{"id":965,"date":"2010-10-04T15:59:18","date_gmt":"2010-10-04T22:59:18","guid":{"rendered":"http:\/\/qpr.ca\/blog\/"},"modified":"2022-01-18T16:25:53","modified_gmt":"2022-01-19T00:25:53","slug":"neater-version","status":"publish","type":"page","link":"https:\/\/qpr.ca\/blogs\/pages\/mathstuff5\/quora\/cute-set-sizes-problem\/neater-version\/","title":{"rendered":"Neater Version"},"content":{"rendered":"

Let k be the number of elements common to all three sets, p be those in just A and B but not C, q in just B and C and r in just C and A, and let \u03b1 be the number in just A, \u03b2 in just B and \u03b3 in just C.
\nIf that is not all clear, then the following diagram might help:\"\"<\/a><\/p>\n

Now the three conditions are:
\n\u03b1+r=p+k, or p+k-r=\u03b1\u22650;
\n\u03b2+p=q+k, or q+k-p=\u03b2\u22650; and
\n\u03b3+q=r+k, or r+k-q=\u03b3\u22650<\/p>\n

We could use the three conditions to express any three of the variables in terms of the other four. It doesn’t matter which four we take as independent but a nice symmetric choice is to use p,q,r and k as we have done above to solve for \u03b1, \u03b2, and \u03b3.<\/p>\n

In terms of these we get
\na=2(k+p)
\nb=2(k+q)
\nc=2(k+r)
\nand our objective is to find extrema of
\nR=a\/c=(k+p)\/(k+r)
\nfor positive integer values of k,p,q,r with p+k-r\u22650, q+k-p\u22650;, and r+k-q\u22650; or equivalently r-p\u2264k, p-q\u2264k, and q-r\u2264k.<\/p>\n

Equivalently, we could take k=1, look for rational values of p,q,r and rescale at the end to get integer solutions.<\/p>\n

So we’re looking ar R=(1+p)\/(1+r) with p,q,r\u22650, and r-p\u22641, p-q\u22641, and q-r\u22641.<\/p>\n

Since r\u22641+p, we get R\u2265(1+p)\/(2+p)\u22651\/2<\/p>\n

And since p-r=(p-q)+(q-r)\u22642, we get p\u22642+r so R\u2264(3+r)\/(1+r)\u22643<\/p>\n

So R is never greater than 3, and the bound is attained when r=0 with q-r=1, and p-q=1.(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, p be those in just A and B but not C, q in just B and C and r in just C and A, and let \u03b1 be … Continue reading →<\/span><\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"parent":949,"menu_order":0,"comment_status":"open","ping_status":"open","template":"","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-965","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/qpr.ca\/blogs\/wp-json\/wp\/v2\/pages\/965","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/qpr.ca\/blogs\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/qpr.ca\/blogs\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/qpr.ca\/blogs\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/qpr.ca\/blogs\/wp-json\/wp\/v2\/comments?post=965"}],"version-history":[{"count":6,"href":"https:\/\/qpr.ca\/blogs\/wp-json\/wp\/v2\/pages\/965\/revisions"}],"predecessor-version":[{"id":4400,"href":"https:\/\/qpr.ca\/blogs\/wp-json\/wp\/v2\/pages\/965\/revisions\/4400"}],"up":[{"embeddable":true,"href":"https:\/\/qpr.ca\/blogs\/wp-json\/wp\/v2\/pages\/949"}],"wp:attachment":[{"href":"https:\/\/qpr.ca\/blogs\/wp-json\/wp\/v2\/media?parent=965"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/qpr.ca\/blogs\/wp-json\/wp\/v2\/categories?post=965"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/qpr.ca\/blogs\/wp-json\/wp\/v2\/tags?post=965"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}