Eliminate a, b, c;
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From: Ken Pledger on 14 Jan 2010 17:58 In article <4b4f92fd(a)news.x-privat.org>, "Amy" <nospam(a)spamless.com> wrote: > "KY" <wkfkh056(a)yahoo.co.jp> wrote in message > news:1082409701.70238.1263485757008.JavaMail.root(a)gallium.mathforum.org... > > Eliminate a, b, c; > > a + b + c = 0, > > ((b - c)/a + (c - a)/b + (a - b)/c)*(a/(b - c) + b/(c - a) + c/(a - b)) = > > k > > 2 equations, 4 unknowns.................... Well, eliminating a, b, c still allows k to stay; but that would usually require 4 equations as you suggest. However, these two equations are cunningly contrived to lead to the conclusion k = 9. Can you work out how? Ken Pledger.
From: preedmont on 14 Jan 2010 23:24
"Ken Pledger" <ken.pledger(a)mcs.vuw.ac.nz> wrote in message news:ken.pledger-748EF1.11584115012010(a)nothing.attdns.com... > In article <4b4f92fd(a)news.x-privat.org>, "Amy" <nospam(a)spamless.com> > wrote: > >> "KY" <wkfkh056(a)yahoo.co.jp> wrote in message >> news:1082409701.70238.1263485757008.JavaMail.root(a)gallium.mathforum.org... >> > Eliminate a, b, c; >> > a + b + c = 0, >> > ((b - c)/a + (c - a)/b + (a - b)/c)*(a/(b - c) + b/(c - a) + c/(a - b)) >> > = >> > k >> >> 2 equations, 4 unknowns.................... > > > > Well, eliminating a, b, c still allows k to stay; but that would > usually require 4 equations as you suggest. However, these two > equations are cunningly contrived to lead to the conclusion k = 9. Can > you work out how? > > Ken Pledger. ez to work out, may take a page, lot of 2nd order terms, |