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Question Number 88708 by ajfour last updated on 12/Apr/20

Commented by ajfour last updated on 12/Apr/20

$$IfAE=BF,findr/R.$$

Commented by mr W last updated on 12/Apr/20

$$AE=FB=k$$$${(2R-k)}^2+r^2={(k+r)}^2$$$$2R^2=(2R+r)k$$$$\Rightarrow k=\frac{2R^2}{2R+r}$$$$\frac{r}{k+r}=\frac{\sqrt{{\left(2R\right)}^2-{(2r+k)}^2}}{2R}=\sqrt{1-{(\frac{r}{R}+\frac{k}{2R})}^2}$$$${\left(\frac{1}{\frac{k}{r}+1}\right)}^2=1-{(\frac{r}{R}+\frac{k}{2R})}^2$$$$with\lambda =\frac{r}{R}$$$${\left(\frac{1}{\frac{2}{(2+\lambda )\lambda }+1}\right)}^2=1-{(\lambda +\frac{1}{2+\lambda })}^2$$$${\lambda }^2{(2+\lambda )}^4={({\lambda }^2+2\lambda +2)}^2[{(\lambda +2)}^2-{(\lambda +1)}^4]$$$${(\lambda +1)}^2({\lambda }^3+3{\lambda }^2+2\lambda -2)({\lambda }^3+3{\lambda }^2+6\lambda +6)=0$$$$\Rightarrow \lambda =0.52138$$

Commented by john santu last updated on 12/Apr/20

Commented by ajfour last updated on 12/Apr/20

$$ThanksforconfirmationSir.$$$$Nicesolution.$$

Commented by john santu last updated on 12/Apr/20

$$letmetry$$$$letAE=BF=x$$$$\sin \theta =\sin \theta$$$$\frac{r}{r+x}=\frac{BD}{2R}\left(i\right)$$$$BD=\sqrt{4R^2-{(2r+x)}^2}$$$$r+x=\sqrt{{(2R-x)}^2-r^2}$$$$\Rightarrow \frac{r}{\sqrt{{(2R-x)}^2-r^2}}=\frac{\sqrt{4R^2-{(2r+x)}^2}}{2R}$$$$\frac{r^2}{{(2R-x)}^2-r^2}=\frac{4R^2-{(2r+x)}^2}{4R^2}$$$$$$

