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In a triangle ABC prove that 2r<=(acotA +bcotB+ccotC)/3<=R.
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.
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
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,
since r = 4RsinA/2sinB/2SinC/2
or to prove that
On solving,
Consider s-b, s-c and use AM GM
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
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