Question-204800

Question Number 204800 by sonukgindia last updated on 27/Feb/24

Answered by mr W last updated on 27/Feb/24

Commented by mr W last updated on 27/Feb/24

r=radius of circles=1 s=side length of square a, b=parameters of ellipse a=2b+r (x^2 /a^2 )+(y^2 /b^2 )=1 C(b+r, 0) (x−b−r)^2 +y^2 =r^2 (x^2 /a^2 )+((r^2 −(x−b−r)^2 )/b^2 )=1 (a^2 −b^2 )x^2 −2a^2 (b+r)x+2a^2 b(b+r)=0 Δ=a^4 (b+r)^2 −2a^2 b(a^2 −b^2 )(b+r)=0 a^2 (r−b)+2b^3 =0 (2b+r)^2 (r−b)+2b^3 =0 ((b/r))^3 −(3/2)((b/r))−(1/2)=0 ⇒(b/r)=(√2) sin (((5π)/(12)))=(((√2)((√2)+(√6)))/4)=((1+(√3))/2) s=2a=2(2b+r) (s/r)=2(1+1+(√3))=2(2+(√3)) s^2 =4(7+4(√3))≈55.7128

$${r}={radius}\:{of}\:{circles}=\mathrm{1} \\ $$$${s}={side}\:{length}\:{of}\:{square} \\ $$$${a},\:{b}={parameters}\:{of}\:{ellipse} \\ $$$${a}=\mathrm{2}{b}+{r} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${C}\left({b}+{r},\:\mathrm{0}\right) \\ $$$$\left({x}−{b}−{r}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{r}^{\mathrm{2}} −\left({x}−{b}−{r}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){x}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} \left({b}+{r}\right){x}+\mathrm{2}{a}^{\mathrm{2}} {b}\left({b}+{r}\right)=\mathrm{0} \\ $$$$\Delta={a}^{\mathrm{4}} \left({b}+{r}\right)^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} {b}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left({b}+{r}\right)=\mathrm{0} \\ $$$${a}^{\mathrm{2}} \left({r}−{b}\right)+\mathrm{2}{b}^{\mathrm{3}} =\mathrm{0} \\ $$$$\left(\mathrm{2}{b}+{r}\right)^{\mathrm{2}} \left({r}−{b}\right)+\mathrm{2}{b}^{\mathrm{3}} =\mathrm{0} \\ $$$$\left(\frac{{b}}{{r}}\right)^{\mathrm{3}} −\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{{b}}{{r}}\right)−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow\frac{{b}}{{r}}=\sqrt{\mathrm{2}}\:\mathrm{sin}\:\left(\frac{\mathrm{5}\pi}{\mathrm{12}}\right)=\frac{\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{6}}\right)}{\mathrm{4}}=\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${s}=\mathrm{2}{a}=\mathrm{2}\left(\mathrm{2}{b}+{r}\right) \\ $$$$\frac{{s}}{{r}}=\mathrm{2}\left(\mathrm{1}+\mathrm{1}+\sqrt{\mathrm{3}}\right)=\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right) \\ $$$${s}^{\mathrm{2}} =\mathrm{4}\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\right)\approx\mathrm{55}.\mathrm{7128} \\ $$

Commented by mr W last updated on 27/Feb/24