Question-143322

Question Number 143322 by enter last updated on 13/Jun/21

Answered by mr W last updated on 13/Jun/21

Commented by mr W last updated on 13/Jun/21

R=radius of big circle r=radius of small circles OC=R−r (R−r)^2 −r^2 =((R/2))^2 ⇒(r/R)=(3/8) sin α=(r/(R−r))=(3/5) BC^2 =R^2 +(R−r)^2 −2R(R−r)sin α =R^2 +((25R^2 )/(64))−2R×((5R)/8)×(3/5) =((41R^2 )/(64)) b^2 =BD^2 =BC^2 −r^2 =((41R^2 )/(64))−((9R^2 )/(64))=(R^2 /2) ⇒b=(((√2)R)/2) AC^2 =R^2 +(R−r)^2 +2R(R−r)sin α =R^2 +((25R^2 )/(64))+2R×((5R)/8)×(3/5) =((137R^2 )/(64)) a^2 =AE^2 =AC^2 −r^2 =((137R^2 )/(64))−((9R^2 )/(64))=2R^2 ⇒a=(√2)R ⇒(a/b)=2

$${R}={radius}\:{of}\:{big}\:{circle} \\ $$$${r}={radius}\:{of}\:{small}\:{circles} \\ $$$${OC}={R}−{r} \\ $$$$\left({R}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} =\left(\frac{{R}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\frac{{r}}{{R}}=\frac{\mathrm{3}}{\mathrm{8}} \\ $$$$\mathrm{sin}\:\alpha=\frac{{r}}{{R}−{r}}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$${BC}^{\mathrm{2}} ={R}^{\mathrm{2}} +\left({R}−{r}\right)^{\mathrm{2}} −\mathrm{2}{R}\left({R}−{r}\right)\mathrm{sin}\:\alpha \\ $$$$\:\:\:\:\:\:\:\:\:={R}^{\mathrm{2}} +\frac{\mathrm{25}{R}^{\mathrm{2}} }{\mathrm{64}}−\mathrm{2}{R}×\frac{\mathrm{5}{R}}{\mathrm{8}}×\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{41}{R}^{\mathrm{2}} }{\mathrm{64}} \\ $$$${b}^{\mathrm{2}} ={BD}^{\mathrm{2}} ={BC}^{\mathrm{2}} −{r}^{\mathrm{2}} =\frac{\mathrm{41}{R}^{\mathrm{2}} }{\mathrm{64}}−\frac{\mathrm{9}{R}^{\mathrm{2}} }{\mathrm{64}}=\frac{{R}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow{b}=\frac{\sqrt{\mathrm{2}}{R}}{\mathrm{2}} \\ $$$${AC}^{\mathrm{2}} ={R}^{\mathrm{2}} +\left({R}−{r}\right)^{\mathrm{2}} +\mathrm{2}{R}\left({R}−{r}\right)\mathrm{sin}\:\alpha \\ $$$$\:\:\:\:\:\:\:\:\:={R}^{\mathrm{2}} +\frac{\mathrm{25}{R}^{\mathrm{2}} }{\mathrm{64}}+\mathrm{2}{R}×\frac{\mathrm{5}{R}}{\mathrm{8}}×\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{137}{R}^{\mathrm{2}} }{\mathrm{64}} \\ $$$${a}^{\mathrm{2}} ={AE}^{\mathrm{2}} ={AC}^{\mathrm{2}} −{r}^{\mathrm{2}} =\frac{\mathrm{137}{R}^{\mathrm{2}} }{\mathrm{64}}−\frac{\mathrm{9}{R}^{\mathrm{2}} }{\mathrm{64}}=\mathrm{2}{R}^{\mathrm{2}} \\ $$$$\Rightarrow{a}=\sqrt{\mathrm{2}}{R} \\ $$$$\Rightarrow\frac{{a}}{{b}}=\mathrm{2} \\ $$

Commented by peter frank last updated on 13/Jun/21