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Manisha Mishra Grade: 11
        








In a triangle ABC prove that 2r<=(acotA +bcotB+ccotC)/3<=R.
 
8 years ago

Answers : (1)

ashish kumar
17 Points
										

First consider : -


( a cotA + b cotB + c cotC ) / 3 R


             2R (cosA + cosB + cosC ) / 3 R


cosA + cosB + cosC 3 / 2


2 cos(A+B)/2 cos(A-B)/2 + 1 - 2 sin2C/2 3 / 2


- 2 sin2C/2 + 2 (sinC/2)cos(A - B)/2 - 1 / 2 0

 


This a quadratic equation in sinC/2 withnegative coefficient of  sin2C/2


D = b2 - 4ac


= 4 cos2(A - B)/2 - 4


since cos is always less than 0 therefore  the eqauation is always valid with a critical point


Therefore,     


( a cotA + b cotB + c cotC ) / 3 R


 


Now, 2r   2R (cosA + cosB + cosc)

 


since r = 4RsinA/2sinB/2SinC/2


or to prove that


4RsinA/2sinB/2SinC/2 R (cosA + cosB + cosC ) / 3

 


On solving,


RHS = ( 4RsinA/2sinB/2SinC/2 + 1 ) / 3

 


Consider s-b, s-c and use AM GM


(s-b + s-c ) / 2 (( s-b )(s-c ))(1/2) / bc

 


a / 2(bc)(1/2) sinA/2

 


similiarly for sinB/2 and sinC/2


therefore,


sinA/2sinB/2SinC/2 1/8


let sinA/2sinB/2SinC/2 = k


To prove 4k ( 4k + 1) / 3


or,   12k 4k + 1


or,         k 1/8 which is always possible.

 


 

8 years ago
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