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Manisha Mishra Grade: 11
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In a triangle ABC prove that 2r<=(acotA +bcotB+ccotC)/3<=R.

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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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