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Question Number 141291 by mr W last updated on 17/May/21

Commented by mr W last updated on 17/May/21

$$OB=\sqrt{{(R-r)}^2-r^2}=\sqrt{R^2-2Rr}$$$$AB=R+\sqrt{R^2-2Rr}=\frac{r}{\tan 22.5°}$$$$BC=R-\sqrt{R^2-2Rr}=1$$$$R-1=\sqrt{R^2-2Rr}$$$$R^2-2R+1=R^2-2Rr$$$$1=2(1-r)R$$$$2R=\frac{1}{1-r}$$$$2R=1+\frac{r}{\tan 22.5°}$$$$\frac{1}{1-r}=1+\frac{r}{\tan 22.5°}$$$$\frac{r}{\tan 22.5°}=\frac{1}{\tan 22.5°}-1$$$$AE=AB=\frac{1}{\tan 22.5°}-1=1+\sqrt{2}-1=\sqrt{2}$$

Commented by mnjuly1970 last updated on 17/May/21

$$gratefulmrW...thanksalot$$

Commented by BHOOPENDRA last updated on 17/May/21

$$greatsir$$

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