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Question Number 124743 by ajfour last updated on 05/Dec/20

Commented by ajfour last updated on 05/Dec/20

$$GivenarclengthAB=1,find$$$$maximumradiusofsmaller$$$$circles.$$

Answered by mr W last updated on 05/Dec/20

Commented by mr W last updated on 05/Dec/20

$$2\theta R=1$$$$\theta =\frac{1}{2R}$$$${(R-r)}^2=r^2+{(r+R\cos \theta )}^2$$$$r^2+2R(1+\cos \theta )r-R^2(1-{\cos }^2\theta )=0$$$$r=R[\sqrt{2(1+\cos \theta )}-(1+\cos \theta )]$$$$r=\frac{1}{2\theta }[\sqrt{2(1+\cos \theta )}-(1+\cos \theta )]$$$$at\theta =\frac{\pi }{2}:$$$$r_{max}=\frac{\sqrt{2}-1}{\pi }$$

Commented by ajfour last updated on 05/Dec/20

sir how did you confirm, i tried grapher, it draws weird graph...

Commented by mr W last updated on 06/Dec/20

$$wegotthesamefunctionforr$$$$intermsof\theta :$$$$r=\frac{\sqrt{2(1+\cos \theta )}-(1+\cos \theta )}{2\theta }$$$$for0<\theta ⩽\pi /2thefunctionis$$$$increasing,therefore{r}_{max}isat\theta =\frac{\pi }{2}.$$$$becausesegmentisdefinedfor$$$$0<\theta ⩽\pi /2.$$

Commented by mr W last updated on 05/Dec/20

Commented by ajfour last updated on 05/Dec/20

$$\frac{\pi }{2}\approx 1.57079$$

Commented by mr W last updated on 06/Dec/20

$${that}^{\prime }swhatimeant.wecannotget$$$$themathematicalmaximumat$$$$\theta =1.645,becausethatisbehind$$$$\theta =\frac{\pi }{2},thepossiblesegment.$$$$theactualmaximumwecanreach$$$$isat\theta =\frac{\pi }{2}:r_{max}=\frac{\sqrt{2}-1}{\pi }whichis$$$$lessthan0.1322.$$

Commented by ajfour last updated on 06/Dec/20

$$Iguess,ifollownow,Sir!$$$$youmakeitquiteclear,\mathcal{T}hanks.$$

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