The discussion revolves around the problem of proving that among ten segments of integer lengths, each greater than 1cm and less than 55cm, it is possible to select three segments that can form a triangle. The conversation touches on the conditions of the lengths and the reasoning behind the proof.
Participants express some agreement on the proof strategy, but there is disagreement regarding the necessity of the strict conditions on segment lengths and whether degenerate triangles should be considered.
The discussion includes assumptions about the properties of triangle formation and the implications of segment length ordering, which may not be universally accepted or resolved.
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.
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
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.
Nice.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