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Question-205289

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Question Number 205289 by mr W last updated on 14/Mar/24
Answered by A5T last updated on 16/Mar/24
Commented by A5T last updated on 16/Mar/24
General idea, there could be a far more simpler solution.
$${General}\:{idea},\:{there}\:{could}\:{be}\:{a}\:{far}\:{more}\:{simpler} \\ $$$${solution}. \\ $$
Commented by A5T last updated on 16/Mar/24
Let O be the centreof the biggest circle,Ω,and let O′ be the centre of the middle circle on diameter of Ω. Then O′P_1 =r_5 +r_1 ,OP_1 =68−r_1 (68−r_1 )^2 =42.5^2 +(25.5+r_1 )^2 −2(42.5)(25.5+r_1 )cosθ ⇒r_1 =−((255(cosθ+1))/(2(5cosθ−11))) O′a=r_5 +25.5,aP_1 =25.5+r_1 ,O′P_1 =r_5 +r_1 (r_5 +r_1 )^2 =(r_5 +25.5)^2 +(r_1 +25.5)^2 −2(r_5 +25.5)(r_1 +25.5)cosθ ⇒(r_5 −((255(cosθ+1))/(2(5cosθ−11))))^2 =(r_5 +25.5)^2 +(25.5−((255(cosθ+1))/(2(5cosθ−11))))^2 −2(r_5 +25.5)(((−255cos(θ+1))/(2(5cosθ−11)))+25.5)cosθ ⇒Then,express r_5 in terms of cosθ 2r_3 +2r_5 =85⇒r_3 +r_5 =42.5⇒r_3 =42.5−r_5 Let β=∠P_2 P_3 O′ Then, (r_5 +r_2 )^2 =(r_3 +r_2 )^2 +42.5^2 −85(r_3 +r_2 )cosβ...(i) ii: (68−r_2 )^2 =(68−r_3 )^2 +(r_3 +r_2 )^2 −2(68−r_3 )(r_3 +r_2 )cosβ ii⇒cosβ=(((68−r_2 )^2 −(68+((204)/(5cosθ−3))−42.5)^2 −(42.5−((204)/(5cosθ−3))+r_2 )^2 )/(−2(68+((204)/(5cosθ−3))−42.5)(42.5−((204)/(5cosθ−3))+r_2 ))) i⇒cosβ=(((r_5 +r_2 )^2 −(r_3 +r_2 )^2 −42.5^2 )/(−85(r_3 +r_2 ))) Express r_2 in terms of cosθ by equating cosβ of (ii) and (i) If we can find θ or cosθ,then the rest would be easy.
$${Let}\:{O}\:{be}\:{the}\:{centreof}\:{the}\:{biggest}\:{circle},\Omega,{and} \\ $$$${let}\:{O}'\:{be}\:{the}\:{centre}\:{of}\:{the}\:{middle}\:{circle}\:{on}\:{diameter}\:{of}\:\Omega. \\ $$$${Then}\:{O}'{P}_{\mathrm{1}} ={r}_{\mathrm{5}} +{r}_{\mathrm{1}} ,{OP}_{\mathrm{1}} =\mathrm{68}−{r}_{\mathrm{1}} \\ $$$$\left(\mathrm{68}−{r}_{\mathrm{1}} \right)^{\mathrm{2}} =\mathrm{42}.\mathrm{5}^{\mathrm{2}} +\left(\mathrm{25}.\mathrm{5}+{r}_{\mathrm{1}} \right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{42}.\mathrm{5}\right)\left(\mathrm{25}.\mathrm{5}+{r}_{\mathrm{1}} \right){cos}\theta \\ $$$$\Rightarrow{r}_{\mathrm{1}} =−\frac{\mathrm{255}\left({cos}\theta+\mathrm{1}\right)}{\mathrm{2}\left(\mathrm{5}{cos}\theta−\mathrm{11}\right)} \\ $$$${O}'{a}={r}_{\mathrm{5}} +\mathrm{25}.\mathrm{5},{aP}_{\mathrm{1}} =\mathrm{25}.\mathrm{5}+{r}_{\mathrm{1}} ,{O}'{P}_{\mathrm{1}} ={r}_{\mathrm{5}} +{r}_{\mathrm{1}} \\ $$$$\left({r}_{\mathrm{5}} +{r}_{\mathrm{1}} \right)^{\mathrm{2}} =\left({r}_{\mathrm{5}} +\mathrm{25}.\mathrm{5}\right)^{\mathrm{2}} +\left({r}_{\mathrm{1}} +\mathrm{25}.\mathrm{5}\right)^{\mathrm{2}} −\mathrm{2}\left({r}_{\mathrm{5}} +\mathrm{25}.\mathrm{5}\right)\left({r}_{\mathrm{1}} +\mathrm{25}.