Since u+v is the number of elements in B but not C, the same number must be in both B and C. k are in all three so there must be u+v-k in B and C but not A.<\/p>\n
Now let w be the number of elements in C only. Then the total in C but not A is (u+v-k)+w=u+v+w-k, and since there are k in the common set, the number in A and C but not B must be u+v+w-2k. <\/p>\n
Since the total number in A must be twice what’s shared with B, that is 2(u+k), the number that are only in A must be (u+k)-(u+v+w-2k)=3k-v-w.<\/p>\n
If that is not all clear, then the following diagram might help: In terms of the independent variables k,u,v, and w, we get For fixed v and w, R->1 as u->infinity and is monotone in the domain so there are no interior critical points. On u+v=1, we have u=1-v≤1, So R is never greater than 3, and the bound is attained when v=h=0 (which corresponds to the example cited earlier)<\/p>\n","protected":false},"excerpt":{"rendered":" Let k be the number of elements common to all three sets, u be the number of elements in both A and B but not C, and v the number of elements that are only in B. (Note: The choice … Continue reading
\n![\"\"](\"http:\/\/qpr.ca\/blog\/wp-content\/uploads\/2010\/10\/venn-300x212.jpg\" "\\"venn\\"")
<\/a><\/p>\n
\na=2(k+u)
\nb=2(u+v)
\nc=2(u+v+w-k)
\nand our objective is to find extrema of
\nR=a\/c=(u+k)\/(u+v+w-k)
\nfor positive integer values of k,u,v,w with u+v-k≥0 , u+v+w-2k≥0, and 3k-v-w≥0.
\nEquivalently, we could take k=1, look for rational values of u,v,w and rescale at the end to get integer solutions.
\nSo we’re looking at R=(u+1)\/(u+v+w-1) with u,v,w≥0, u+v+w≥2, and 3≥u+v≥1<\/p>\n
\nWe can therefore restrict our attention to the boundary.<\/p>\n
\nso R=(u+1)\/w=(2-v)\/w≤2 (since w≥1 since u+v+w≥2);
\nOn u+v=3, R=(4-v)\/(2+w)≤2; and finally
\nOn u+v+w=2, u=2-v-w, and so
\nR=(3-v-h)\/(2-1)≤3<\/p>\n