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Question Number 198479 by ajfour last updated on 20/Oct/23

Commented by ajfour last updated on 20/Oct/23

$$Findrif\alpha beknown.Outercircle$$$$hasunitradius.$$

Commented by dangduomg last updated on 21/Oct/23

$$doestheinnercirclehavwtobetangentialtooutercircle?$$

Answered by mr W last updated on 21/Oct/23

$${(\frac{r}{\tan \alpha }-1)}^2+r^2={(1-r)}^2$$$$\Rightarrow r=2(1-\tan \alpha )\tan \alpha$$

Commented by mr W last updated on 21/Oct/23

Commented by ajfour last updated on 21/Oct/23

yes sir, short correct and beautiful solution.

Answered by a.lgnaoui last updated on 21/Oct/23

$$△\mathrm{MIF}\sin \mathit{\alpha }=\frac{\mathbf{MI}}{\mathrm{I}\mathbf{F}}=\frac{\mathbf{BC}}{\mathbf{BF}}\left(\mathbf{i}\right)$$$$△\mathrm{OMI}\measuredangle \mathbf{MOI}=\mathit{\theta }\sin \mathit{\theta }=\frac{\mathbf{MI}}{\mathbf{OI}}=\frac{\mathbf{r}}{1-\mathbf{r}}$$$$\Rightarrow \sin \mathit{\theta }=\frac{\mathbf{r}}{1-\mathbf{r}}\cos \mathit{\theta }=\frac{\sqrt{1-2\mathbf{r}}}{1-\mathbf{r}}$$$$\Rightarrow \mathbf{MF}=1+\sqrt{1-2\mathbf{r}}$$$$$$$$\mathbf{BC}=1\times \sin 2\mathit{\alpha }\mathbf{CF}=1+\cos 2\mathit{\alpha }$$$$\tan \mathit{\alpha }=\frac{\mathbf{BC}}{\mathbf{CF}}=\frac{\mathbf{MI}}{\mathbf{IF}}=\frac{\mathbf{r}}{1+\sqrt{1-2\mathbf{r}}}$$$$\mathrm{posons}\mathbf{t}=\tan \mathit{\alpha }$$$$\mathbf{t}(1+\sqrt{1-2\mathbf{r}})=\mathbf{r}$$$$\mathbf{t}\sqrt{1-2\mathbf{r}}=\mathbf{r}-\mathbf{t}$$$${\mathbf{t}}^2(1-2\mathbf{r})={\mathbf{r}}^2+{\mathbf{t}}^2-2\mathbf{rt}$$$$$$$$\mathbf{r}=2\mathbf{t}(1-\mathbf{t})$$$$$$

Commented by a.lgnaoui last updated on 21/Oct/23

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