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
The ten segments represent the sides or edges of a shape or figure.
A minimum of three segments are needed to form a triangle.
No, not all ten segments can be used to form a triangle. Only three segments can be used at a time to form a triangle.
The number of segments used will determine the type of triangle formed. For example, using three segments will form a triangle, while using four segments will form a quadrilateral.
Yes, there are other shapes that can be formed with ten segments, such as a pentagon, hexagon, or decagon.