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A stick is broken in three parts of any length , if we join the pieces then what's the probability of getting a acute angled triangle?

aniruddha sharma
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    This problem has been answered several times on Math.SE. Since you ask for a probability, it is necessary to work with a probability distribution of the lengths into which the stick is broken. There is more than one reasonable interpretation of how to proceed in breaking the stick. – hardmath Dec 27 '16 at 18:25
  • To characterize acuteness, you can use the Pythagorean Theorem. That is, If $a \le b \le c$ are the stick lengths in increasing order, then you need both $a+b > c$ for the sticks to form a triangle, and $a^2+b^2 > c^2$ for the triangle to be acute. – angryavian Dec 27 '16 at 18:25
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    Possible duplicate of Calculating probability for forming a triangle – amd Dec 27 '16 at 18:27
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    I don't think this is a duplicate. It sure is related to the linked question. – Shraddheya Shendre Dec 27 '16 at 18:30
  • I think this question is the same as http://math.stackexchange.com/questions/1962301/acute-triangle-from-a-stick, but that question was not answered. – David K Dec 27 '16 at 18:34
  • The interested Reader may want to look at the write-up here which completes the discussion of chance of forming a triangle with details regarding the acute and obtuse outcomes. – hardmath Dec 27 '16 at 20:59
  • See [this post] for a relevant simulation. – BruceET Jan 20 '17 at 03:52

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WLOG the stick has length 1, here the experience can be modeled e.g. by choosing two cutting points $x_1 \le x_2$ uniformly at random in $[0, 1]$. Then we have three sticks with length $l_1=x_1$, $l_2=x_2-x_1$ and $l_3=1-x_2$.

Then you can notice geometrically that the triangle formed with them is acute iff. $\sqrt{l_2^2-l_1^2} \le l_3 \le \sqrt{l_1^2+l_2^2}$ (the two extreme cases are obtained for right triangles).

md5
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  • The condition in the last paragraph assumes that $l_2>l_1$. But I think putting an absolute value around the difference would fix that. – David K Jan 18 '17 at 19:06