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Seven line segments, with lengths no greater than 10 inches, and no shorter than 1 inch, are given. Show that one can choose three of them to represent the sides of a triangle.

Pedro
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Aditya Kumar
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    It is my opinion that closing such old Questions merely for lack of context is not in the best interest of the Math.SE site. This falls into the category of reasonable mathematical problems stated without evidence of research or motivation. No one has bothered to comment on the Question to suggest that the original poster should provide context, I'm voting to leave open. – hardmath Jan 29 '22 at 13:46

4 Answers4

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Let the lengths of seven of the line segments be $l_1, l_2, l_3, \dotsc, l_7$ in (weakly) ascending order. Now suppose that no triangle can be formed from these lengths. Furthermore, let $l_2 ≥l_1 ≥1$ (the minimum length,suggested by meelo).

Then $l_3 > 2$, $l_4 > 3$, $l_5 > 5$, $l_6 > 8$, $l_7 > 13$ (because if $l_i + l_{i+1} \geq l_{i+2}$, then we can form a triangle). But we know that $l_i \leq 10$, so thus a triangle has to be formed.

avz2611
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    It would be better to phrase this as letting $l_2\geq l_1\geq 1$ - it stills follows that $l_3> l_2+l_1 \geq 2$ and so on, and you don't need to make any unwarranted assumption about $l_1$ and $l_2$ being equal to anything. – Milo Brandt Feb 24 '15 at 23:31
  • @Meelo yes you are right ,edited . thanks for your feedback :) – avz2611 Feb 25 '15 at 04:13
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Hint: Suppose $a < b < c$ are three lengths that do not form a triangle. Then $c > a+b > 2a$.

MJD
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Hint given by @MJD was really helpful

If you number the line segments from 1 to 7, and suppose no three of them can be used to make a triangle.

Start from seg. 1 and seg.2 . As you cannot use seg. 3 to make a triangle from seg.1 and seg. 2 so length of seg. 3 must be strictly more than 2 inches.

Continue same way you shall get (strict) lower bounds on lengths for

seg. 1 ------ 1 inch

seg 2 --------1 inch

seg 3---------2 inch

seg 4---------3 inch

seg 5---------5 inch

seg 6---------8 inch

Seventh segment cannot be greater than 10 inches, then it can be used to make a triangle with two of the previous 6 segments. (seg. 5 and seg. 6 for example)

Now question how to use Pigeonhole principle in here...? I am sorry I did not answer your question completely

N. F. Taussig
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Harish
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Minimum sequence (degenerate triangles):

1, 1, 2, 3, 5, 8, 13

If any number were reduced a triangle would form. Therefore to make no triangles with 7, the upper bound must be 13.

I don't see how the pigeonhole principle even applies.

Joshua
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