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- #1

- Mar 10, 2012

- 834

- Thread starter caffeinemachine
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- Thread starter
- #1

- Mar 10, 2012

- 834

- Jan 26, 2012

- 890

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

- Mar 10, 2012

- 834

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

- Jan 26, 2012

- 890

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

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

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:

- Thread starter
- #5

- Mar 10, 2012

- 834

Nice.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