wu :: forums
wu :: forums - (tanA)(tanB)(tanC)

Welcome, Guest. Please Login or Register.
Oct 27th, 2021, 8:49am

RIDDLES SITE WRITE MATH! Home Home Help Help Search Search Members Members Login Login Register Register
   wu :: forums
   riddles
   putnam exam (pure math)
(Moderators: SMQ, towr, william wu, Icarus, Grimbal, Eigenray)
   (tanA)(tanB)(tanC)
« Previous topic | Next topic »
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print
   Author  Topic: (tanA)(tanB)(tanC)  (Read 9905 times)
ThudnBlunder
Uberpuzzler
*****




The dewdrop slides into the shining Sea

   


Gender: male
Posts: 4489
(tanA)(tanB)(tanC)  
« on: May 17th, 2008, 12:09pm »
Quote Quote Modify Modify

Prove that for any acute triangle (tanA)(tanB)(tanC) 33
« Last Edit: May 17th, 2008, 9:00pm by ThudnBlunder » IP Logged

THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
Benny
Uberpuzzler
*****





   


Gender: male
Posts: 1024
Re: (tanA)(tanB)(tanC)  
« Reply #1 on: May 17th, 2008, 12:40pm »
Quote Quote Modify Modify

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.
IP Logged

If we want to understand our world or how to change it we must first understand the rational choices that shape it.
ThudnBlunder
Uberpuzzler
*****




The dewdrop slides into the shining Sea

   


Gender: male
Posts: 4489
Re: (tanA)(tanB)(tanC)  
« Reply #2 on: May 17th, 2008, 1:07pm »
Quote Quote Modify Modify

on May 17th, 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  Wink
 
HINT:Once you have proved tanA + tanB + tanC = tanA * tanB * tanC you can use a well-known inequality for the rest.
« Last Edit: May 17th, 2008, 1:16pm by ThudnBlunder » IP Logged

THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
william wu
wu::riddles Administrator
*****





   
WWW

Gender: male
Posts: 1291
Re: (tanA)(tanB)(tanC)  
« Reply #3 on: May 20th, 2008, 2:28pm »
Quote Quote Modify Modify


 
For notational brevity, let a = tan A, b = tan B, and c = tan C.
Applying the AM-GM 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 20th, 2008, 2:29pm by william wu » IP Logged


[ wu ] : http://wuriddles.com / http://forums.wuriddles.com
l4z3r
Newbie
*





   


Posts: 10
Re: (tanA)(tanB)(tanC)  
« Reply #4 on: Aug 30th, 2008, 6:05am »
Quote Quote Modify Modify

on May 17th, 2008, 1:07pm, ThudanBlunder wrote:

HINT:well-known 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 Smiley WOOT!
 
my first solution on this site, i think.
IP Logged
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print

« Previous topic | Next topic »

Powered by YaBB 1 Gold - SP 1.4!
Forum software copyright 2000-2004 Yet another Bulletin Board