This is the problem link.

The first two parts build the tempo. Once you solve these two(please don't see the solution), move to the last part.

**If six numbers are chosen at random, uniformly and independently, from the interval [0,1], what is the probability that they are the lengths of the edges of a tetrahedron?**

My attempt -

Once I have chosen such \(x,y,z\) which form a triangle, number of points lying inside that triangle will be proportional to the number of tetrahedrons.

So, the \(\text{Probability(x,y,z, a,b,c forming a tetrahedron | x,y,z form a triangle)} \propto \text{Area of the triangle}\)

\(= \frac{k}{4} \sqrt{(x+y+z) (x+y-z) (x-y+z) (-x+y+z)} \)

\(= \iiint \frac{k}{4} \sqrt{(x+y+z) (x+y-z) (x-y+z) (-x+y+z)} \ dx \ dy \ dz\)

\(\displaystyle = Re\left( \int_{z=0}^{z=1} {\int_{y=0}^{y=1}{\int_{x=0}^{x=1}{ \frac{k}{4} \sqrt{(x+y+z) (x+y-z) (x-y+z) (-x+y+z)} \ dx} \ dy }\ dz} \right)\)

I don't really know how to follow next, or if what I did until now is any helpful or not.

## Comments

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TopNewestThis is a complicated matter. The third vertex of a tetrahedron does not have to lie above the base. For that matter, the uniform distribution of the side lengths will not convert into a uniform distribution of the (projected) location of the fourth vertex onto the base.

You might like to look at the article by Wirth & Dreiding that can be found here. – Mark Hennings · 5 months, 3 weeks ago

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– Kartik Sharma · 5 months, 3 weeks ago

Oh I see. It's really a complicated matter. And indeed out of my abilities as well.Log in to reply

@Mark Hennings @Michael Mendrin – Kartik Sharma · 5 months, 3 weeks ago

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