Geometry Probability – 1

Solution: This problem is equivalent to the following:

Two cuts are randomly made on a line of length $180$, dividing the line into three segments. The lengths of the three segments are taken as the value in degrees of $\angle{A}$, $\angle{B}$, and $\angle{C}$ of $\triangle{ABC}$. What is the probability that $\triangle{ABC}$ is acute?

$$\angle{A}+\angle{B}+\angle{C}=180, \angle{A}<90, \angle{B}<90, \angle{C}<90$$

Let $x=\dfrac{\angle{A}}{180}$, $y=\dfrac{\angle{B}}{180}$, $z=\dfrac{\angle{C}}{180}$, we have:

$$x+y+z=1, x<\dfrac{1}{2}, y<\dfrac{1}{2}, z<\dfrac{1}{2}$$

Therefore $$x+y>\dfrac{1}{2}, x+y<1, x<\dfrac{1}{2}, y<\dfrac{1}{2} \tag{1}$$

In the above diagram, $\triangle{ODE}$ represents all possible $x$ and $y$ values. And $\triangle{ABC}$ represents all $x$ and $y$ values satisfying $(1)$. Therefore, the probability to the original question is $\dfrac{1}{4}$.