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Question Number 63176 by ajfour last updated on 30/Jun/19

Commented by ajfour last updated on 30/Jun/19

$$Findmaximumcolouredarea$$$$iflinesegmentsBPandAP$$$$lieentirelyoutsidethecircle.$$

Answered by mr W last updated on 30/Jun/19

$$let\theta =\mathrm{\angle }APB,\phi =\mathrm{\angle }ACB$$$$AB=\sqrt{a^2+b^2-2ab\cos \theta }$$$$AB=2R\sin \frac{\phi }{2}$$$$\Rightarrow \phi =2{\sin }^{-1}\frac{\sqrt{a^2+b^2-2ab\cos \theta }}{2R}$$$$\frac{d\phi }{d\theta }=\frac{2}{\sqrt{1-\frac{a^2+b^2-2ab\cos \theta }{4R^2}}}\times \frac{ab\sin \theta }{2R\sqrt{a^2+b^2-2ab\cos \theta }}$$$$=\frac{2ab\sin \theta }{\sqrt{(4R^2-a^2-b^2+2ab\cos \theta )(a^2+b^2-2ab\cos \theta )}}$$$$A_{blue}=\frac{1}{2}ab\sin \theta -\frac{R^2}{2}(\phi -\sin \phi )$$$$\frac{{dA}_{bue}}{d\theta }=0$$$$ab\cos \theta -R^2(1-\cos \phi )\frac{d\phi }{d\theta }=0$$$$ab\cos \theta -R^2\left(2{\sin }^2\frac{\phi }{2}\right)\frac{2ab\sin \theta }{\sqrt{(4R^2-a^2-b^2+2ab\cos \theta )(a^2+b^2-2ab\cos \theta )}}=0$$$$ab\cos \theta -R^2\left(\frac{2\sqrt{a^2+b^2-2ab\cos \theta }}{2R}\right)\frac{2ab\sin \theta }{\sqrt{(4R^2-a^2-b^2+2ab\cos \theta )(a^2+b^2-2ab\cos \theta )}}=0$$$$\cos \theta -\frac{2R\sin \theta }{\sqrt{(4R^2-a^2-b^2+2ab\cos \theta )}}=0$$$$4R^2{\tan }^2\theta =4R^2-a^2-b^2+2ab\cos \theta$$$$4R^2=(4R^2-a^2-b^2+2ab\cos \theta )(1+{\cos }^2\theta )$$$$1=(1-\frac{a^2+b^2}{4R^2}+\frac{ab}{2R^2}\cos \theta )(1+{\cos }^2\theta )$$$$1=(1-\lambda +\mu \cos \theta )(1+{\cos }^2\theta )$$$$0=-\lambda +\mu \cos \theta +(1-\lambda ){\cos }^2\theta +\mu {\cos }^3\theta$$$$\Rightarrow {\cos }^3\theta +\frac{1-\lambda }{\mu }{\cos }^2\theta +\cos \theta -\frac{\lambda }{\mu }=0$$$$\Rightarrow \cos \theta =.....$$

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