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Question Number 165600 by mr W last updated on 05/Feb/22

Commented by mr W last updated on 05/Feb/22

$$attempttoQ165565$$

Commented by mr W last updated on 05/Feb/22

Commented by mr W last updated on 06/Feb/22

$$let\mu =\frac{b}{a}⩽1$$$$A(\frac{a}{\tan \theta },a)=(ma,a)$$$$B(b,y_B)$$$${(y_B-a)}^2+{(\frac{a}{\tan \theta }-b)}^2={(a+b)}^2$$$${(y_B-a)}^2=a^2(1+\frac{1}{\tan \theta })(2\mu +1-\frac{1}{\tan \theta })$$$$letm=\frac{1}{\tan \theta }$$$$y_B=a+a\sqrt{(1+m)(2\mu +1-m)}$$$$\tan \theta ⩾\frac{a}{a+2b}=\frac{1}{2\mu +1}$$$$\theta ⩽\frac{\pi }{4}$$$$\Rightarrow 1⩽m⩽2\mu +1$$$$$$$$x_P=\frac{a}{\tan \theta }+a\sin \alpha =a(m+\sin \alpha )$$$$y_P=a-a\cos \alpha =a(1-\cos \alpha )$$$$x_Q=b-b\cos \beta =a\mu (1-\cos \beta )$$$$y_Q=y_B+b\sin \beta =a[1+\sqrt{(1+m)(2\mu +1-m)}+\mu \sin \beta ]$$$$$$$$areaoftriangleOPQ:$$$$\mathrm{\Delta }=\frac{x_Q{y}_Q}{2}+\frac{(x_P-x_Q)(y_P+y_Q)}{2}-\frac{x_P{y}_P}{2}$$$$2\mathrm{\Delta }=x_P{y}_Q-x_Q{y}_P$$$$2\mathrm{\Delta }=a^2(m+\sin \alpha )(1+\sqrt{(1+m)(2\mu +1-m)}+\mu \sin \beta )-a^2\mu (1-\cos \beta )(1-\cos \alpha )$$$$\frac{2\mathrm{\Delta }}{\mu a^2}=(m+\sin \alpha )(\frac{1+\sqrt{(1+m)(2\mu +1-m)}}{\mu }+\sin \beta )-(1-\cos \beta )(1-\cos \alpha )$$$$let\mathrm{\Phi }=\frac{2\mathrm{\Delta }}{\mu a^2},\lambda =\frac{1+\sqrt{(1+m)(2\mu +1-m)}}{\mu }$$$$\mathrm{\Phi }=(m+\sin \alpha )(\lambda +\sin \beta )-(1-\cos \beta )(1-\cos \alpha )$$$$\mathrm{\Phi }=m\lambda -1+\lambda \sin \alpha +\cos \alpha +m\sin \beta +\cos \beta -\cos (\alpha +\beta )$$$$\mathrm{\Phi }=m\lambda -1+\sqrt{1+{\lambda }^2}\cos (\alpha -{\tan }^{-1}\lambda )+\sqrt{1+m^2}\cos (\beta -{\tan }^{-1}m)-\cos (\alpha +\beta )$$$$\mathrm{\Phi }isafunctionofthreevariables:$$$$m,\alpha ,\beta .