What is the probability for a wood stick of real number length breaking in three piece that can forming precisely a triangle? What is the probability for a wood stick of any real number length breaking in three piece that can forming precisely a triangle without any extra segment extend  from the side of the triangle?
My friend told me it can be solve in a calculus way and a algebra-precalculus way, what is the calculus way?
I found a relevant of this problem, but it is not the same to me : Broken stick probability problem
Last but not least: happy puzzle solving!
 A: Picking two numbers independently that are uniformly distributed in $[0,1]$ is the same as picking a uniformly distributed point in the unit square.  Call the smaller one $x$ and the bigger one $y$.  The graph of $0<x<y<1$ is a triangular region and we are choosing a point within it.  The lengths of the three sides will then be $x$, $y-x$, and $1-y$.  We need three instances of the triangle inequality:
\begin{align}
x + (y-x) & > 1-y, \\
x + (1-y) & > y-x, \\
(y-x)+(1-y) & > x.
\end{align}
Draw the graph of the set of inequalities above within the aforementioned triangular region and you'll see the answer.
A: In the comments you asked me by name for a calculus way. Personally I would not do it this way.
Let us assume that the stick is broken at two random points $X, Y$ where the two points are independently uniformly distributed on $[0,l]$, so pieces with lengths $\min(X,Y), \max(X,Y)-\min(X,Y), l-\max(X,Y)$ and a triangle cannot be constructed if any of the lengths is $l/2$ or more (almost certainly mutually exclusive events).
Then $\displaystyle \Pr(\min(X,Y) \ge l/2)=\int_{x=l/2}^{l} \int_{y=l/2}^{l} \frac1{l^2} \,dy \,dx = \frac14$
$\displaystyle \Pr\left(\max(X,Y)-\min(X,Y) \ge l/2\right)=\int_{x=l/2}^{l}  \int_{y=0}^{x-l/2} \frac1{l^2}  \,dy \,dx + \int_{x=0}^{l/2}  \int_{y=x+l/2}^{l} \frac1{l^2}  \,dy \,dx = \frac14$
$\displaystyle \Pr(l-\max(X,Y) \ge l/2)=\int_{x=0}^{l/2}  \int_{y=0}^{l/2} \frac1{l^2} \,dy \,dx = \frac14$.
The probability you want is $1-\dfrac14-\dfrac14-\dfrac14=\dfrac14$. 
