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Question Number 160719 by mnjuly1970 last updated on 05/Dec/21

Commented by kowalsky78 last updated on 05/Dec/21

$$Igot\frac{3}{5}.MaybeImadesomemistake.$$

Commented by mr W last updated on 05/Dec/21

$$\frac{3}{5}isthecorrectanswer.$$

Answered by mr W last updated on 05/Dec/21

Commented by mr W last updated on 05/Dec/21

$$AB=AC=BC=x+y$$$${AD}^2=x^2+{(x+y)}^2-2x(x+y)\cos \frac{\pi }{3}$$$${AD}^2=x^2+y^2+xy$$$$\Rightarrow AD=\sqrt{x^2+y^2+xy}$$$$$$$$AD=AG+GD=AF+ED$$$$AD=(x+y-\sqrt{3}{r}_2)+(y-\sqrt{3}{r}_2)$$$$AD=x+2y-2\sqrt{3}{r}_2$$$$\Rightarrow r_2=\frac{x+2y-AD}{2\sqrt{3}}$$$$\Rightarrow r_2=\frac{x+2y-\sqrt{x^2+y^2+xy}}{2\sqrt{3}}$$$$similarly$$$$\Rightarrow r_1=\frac{2x+y-\sqrt{x^2+y^2+xy}}{2\sqrt{3}}$$$$\frac{r_1}{r_2}=\frac{2x+y-\sqrt{x^2+y^2+xy}}{x+2y-\sqrt{x^2+y^2+xy}}$$$$let\lambda =\frac{x}{y},\mu =\frac{r_1}{r_2}⩽1$$$$\Rightarrow \mu =\frac{2\lambda +1-\sqrt{{\lambda }^2+\lambda +1}}{\lambda +2-\sqrt{{\lambda }^2+\lambda +1}}$$$$(2-\mu )\lambda +1-2\mu =(1-\mu )\sqrt{{\lambda }^2+\lambda +1}$$$${(2-\mu )}^2{\lambda }^2+{(1-2\mu )}^2+2(2-\mu )(1-2\mu )\lambda ={(1-\mu )}^2({\lambda }^2+\lambda +1)$$$$(3-2\mu ){\lambda }^2+(3{\mu }^2-8\mu +3)\lambda +(3\mu -2)\mu =0$$$$\lambda =\frac{-(3{\mu }^2-8\mu +3)+\sqrt{{(3{\mu }^2-8\mu +3)}^2-4(3-2\mu )(3\mu -2)\mu }}{2(3-2\mu )}$$$$\Rightarrow \lambda =\frac{8\mu -3(1+{\mu }^2)+(1-\mu )\sqrt{3(3{\mu }^2-2\mu +3)}}{2(3-2\mu )}$$$$with\mu =\frac{r_1}{r_2}=\frac{2}{3}$$$$\Rightarrow \lambda =\frac{x}{y}=\frac{3}{5}$$

Commented by mnjuly1970 last updated on 05/Dec/21

$$gratefulsirW$$

Commented by mr W last updated on 05/Dec/21

Commented by Tawa11 last updated on 05/Dec/21

$$\mathrm{Great}\mathrm{sir}.$$

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