An iron rod of $1$ foot long is cut into three segments with random length. (1) What is the probability that the three segments form a triangle? (2) What is the probability that the three segments form an acute triangle? Click here for the solution.
Solution (1): Let $x$, $y$, $z$ be the lengths of three segments. Therefore $$x+y+z=1 \tag{1}$$ In order to form a triangle with these three segments, the following conditions must be met: $$x+y>z,y+z>x,z+x>y \tag{2}$$
Combining $(1)$ and $(2)$, we have $$x+y>\dfrac{1}{2},x+y<1,x<\dfrac{1}{2},y<\dfrac{1}{2}\tag{3}$$
In the above diagram, $\triangle{ODE}$ represents all possible $x$ and $y$ values. And $\triangle{ABC}$ represents all $x$ and $y$ values satisfying $(3)$. Therefore, the probability to the original question is $\boxed{\dfrac{1}{4}}$.
Solution (2): Following Solution (1), the following additional conditions must be satisfied for the triangle to be acute: $$x^2+y^2>z^2, x^2+z^2>y^2,y^2+z^2>x^2\tag{4}$$
Replacing $z=1-(x+y)$ in $(4)$, we have: $$(1-x)(1-y)>\dfrac{1}{2}, (x+y)(1-x)<\dfrac{1}{2}, (x+y)(1-y)<\dfrac{1}{2}\tag{5}$$
In the above diagram, the area bounded by three hyperbola curves contains all $x$ and $y$ values that form an acute triangle.
The area bounded by line $AB$ and hyperbola $AB$ is $$Area(AB)=\int_{0}^{\frac{1}{2}}\left\{\left(1-\dfrac{1}{2(1-x)} \right)-\left(\dfrac{1}{2}-x\right)\right\} \,dx=\int_{0}^{\frac{1}{2}}\left(\frac{1}{2}+x-\dfrac{1}{2(1-x)}\right)\,dx$$ $$=\dfrac{1}{2}\left(x+x^2+\log{(1-x)}\right)\Big\rvert_{0}^{\frac{1}{2}}=\dfrac{3}{8}-\dfrac{1}{2}\log{2}$$
The area bounded by line $AC$ and hyperbola $AC$, which is same as the area bounded by line $BC$ and hyperbola $BC$, is $$Area(AC)=Area(BC)=\int_{0}^{\frac{1}{2}}\left\{\dfrac{1}{2}-\left(\dfrac{1}{2(1-x)}-x\right)\right\}\,dx=$$ $$=\int_{0}^{\frac{1}{2}}\left(\frac{1}{2}+x-\dfrac{1}{2(1-x)}\right)\,dx=\dfrac{3}{8}-\dfrac{1}{2}\log{2}=Area(AB)$$
The area bounded by three hyperbola curves is $$\dfrac{1}{8}-3\left(\dfrac{3}{8}-\dfrac{1}{2}\log{2}\right)=\dfrac{3}{2}\log{2}-1$$
Therefore, the probability of three segments that form an acute triangle is $$2\cdot\left(\dfrac{3}{2}\log{2}-1\right)=\boxed{3\log{2}-2\approx 0.08}$$