Problem:
Three numbers in the interval are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?
Answer Choices:
A.
B.
C.
D.
E.
Solution:
Let the chosen numbers be , and . The set of possible ordered triples forms a solid unit cube, two of whose vertices are and . 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 lie in the region on and above the plane . The intersection of this region with the solid cube is a solid tetrahedron with vertices , and . The volume of this tetrahedron is . The intersections of the solid cube with the regions defined by the inequalities and are solid tetrahedra with the same volume. Because at most one of the inequalities , and can be true for any choice of , and , the three tetrahedra have disjoint interiors. Thus the required probability is .
OR
As in the first solution, the set of possible ordered triples forms a solid unit cube. First consider only the points for which and . These points form a square pyramid whose vertex is and whose base has vertices at , and . Such an ordered triple corresponds to the side lengths of a triangle if and only if . The plane passes through the vertex of the pyramid and bisects its base, so it bisects the volume of the pyramid. The probability of forming a triangle is the same as the probability of not forming a triangle. The same argument applies when or is the largest element in the triple. The probability of any two of , and being equal is 0 , so this case can be ignored. Thus this event and its complement are equally likely; the probability is .
The problems on this page are the property of the MAA's American Mathematics Competitions