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Question Number 51088 by ajfour last updated on 23/Dec/18

Commented by ajfour last updated on 23/Dec/18

$$Findmaximumareaof△ABC.$$

Answered by ajfour last updated on 23/Dec/18

$$Formax△,AB=AC$$$$△=\frac{(\sqrt{a^2+R^2}+x)}{2}\times 2\sqrt{R^2-x^2}$$$$letx=R\sin \theta ,then$$$$△=R\cos \theta (R\sin \theta +d)$$$$whered=\sqrt{a^2+R^2}$$$$\frac{d△}{d\theta }=R^2(1-2\sin {}^2\theta )-Rd\sin \theta$$$$\frac{d△}{d\theta }=0\Rightarrow$$$$2R\sin {}^2\theta +d\sin \theta -R=0$$$$\Rightarrow \sin \theta =-\frac{d}{4R}+\frac{\sqrt{d^2+8R^2}}{4R}$$$${△}_{max}=R^2\sqrt{1-\sin {}^2\theta }(\sin \theta +\frac{d}{R})$$$$...$$

Answered by mr W last updated on 24/Dec/18

$$O=centerofcircle$$$$AO=d=\sqrt{a^2+R^2}$$$$\mathrm{\angle }BOC=2\theta$$$$A=\frac{1}{2}\left(2R\sin \theta \right)(d+R\cos \theta )=R^2\sin \theta (\frac{d}{R}+\cos \theta )$$$$A=R^2\sin \theta (\lambda +\cos \theta )with\lambda =\frac{d}{R}=\sqrt{1+\frac{a^2}{R^2}}$$$$\frac{dA}{d\theta }=0$$$$\cos \theta (\lambda +\cos \theta )-{\sin }^2\theta =0$$$${2\cos }^2\theta +\lambda \cos \theta -1=0$$$$\Rightarrow \cos \theta =\frac{\sqrt{{\lambda }^2+8}-\lambda }{4}$$$$\Rightarrow \sin \theta =\frac{\sqrt{8-2{\lambda }^2+2\lambda \sqrt{{\lambda }^2+8}}}{4}$$$$\Rightarrow A_{max}=\frac{R^2(3\lambda +\sqrt{{\lambda }^2+8})\sqrt{8-2{\lambda }^2+2\lambda \sqrt{{\lambda }^2+8}}}{16}$$$$$$$$ifpointAisonthecircle,\Rightarrow d=R,\lambda =1$$$$A_{max}=\frac{R^2(3+\sqrt{9})\sqrt{8-2+2\sqrt{9}})}{16}=\frac{R^{2}6\sqrt{12}}{16}=\frac{3\sqrt{3}{R}^2}{4}$$$$\Rightarrow areaofinscribedequilateraltriangle$$

Commented by ajfour last updated on 24/Dec/18

$$ThanksSir!$$

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