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

Commented by kowalsky78 last updated on 05/Dec/21

I got (3/5). Maybe I made some mistake.

$${I}\:{got}\:\frac{\mathrm{3}}{\mathrm{5}}.\:{Maybe}\:{I}\:{made}\:{some}\:{mistake}. \\ $$

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

(3/5) is the correct answer.

$$\frac{\mathrm{3}}{\mathrm{5}}\:{is}\:{the}\:{correct}\:{answer}. \\ $$

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 (π/3) AD^2 =x^2 +y^2 +xy ⇒AD=(√(x^2 +y^2 +xy)) AD=AG+GD=AF+ED AD=(x+y−(√3)r_2 )+(y−(√3)r_2 ) AD=x+2y−2(√3)r_2 ⇒r_2 =((x+2y−AD)/(2(√3))) ⇒r_2 =((x+2y−(√(x^2 +y^2 +xy)))/(2(√3))) similarly ⇒r_1 =((2x+y−(√(x^2 +y^2 +xy)))/(2(√3))) (r_1 /r_2 )=((2x+y−(√(x^2 +y^2 +xy)))/(x+2y−(√(x^2 +y^2 +xy)))) let λ=(x/y), μ=(r_1 /r_2 )≤1 ⇒μ=((2λ+1−(√(λ^2 +λ+1)))/(λ+2−(√(λ^2 +λ+1)))) (2−μ)λ+1−2μ=(1−μ)(√(λ^2 +λ+1)) (2−μ)^2 λ^2 +(1−2μ)^2 +2(2−μ)(1−2μ)λ=(1−μ)^2 (λ^2 +λ+1) (3−2μ)λ^2 +(3μ^2 −8μ+3)λ+(3μ−2)μ=0 λ=((−(3μ^2 −8μ+3)+(√((3μ^2 −8μ+3)^2 −4(3−2μ)(3μ−2)μ)))/(2(3−2μ))) ⇒λ=((8μ−3(1+μ^2 )+(1−μ)(√(3(3μ^2 −2μ+3))))/(2(3−2μ))) with μ=(r_1 /r_2 )=(2/3) ⇒λ=(x/y)=(3/5)

$${AB}={AC}={BC}={x}+{y} \\ $$$${AD}^{\mathrm{2}} ={x}^{\mathrm{2}} +\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{x}\left({x}+{y}\right)\mathrm{cos}\:\frac{\pi}{\mathrm{3}} \\ $$$${AD}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy} \\ $$$$\Rightarrow{AD}=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}} \\ $$$$ \\ $$$${AD}={AG}+{GD}={AF}+{ED} \\ $$$${AD}=\left({x}+{y}−\sqrt{\mathrm{3}}{r}_{\mathrm{2}} \right)+\left({y}−\sqrt{\mathrm{3}}{r}_{\mathrm{2}} \right) \\ $$$${AD}={x}+\mathrm{2}{y}−\mathrm{2}\sqrt{\mathrm{3}}{r}_{\mathrm{2}} \\ $$$$\Rightarrow{r}_{\mathrm{2}} =\frac{{x}+\mathrm{2}{y}−{AD}}{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{r}_{\mathrm{2}} =\frac{{x}+\mathrm{2}{y}−\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}}}{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$${similarly} \\ $$$$\Rightarrow{r}_{\mathrm{1}} =\frac{\mathrm{2}{x}+{y}−\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}}}{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$$\frac{{r}_{\mathrm{1}} }{{r}_{\mathrm{2}} }=\frac{\mathrm{2}{x}+{y}−\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}}}{{x}+\mathrm{2}{y}−\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}}} \\ $$$${let}\:\lambda=\frac{{x}}{{y}},\:\mu=\frac{{r}_{\mathrm{1}} }{{r}_{\mathrm{2}} }\leqslant\mathrm{1} \\ $$$$\Rightarrow\mu=\frac{\mathrm{2}\lambda+\mathrm{1}−\sqrt{\lambda^{\mathrm{2}} +\lambda+\mathrm{1}}}{\lambda+\mathrm{2}−\sqrt{\lambda^{\mathrm{2}} +\lambda+\mathrm{1}}} \\ $$$$\left(\mathrm{2}−\mu\right)\lambda+\mathrm{1}−\mathrm{2}\mu=\left(\mathrm{1}−\mu\right)\sqrt{\lambda^{\mathrm{2}} +\lambda+\mathrm{1}} \\ $$$$\left(\mathrm{2}−\mu\right)^{\mathrm{2}} \lambda^{\mathrm{2}} +\left(\mathrm{1}−\mathrm{2}\mu\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{2}−\mu\right)\left(\mathrm{1}−\mathrm{2}\mu\right)\lambda=\left(\mathrm{1}−\mu\right)^{\mathrm{2}} \left(\lambda^{\mathrm{2}} +\lambda+\mathrm{1}\right) \\ $$$$\left(\mathrm{3}−\mathrm{2}\mu\right)\lambda^{\mathrm{2}} +\left(\mathrm{3}\mu^{\mathrm{2}} −\mathrm{8}\mu+\mathrm{3}\right)\lambda+\left(\mathrm{3}\mu−\mathrm{2}\right)\mu=\mathrm{0} \\ $$$$\lambda=\frac{−\left(\mathrm{3}\mu^{\mathrm{2}} −\mathrm{8}\mu+\mathrm{3}\right)+\sqrt{\left(\mathrm{3}\mu^{\mathrm{2}} −\mathrm{8}\mu+\mathrm{3}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{3}−\mathrm{2}\mu\right)\left(\mathrm{3}\mu−\mathrm{2}\right)\mu}}{\mathrm{2}\left(\mathrm{3}−\mathrm{2}\mu\right)} \\ $$$$\Rightarrow\lambda=\frac{\mathrm{8}\mu−\mathrm{3}\left(\mathrm{1}+\mu^{\mathrm{2}} \right)+\left(\mathrm{1}−\mu\right)\sqrt{\mathrm{3}\left(\mathrm{3}\mu^{\mathrm{2}} −\mathrm{2}\mu+\mathrm{3}\right)}}{\mathrm{2}\left(\mathrm{3}−\mathrm{2}\mu\right)} \\ $$$${with}\:\mu=\frac{{r}_{\mathrm{1}} }{{r}_{\mathrm{2}} }=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow\lambda=\frac{{x}}{{y}}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$

Commented by mnjuly1970 last updated on 05/Dec/21

grateful sir W

$${grateful}\:{sir}\:\:{W} \\ $$

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

Commented by Tawa11 last updated on 05/Dec/21

Great sir.

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

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