Question Number 59181 by ajfour last updated on 05/May/19
Commented by mr W last updated on 05/May/19
$${r}\leqslant\frac{{R}}{\mathrm{2}}\:! \\ $$$${so}\:{let}\:{r}=\mathrm{2}\:{instead}\:{of}\:\mathrm{3}. \\ $$
Answered by mr W last updated on 05/May/19
![(√((R−r)^2 −r^2 ))+(√((R−x)^2 −x^2 ))=(√((r+x)^2 −(r−x)^2 )) (√(R(R−2r)))+(√(R(R−2x)))=2(√(rx)) let λ=(x/R), μ=(r/R) (√(1−2μ))+(√(1−2λ))=2(√(μλ)) 1−2μ+1−2λ+2(√((1−2μ)(1−2λ)))=4μλ (√((1−2μ)(1−2λ)))=(2μ+1)λ−(1−μ) (1−2μ)(1−2λ)=(2μ+1)^2 λ^2 −2(1−μ)(2μ+1)λ+(1−μ)^2 (2μ+1)^2 λ^2 −2μ(3−2μ)λ+μ^2 =0 ⇒λ=((μ[(3−2μ)±2(√(2(1−2μ)))])/((2μ+1)^2 )) with μ=r/R=2/5=0.4 ⇒λ=((2(11±2(√(10))))/(81))= { ((0.094453)),((0.349991)) :}](https://www.tinkutara.com/question/Q59189.png)
$$\sqrt{\left({R}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }+\sqrt{\left({R}−{x}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} }=\sqrt{\left({r}+{x}\right)^{\mathrm{2}} −\left({r}−{x}\right)^{\mathrm{2}} } \\ $$$$\sqrt{{R}\left({R}−\mathrm{2}{r}\right)}+\sqrt{{R}\left({R}−\mathrm{2}{x}\right)}=\mathrm{2}\sqrt{{rx}} \\ $$$${let}\:\lambda=\frac{{x}}{{R}},\:\mu=\frac{{r}}{{R}} \\ $$$$\sqrt{\mathrm{1}−\mathrm{2}\mu}+\sqrt{\mathrm{1}−\mathrm{2}\lambda}=\mathrm{2}\sqrt{\mu\lambda} \\ $$$$\mathrm{1}−\mathrm{2}\mu+\mathrm{1}−\mathrm{2}\lambda+\mathrm{2}\sqrt{\left(\mathrm{1}−\mathrm{2}\mu\right)\left(\mathrm{1}−\mathrm{2}\lambda\right)}=\mathrm{4}\mu\lambda \\ $$$$\sqrt{\left(\mathrm{1}−\mathrm{2}\mu\right)\left(\mathrm{1}−\mathrm{2}\lambda\right)}=\left(\mathrm{2}\mu+\mathrm{1}\right)\lambda−\left(\mathrm{1}−\mu\right) \\ $$$$\left(\mathrm{1}−\mathrm{2}\mu\right)\left(\mathrm{1}−\mathrm{2}\lambda\right)=\left(\mathrm{2}\mu+\mathrm{1}\right)^{\mathrm{2}} \lambda^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}−\mu\right)\left(\mathrm{2}\mu+\mathrm{1}\right)\lambda+\left(\mathrm{1}−\mu\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{2}\mu+\mathrm{1}\right)^{\mathrm{2}} \lambda^{\mathrm{2}} −\mathrm{2}\mu\left(\mathrm{3}−\mathrm{2}\mu\right)\lambda+\mu^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\lambda=\frac{\mu\left[\left(\mathrm{3}−\mathrm{2}\mu\right)\pm\mathrm{2}\sqrt{\mathrm{2}\left(\mathrm{1}−\mathrm{2}\mu\right)}\right]}{\left(\mathrm{2}\mu+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${with}\:\mu={r}/{R}=\mathrm{2}/\mathrm{5}=\mathrm{0}.\mathrm{4} \\ $$$$\Rightarrow\lambda=\frac{\mathrm{2}\left(\mathrm{11}\pm\mathrm{2}\sqrt{\mathrm{10}}\right)}{\mathrm{81}}=\begin{cases}{\mathrm{0}.\mathrm{094453}}\\{\mathrm{0}.\mathrm{349991}}\end{cases} \\ $$
Commented by ajfour last updated on 06/May/19
$$\mathrm{Rare}\:\mathrm{and}\:\mathcal{E}{legant}\:{S}\mathrm{ir}! \\ $$$$\mathrm{Thank}\:\mathrm{you},\:\mathrm{so}\:\mathrm{nicely}\:\mathrm{presented}. \\ $$