There is a rope that can be used to form a square with a side length of 1. Three points are randomly selected on the rope, and the longest segment is chosen. What is the probability that the top-left vertex of the square is in that segment? (We may assume that we start at the top left and work clockwise).
My attempt solution:
Assume the three segment points are A B C. The four vertices are EFGH. The length of the longest segment is $x$.
W.L.O.G, let line AB be the longest segment. Therefore, $|AB|\geq 1/3$.
Our goal is to solve the following objective: $\int^1_{x=\frac{1}{3}} p(\text{top-left vertex is in longest segment} |x)p(x)dx$
(1) When $1/2>|AB|\geq 1/3$, there are must be one vertex lies in |AB|. The probability is $\int^{\frac{1}{2}}_{x=\frac{1}{3}} p(\text{top-left vertex is in longest segment} |x)p(x)dx=\frac{1}{4}\times\frac{(\frac{1}{2}-\frac{1}{3})}{\frac{2}{3}}$.
(2)When $3/4>|AB|\geq 1/2$, there are must be two vertices lies in |AB|. the probability are $\int^{\frac{3}{4}}_{x=\frac{1}{2}} p(\text{top-left vertex is in longest segment} |x)p(x)dx=\frac{1}{2}\times\frac{(\frac{3}{4}-\frac{1}{2})}{\frac{2}{3}}$
(2)When $|AB|\geq 3/4$, there are must be three vertices lies in |AB|. the probability are $\int^{1}_{x=\frac{3}{4}} p(x)p(\text{top-left vertex is in longest segment} |x)dx=\frac{3}{4}\times\frac{(1-\frac{3}{4})}{\frac{2}{3}}$.
Sum them together and we get the final result.
Is my answer correct?
Or is there a simpler method to solve this problem?
Thank you in advance!
