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Question Number 201707 by ajfour last updated on 10/Dec/23

Answered by mr W last updated on 11/Dec/23

Commented by mr W last updated on 11/Dec/23

$$let\frac{r}{R}=\lambda$$$$\cos 2\alpha =\frac{R-2r}{R}=1-2\lambda$$$$\tan \alpha =\frac{\sqrt{\lambda }}{\sqrt{1-\lambda }}$$$$ED=\frac{r}{\tan \alpha }=\frac{r\sqrt{1-\lambda }}{\sqrt{\lambda }}$$$$AD=2AB=2\sqrt{R^2-{(R-2r)}^2}=4\sqrt{(R-r)r}$$$$\cos 2\alpha =\frac{ED}{AD}=\frac{r}{4\tan \alpha \sqrt{(R-r)r}}=\frac{\lambda \sqrt{1-\lambda }}{4\sqrt{\lambda }\sqrt{(1-\lambda )\lambda }}=\frac{1}{4}$$$$1-2\lambda =\frac{1}{4}$$$$\Rightarrow \lambda =\frac{3}{8}$$

Commented by ajfour last updated on 10/Dec/23

$$letR=1$$$$AB=\sqrt{1-{(1-2r)}^2}=2\sqrt{r(1-r)}$$$$\frac{p}{1}=\frac{r}{1-2r}\Rightarrow p+r=\frac{2r(1-r)}{1-2r}$$$$\frac{2AB-s}{AB}=\frac{r}{1-2r}$$$$2-\frac{s}{2\sqrt{r(1-r)}}=\frac{r}{1-2r}......\left(i\right)$$$$s^2=(p+r)(2-p-r)$$$$2-(p+r)=2-\frac{2r(1-r)}{1-2r}=\frac{2-6r+2r^2}{1-2r}$$$$s^2=\frac{2r(1-r)}{(1-2r)}\times \frac{2(1-3r+r^2)}{(1-2r)}$$$$butfrom\left(i\right)$$$$\frac{s^2}{4r(1-r)}={(2-\frac{r}{1-2r})}^2$$$$\Rightarrow (1-3r+r^2)={(2-5r)}^2$$$$24r^2-17r+3=0$$$$r=\frac{17\pm \sqrt{289-4\times 3\times 24}}{48}$$$$r=\frac{17\pm 1}{48}\Rightarrow r=\frac{1}{3}orr=\frac{3}{8}$$$$$$

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