i'll do it in the framework of partitions of 1 . define the mesh of a partition to be its smallest part. rich's partition is 5/84 + 5/84 + 5/84 + 5/84 + 5/84 + 5/84 + 1/14 + 1/14 + 1/12 + 1/12 + 1/12 + 1/12 + 1/12 + 1/12 , which has mesh 5/84 , and is simultaneously a refinement of 1/4 + 1/4 + 1/4 + 1/4 and of 1/7 + 1/7 + 1/7 + 1/7 + 1/7 + 1/7 + 1/7 . if some 1/4 is split into 5 or more parts, then the mesh is <= 1/20 < 5/84 . since 1/4 cannot be a part of 1/7 , each 1/4 must be split into at least 2 parts. suppose some 1/4 is split into exactly 2 parts, say 1/4 = a + b , where a <= b . we have 1/8 <= b <= 1/7 , so 3/28 <= a <= 1/8 . now a occurs as a part of 1/7 , so it has another part at most 1/7 - a <= 1/7 - 3/28 = 1/28 < 5/84 . thus, in an optimal partition, each 1/4 is split into 3 or 4 parts. if some 1/7 is split into 3 or more parts, then the mesh is <= 1/21 < 5/84 . if some 1/7 is not split, then it must occur as a part of some 1/4 . as above, this 1/4 must be split into at least 3 parts, so one of the other parts is at most (1/4 - 1/7) / 2 = 3/56 < 5/84 . thus, in an optimal partition, each 1/7 is split into exactly 2 parts. therefore, there are exactly 14 parts, and 2 of the 1/4 's are split into 3 parts, and the other 2 split into 4 parts. consider a 1/4 that is split into 3 parts. we then have 1/4 = a + b + c , where a <= b <= c , so c >= 1/12 . this occurs as a part of 1/7 , so the other part is 1/7 - c <= 5/84 . this proves that the mesh is at most 5/84 in all cases. moreover, if the mesh is indeed 5/84 , then the 1/4 's that are split into 3 parts must both split as 1/12 + 1/12 + 1/12 . each of these six 1/12 's must be part of 1/7 , which each split as 1/12 + 5/84 . so now there are six parts of 5/84 , and this shows that the two 1/4 's that split into 4 parts must both split as 5/84 + 5/84 + 5/84 + 1/14 . this proves that the partition is the same as rich's. mike