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Question Number 28516 by beh.i83417@gmail.com last updated on 26/Jan/18

Commented by beh.i83417@gmail.com last updated on 26/Jan/18

(r/s)=?

$$\frac{\boldsymbol{{r}}}{\boldsymbol{{s}}}=? \\ $$

Answered by mrW2 last updated on 26/Jan/18

Commented by mrW2 last updated on 26/Jan/18

sin α=(((√2)s)/(r+s)) ⇒cos α=(√(1−((((√2)s)/(r+s)))^2 ))=((√(r^2 +2rs−s^2 ))/(r+s)) (r+s) cos ((π/4)−α)=r−s (r+s) (1/(√2))(cos α+sin α)=r−s (r+s) (1/(√2))((((√(r^2 +2rs−s^2 ))+(√2)s)/(r+s)))=r−s (√(r^2 +2rs−s^2 ))=(√2)(r−2s) r^2 +2rs−s^2 =2(r^2 −4rs+4s^2 ) 9s^2 −10rs+r^2 =0 (9s−r)(s−r)=0 ⇒s=(r/9) or s=r (not suitable) ⇒(r/s)=9

$$\mathrm{sin}\:\alpha=\frac{\sqrt{\mathrm{2}}{s}}{{r}+{s}} \\ $$$$\Rightarrow\mathrm{cos}\:\alpha=\sqrt{\mathrm{1}−\left(\frac{\sqrt{\mathrm{2}}{s}}{{r}+{s}}\right)^{\mathrm{2}} }=\frac{\sqrt{{r}^{\mathrm{2}} +\mathrm{2}{rs}−{s}^{\mathrm{2}} }}{{r}+{s}} \\ $$$$\left({r}+{s}\right)\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}−\alpha\right)={r}−{s} \\ $$$$\left({r}+{s}\right)\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\left(\mathrm{cos}\:\alpha+\mathrm{sin}\:\alpha\right)={r}−{s} \\ $$$$\left({r}+{s}\right)\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\left(\frac{\sqrt{{r}^{\mathrm{2}} +\mathrm{2}{rs}−{s}^{\mathrm{2}} }+\sqrt{\mathrm{2}}{s}}{{r}+{s}}\right)={r}−{s} \\ $$$$\sqrt{{r}^{\mathrm{2}} +\mathrm{2}{rs}−{s}^{\mathrm{2}} }=\sqrt{\mathrm{2}}\left({r}−\mathrm{2}{s}\right) \\ $$$${r}^{\mathrm{2}} +\mathrm{2}{rs}−{s}^{\mathrm{2}} =\mathrm{2}\left({r}^{\mathrm{2}} −\mathrm{4}{rs}+\mathrm{4}{s}^{\mathrm{2}} \right) \\ $$$$\mathrm{9}{s}^{\mathrm{2}} −\mathrm{10}{rs}+{r}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left(\mathrm{9}{s}−{r}\right)\left({s}−{r}\right)=\mathrm{0} \\ $$$$\Rightarrow{s}=\frac{{r}}{\mathrm{9}}\:{or}\:{s}={r}\:\left({not}\:{suitable}\right) \\ $$$$\Rightarrow\frac{{r}}{{s}}=\mathrm{9} \\ $$

Answered by ajfour last updated on 26/Jan/18

line through centre of two small circles (that touch the bigger circle) be S_2 S_3 let midpoint of S_2 S_3 be M distance of centre of big circle C from this line is CM= r(√2)−2(√2)s , MS_3 =s(√2) MS_3 ^( 2) +CM^( 2) =CS_2 ^2 =CS_3 ^2 ⇒ 2s^2 +2(r−2s)^2 =(r+s)^2 ⇒ 2s^2 +2r^2 −8rs+8s^2 =r^2 +s^2 +2rs r^2 −10rs+9s^2 =0 (r−9s)(r−s)=0 ⇒ (r/s) = 9 .

$$\:{line}\:{through}\:{centre}\:{of} \\ $$$${two}\:{small}\:{circles}\:\left({that}\:{touch}\:{the}\right. \\ $$$$\left.{bigger}\:{circle}\right)\:\:{be}\:{S}_{\mathrm{2}} {S}_{\mathrm{3}} \\ $$$${let}\:{midpoint}\:{of}\:{S}_{\mathrm{2}} {S}_{\mathrm{3}} \:{be}\:{M} \\ $$$${distance}\:{of}\:{centre}\:{of}\:{big}\:{circle}\:{C} \\ $$$${from}\:{this}\:{line}\:{is} \\ $$$$\:\:\:\:{CM}=\:{r}\sqrt{\mathrm{2}}−\mathrm{2}\sqrt{\mathrm{2}}{s}\:\:,\:{MS}_{\mathrm{3}} ={s}\sqrt{\mathrm{2}} \\ $$$${MS}_{\mathrm{3}} ^{\:\mathrm{2}} +{CM}^{\:\mathrm{2}} ={CS}_{\mathrm{2}} ^{\mathrm{2}} \:={CS}_{\mathrm{3}} ^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{2}{s}^{\mathrm{2}} +\mathrm{2}\left({r}−\mathrm{2}{s}\right)^{\mathrm{2}} =\left({r}+{s}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{2}{s}^{\mathrm{2}} +\mathrm{2}{r}^{\mathrm{2}} −\mathrm{8}{rs}+\mathrm{8}{s}^{\mathrm{2}} ={r}^{\mathrm{2}} +{s}^{\mathrm{2}} +\mathrm{2}{rs} \\ $$$${r}^{\mathrm{2}} −\mathrm{10}{rs}+\mathrm{9}{s}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({r}−\mathrm{9}{s}\right)\left({r}−{s}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\frac{{r}}{{s}}\:=\:\mathrm{9}\:. \\ $$

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