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# Another Missouri state problem!

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.

Note by Kartik Sharma
5 months, 3 weeks ago

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This 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. · 5 months, 3 weeks ago

Oh I see. It's really a complicated matter. And indeed out of my abilities as well. · 5 months, 3 weeks ago