MHB Ten segments. One can form a triangle.

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The discussion focuses on proving that among ten segments with integer lengths between 1cm and 55cm, three can be selected to form a triangle. It suggests that the conditions can be relaxed to include lengths greater than or equal to 1cm. The argument involves sorting the lengths and assuming the claim is false, leading to a super Fibonacci sequence where the maximum length contradicts the upper limit of 55cm. The conclusion drawn is that it's impossible to select three segments that do not form a triangle under the given conditions. The reasoning highlights the importance of the triangle inequality in this context.
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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.
 
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caffeinemachine said:
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
 
CaptainBlack said:
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.
 
caffeinemachine said:
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:
CaptainBlack said:
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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