Geometric Probability Problem, Random Numbers $0$-$1+$ Triangles. Randy presses RANDOM 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 $p$.
At first, I thought that the answer would be $.25$, because the probability of both of the numbers being above $.5$ is $.5$. But I realized that this is not true, for cases such as $.99, .99, 1,$ etc. This problem has me a bit stumped. I'm fairly certain I need to use a graph, but I don't know how.
 A: Let $a$ and $b$ be your two independent random numbers.
First of all, we need to check whether the three given sides even form a triangle. For that, $a, b, 1$ need to satisfy the triangle equation, i.e. $a + b \geq 1$ (the other two permutations are always satisfied anyway).
Now when is this triangle obtuse? Well, we know that the side of length $1$ is the longest side, so it must be opposite the largest angle $\gamma$.
When is that angle $\gamma$ obtuse? For that, remember Pythagoras. $\gamma$ is a right angle exactly when $a^2+b^2 = 1^2$, similarly $\gamma$ is obtuse exactly when $a^2+b^2 < 1^2$.
Now what is the probability of $a + b \geq 1 \wedge a^2+b^2 < 1^2$? Let $a$ and $b$ denote coordinates of a point $(a, b)$. This point is, by (assumed) independence of $a$ and $b$, uniformly random in the unit square $[0,1]^2$.
Now $a$ and $b$ satisfy the property $a^2+b^2 < 1^2$ exactly when $(a, b)$ lies in the unit sphere. The part of the unit sphere that lies in the unit square is a unit quarter-sphere, of area $\pi/4$.
Furthermore, $a + b \geq 1$ is the case exactly when $(a, b)$ lies above the diagonal between $(0, 1)$ and $(1, 0)$. This diagonal divides the circular sector into a segment and a triangle. The triangle has area $1/2$, so the segment has area $\pi/4 - 1/2$.
Two numbers $a, b$ now satisfy the property given in the problem exactly when $(a, b)$ lies in this segment. Since $(a, b)$ is uniformly distributed over the unit square, the probability that this happens is exactly the area of the segment divided by the area of the square (which is $1$). The probability you are looking for is exactly $\pi/4 - 1/2$.
A: I bet they are looking for you to use the Pythagorean Theorem. Since the random number will be less than 1, you can assume 1 is the largest side length. For obtuse triangles, $a^2+b^2<c^2$. In this case, because c is 1, you can simplify this to $a^2+b^2<1.$ This is a two-variable probability problem, so the easiest way to do it is integration, but you needn't integrate. If you set $a=x$ and $b=y$, you get the unit circle equation $x^2+y^2<1$. The area of this circle inside the parameters $0<x<1$ and $0<y<1$ is $[\pi(r^2)]/4$, or $\pi/4$. The total area is the unit square. Thus, the probability of an obtuse triangle is $(\pi/4)/1$, or just $\pi/4.$
