Question-172763

Question Number 172763 by mnjuly1970 last updated on 01/Jul/22

Answered by aleks041103 last updated on 01/Jul/22

Commented by aleks041103 last updated on 01/Jul/22

1) (R+r)^2 =(R−r)^2 +(a−(R+r))^2 4Rr=a^2 −2a(R+r)+R^2 +r^2 +2Rr ⇒a^2 +(R−r)^2 −2a(R+r)=0 2) (r/x)=(R/a) 3) (1/2)(x+a+(√(x^2 +a^2 )))r=(1/2)ax ⇒x+a+(√(x^2 +a^2 ))=((ax)/r) Without loss of generalization a=1 ⇒1+R^2 +r^2 −2rR−2R−2r=0 ⇒x=((ar)/R)=(r/R) ⇒(r/R)+1+(√(((r/R))^2 +1))=(1/R) ⇒r+R+(√(r^2 +R^2 ))=1 ⇒r^2 +R^2 =1+r^2 +R^2 −2r−2R+2Rr ⇒1−2r−2R+2Rr=0 ⇒−2r−2R=−1−2Rr ⇒1+R^2 +r^2 −2Rr−1−2Rr=0 R^2 +r^2 −4Rr=0 ⇒((R/r))^2 −4((R/r))+1=0 ⇒(R/r)=((4±(√(16−4)))/2)=((4±(√(12)))/2)=2±(√3) R>r ⇒(R/r)=2+(√3)

$$\left.\mathrm{1}\right) \\ $$$$\left({R}+{r}\right)^{\mathrm{2}} =\left({R}−{r}\right)^{\mathrm{2}} +\left({a}−\left({R}+{r}\right)\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{Rr}={a}^{\mathrm{2}} −\mathrm{2}{a}\left({R}+{r}\right)+{R}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}{Rr} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +\left({R}−{r}\right)^{\mathrm{2}} −\mathrm{2}{a}\left({R}+{r}\right)=\mathrm{0} \\ $$$$\left.\mathrm{2}\right) \\ $$$$\frac{{r}}{{x}}=\frac{{R}}{{a}} \\ $$$$\left.\mathrm{3}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left({x}+{a}+\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }\right){r}=\frac{\mathrm{1}}{\mathrm{2}}{ax} \\ $$$$\Rightarrow{x}+{a}+\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }=\frac{{ax}}{{r}} \\ $$$$ \\ $$$${Without}\:{loss}\:{of}\:{generalization} \\ $$$${a}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{1}+{R}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{rR}−\mathrm{2}{R}−\mathrm{2}{r}=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{{ar}}{{R}}=\frac{{r}}{{R}} \\ $$$$\Rightarrow\frac{{r}}{{R}}+\mathrm{1}+\sqrt{\left(\frac{{r}}{{R}}\right)^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{1}}{{R}} \\ $$$$\Rightarrow{r}+{R}+\sqrt{{r}^{\mathrm{2}} +{R}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\Rightarrow{r}^{\mathrm{2}} +{R}^{\mathrm{2}} =\mathrm{1}+{r}^{\mathrm{2}} +{R}^{\mathrm{2}} −\mathrm{2}{r}−\mathrm{2}{R}+\mathrm{2}{Rr} \\ $$$$\Rightarrow\mathrm{1}−\mathrm{2}{r}−\mathrm{2}{R}+\mathrm{2}{Rr}=\mathrm{0} \\ $$$$\Rightarrow−\mathrm{2}{r}−\mathrm{2}{R}=−\mathrm{1}−\mathrm{2}{Rr} \\ $$$$\Rightarrow\mathrm{1}+{R}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{Rr}−\mathrm{1}−\mathrm{2}{Rr}=\mathrm{0} \\ $$$${R}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{4}{Rr}=\mathrm{0} \\ $$$$\Rightarrow\left(\frac{{R}}{{r}}\right)^{\mathrm{2}} −\mathrm{4}\left(\frac{{R}}{{r}}\right)+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\frac{{R}}{{r}}=\frac{\mathrm{4}\pm\sqrt{\mathrm{16}−\mathrm{4}}}{\mathrm{2}}=\frac{\mathrm{4}\pm\sqrt{\mathrm{12}}}{\mathrm{2}}=\mathrm{2}\pm\sqrt{\mathrm{3}} \\ $$$${R}>{r} \\ $$$$\Rightarrow\frac{{R}}{{r}}=\mathrm{2}+\sqrt{\mathrm{3}} \\ $$

Commented by Tawa11 last updated on 01/Jul/22

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

Answered by mr W last updated on 01/Jul/22

Commented by mr W last updated on 01/Jul/22

α+β=(π/4) side length of square: r+(r/(tan α))=R+(R/(tan β))=R+r+(√((R+r)^2 −(R−r)^2 )) ⇒(1/(tan α))=(R/r)+2(√(R/r))=λ+2(√λ) ⇒(1/(tan β))=(r/R)+2(√(r/R))=(1/λ)+(2/( (√λ))) with λ=(R/r) tan (α+β)=(((1/(tan α))+(1/(tan β)))/((1/(tan α tan β))−1))=1 ((λ+2(√λ)+(1/λ)+(2/( (√λ))))/((λ+2(√λ))((1/λ)+(2/( (√λ))))−1))=1 λ+(1/λ)=4 λ^2 −4λ+1=0 ⇒λ=2+(√3)

$$\alpha+\beta=\frac{\pi}{\mathrm{4}} \\ $$$${side}\:{length}\:{of}\:{square}: \\ $$$${r}+\frac{{r}}{\mathrm{tan}\:\alpha}={R}+\frac{{R}}{\mathrm{tan}\:\beta}={R}+{r}+\sqrt{\left({R}+{r}\right)^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}=\frac{{R}}{{r}}+\mathrm{2}\sqrt{\frac{{R}}{{r}}}=\lambda+\mathrm{2}\sqrt{\lambda} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{tan}\:\beta}=\frac{{r}}{{R}}+\mathrm{2}\sqrt{\frac{{r}}{{R}}}=\frac{\mathrm{1}}{\lambda}+\frac{\mathrm{2}}{\:\sqrt{\lambda}} \\ $$$${with}\:\lambda=\frac{{R}}{{r}} \\ $$$$\mathrm{tan}\:\left(\alpha+\beta\right)=\frac{\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}+\frac{\mathrm{1}}{\mathrm{tan}\:\beta}}{\frac{\mathrm{1}}{\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta}−\mathrm{1}}=\mathrm{1} \\ $$$$\frac{\lambda+\mathrm{2}\sqrt{\lambda}+\frac{\mathrm{1}}{\lambda}+\frac{\mathrm{2}}{\:\sqrt{\lambda}}}{\left(\lambda+\mathrm{2}\sqrt{\lambda}\right)\left(\frac{\mathrm{1}}{\lambda}+\frac{\mathrm{2}}{\:\sqrt{\lambda}}\right)−\mathrm{1}}=\mathrm{1} \\ $$$$\lambda+\frac{\mathrm{1}}{\lambda}=\mathrm{4} \\ $$$$\lambda^{\mathrm{2}} −\mathrm{4}\lambda+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\lambda=\mathrm{2}+\sqrt{\mathrm{3}} \\ $$

Commented by mnjuly1970 last updated on 01/Jul/22