1

I wonder if anybody can help me solve the following problem:

Three numbers are chosen from the first $n$ natural numbers. What is the probability that they can be sides of a triangle?

NOTE: The sides can be equal.

N. F. Taussig
  • 76,571
user512347
  • 11
  • 1
    Any thoughts? Have you, say, worked it out for small $n$? My first thought would be to do it for small $n$ and then work recursively. – lulu Apr 29 '18 at 20:38
  • 1
    Also you should specify whether or not you allow "degenerate" triangles, like the one with sides ${1,1,2}$. – lulu Apr 29 '18 at 20:40
  • You're sure about that? The "triangle" I wrote down is just a line segment. Up to you, of course. What about my first question? What have you tried so far? What is the answer for the first few $n$? – lulu Apr 29 '18 at 21:15
  • I don't think this is a duplicate. The other questions considered real lengths in an interval while this is nature number lengths. – fleablood Apr 30 '18 at 20:01
  • @fleablood It seems that you've nominated this post for reopening, but where's OP's own thoughts? – GNUSupporter 8964民主女神 地下教會 Apr 30 '18 at 20:36
  • I voted to reopen because it is not a duplicate IMO. If it is voted to close for lack of showing thought, I would not vote to reopen. Either way, I am not going to lose sleep if it is not reopened. And either way, I advise the OP to add his/her thoughts. – fleablood Apr 30 '18 at 20:46
  • @fleablood I appreciate your insistence on procedural righteousness, but unluckily, none of the reviewers voted to reopen. It seems that nobody would reopen and then close it for another reason. – GNUSupporter 8964民主女神 地下教會 May 01 '18 at 08:46

1 Answers1

0

If A=B, then C is a multiple of sqrt(2) for a right triangle. Ignore right triangles for now. If A+B > C, you have 100% chance of making a triangle. If A+B=C, the triangle is 2d and does not qualify. If A+B < C, you have three joined line segments that do not meet at two ends.

If you want Pythagorean triplets, you will need formulas for A, B, and C that use two natural numbers to specify each. (I have developed such formulas for a paper I am writing if you want them.) From there you will have to combine the chances of each of the three points as in regular probability calculations.

poetasis
  • 6,338
  • 1
    " Ignore right triangles for now. " Ignore them forever. The triangles don't need to be right. So being a right triangle has utterly no relevance. "If A+B > C you have 100% chance of making a triangle." What if $A = 4$, $B = 1$ and $C=2$ then $A + B > C$. Can you make a triangle? Also if you add the two larger numbers they will always be greater than the smallest. Are you saying all three numbers will be triangles? – fleablood Apr 30 '18 at 02:10
  • If two of the numbers add up to less than the third number, there is no trtiangle, merely three joined line segments with dangling ends. – poetasis Apr 30 '18 at 02:16
  • "If two of the numbers add up to less than the third number, there is no trtiangle, merely three joined line segments with dangling ends." Which is something you completely ignored in your post. You claimed if $A + B >C $ you will always have a triangle. That is obviously false as $A=4; B=1; C=2$ can not be a triangle. – fleablood Apr 30 '18 at 02:21
  • The assignment of A, B, and C is arbitrary and it is obviously useful to assign 4 to C in your example. 1+2<4 so no triangle. – poetasis Apr 30 '18 at 17:23
  • Yes, but your post does no such thing and makes absolutely no statement that such would need to be done. And you give utterly no hint as to how to figure the probability that given three sides whether the third is less than the sum of the other two. – fleablood Apr 30 '18 at 19:58
  • It just makes sense to assign the largest number(s) to C. I didn't feel the need to point it out because it seemed the obvious choice for a thinking person. – poetasis May 01 '18 at 18:46