### Instruction: Randy presses RAND on his calculator twice to obtain two random numbers between 0 and 1. Let $p$ be the probability that these two numbers and 1 form the sides of an obtuse triangle.  Find $4p$.### Response: An obtuse triangle is a triangle that has one angle greater than 90 degrees.
I can use the law of cosines to relate the sides and angles of a triangle.
If I let $a, b, c$ be the sides of the triangle, and $C$ be the angle opposite $c$, then the law of cosines says that $c^2 = a^2 + b^2 - 2ab \cos C$.
I want to find the probability that $c^2 > a^2 + b^2$, since that would mean that $C$ is obtuse.
Since $c = 1$, this is equivalent to finding the probability that $a^2 + b^2 < 1$.
This is a geometric probability problem, where I need to find the fraction of the unit square that lies inside the unit circle.
The unit circle has area $\pi$, and the unit square has area 4, so the fraction is $\frac{\pi}{4}$.
Therefore, $p = \frac{\pi}{4}$, and $4p = \pi$.