AMC10 2003 10A, Problem 12

A point $(x,y)$ is randomly picked from inside the rectangle with  vertices $(0,0)$, $(4,0)$, $(4,1)$, and $(0,1)$. What is the probability that $x<y$?

Solution: The rectangle has a width of $4$ and a height of $1$. The area of this rectangle is $4\cdot1=4$. The line $x=y$ intersects the rectangle at $(0,0)$ and $(1,1)$. The area which $x<y$ is the right isosceles triangle with side length $1$ that has vertices at $(0,0)$, $(1,1)$, and $(0,1)$. The area of this triangle is $\frac{1}{2}\cdot1^{2}=\frac{1}{2}$ Therefore, the probability that $x<y$ is $\frac{\frac{1}{2}}{4}=\frac{1}{8} \Rightarrow \fbox{\frac{1}{8}}$

Answer: \frac{1}{8}