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For a triangle A B C, the value of \cos 2 A+\cos 2 B+\cos 2 C is least. If its inradius is 3 and incentre is M, then which of the following is NOT correct?

Option: 1

perimeter of \triangle A B C  is 18 \sqrt{3}


Option: 2

\sin 2 A+\sin 2 B+\sin 2 C=\sin A+\sin B+\sin C


Option: 3

\overrightarrow{M A} \cdot \overrightarrow{M B}=-18


Option: 4

area of  \triangle A B C  is \frac{27 \sqrt{3}}{2}


Answers (1)

best_answer

Let \mathrm{P}
$$ \begin{aligned} & =\cos 2 \mathrm{~A}+\cos 2 \mathrm{~B}+\cos 2 \mathrm{C} \\ & =2 \cos (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})+2 \cos ^2 \mathrm{C}-1 \\ & =2 \cos (\pi-\mathrm{C}) \cos (\mathrm{A}-\mathrm{B})+2 \cos ^2 \mathrm{C}-1 \\ & =-2 \cos \mathrm{C}[\cos (\mathrm{A}-\mathrm{B})+\cos (\mathrm{A}+\mathrm{B})]-1 \\ & =-1-4 \cos \mathrm{A} \cos \mathrm{B} \cos \mathrm{C} \end{aligned}
for P to be minimum

\cos \mathrm{A} \cos \mathrm{B} \cos \mathrm{C} must be maximum

\Rightarrow \triangle \mathrm{ABC} is equilateral triangle.

Let side length of triangle is a

\begin{aligned} & \tan \theta=\frac{3}{a / 2} \\ & \Rightarrow \frac{1}{\sqrt{3}}=\frac{6}{a} \Rightarrow a=6 \sqrt{3} \\ & \text { area of triangle }=\frac{\sqrt{3}}{4} \mathrm{a}^2 \\ & =\frac{\sqrt{3}}{4}(108)=27 \sqrt{3} \end{aligned}

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