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 $4p$.
Let the two random numbers be $x$ and $y$. In order to form an obtuse triangle, since 1 will be the longest side, we must simultaneously satisfy the following inequalities: $$x+y>1\text{ and }x^2+y^2<1.$$The first is the triangle inequality and the second guarantees that the triangle is obtuse. Graphing these in the $xy$-plane, we get the following shaded region. [asy]
draw(unitsquare);
draw((0,0)--(1,0),EndArrow);
draw((0,0)--(0,1),EndArrow);
label("0",(0,0),SW);
label("1",(1,0),S);
label("1",(0,1),W);
label("$x$",(.5,0),S);
label("$y$",(0,.5),W);
fill((1,0)--(0,1)..(3/5,4/5)..cycle,gray(.7));
[/asy] The curve is an arc of the unit circle centered at the origin. This area is then equal to that sector minus the right isosceles triangle within it, or $\frac{\pi}{4}-\frac{1}{2}=\frac{\pi-2}{4}.$ And since the area of the square is $1,$ $p = \frac{\pi-2}{4}.$

Four times $p$ is $\pi-2$.