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Question Number 58986 by ajfour last updated on 02/May/19

Commented by mr W last updated on 02/May/19

$$doyoumeanmaximumarea?$$$$ithinkthereisnomaximumor$$$$minimumperimeter,since$$$$max.p\to 4r,whenC\to MandB\to N.$$$$min.p\to 2(r-a),whenC\to B\to N.$$$$$$$$max.Area=\frac{r(r+a)}{2}$$

Commented by ajfour last updated on 02/May/19

$$\mathrm{I}\mathrm{actually}\mathrm{was}\mathrm{curious}\mathrm{about}$$$$\max \mathrm{perimeter}\mathrm{itself},4\mathrm{r}\mathrm{dont}\mathrm{seem}$$$$\mathrm{obvious}\mathrm{to}\mathrm{me}\mathrm{Sir}.$$

Commented by mr W last updated on 02/May/19

$$AC⩽AM$$$$AB+BC⩽AN+NM$$$$\Rightarrow p=AC+AB+BC⩽AM+AN+NM=4r$$

Answered by mr W last updated on 02/May/19

Commented by mr W last updated on 02/May/19

$$AB=\sqrt{a^2+r^2-2ar\cos \alpha }$$$$AC=\sqrt{a^2+r^2+2ar\cos \beta }$$$$BC=\sqrt{2r^2+2r^2\cos (\alpha +\beta )}=r\sqrt{2[1+\cos (\alpha +\beta )]}$$$$p=\sqrt{a^2+r^2-2ar\cos \alpha }+\sqrt{a^2+r^2+2ar\cos \beta }+r\sqrt{2[1+\cos (\alpha +\beta )]}$$$$\frac{\partial p}{\partial \alpha }=\frac{ar\sin \alpha }{\sqrt{a^2+r^2-2ar\cos \alpha }}-r\frac{\sin (\alpha +\beta )}{\sqrt{2[1+\cos (\alpha +\beta )]}}=0$$$$\Rightarrow \frac{\sin \alpha }{\sqrt{1+{\left(\frac{r}{a}\right)}^2-2\left(\frac{r}{a}\right)\cos \alpha }}=\frac{\sin (\alpha +\beta )}{\sqrt{2[1+\cos (\alpha +\beta )]}}$$$$\Rightarrow \frac{\sin \alpha }{\sqrt{1+{\lambda }^2-2\lambda \cos \alpha }}=\frac{\sin (\alpha +\beta )}{\sqrt{2[1+\cos (\alpha +\beta )]}}$$$$similarly$$$$\Rightarrow \frac{-\sin \beta }{\sqrt{1+{\lambda }^2+2\lambda \cos \beta }}=\frac{\sin (\alpha +\beta )}{\sqrt{2[1+\cos (\alpha +\beta )]}}$$$$$$$$\Rightarrow \frac{\sin \alpha }{\sqrt{1+{\lambda }^2-2\lambda \cos \alpha }}=\frac{-\sin \beta }{\sqrt{1+{\lambda }^2+2\lambda \cos \beta }}$$$$with\mu =\frac{1+{\lambda }^2}{2\lambda }$$$$\Rightarrow \frac{\sin \alpha }{\sqrt{\mu -\cos \alpha }}=\frac{-\sin \beta }{\sqrt{\mu +\cos \beta }}...\left(i\right)$$$$$$$$\alpha =\beta =0\left(180°\right)fulfilltheequations.$$$$thisistheonlysolution.if\alpha \ne 0,\beta \ne 0,$$$$from\left(i\right)weget\sin \alpha or\sin \beta mustbe$$$$negative,butthisisnotpossible,$$$$since0⩽\alpha ,\beta ⩽180°,\sin \alpha and\sin \beta$$$$arealwayspositive.$$$$\Rightarrow min.ormax.piswhenB,Care$$$$onthediameterofthecircle.$$

Commented by ajfour last updated on 03/May/19

$$Thank\mathrm{you}\mathrm{sir}.(\mathrm{you}\mathrm{knew}\mathrm{it}\mathrm{before}$$$$\mathrm{proving},\mathrm{and}\mathrm{you}\mathrm{proved}\mathrm{it}\mathrm{too}).$$

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