Triangle forming probability for area Say you have a stick which breaks randomly into three pieces (we can choose the points randomly). What is the probability that the area is greater than or equal to $0.4$? 
I can see it has something to do with Heron's formula but I just can't put t together. 
 A: We will assume the length of the stick is $1$ and 
we break the stick by picking two points $u, v$ uniformly and independently from $(0,1)$.
Let $a = \min(u,v)$, $b = 1 - \max(u,v)$ and $c = |u-v| = 1 - a - b$.  $a, b, c$ forms the sides of a triangle when and only when
$$a \le b+c,\;b \le c+a,\;c \le a+b\;\iff\;a \le \frac12,\;b \le \frac12,\;a+b \ge \frac12$$
Change variable to $x = 1-2a, y = 1-2b$, the lengths $a,b,c$ forms a triangle when $(x,y)$ falls inside another triangle 
$$\Delta = \big\{ (x,y) : x \ge 0, y \ge 0, x+y \le 1 \big\}$$
with area $\frac12$. It is clear conditional to $a,b,c$ forming a triangle, the probability "density" of picking a particular $(x,y)$ is $2dxdy$.
Let $A$ be the area of a triangle with sides $a,b,c$ and $A_0 = 0.04$.
By Heron's formula, we have
$$\begin{align}
          A &= \sqrt{s(s-a)(s-b)(s-c)}\\
\iff  16A^2 & = 1(1-2a)(1-2b)(1-2c) = xy(1-x-y)\\
\iff  64A^2 & = ((x+y)^2 - (x-y)^2)(1-x-y)
\end{align}
$$
Change variable once again to $p = (x+y), q = (x-y)$, the triangle $\Delta$ becomes
$$\Delta' = \big\{ (p,q) : 0 \le |q| \le p \le 1 \big\}$$
Furthermore,
$$A \ge A_0 
\quad\iff\quad
(p^2 - q^2)(1-p) \ge (8A_0)^2
\quad\iff\quad q^2 \le p^2 - \frac{(8A_0)^2}{1-p}
$$
Let $\displaystyle\;f(p) = \sqrt{p^2 - \frac{(8A_0)^2}{1-p}}\;$ 
and $\lambda_1, \lambda_2$ be the two roots of $f(p)$ in $(0,1)$. The condition above is equivalent to $\lambda_1 \le p \le \lambda_2$ and $|q| \le f(p)$. 
Since $dpdq = 2dxdy$, the probability we seek equals to
$$\mathbb{P}[A\ge A_0] = \int_{\lambda_1}^{\lambda_2} \int_{-f(p)}^{f(p)} dqdp = 2\int_{\lambda_1}^{\lambda_2} f(p) dp$$
Change variable to $t = \sqrt{1-p}$, this becomes
$$\mathbb{P}[A\ge A_0] = 4\int_{\mu_1}^{\mu_2} \sqrt{t^2(1-t^2)^2 - (8A_0)^2} dt\tag{*1}$$
where $\mu_1, \mu_2$ are now the roots of the polynomial $\;t(1-t^2) - 8A_0\;$ in $(0,1)$.
For the problem at hand where $A_0 = 0.04$, we have
$$\mu_1 \approx 0.3711104191979701,\; \mu_2 \approx 0.7613913530813122$$
and $(*1)$ evaluates numerically to
$$\mathbb{P}[A\ge A_0] \approx 0.2586458039398669$$
This is compatible with what another user @Henry obtained through simulation.
