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Question Number 59181 by ajfour last updated on 05/May/19

Commented by mr W last updated on 05/May/19

$r ⩽ \frac{R}{2}!$ $s o l e t r = 2 i n s t e a d o f 3 .$
	Answered by mr W last updated on 05/May/19
	$(√((R−r)^2 −r^2 ))+(√((R−x)^2 −x^2 ))=(√((r+x)^2 −(r−x)^2 )) (√(R(R−2r)))+(√(R(R−2x)))=2(√(rx)) let λ=(x/R), μ=(r/R) (√(1−2μ))+(√(1−2λ))=2(√(μλ)) 1−2μ+1−2λ+2(√((1−2μ)(1−2λ)))=4μλ (√((1−2μ)(1−2λ)))=(2μ+1)λ−(1−μ) (1−2μ)(1−2λ)=(2μ+1)^2 λ^2 −2(1−μ)(2μ+1)λ+(1−μ)^2 (2μ+1)^2 λ^2 −2μ(3−2μ)λ+μ^2 =0 ⇒λ=((μ[(3−2μ)±2(√(2(1−2μ)))])/((2μ+1)^2 )) with μ=r/R=2/5=0.4 ⇒λ=((2(11±2(√(10))))/(81))= { ((0.094453)),((0.349991)) :}$
	$\sqrt{{(R - r)}^{2} - r^{2}} + \sqrt{{(R - x)}^{2} - x^{2}} = \sqrt{{(r + x)}^{2} - {(r - x)}^{2}}$ $\sqrt{R (R - 2 r)} + \sqrt{R (R - 2 x)} = 2 \sqrt{r x}$ $l e t λ = \frac{x}{R}, μ = \frac{r}{R}$ $\sqrt{1 - 2 μ} + \sqrt{1 - 2 λ} = 2 \sqrt{μ λ}$ $1 - 2 μ + 1 - 2 λ + 2 \sqrt{(1 - 2 μ) (1 - 2 λ)} = 4 μ λ$ $\sqrt{(1 - 2 μ) (1 - 2 λ)} = (2 μ + 1) λ - (1 - μ)$ $(1 - 2 μ) (1 - 2 λ) = {(2 μ + 1)}^{2} λ^{2} - 2 (1 - μ) (2 μ + 1) λ + {(1 - μ)}^{2}$ ${(2 μ + 1)}^{2} λ^{2} - 2 μ (3 - 2 μ) λ + μ^{2} = 0$ $\Rightarrow λ = \frac{μ [(3 - 2 μ) \pm 2 \sqrt{2 (1 - 2 μ)}]}{{(2 μ + 1)}^{2}}$ $w i t h μ = r / R = 2 / 5 = 0.4$ $\Rightarrow λ = \frac{2 (11 \pm 2 \sqrt{10})}{81} = {\begin{cases} 0.094453 \\ 0.349991 \end{cases}$
	Commented by ajfour last updated on 06/May/19

	$Rare and E l e g a n t S ir!$ $Thank you, so nicely presented .$
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