AMC12 2016 A

Answer (C): Let the chosen numbers be $x$, $y$, and $z$. The set of possible ordered triples $(x,y,z)$ forms a solid unit cube, two of whose vertices are $(0,0,0)$ and $(1,1,1)$. The numbers fail to be the side lengths of a triangle with positive area if and only if one of the numbers is at least as great as the sum of the other two. The ordered triples that satisfy $z\ge x+y$ lie in the region on and above the plane $z=x+y$. The intersection of this region with the solid cube is a solid tetrahedron with vertices $(0,0,0)$, $(0,0,1)$, $(0,1,1)$, and $(1,0,1)$. The volume of this tetrahedron is $\frac16$. The intersections of the solid cube with the regions defined by the inequalities $y\ge x+z$ and $x\ge y+z$ are solid tetrahedra with the same volume. Because at most one of the inequalities $z>x+y$, $y>x+z$, and $x>y+z$ can be true for any choice of $x$, $y$, and $z$, the three tetrahedra have disjoint interiors. Thus the required probability is $1-3\cdot\frac16=\frac12$.