of a triangle among the segments.

  2. MHB Journeyman
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    #2
    Quote Originally Posted by caffeinemachine View Post
    Each of ten segments has integer length and each one's length is greater than 1cm and less than 55cm. Prove that you can select three sides of a triangle among the segments.
    Is the wording right? It looks like you can relax the conditions to the lengths being greater than or equal 1cm and less than 55cm.

    CB

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    #3
    Quote Originally Posted by CaptainBlack View Post
    Is the wording right? It looks like you can relax the conditions to the lengths being greater than or equal 1cm and less than 55cm.

    CB
    I think you are right. I've taken this from a book and in the book they have the conditions I have posted.

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    #4
    Quote Originally Posted by caffeinemachine View Post
    I think you are right. I've taken this from a book and in the book they have the conditions I have posted.
    Sorting the lengths into non-decreasing order, and assume that the claim is false, then the ordered sequence of lengths is a super Fibonacci sequence, meaning that:

    \[ l_{k} \ge l_{k-1}+l_{k-2}, k=3, .., 10 \]

    Thus the set of lengths which has the smallest maximum value such that three may not be selected to form a triangle is the Fibonacci sequence, but then \(l_{10}\ge 55\), a contradiction.

    (note I regard the degenerate triangle with two zero angles as a non-triangle, the argument is easily modified if you want this to count as a triangle)

    CB
    Last edited by CaptainBlack; Mar-23 at 15:57. Reason: fix LaTeX

  5. MHB Journeyman
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    #5
    Quote Originally Posted by CaptainBlack View Post
    Sorting the lengths into non-decreasing order, and assume that the claim is false, then the ordered sequence of lengths is a super Fibonacci sequence, meaning that:

    \[ l_{k} \ge l_{k-1}+l_{k-2}, k=3, .., 10 \]

    Thus the set of lengths which has the smallest maximum value such that three may not be selected to form a triangle is the Fibonacci sequence, but then \(l_10\ge 55\), a contradiction.

    (note I regard the degenerate triangle with two zero angles as a non-triangle, the argument is easily modified if you want this to count as a triangle)

    CB
    Nice.

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