2012 AMC 10A Problem 25

This problem lends itself to geometric probability since we can view the interval as a range on an axis. WLOG, let $$n \geq x \geq y \geq z \geq 0.$$ Then we have that the points $(x, y, z)$ which satisfy this restriction form a tetrahedron. The height of this tetrahedron is $n,$ and the base has an area of $\dfrac{n^2}{2}.$ This makes the volume $$\dfrac{1}{3} \cdot \dfrac{n^2}{2} \cdot n = \dfrac{n^3}{6}.$$ Now we have to apply the restrictions from the problem statement. We need to find the region where $$|x - y|, |x - z|, |y - z| \geq 1.$$ From our ordering condition that we imposed, these inequalities reduce to $$x - y \geq 1 \text{ and } y - z \geq 1.$$ These two restrictions form another tetrahedron as shown below. Note that in the new tetrahedron, all the dimensions have been reduced by $2.$ This makes the height $n - 2$ and the base $\dfrac{(n - 2)^2}{2}.$ The volume is then $$\dfrac{1}{3} \cdot \dfrac{(n - 2)^2}{2} \cdot (n - 2)$$$$= \dfrac{(n - 2)^3}{6}.$$ The desired probability is then $$\dfrac{(n - 2)^3}{6} \div \dfrac{n^3}{6} = \dfrac{(n - 2)^3}{n^3}.$$ Plugging in all the answer choices, we get that the smallest value such that this fraction is greater than $\dfrac{1}{2}$ is $10.$ Thus, D is the correct answer.