$$$$$$$$i^{\prime }lltryingeometricway:$$$$formaiximumareaoftriangleOPQ,$$$$tangentatQmustbe//OP:$$$$\frac{1}{\tan \beta }=\frac{y_P}{x_P}=\frac{1-\cos \alpha }{m+\sin \alpha }$$$$\Rightarrow \frac{\cos \beta }{\sin \beta }=\frac{1-\cos \alpha }{m+\sin \alpha }$$$$m\cos \beta -\sin \beta +\sin \alpha \cos \beta +\cos \alpha \sin \beta =0$$$$m\cos \beta -\sin \beta +\sin (\alpha +\beta )=0$$$$\Rightarrow \sqrt{1+m^2}\sin (\beta -{\tan }^{-1}m)=\sin (\alpha +\beta )$$$$(wecangetthesamealsofrom\frac{\partial \mathrm{\Phi }}{\partial \beta }=0)$$$$$$$$tangentatPmustbe//OQ:$$$$\tan \alpha =\frac{y_Q}{x_Q}=\frac{1+\sqrt{(1+m)(2\mu +1-m)}+\mu \sin \beta }{\mu (1-\cos \beta )}$$$$\Rightarrow \frac{\sin \alpha }{\cos \alpha }=\frac{\lambda +\sin \beta }{1-\cos \beta }$$$$\lambda \cos \alpha -\sin \alpha +\cos \alpha \sin \beta +\sin \alpha \cos \beta =0$$$$\lambda \cos \alpha -\sin \alpha +\sin (\alpha +\beta )=0$$$$\Rightarrow \sqrt{1+{\lambda }^2}\sin (\alpha -{\tan }^{-1}\lambda )=\sin (\alpha +\beta )$$$$(wecangetthesamealsofrom\frac{\partial \mathrm{\Phi }}{\partial \alpha }=0)$$$$$$$$wecanseethat\frac{\pi }{2}<\alpha +\beta <\pi incaseof$$$$that\mathrm{\Delta }ismaximum.$$$$$$$$say\sqrt{1+{\lambda }^2}\sin (\alpha -{\tan }^{-1}\lambda )=\sqrt{1+m^2}\sin (\beta -{\tan }^{-1}m)=\sin (\alpha +\beta )=\xi$$$$\Rightarrow \alpha ={\tan }^{-1}\lambda +{\sin }^{-1}\frac{\xi }{\sqrt{1+{\lambda }^2}}$$$$\Rightarrow \beta ={\tan }^{-1}m+{\sin }^{-1}\frac{\xi }{\sqrt{1+m^2}}$$$$\Rightarrow \alpha +\beta =\pi -{\sin }^{-1}\xi$$$$\overline{)\begin{array}{c}{\sin }^{-1}\xi +{\sin }^{-1}\frac{\xi }{\sqrt{1+{\lambda }^2}}+{\sin }^{-1}\frac{\xi }{\sqrt{1+m^2}}=\pi -{\tan }^{-1}\lambda -{\tan }^{-1}m\end{array}}...\left(I\right)$$$$$$$$\sqrt{1+{\lambda }^2}\sin (\alpha -{\tan }^{-1}\lambda )=\xi$$$$\Rightarrow \cos (\alpha -{\tan }^{-1}\lambda )=\sqrt{1-\frac{{\xi }^2}{1+{\lambda }^2}}$$$$\sqrt{1+m^2}\sin (\beta -{\tan }^{-1}m)=\xi$$$$\Rightarrow \cos (\beta -{\tan }^{-1}m)=\sqrt{1-\frac{{\xi }^2}{1+m^2}}$$$$\sin (\alpha +\beta )=\xi$$$$\Rightarrow \cos (\alpha +\beta )=-\sqrt{1-{\xi }^2}$$$$\mathrm{\Phi }=m\lambda -1+\sqrt{1+{\lambda }^2}\cos (\alpha -{\tan }^{-1}\lambda )+\sqrt{1+m^2}\cos (\beta -{\tan }^{-1}m)-\cos (\alpha +\beta )$$$$\overline{)\begin{array}{c}\mathrm{\Phi }=m\lambda -1+\sqrt{1+{\lambda }^2-{\xi }^2}+\sqrt{1+m^2-{\xi }^2}+\sqrt{1-{\xi }^2}\end{array}}...\left(II\right)$$

