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Question Number 48560 by behi83417@gmail.com last updated on 25/Nov/18

Commented by behi83417@gmail.com last updated on 25/Nov/18

$$\mathbf{out}\mathbf{circle}\mathbf{radi}=1$$$$\mathbf{a},\mathbf{b}=\mathbf{distance}\mathbf{of}\mathbf{midpoint}\mathbf{of}:\mathbf{AB},\mathbf{BC}\mathbf{from}$$$$\mathbf{out}\mathbf{circle}.$$$$\mathbf{find}:\mathbf{r},\mathbf{in}\mathbf{terms}\mathbf{of}:\mathbf{a},\mathbf{b}.$$

Answered by mr W last updated on 25/Nov/18

$$let$$$$R=radiusofbigcircle=1$$$$2c=AB$$$$2d=BC$$$$R^2={(R-a)}^2+c^2$$$$\Rightarrow c=\sqrt{a(2R-a)}$$$$R^2={(R-b)}^2+d^2$$$$\Rightarrow d=\sqrt{b(2R-b)}$$$${\mathrm{\Delta }}_{ABC}=\frac{1}{2}(2R+2c+2d)r=\frac{2c\times 2d}{2}$$$$\Rightarrow r=\frac{2cd}{R+c+d}=\frac{2\sqrt{ab(2R-a)(2R-b)}}{R+\sqrt{a(2R-a)}+\sqrt{b(2R-b)}}$$$$\Rightarrow r=\frac{2\sqrt{ab(2-a)(2-b)}}{1+\sqrt{a(2-a)}+\sqrt{b(2-b)}}$$

Commented by behi83417@gmail.com last updated on 25/Nov/18

$$dearmaster!ithink:$$$$2c=BC,2d=AB,youaresay.amiright?$$$$ofcoursethisisnotimportantinfinal$$$$resault.$$

Commented by mr W last updated on 25/Nov/18

$$yes.Ilet2c=BCand2d=AB,since$$$$2R=AC,butthis{doesn}^{\prime }tmeanmuch.$$

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