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Question Number 182600 by mnjuly1970 last updated on 11/Dec/22

Answered by mr W last updated on 11/Dec/22

$$\mathit{M}\mathit{e}\mathit{t}\mathit{h}\mathit{o}\mathit{d}\mathit{I}$$$$R=2$$$$r_1=1$$$$r_2=r_3=r=?$$$${(\frac{2}{r}+\frac{1}{1}-\frac{1}{2})}^2=2(\frac{2}{r^2}+\frac{1}{1^2}+\frac{1}{2^2})$$$$\frac{1}{r}-\frac{9}{8}=0$$$$\Rightarrow r=\frac{8}{9}$$

Commented by Acem last updated on 11/Dec/22

$$r=\frac{8}{9}$$

Answered by mr W last updated on 11/Dec/22

Commented by mr W last updated on 11/Dec/22

$$\mathit{M}\mathit{e}\mathit{t}\mathit{h}\mathit{o}\mathit{d}\mathit{I}\mathit{I}$$$$\sqrt{{(1+r)}^2-r^2}-1=\sqrt{{(2-r)}^2-r^2}$$$$3r-1=\sqrt{1+2r}$$$$9r^2-8r=0$$$$\Rightarrow r=\frac{8}{9}$$

Commented by mnjuly1970 last updated on 12/Dec/22

$$thxsirW$$

Answered by Acem last updated on 11/Dec/22

Commented by Acem last updated on 11/Dec/22

$$\bullet {(1+r)}^2=r^2+{(1+x)}^2\Rightarrow r=x(\frac{1}{2}x+1)...i$$$$\bullet {(y+r)}^2=r^2+x^2\Rightarrow y^2=x^2-2ry...ii$$$$\bullet y=2-2r.....iii$$$$iii\wedge ii\wedge i\Rightarrow 3x^2+4x-4=0$$$$x=\frac{2}{3}\Rightarrow r=\frac{8}{9}$$$$$$

Commented by mnjuly1970 last updated on 12/Dec/22

$$verynice.thxsirAcm$$

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