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Question Number 107415 by ajfour last updated on 10/Aug/20

Commented by ajfour last updated on 10/Aug/20

$$Findr/R.$$

Answered by mr W last updated on 10/Aug/20

Commented by mr W last updated on 10/Aug/20

$$BA=2R-r$$$$OD=R-2r$$$$\frac{OD}{BC}=\frac{OA}{BA}$$$$\Rightarrow \frac{R-2r}{r}=\frac{R}{2R-r}$$$$(\frac{R}{r}-2)(2-\frac{r}{R})=1$$$$let\lambda =\frac{R}{r}$$$$(\lambda -2)(2\lambda -1)=\lambda$$$$\Rightarrow {\lambda }^2-3\lambda +1=0$$$$\Rightarrow \lambda =\frac{3+\sqrt{5}}{2}\approx 2.618$$

Commented by ajfour last updated on 10/Aug/20

$$VeryverynicesolutionSir,thanks!$$

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