In a triangle ABC prove that 2r&lt;=(acotA +bcotB+ccotC)/3&lt;=R.

Manisha Mishra Grade: 11

        









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

11 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.