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Question Number 177832 by HeferH last updated on 09/Oct/22

Answered by mr W last updated on 10/Oct/22

Commented by mr W last updated on 10/Oct/22

$$r_2=\frac{R}{4}$$$${(r_1+r_2)}^2-{(\frac{3R}{4}-r_1)}^2={(R-r_1)}^2-r_1^2$$$$\Rightarrow r_1=\frac{3R}{8}$$$${(-\frac{1}{R}+\frac{8}{3R}+\frac{4}{R}+\frac{1}{r})}^2=2(\frac{1}{R^2}+\frac{64}{9R^2}+\frac{16}{R^2}+\frac{1}{r^2})$$$${(\frac{17}{3R}+\frac{1}{r})}^2=2(\frac{217}{9R^2}+\frac{1}{r^2})$$$$\frac{9}{r^2}-\frac{102}{Rr}+\frac{145}{R^2}=0$$$$\frac{R}{r}=\frac{1}{9}(51+\sqrt{{51}^2-9\times 145})=\frac{87}{9}=\frac{29}{3}$$$$\Rightarrow \frac{r}{R}=\frac{3}{29}✓$$

Commented by Tawa11 last updated on 10/Oct/22

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

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