\mathrm{5}\right){cos}\theta \\ $$$$\Rightarrow\left({r}_{\mathrm{5}} −\frac{\mathrm{255}\left({cos}\theta+\mathrm{1}\right)}{\mathrm{2}\left(\mathrm{5}{cos}\theta−\mathrm{11}\right)}\right)^{\mathrm{2}} =\left({r}_{\mathrm{5}} +\mathrm{25}.\mathrm{5}\right)^{\mathrm{2}} +\left(\mathrm{25}.\mathrm{5}−\frac{\mathrm{255}\left({cos}\theta+\mathrm{1}\right)}{\mathrm{2}\left(\mathrm{5}{cos}\theta−\mathrm{11}\right)}\right)^{\mathrm{2}} \\ $$$$−\mathrm{2}\left({r}_{\mathrm{5}} +\mathrm{25}.\mathrm{5}\right)\left(\frac{−\mathrm{255}{cos}\left(\theta+\mathrm{1}\right)}{\mathrm{2}\left(\mathrm{5}{cos}\theta−\mathrm{11}\right)}+\mathrm{25}.\mathrm{5}\right){cos}\theta \\ $$$$\Rightarrow{Then},{express}\:{r}_{\mathrm{5}} \:{in}\:{terms}\:{of}\:{cos}\theta \\ $$$$\mathrm{2}{r}_{\mathrm{3}} +\mathrm{2}{r}_{\mathrm{5}} =\mathrm{85}\Rightarrow{r}_{\mathrm{3}} +{r}_{\mathrm{5}} =\mathrm{42}.\mathrm{5}\Rightarrow{r}_{\mathrm{3}} =\mathrm{42}.\mathrm{5}−{r}_{\mathrm{5}} \\ $$$${Let}\:\beta=\angle{P}_{\mathrm{2}} {P}_{\mathrm{3}} {O}' \\ $$$${Then},\:\left({r}_{\mathrm{5}} +{r}_{\mathrm{2}} \right)^{\mathrm{2}} =\left({r}_{\mathrm{3}} +{r}_{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{42}.\mathrm{5}^{\mathrm{2}} −\mathrm{85}\left({r}_{\mathrm{3}} +{r}_{\mathrm{2}} \right){cos}\beta…\left({i}\right) \\ $$$${ii}:\:\left(\mathrm{68}−{r}_{\mathrm{2}} \right)^{\mathrm{2}} =\left(\mathrm{68}−{r}_{\mathrm{3}} \right)^{\mathrm{2}} +\left({r}_{\mathrm{3}} +{r}_{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{68}−{r}_{\mathrm{3}} \right)\left({r}_{\mathrm{3}} +{r}_{\mathrm{2}} \right){cos}\beta \\ $$$${ii}\Rightarrow{cos}\beta=\frac{\left(\mathrm{68}−{r}_{\mathrm{2}} \right)^{\mathrm{2}} −\left(\mathrm{68}+\frac{\mathrm{204}}{\mathrm{5}{cos}\theta−\mathrm{3}}−\mathrm{42}.\mathrm{5}\right)^{\mathrm{2}} −\left(\mathrm{42}.\mathrm{5}−\frac{\mathrm{204}}{\mathrm{5}{cos}\theta−\mathrm{3}}+{r}_{\mathrm{2}} \right)^{\mathrm{2}} }{−\mathrm{2}\left(\mathrm{68}+\frac{\mathrm{204}}{\mathrm{5}{cos}\theta−\mathrm{3}}−\mathrm{42}.\mathrm{5}\right)\left(\mathrm{42}.\mathrm{5}−\frac{\mathrm{204}}{\mathrm{5}{cos}\theta−\mathrm{3}}+{r}_{\mathrm{2}} \right)} \\ $$$${i}\Rightarrow{cos}\beta=\frac{\left({r}_{\mathrm{5}} +{r}_{\mathrm{2}} \right)^{\mathrm{2}} −\left({r}_{\mathrm{3}} +{r}_{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{42}.\mathrm{5}^{\mathrm{2}} }{−\mathrm{85}\left({r}_{\mathrm{3}} +{r}_{\mathrm{2}} \right)} \\ $$$${Express}\:{r}_{\mathrm{2}} \:{in}\:{terms}\:{of}\:{cos}\theta\:\:{by}\:{equating}\:{cos}\beta\: \\ $$$${of}\:\left({ii}\right)\:{and}\:\left({i}\right) \\ $$$${If}\:{we}\:{can}\:{find}\:\theta\:{or}\:{cos}\theta,{then}\:{the}\:{rest}\:{would}\:{be}\:{easy}. \\ $$

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