AD is altitude of isosceles triangle,
AB=ACOD=OE as it is isosceles triangle
AE=AF,BD=BE and CD=CE as tangents drawn from exterior point is equal
So, AB+BC+CA= perimeter
AF+BF+BD+CD+AE+EC=PAAF+BF+(BF+BD+CD+FC)=2AF+4BD
In △AFO||AF=OFtanθ=rtanθBD=ADtanθ=(AO+OD)tanθ=(rsinθ+r)tanθ
So, P(θ)=2rtanθ+4(r+rsinθ)sinθsinθcosθP(θ)=2rcotθ+4secθ+4rtanθP(θ)=0 for maximum/ minimum
P(θ)=r(−2cosec2θ+4secθtanθ+4sec2θ)=0
2(−1sin2θ+2sinθcos2θ+2cos2θ)=0or,−cos2θ+2sin3θ+2sin2θ=0or,−1−sin2θ+2sin3θ+2sin2θ=02sin3θ+3sin2θ−=0(sinθ+1)(2sin2θ−sinθ−1)=0θ≠−1asθ>90°2sin2θ−2sinθ+sinθ−1=0(2sinθ−1)(sinθ+1)=0sinθ≠1
So, sinθ=12θ=π6
If f′(x) changes sign so θ=π6 is maxima
r=[2√3+4×2√3+4×1√3]=18√3r=6√3r