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Question-63176

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Question Number 63176 by ajfour last updated on 30/Jun/19
Commented by ajfour last updated on 30/Jun/19
Find maximum coloured area if line segments BP and AP lie entirely outside the circle.
$${Find}\:{maximum}\:{coloured}\:{area} \\ $$$${if}\:{line}\:{segments}\:{BP}\:{and}\:{AP}\: \\ $$$${lie}\:{entirely}\:{outside}\:{the}\:{circle}. \\ $$
Answered by mr W last updated on 30/Jun/19
let θ=∠APB, ϕ=∠ACB AB=(√(a^2 +b^2 −2ab cos θ)) AB=2R sin (ϕ/2) ⇒ϕ=2 sin^(−1) ((√(a^2 +b^2 −2ab cos θ))/(2R)) (dϕ/dθ)=(2/( (√(1−((a^2 +b^2 −2ab cos θ)/(4R^2 ))))))×((ab sin θ)/(2R(√(a^2 +b^2 −2ab cos θ)))) =((2ab sin θ)/( (√((4R^2 −a^2 −b^2 +2ab cos θ)(a^2 +b^2 −2ab cos θ))))) A_(blue) =(1/2)ab sin θ−(R^2 /2)(ϕ−sin ϕ) (dA_(bue) /dθ)=0 ab cos θ−R^2 (1−cos ϕ)(dϕ/dθ)=0 ab cos θ−R^2 (2 sin^2 (ϕ/2))((2ab sin θ)/( (√((4R^2 −a^2 −b^2 +2ab cos θ)(a^2 +b^2 −2ab cos θ)))))=0 ab cos θ−R^2 (((2(√(a^2 +b^2 −2ab cos θ)))/(2R)))((2ab sin θ)/( (√((4R^2 −a^2 −b^2 +2ab cos θ)(a^2 +b^2 −2ab cos θ)))))=0 cos θ−((2Rsin θ)/( (√((4R^2 −a^2 −b^2 +2ab cos θ)))))=0 4R^2 tan^2 θ=4R^2 −a^2 −b^2 +2ab cos θ 4R^2 =(4R^2 −a^2 −b^2 +2ab cos θ)(1+cos^2 θ) 1=(1−((a^2 +b^2 )/(4R^2 ))+((ab)/(2R^2 )) cos θ)(1+cos^2 θ) 1=(1−λ+μ cos θ)(1+cos^2 θ) 0=−λ+μ cos θ+(1−λ)cos^2 θ+μ cos^3 θ ⇒cos^3 θ+((1−λ)/μ) cos^2 θ+cos θ−(λ/μ)=0 ⇒cos θ=.....
$${let}\:\theta=\angle{APB},\:\varphi=\angle{ACB} \\ $$$${AB}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\:\theta} \\ $$$${AB}=\mathrm{2}{R}\:\mathrm{sin}\:\frac{\varphi}{\mathrm{2}} \\ $$$$\Rightarrow\varphi=\mathrm{2}\:\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\:\theta}}{\mathrm{2}{R}} \\ $$$$\frac{{d}\varphi}{{d}\theta}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\:\theta}{\mathrm{4}{R}^{\mathrm{2}} }}}×\frac{{ab}\:\mathrm{sin}\:\theta}{\mathrm{2}{R}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\:\theta}} \\ $$$$=\frac{\mathrm{2}{ab}\:\mathrm{sin}\:\theta}{\:\sqrt{\left(\mathrm{4}{R}^{\mathrm{2}} −{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +\mathrm{2}{ab}\:\mathrm{cos}\:\theta\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\:\theta\right)}} \\ $$$${A}_{{blue}} =\frac{\mathrm{1}}{\mathrm{2}}{ab}\:\mathrm{sin}\:\theta−\frac{{R}^{\mathrm{2}} }{\mathrm{2}}\left(\varphi−\mathrm{sin}\:\varphi\right) \\ $$$$\frac{{dA}_{{bue}} }{{d}\theta}=\mathrm{0} \\ $$$${ab}\:\mathrm{cos}\:\theta−{R}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\varphi\right)\frac{{d}\varphi}{{d}\theta}=\mathrm{0} \\ $$$${ab}\:\mathrm{cos}\:\theta−{R}^{\mathrm{2}} \left(\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\frac{\varphi}{\mathrm{2}}\right)\frac{\mathrm{2}{ab}\:\mathrm{sin}\:\theta}{\:\sqrt{\left(\mathrm{4}{R}^{\mathrm{2}} −{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +\mathrm{2}{ab}\:\mathrm{cos}\:\theta\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\:\theta\right)}}=\mathrm{0} \\ $$$${ab}\:\mathrm{cos}\:\theta−{R}^{\mathrm{2}} \left(\frac{\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\:\theta}}{\mathrm{2}{R}}\right)\frac{\mathrm{2}{ab}\:\mathrm{sin}\:\theta}{\:\sqrt{\left(\mathrm{4}{R}^{\mathrm{2}} −{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +\mathrm{2}{ab}\:\mathrm{cos}\:\theta\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\:\theta\right)}}=\mathrm{0} \\ $$$$\mathrm{cos}\:\theta−\frac{\mathrm{2}{R}\mathrm{sin}\:\theta}{\:\sqrt{\left(\mathrm{4}{R}^{\mathrm{2}} −{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +\mathrm{2}{ab}\:\mathrm{cos}\:\theta\right)}}=\mathrm{0} \\ $$$$\mathrm{4}{R}^{\mathrm{2}} \:\mathrm{tan}^{\mathrm{2}} \:\theta=\mathrm{4}{R}^{\mathrm{2}} −{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +\mathrm{2}{ab}\:\mathrm{cos}\:\theta \\ $$$$\mathrm{4}{R}^{\mathrm{2}} =\left(\mathrm{4}{R}^{\mathrm{2}} −{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +\mathrm{2}{ab}\:\mathrm{cos}\:\theta\right)\left(\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:\theta\right) \\ $$$$\mathrm{1}=\left(\mathrm{1}−\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{4}{R}^{\mathrm{2}} }+\frac{{ab}}{\mathrm{2}{R}^{\mathrm{2}} }\:\mathrm{cos}\:\theta\right)\left(\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:\theta\right) \\ $$$$\mathrm{1}=\left(\mathrm{1}−\lambda+\mu\:\mathrm{cos}\:\theta\right)\left(\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:\theta\right) \\ $$$$\mathrm{0}=−\lambda+\mu\:\mathrm{cos}\:\theta+\left(\mathrm{1}−\lambda\right)\mathrm{cos}^{\mathrm{2}} \:\theta+\mu\:\mathrm{cos}^{\mathrm{3}} \:\theta \\ $$$$\Rightarrow\mathrm{cos}^{\mathrm{3}} \:\theta+\frac{\mathrm{1}−\lambda}{\mu}\:\mathrm{cos}^{\mathrm{2}} \:\theta+\mathrm{cos}\:\theta−\frac{\lambda}{\mu}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{cos}\:\theta=….. \\ $$

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