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Topic: (tanA)(tanB)(tanC) (Read 9883 times) 

ThudnBlunder
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(tanA)(tanB)(tanC)
« on: May 17^{th}, 2008, 12:09pm » 
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Prove that for any acute triangle (tanA)(tanB)(tanC) 33

« Last Edit: May 17^{th}, 2008, 9:00pm by ThudnBlunder » 
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Benny
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Re: (tanA)(tanB)(tanC)
« Reply #1 on: May 17^{th}, 2008, 12:40pm » 
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A+B+C = pi C = pi  (A+B) tanC = tan(A+B) tanC = (tanA + tanB)/(1  tanA tanB) tanA + tanB + tanC = tanA + tanB  (tanA + tanB)/(1  tanA tanB) = {tanA(1  tanA tanB) + tanB(1  tanA tanB)  tanA  tanB}/(1  tanA tanB) = {tanA  tanA*tanA*tanB + tanB  tanA*tanB*tanB  tanA  tanB}/(1  tanA tanB) = tanA*tanB(tanA + tanB)/(1  tanA tanB) = tanA * tanB * tanC So we have tanA + tanB + tanC = tanA * tanB * tanC or by dividing by tanA tanB tanC cotB cotC + cotA cotC + cotA cotB = 1. By squaring both sides of S = cotA + cotB + cotC we find S^2 = (cotA)^2 + (cotB)^2 + (cotC)^2 + 2. Now we know that (cotA  cotB)^2 + (cotB  cotC)^2 + (cotC  cotA)^2 >= 0 and thus 2((cotA)^2 + (cotB)^2 + (cotC)^2)  2(cotB cotC + cotA cotC + cotA cotB) >=0 or 2(S^2  2)  2 >= 0 2S^2  6 >= 0 S^2  3 >= 0 and we find that S>= sqrt(3) (as it must be positive), which is exactly what you were looking for.


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ThudnBlunder
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Re: (tanA)(tanB)(tanC)
« Reply #2 on: May 17^{th}, 2008, 1:07pm » 
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on May 17^{th}, 2008, 12:40pm, BenVitale wrote: So we have tanA + tanB + tanC = tanA * tanB * tanC <snipped> and we find that S>= sqrt(3) (as it must be positive), which is exactly what you were looking for. 
 Looks good, except 3 is not exactly what I was looking for. That would be 33 HINT:Once you have proved tanA + tanB + tanC = tanA * tanB * tanC you can use a wellknown inequality for the rest.

« Last Edit: May 17^{th}, 2008, 1:16pm by ThudnBlunder » 
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william wu
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Re: (tanA)(tanB)(tanC)
« Reply #3 on: May 20^{th}, 2008, 2:28pm » 
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For notational brevity, let a = tan A, b = tan B, and c = tan C. Applying the AMGM inequality yields (a + b + c)/3 (abc)^{1/3}. As Ben showed, a b c = a + b + c in this case. Substituting this relation into the left hand side yields abc/3 (abc)^{1/3}, which, after some algebra, yields abc >= 3 {3}. Hence, a + b + c = abc 3 {3}.

« Last Edit: May 20^{th}, 2008, 2:29pm by william wu » 
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l4z3r
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Re: (tanA)(tanB)(tanC)
« Reply #4 on: Aug 30^{th}, 2008, 6:05am » 
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on May 17^{th}, 2008, 1:07pm, ThudanBlunder wrote: HINT:wellknown inequality for the rest. 
 Jensen's Inequality? When A+B+C=n then (tan A) = (tan A). Let f(x)=tan x. 0<x< /2 The second differential of f(x) is always positive between o and /2. Hence, the function is convex. (It would have sufficed if the graph of tan x was observed, actually). Jensen's inequality gives us: (tan A + tan B + tan C)/3 tan ((A+B+C)/3) A + B + C = . Hence Proved WOOT! my first solution on this site, i think.


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