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Question Number 173032 by mnjuly1970 last updated on 05/Jul/22

Answered by mr W last updated on 05/Jul/22

Commented by mr W last updated on 05/Jul/22

$$AC=\sqrt{{(r+R)}^2-R^2}=\sqrt{r(r+2R)}$$$$OC=r+\sqrt{r(r+2R)}$$$${OB}^2=R^2+{(\frac{r}{\sqrt{2}}+\sqrt{{(r+R)}^2-{(R-\frac{r}{\sqrt{2}})}^2})}^2=R^2+{(r+\sqrt{r(r+2R)})}^2$$$$\frac{r}{\sqrt{2}}+\sqrt{(2+\sqrt{2})rR+\frac{r^2}{2}}=r+\sqrt{r(r+2R)}$$$$\sqrt{2(2+\sqrt{2})rR+r^2}=(\sqrt{2}-1)r+\sqrt{2r(r+2R)}$$$$2(2+\sqrt{2})rR+r^2={(\sqrt{2}-1)}^2{r}^2+2r(r+2R)+2(\sqrt{2}-1)r\sqrt{2r(r+2R)}$$$$\sqrt{2}R-(2-\sqrt{2})r=(\sqrt{2}-1)\sqrt{2r(r+2R)}$$$$R=2(2-\sqrt{2})r$$$$\frac{R}{r}=2(2-\sqrt{2})\approx 1.172$$

Commented by Tawa11 last updated on 06/Jul/22

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

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