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Question Number 37804 by ajfour last updated on 17/Jun/18

Commented by ajfour last updated on 17/Jun/18

$$Findmaximumareaof$$$$quadrilateralMNOPinterms$$$$ofradius\mathit{R}ofcircle.$$$$\left(Oisthecenterofcircle\right).$$

Answered by MrW3 last updated on 18/Jun/18

Commented by MrW3 last updated on 18/Jun/18

$$OQ=R\sin \phi ,QD=R\cos \phi$$$$OR=R\sin \theta ,RA=R\cos \theta$$$$\frac{RM}{RA}=\frac{MQ}{QD}$$$$\frac{RM}{RA}=\frac{RQ-RM}{QD}$$$$\frac{RM}{RA}=\frac{OR+OQ-RM}{QD}$$$$\frac{RM}{R\cos \theta }=\frac{R(\sin \theta +\sin \phi )-RM}{R\cos \phi }$$$$(\cos \theta +\cos \phi )RM=\cos \theta (\sin \theta +\sin \phi )R$$$$\Rightarrow RM=\frac{\cos \theta (\sin \theta +\sin \phi )R}{(\cos \theta +\cos \phi )}$$$$OM=OR-RM=R\sin \theta -\frac{\cos \theta (\sin \theta +\sin \phi )R}{(\cos \theta +\cos \phi )}$$$$OM=R\frac{\sin \theta \cos \theta +\sin \theta \cos \phi -\cos \theta \sin \theta -\cos \theta \sin \phi }{(\cos \theta +\cos \phi )}$$$$\Rightarrow OM=\frac{\sin (\theta -\phi )R}{(\cos \theta +\cos \phi )}$$$$$$$$...$$

Commented by ajfour last updated on 18/Jun/18

$$Sir,willitbesimply{A}_{max}=R^2?$$

Answered by MJS last updated on 19/Jun/18

$$A\mathrm{and}B\mathrm{must}\mathrm{be}\mathrm{above}C\mathrm{and}D$$$$A=\left(\begin{array}{c}r\sqrt{1-p^2}\\ rp\end{array}\right)B=\left(\begin{array}{c}-r\sqrt{1-p^2}\\ rp\end{array}\right)$$$$C=\left(\begin{array}{c}-r\sqrt{1-q^2}\\ rq\end{array}\right)D=\left(\begin{array}{c}r\sqrt{1-q^2}\\ rq\end{array}\right)$$$$0⩽p<rp$$$$-p<q⩽p$$$$(\mathrm{think}\mathrm{it}\mathrm{over},{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}\mathrm{to}\mathrm{see}\mathrm{why})$$$$\Rightarrow \mathrm{if}C=B=N\mathrm{and}D=A=P\mathrm{we}\mathrm{get}\mathrm{a}\mathrm{triangle}$$$$\mathrm{with}NMP\mathrm{as}\mathrm{base}\Rightarrow \mathrm{area}\left(ABO\right)\mathrm{must}\mathrm{be}$$$$\mathrm{maximal}.\mathrm{so}\mathrm{the}\mathrm{problem}\mathrm{is}\mathrm{reduced}\mathrm{to}:$$$$\mathrm{find}\mathrm{triangle}\mathrm{of}\mathrm{greatest}\mathrm{area}\mathrm{in}\mathrm{a}\mathrm{half}\mathrm{circle}$$$$\Rightarrow \mathrm{solution}:p=q=\frac{\sqrt{2}}{2},\mathrm{maximal}\mathrm{area}=\frac{r^2}{2}$$

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