Problem:
Let T be the set of ordered triples (x,y,z) of nonnegative real numbers that lie in the plane x+y+z=1. Let us say that (x,y,z) supports (a,b,c) when exactly two of the following are true: x≥a,y≥b,z≥c. Let S consist of those triples in T that support (21​,31​,61​). The area of S divided by the area of T is m/n, where m and n are relatively prime positive integers. Find m+n.
Solution:
Notice that T is a first-octant equilateral triangle, whose vertices are (1,0,0),(0,1,0), and (0,0,1). The planes x=21​,y=31​, and z=61​ intersect T along line segments that are parallel to the sides of T. Let A consist of those points of T that satisfy x≥21​ and y≥31​. Notice that z≤61​ for any point in A, so the points of A support (21​,31​,61​), with the exception of (21​,31​,61​) itself. In a similar fashion, let B consist of those points of T that satisfy x≥21​ and z≥61​, and let C consist of those points of T that satisfy y≥31​ and z≥61​. Except for the point (21​,31​,61​),S is the union of the equilateral triangles A,B, and C, whose sides are 61​,31​, and 21​ times as long as the sides of T, and whose areas are 361​,91​, and 41​ times the area of T, respectively. It follows that the area of S divided by the area of T is 361​+91​+41​=187​. Thus m+n=25​.
The problems on this page are the property of the MAA's American Mathematics Competitions