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You are given three unique line segments randomly (all outcomes with equal probability) with integer lengths ranging from 1-100.

Question:

What is the probability, given the line segments you got, that you're able to create a triangle?


Without loss of generality, let our three lengths be $A,B,C$ with $A>B>C$. If it is a triangle we must have $B>A/2$

Here is an image illustrating my approach:

enter image description here


Black line is A. Blue line is B. Green line is C. B must cross the mid section. C must cross B but not be longer than B.


But here is where i'm stuck... Is there a nice (more towards easy rather than advanced) approach to solve this problem? If you have another approach, feel free to demonstrate it.

JMoravitz
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Prim3numbah
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  • What is the problem with С crossing the red mark? Segments of lengths $28, 27, 26$ do form a triangle despite $26 > 14$. – Abstraction Oct 26 '23 at 11:58
  • Ahh you're right. I'll make an edit. – Prim3numbah Oct 26 '23 at 12:00
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    More importantly, the question seems to imply that lengths are integers (you can't have any real length, otherwise the "all outcomes with equal probability" clause is impossible). In which case you have $161 700$ possible combinations and the answer must have the form of $n / 161 700$. Is that correct? – Abstraction Oct 26 '23 at 12:07
  • @Abstraction Yes, that's what I meant. Edited now. – Prim3numbah Oct 26 '23 at 12:14
  • If $~a < b,~$ can $~c~$ be $~\leq (b-a)~?~$ Further, can $~c~$ be $~\geq (b+a)~?$ – user2661923 Oct 26 '23 at 13:44
  • Your posting is missing several details. Please see this article on MathSE protocol. – user2661923 Oct 26 '23 at 13:56
  • You should not have needed an image to write something simple like $A>B$. Use MathJax and $\LaTeX$ to type it. – JMoravitz Oct 26 '23 at 14:03
  • All of the work shown so far can be rewritten into the more concise and complete statements: $A<B+C,~ B<A+C,~ C<A+B$... a specific application of the well known triangle inequality which should have been prior knowledge and not the focus of this problem assuming this were presented to you in a course on probability. – JMoravitz Oct 26 '23 at 14:06
  • Related Probability that a triangle can be formed... except that the analysis there allows for lengths to repeat. Their approaches however can be modified if careful to talk about the cases where lengths may not repeat. The end result is that the probability will be near $\frac{1}{2}$. – JMoravitz Oct 26 '23 at 14:18
  • @JMoravitz If triangles of area $0$ are prohibited the answer will be less than $\frac12$ and you get $\frac{79625}{161700} = \frac{65}{132} \approx 0.4924$. The probability of "A > B+C" is also $\frac{65}{132}$ and of "A = B+C" is the remaining $\frac{2}{132}$ – Henry Oct 26 '23 at 15:00
  • I apparently read the entry for https://oeis.org/A002623 incorrectly, the oeis I found was for allowing same-length. I've deleted the incorrect comment and replacing it with this one. I'm failing to find one more specific to the situation at hand, but the oeis entry contained the following link https://babel.hathitrust.org/cgi/pt?id=nyp.33433062732734&view=1up&seq=308 which does cover the case of no repeats, confirming your answer of $\dfrac{65}{132}$. – JMoravitz Oct 26 '23 at 15:12

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