# 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.

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.
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Geovanni Shelton
Step 1
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, ${a}^{2}+{b}^{2}<{c}^{2}$.
Step 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 and $0 is $\left[\pi \left({r}^{2}\right)\right]/4$, or $\pi /4$. The total area is the unit square. Thus, the probability of an obtuse triangle is $\left(\pi /4\right)/1$, or just $\pi /4.$.
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Bayobusalue
Step 1
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. $a+b\ge 1$ (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 ${a}^{2}+{b}^{2}<{1}^{2}$.
Step 2
Now what is the probability of $a+b\ge 1\wedge {a}^{2}+{b}^{2}<{1}^{2}$? 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 $\left[0,1{\right]}^{2}$.
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, $a+b\ge 1$ 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 $\pi /4-1/2$.