Question-224529

Question Number 224529 by ajfour last updated on 17/Sep/25

Commented by ajfour last updated on 17/Sep/25

Find r/R if r_1 =r_2 =r_3 and R_2 =R_1 .

$${Find}\:{r}/{R}\:{if}\:{r}_{\mathrm{1}} ={r}_{\mathrm{2}} ={r}_{\mathrm{3}} \:{and}\:{R}_{\mathrm{2}} ={R}_{\mathrm{1}} . \\ $$

Answered by mahdipoor last updated on 17/Sep/25

circle ⇒ c=(0,r),r c_+ ,c_− =(±b,r),r C_+ ,C_− =(±a,0),R C_+ c_− =C_− c_+ =R+r=(√((b+a)^2 +(r)^2 )) C_+ c=C_− c=R−r=(√((a)^2 +(r)^2 )) C_+ c_+ =C_− c_− =R−r=(√((b−a)^2 +(r)^2 )) { ((R^2 +2Rr=b^2 +a^2 +2ab 1)),((R^2 −2Rr=a^2 2 )),((R^2 −2Rr=b^2 +a^2 −2ba 3)) :} 2,3⇒b^2 −2ba=0=b(b−2a)⇒b=2a 1,2⇒ { ((R^2 +2Rr=9a^2 )),((R^2 −2Rr=a^2 )) :}⇒R^2 +2Rr=9R^2 −18Rr ⇒8R^2 −20Rr=0=4R(2R−5r)=0 ⇒(r/R)=(2/5)

$${circle}\:\Rightarrow \\ $$$${c}=\left(\mathrm{0},{r}\right),{r} \\ $$$${c}_{+} ,{c}_{−} =\left(\pm{b},{r}\right),{r} \\ $$$${C}_{+} ,{C}_{−} =\left(\pm{a},\mathrm{0}\right),{R} \\ $$$${C}_{+} {c}_{−} ={C}_{−} {c}_{+} ={R}+{r}=\sqrt{\left({b}+{a}\right)^{\mathrm{2}} +\left({r}\right)^{\mathrm{2}} } \\ $$$${C}_{+} {c}={C}_{−} {c}={R}−{r}=\sqrt{\left({a}\right)^{\mathrm{2}} +\left({r}\right)^{\mathrm{2}} } \\ $$$${C}_{+} {c}_{+} ={C}_{−} {c}_{−} ={R}−{r}=\sqrt{\left({b}−{a}\right)^{\mathrm{2}} +\left({r}\right)^{\mathrm{2}} } \\ $$$$\begin{cases}{{R}^{\mathrm{2}} +\mathrm{2}{Rr}={b}^{\mathrm{2}} +{a}^{\mathrm{2}} +\mathrm{2}{ab}\:\:\:\mathrm{1}}\\{{R}^{\mathrm{2}} −\mathrm{2}{Rr}={a}^{\mathrm{2}} \:\:\:\mathrm{2}\:\:}\\{{R}^{\mathrm{2}} −\mathrm{2}{Rr}={b}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{2}{ba}\:\:\:\mathrm{3}}\end{cases} \\ $$$$\mathrm{2},\mathrm{3}\Rightarrow{b}^{\mathrm{2}} −\mathrm{2}{ba}=\mathrm{0}={b}\left({b}−\mathrm{2}{a}\right)\Rightarrow{b}=\mathrm{2}{a} \\ $$$$\mathrm{1},\mathrm{2}\Rightarrow\begin{cases}{{R}^{\mathrm{2}} +\mathrm{2}{Rr}=\mathrm{9}{a}^{\mathrm{2}} }\\{{R}^{\mathrm{2}} −\mathrm{2}{Rr}={a}^{\mathrm{2}} }\end{cases}\Rightarrow{R}^{\mathrm{2}} +\mathrm{2}{Rr}=\mathrm{9}{R}^{\mathrm{2}} −\mathrm{18}{Rr} \\ $$$$\Rightarrow\mathrm{8}{R}^{\mathrm{2}} −\mathrm{20}{Rr}=\mathrm{0}=\mathrm{4}{R}\left(\mathrm{2}{R}−\mathrm{5}{r}\right)=\mathrm{0} \\ $$$$\Rightarrow\frac{{r}}{{R}}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$

Commented by ajfour last updated on 17/Sep/25

$${yes},\:{thanks}.\:{Thats}\:{correct}. \\ $$

Answered by ajfour last updated on 18/Sep/25

Commented by ajfour last updated on 18/Sep/25

E is center of left semicircle while A is center of right one. so OE=OA=AB=t EC=R+r AC=R−r t^2 =(R−r)^2 −r^2 (3t)^2 =(R+r)^2 −r^2 ⇒ 9(R−r)^2 −9r^2 =(R+r)^2 −r^2 ⇒ 8R^2 =20Rr (r/R)=(2/5)

$${E}\:{is}\:{center}\:{of}\:{left}\:{semicircle} \\ $$$${while}\:{A}\:{is}\:{center}\:{of}\:{right}\:{one}. \\ $$$${so}\:\:{OE}={OA}={AB}={t} \\ $$$${EC}={R}+{r} \\ $$$${AC}={R}−{r} \\ $$$${t}^{\mathrm{2}} =\left({R}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} \\ $$$$\left(\mathrm{3}{t}\right)^{\mathrm{2}} =\left({R}+{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{9}\left({R}−{r}\right)^{\mathrm{2}} −\mathrm{9}{r}^{\mathrm{2}} =\left({R}+{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{8}{R}^{\mathrm{2}} =\mathrm{20}{Rr} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\frac{{r}}{{R}}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$

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