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# Ten segments. One can form a triangle.

This is a discussion on Ten segments. One can form a triangle. within the Challenge Questions and Puzzles forums, part of the MHB Lounge category; Each of ten segments has integer length and each one's length is greater than 1cm and less than 55cm. Prove ...

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. Originally Posted by caffeinemachine
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

3. Originally Posted by CaptainBlack
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.

4. Originally Posted by caffeinemachine
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

5. Originally Posted by CaptainBlack
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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