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        riddles >> putnam exam (pure math) >> (tanA)(tanB)(tanC)
        
(Message started by: ThudanBlunder on May 17^th, 2008, 12:09pm)

Title: (tanA)(tanB)(tanC)
Post by ThudanBlunder on May 17^th, 2008, 12:09pm

Prove that for any acute triangle (tanA)(tanB)(tanC) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/geqslant.gif 3http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3

Title: Re: (tanA)(tanB)(tanC)
Post by BenVitale on May 17^th, 2008, 12:40pm

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.

Title: Re: (tanA)(tanB)(tanC)
Post by ThudanBlunder on May 17^th, 2008, 1:07pm

on 05/17/08 at 12:40:20, 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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3 is not exactly what I was looking for. That would be 3http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3 ;)

HINT:[hide]Once you have proved tanA + tanB + tanC = tanA * tanB * tanC you can use a well-known inequality for the rest.[/hide]

Title: Re: (tanA)(tanB)(tanC)
Post by william wu on May 20^th, 2008, 2:28pm

[hide]

For notational brevity, let a = tan A, b = tan B, and c = tan C.
Applying the AM-GM inequality yields
(a + b + c)/3 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif(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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif(abc)^1/3, which, after some algebra, yields abc >= 3 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{3}.

Hence, a + b + c = abc http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif3 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{3}.

[/hide]

Title: Re: (tanA)(tanB)(tanC)
Post by l4z3r on Aug 30^th, 2008, 6:05am

on 05/17/08 at 13:07:10, ThudanBlunder wrote:

HINT:[hide]well-known inequality for the rest.[/hide]

[hide]Jensen's Inequality?

When A+B+C=n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif then http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cpi.gif(tan A) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/csigma.gif(tan A).

Let f(x)=tan x. 0<x<http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif /2

The second differential of f(x) is always positive between o and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif tan ((A+B+C)/3)

A + B + C = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif.[/hide]

Hence Proved :) WOOT!

my first solution on this site, i think.