Commented by mr W last updated on 05/Feb/22

Commented by mr W last updated on 05/Feb/22

Commented by mr W last updated on 05/Feb/22

Commented by mr W last updated on 05/Feb/22

$$thepositionsofthecirclesaredefined$$$$bytheparameterm=\frac{1}{\tan \theta }with$$$$1⩽m⩽2\mu +1.$$$$foragivenmwecandeterminethe$$$$parameter\xi whichdefinesthe$$$$positionsofpointsPandQonthe$$$$circles,i.e.\alpha and\beta ,suchthat{\mathrm{\Delta }}_{OPQ}is$$$$maximumrelativelytothegivenm.$$$$butsincetheequationfor\xi cannotbe$$$$solvedanalytically,ihavenotfound$$$$awaytodeterminethemforthe$$$$absolutemaximum{\mathrm{\Delta }}_{OPQ}.$$

Commented by aleks041103 last updated on 05/Feb/22

$$Inaverycutevectorialwayyoucan$$$$provethatformaxarea$$$$OP\mathrm{\perp }QBandOQ\mathrm{\perp }PA$$$$Itispossibletoprovethatformaxarea$$$$\frac{x_P}{y_Q}=\frac{y_B-y_A}{x_A-x_B}$$$$later{i}^{\prime }llpostaproof!Ihopethathelps!$$

Commented by mr W last updated on 05/Feb/22

$${that}^{\prime }srightsir.thanks!$$$$thisisalsothatwhatmydiagram$$$$atthebeginningshows:$$$$tangentatQ//AP,i.e.BQ\mathrm{\perp }OP$$$$tangentatP//AQ,i.e.AP\mathrm{\perp }OQ$$

Commented by Tawa11 last updated on 05/Feb/22

$$\mathrm{Great}\mathrm{sir}$$

Commented by mr W last updated on 06/Feb/22

$$Nowihavefoundawaytofind$$$$theabsolutemaximumareaofthe$$$$triangleandthecorresponding$$$$parameterm.$$$$Therelationshipbetweentheparamters$$$$\xi andmistheequationI:$$$$\overline{)\begin{array}{c}{\sin }^{-1}\xi +{\sin }^{-1}\frac{\xi }{\sqrt{1+{\lambda }^2}}+{\sin }^{-1}\frac{\xi }{\sqrt{1+m^2}}=\pi -{\tan }^{-1}\lambda -{\tan }^{-1}m\end{array}}$$$$theareaofthetriangleisexpressed$$$$bytheequationII:$$$$\overline{)\begin{array}{c}\mathrm{\Phi }=m\lambda -1+\sqrt{1+{\lambda }^2-{\xi }^2}+\sqrt{1+m^2-{\xi }^2}+\sqrt{1-{\xi }^2}\end{array}}$$$$with\mathrm{\Phi }=\frac{2\mathrm{\Delta }}{\mu a^2}.$$$$themaximum\mathrm{\Phi }iswhenthecurves$$$$ofbothequationsjusttoucheach$$$$other,asfollowingdiagramshows.$$$$$$$$\underset{―}{example:}$$$$\mu =\frac{b}{a}=0.75$$$${\mathrm{\Phi }}_{max}=11.125757or$$$${\mathrm{\Delta }}_{max}=\frac{\mu a^2{\mathrm{\Phi }}_{max}}{2}=4.172a^2$$$$atm=1.9858or\theta ={\tan }^{-1}\left(\frac{1}{m}\right)$$

Commented by mr W last updated on 06/Feb/22

Commented by mr W last updated on 06/Feb/22

Commented by mr W last updated on 06/Feb/22

Commented by mr W last updated on 06/Feb/22

Commented by mr W last updated on 06/Feb/22

Commented by mr W last updated on 06/Feb/22

Commented by mr W last updated on 06/Feb/22

Commented by mr W last updated on 06/Feb/22

$$furtherexamples:$$$$\mu =\frac{b}{a}=0.4$$$${\mathrm{\Phi }}_{max}=13.51418or$$$${\mathrm{\Delta }}_{max}=\frac{\mu a^2{\mathrm{\Phi }}_{max}}{2}=2.703a^2$$$$atm=1.3871$$

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