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Question Number 214409 by ajfour last updated on 07/Dec/24

Answered by A5T last updated on 07/Dec/24

Commented by A5T last updated on 07/Dec/24

$$AB=\sqrt{{(r+R)}^2-{(R-r)}^2}=2\sqrt{rR}$$$$\Rightarrow tan2\theta =\frac{2tan\theta }{1-{tan}^2\theta }=\frac{R}{2\sqrt{rR}}=\frac{1}{2}\sqrt{\frac{R}{r}}=p$$$$\Rightarrow {ptan}^2\theta +2tan\theta -p=0$$$$\Rightarrow tan\theta =\frac{-2\underset{-}{+}\sqrt{4+4p^2}}{2p}=\frac{-2\sqrt{r}\underset{-}{+}\sqrt{4r+R}}{\sqrt{R}}$$$$For\theta <90;tan\theta =\frac{\sqrt{4r+R}-2\sqrt{r}}{\sqrt{R}}=\frac{r}{AC}$$$$\Rightarrow AC=\frac{r\sqrt{R}}{\sqrt{4r+R}-2\sqrt{r}}$$$$\frac{R}{r}=\frac{BC=BA+AC}{AC}=\frac{BA}{AC}+1=\frac{2\sqrt{4r^2+Rr}-3r}{r}$$$$\Rightarrow {(R+3r)}^2=4(4r^2+Rr)\Rightarrow R^2+2Rr=7r^2$$$$\stackrel{/r^2}{\Rightarrow }{\left(\frac{R}{r}\right)}^2+2\left(\frac{R}{r}\right)-7=0\Rightarrow \frac{R}{r}=\frac{-2\underset{-}{+}\sqrt{4+28}}{2}$$$$\frac{R}{r}>0\Rightarrow \frac{R}{r}=\frac{4\sqrt{2}-2}{2}=2\sqrt{2}-1$$

Commented by ajfour last updated on 07/Dec/24

$$wowsupergood!$$

Answered by mr W last updated on 07/Dec/24

Commented by mr W last updated on 07/Dec/24

$$let\lambda =\frac{R}{r}$$$$\sin \alpha =\frac{R-r}{R+r}=\frac{\lambda -1}{\lambda +1}$$$$\cos \alpha =\frac{{(R+r)}^2+4Rr+R^2-r^2}{2(R+r)\sqrt{4Rr+R^2}}$$$$=\frac{R^2+3Rr}{(R+r)\sqrt{4Rr+R^2}}=\frac{(\lambda +3)\lambda }{(\lambda +1)\sqrt{{\lambda }^2+4\lambda }}$$$${\left(\frac{\lambda -1}{\lambda +1}\right)}^2+{\left[\frac{(\lambda +3)\lambda }{(\lambda +1)\sqrt{{\lambda }^2+4\lambda }}\right]}^2=1$$$${\left[\frac{(\lambda +3)\lambda }{\sqrt{{\lambda }^2+4\lambda }}\right]}^2={(\lambda +1)}^2-{(\lambda -1)}^2=4\lambda$$$${\lambda }^2+2\lambda -7=0$$$$\Rightarrow \lambda =2\sqrt{2}-1✓$$

Commented by ajfour last updated on 08/Dec/24

$$greatway!Sir.$$

Commented by ajfour last updated on 08/Dec/24

$$\tan \theta =\frac{R-r}{2\sqrt{Rr}}$$$$\tan 2\theta =\frac{R}{2\sqrt{Rr}}$$$$\Rightarrow \frac{\tan 2\theta }{\tan \theta }=\frac{2}{1-\tan {}^2\theta }=\frac{R}{R-r}=\frac{1}{1-p}$$$$1-\tan {}^2\theta =1-\frac{{(R-r)}^2}{4rR}$$$$2=\left(\frac{1}{1-p}\right)\{1-\frac{{(1-p)}^2}{4p}\}$$$$8p(1-p)=4p-{(1-p)}^2$$$$7p^2-2p-1=0$$$$p=\frac{1+2\sqrt{2}}{7}$$$$\frac{1}{p}=\frac{R}{r}=2\sqrt{2}-1$$

Answered by MathematicalUser2357 last updated on 12/Dec/24

$$\mathrm{Oh}\mathrm{this}\mathrm{might}\mathrm{be}\mathrm{it}$$$$$$

Commented by mr W last updated on 12/Dec/24

$$thisappispowerfulforwritting$$$$formulae,butweakfordrawing$$$$diagrams.$$

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