Answered by mahdi last updated on 12/Apr/20

$$2\mathrm{R}=\mathrm{AF}+\mathrm{FB}=\sqrt{{\mathrm{AC}}^2-{\mathrm{CF}}^2}+\mathrm{FB}$$$$=\sqrt{{(\mathrm{AC}+\mathrm{EC})}^2-{\mathrm{CF}}^2}+\mathrm{FB}\Rightarrow$$$$(\mathrm{AE}=\mathrm{FB}=\mathrm{a},\mathrm{EC}=\mathrm{CF}=\mathrm{r})$$$$2\mathrm{R}=\sqrt{{(\mathrm{r}+\mathrm{a})}^2-{\mathrm{r}}^2}+\mathrm{a}\Rightarrow 2\mathrm{R}-\mathrm{a}=\sqrt{{\mathrm{a}}^2+2\mathrm{ar}}$$$$\Rightarrow {4\mathrm{R}}^2+{\mathrm{a}}^2-4\mathrm{Ra}={\mathrm{a}}^2+2\mathrm{ar}\Rightarrow$$$${2\mathrm{R}}^2-2\mathrm{Ra}=\mathrm{ar}\Rightarrow \mathrm{a}=\frac{{2\mathrm{R}}^2}{2\mathrm{R}+\mathrm{r}}\left(\mathrm{i}\right)$$$$\mathrm{A}\stackrel{\mathrm{\Delta }}{\mathrm{O}D}:{\mathrm{OD}}^2={\mathrm{AO}}^2+{\mathrm{AD}}^2-2.\mathrm{AD}.\mathrm{AO}.\cos \hat{\mathrm{A}}$$$$\Rightarrow (\cos \hat{\mathrm{A}}=\frac{\mathrm{AF}}{\mathrm{AC}}=\frac{2\mathrm{R}-\mathrm{a}}{\mathrm{a}+\mathrm{r}})\Rightarrow$$$${\mathrm{R}}^2={\mathrm{R}}^2+{(\mathrm{a}+2\mathrm{r})}^2-2.\mathrm{R}.(\mathrm{a}+2\mathrm{r}).\frac{2\mathrm{R}-\mathrm{a}}{\mathrm{a}+\mathrm{r}}$$$$\Rightarrow (\mathrm{a}+\mathrm{r})(\mathrm{a}+2\mathrm{r})=2\mathrm{R}(2\mathrm{R}-\mathrm{a})\left(\mathrm{ii}\right)$$$$\Rightarrow \mathrm{paste}\mathrm{a}=\frac{{2\mathrm{R}}^2}{2\mathrm{R}+\mathrm{r}}\mathrm{in}\left(\mathrm{ii}\right)\Rightarrow$$$$$$$$\frac{{2\mathrm{R}}^2+{\mathrm{r}}^2+2\mathrm{Rr}}{2\mathrm{R}+\mathrm{r}}\times \frac{{2\mathrm{R}}^2+{2\mathrm{r}}^2+4\mathrm{Rr}}{2\mathrm{R}+\mathrm{r}}=2\mathrm{R}\frac{{2\mathrm{R}}^2+2\mathrm{Rr}}{2\mathrm{R}+\mathrm{r}}$$$$({2\mathrm{R}}^2+{\mathrm{r}}^2+2\mathrm{Rr})(\mathrm{R}+\mathrm{r})={2\mathrm{R}}^2(2\mathrm{R}+\mathrm{r})$$$$\Rightarrow \mathrm{get}\mathrm{c}=\frac{\mathrm{r}}{\mathrm{R}}\Rightarrow$$$$(2+{\mathrm{c}}^2+2\mathrm{c})(1+\mathrm{c})=2(2+\mathrm{c})\Rightarrow$$$${(1+\mathrm{c})}^3-(1+\mathrm{c})-2=0\Rightarrow 1+\mathrm{c}\backsimeq 1.5213$$$$\mathrm{c}=\frac{\mathrm{r}}{\mathrm{R}}\backsimeq .5213$$

Answered by ajfour last updated on 12/Apr/20

$$let\mathrm{\angle }DAB=\theta$$$$AE=BF=p=BD$$$$2R\cos \theta =p+2r...\left(i\right)$$$$2R\sin \theta =p...\left(ii\right)$$$$\cos \theta =\frac{2R-p}{p+r}...\left(iii\right)$$$$\sin \theta =\frac{r}{p+r}...\left(iv\right)$$$$let\frac{p}{R}=\mu ,\frac{r}{R}=\lambda$$$$\Rightarrow 2\cos \theta =\mu +2\lambda ...\left(I\right)$$$$2\sin \theta =\mu ...\left(II\right)$$$$2\cos \theta =\frac{4-2\mu }{\mu +\lambda }...\left(III\right)$$$$2\sin \theta =\frac{2\lambda }{\mu +\lambda }...\left(IV\right)$$$$\Rightarrow \mu +2\lambda =\frac{4-2\mu }{\mu +\lambda }\mathrm{\&}$$$$\mu (\mu +\lambda )=2\lambda$$$$\Rightarrow \lambda =\frac{{\mu }^2}{2-\mu },hence$$$${\mu }^2+\frac{3{\mu }^3}{2-\mu }+\frac{2{\mu }^4}{{(2-\mu )}^2}+2\mu -4=0$$$$2{\mu }^4+3{\mu }^3(2-\mu )+{(2-\mu )}^2({\mu }^2+2\mu -4)=0$$$$$$$$\Rightarrow 2{\mu }^4+6{\mu }^3-3{\mu }^4+{\mu }^4+2{\mu }^3-4{\mu }^2$$$$-4{\mu }^3-8{\mu }^2+16\mu +4{\mu }^2+8\mu -16=0$$$$$$$$\Rightarrow 4{\mu }^3-8{\mu }^2+24\mu -16=0$$$$or{\mu }^3-2{\mu }^2+6\mu -4=0$$$$\Rightarrow \mu \approx 0.79321651$$$$\lambda =\frac{r}{R}=\frac{{\mu }^2}{2-\mu }\approx 0.52138.$$$$$$

Commented by mr W last updated on 12/Apr/20

$$nicelysolvedsir!igotsameresult.$$

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