Question Number 205507 by mr W last updated on 23/Mar/24
Commented by mr W last updated on 23/Mar/24
$${an}\:{unsolved}\:{old}\:{question} \\ $$
Answered by mr W last updated on 25/Mar/24
Commented by mr W last updated on 23/Mar/24
![R=((85+51)/2)=68 r_1 =((51)/2)=25.5 R+d=s+2r_1 ⇒d=s−17 λ=((R^2 +s^2 −d^2 )/(2Rs))=((68^2 +s^2 −d^2 )/(136s))=((68^2 +s^2 −(s−17)^2 )/(136s))=((34s+4335)/(136s)) for 6 circles between the outer and the inner circles the inversive distance from them should be δ=cosh^(−1) λ =2 ln [tan ((π/4)+(π/(2n)))] with n=6 cosh^(−1) λ=2 ln [tan ((π/4)+(π/(2×6)))] cosh^(−1) λ=2 ln (tan (π/3)) cosh^(−1) λ=ln 3 ln (λ+(√(λ^2 −1)))=ln 3 λ+(√(λ^2 −1))=3 ⇒λ=(5/3)=((34s+4335)/(136s)) ⇒s=((45)/2)=22.5 ⇒d=22.5−17=5.5 ⇒r_4 =R−s−r_1 =68−((51)/2)−((45)/2)=20 (((s+r_2 )^2 +(s+r_1 )^2 −(r_1 +r_2 )^2 )/(2(s+r_2 )(s+r_1 )))=(((s+r_2 )^2 +d^2 −(R−r_2 )^2 )/(2(s+r_2 )d)) ((s^2 +sr_1 +(s−r_1 )r_2 )/(s+r_1 ))=((s^2 +2(R+s)r_2 +d^2 −R^2 )/(2d)) (((((45)/2))^2 +((45)/2)×((51)/2)+(((45)/2)−((51)/2))r_2 )/(((45)/2)+((51)/2)))=(((((45)/2))^2 +2(68+((45)/2))r_2 +(((11)/2))^2 −68^2 )/(2×((11)/2))) ⇒r_2 =((1360)/(57))≈23.859649 similarly ((2s^2 +2sr_4 +2(s−r_4 )r_3 )/(s+r_4 ))=((s^2 +2(R+s)r_3 +d^2 −R^2 )/(−d)) (((((45)/2))^2 +((45)/2)×20+(((45)/2)−20)r_3 )/(((45)/2)+20))=(((((45)/2))^2 +2(68+((45)/2))r_3 +(((11)/2))^2 −68^2 )/(−2×((11)/2))) ⇒r_3 =((4080)/(193))≈21.139 896 c=r_2 +r_3 =((1360)/(57))+((4080)/(193))=((495040)/(11001))≈44.999 545 496](https://www.tinkutara.com/question/Q205510.png)
$${R}=\frac{\mathrm{85}+\mathrm{51}}{\mathrm{2}}=\mathrm{68} \\ $$$${r}_{\mathrm{1}} =\frac{\mathrm{51}}{\mathrm{2}}=\mathrm{25}.\mathrm{5} \\ $$$${R}+{d}={s}+\mathrm{2}{r}_{\mathrm{1}} \\ $$$$\Rightarrow{d}={s}−\mathrm{17} \\ $$$$\lambda=\frac{{R}^{\mathrm{2}} +{s}^{\mathrm{2}} −{d}^{\mathrm{2}} }{\mathrm{2}{Rs}}=\frac{\mathrm{68}^{\mathrm{2}} +{s}^{\mathrm{2}} −{d}^{\mathrm{2}} }{\mathrm{136}{s}}=\frac{\mathrm{68}^{\mathrm{2}} +{s}^{\mathrm{2}} −\left({s}−\mathrm{17}\right)^{\mathrm{2}} }{\mathrm{136}{s}}=\frac{\mathrm{34}{s}+\mathrm{4335}}{\mathrm{136}{s}} \\ $$$${for}\:\mathrm{6}\:{circles}\:{between}\:{the}\:{outer} \\ $$$${and}\:{the}\:{inner}\:{circles}\:{the}\:{inversive}\: \\ $$$${distance}\:{from}\:{them}\:{should}\:{be} \\ $$$$\delta=\mathrm{cosh}^{−\mathrm{1}} \lambda\:=\mathrm{2}\:\mathrm{ln}\:\left[\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{2}{n}}\right)\right] \\ $$$${with}\:{n}=\mathrm{6} \\ $$$$\mathrm{cosh}^{−\mathrm{1}} \:\lambda=\mathrm{2}\:\mathrm{ln}\:\left[\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{2}×\mathrm{6}}\right)\right] \\ $$$$\mathrm{cosh}^{−\mathrm{1}} \:\lambda=\mathrm{2}\:\mathrm{ln}\:\left(\mathrm{tan}\:\frac{\pi}{\mathrm{3}}\right) \\ $$$$\mathrm{cosh}^{−\mathrm{1}} \:\lambda=\mathrm{ln}\:\mathrm{3} \\ $$$$\mathrm{ln}\:\left(\lambda+\sqrt{\lambda^{\mathrm{2}} −\mathrm{1}}\right)=\mathrm{ln}\:\mathrm{3} \\ $$$$\lambda+\sqrt{\lambda^{\mathrm{2}} −\mathrm{1}}=\mathrm{3} \\ $$$$\Rightarrow\lambda=\frac{\mathrm{5}}{\mathrm{3}}=\frac{\mathrm{34}{s}+\mathrm{4335}}{\mathrm{136}{s}} \\ $$$$\Rightarrow{s}=\frac{\mathrm{45}}{\mathrm{2}}=\mathrm{22}.\mathrm{5} \\ $$$$\Rightarrow{d}=\mathrm{22}.\mathrm{5}−\mathrm{17}=\mathrm{5}.\mathrm{5} \\ $$$$\Rightarrow{r}_{\mathrm{4}} ={R}−{s}−{r}_{\mathrm{1}} =\mathrm{68}−\frac{\mathrm{51}}{\mathrm{2}}−\frac{\mathrm{45}}{\mathrm{2}}=\mathrm{20} \\ $$$$\frac{\left({s}+{r}_{\mathrm{2}} \right)^{\mathrm{2}} +\left({s}+{r}_{\mathrm{1}} \right)^{\mathrm{2}} −\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{2}\left({s}+{r}_{\mathrm{2}} \right)\left({s}+{r}_{\mathrm{1}} \right)}=\frac{\left({s}+{r}_{\mathrm{2}} \right)^{\mathrm{2}} +{d}^{\mathrm{2}} −\left({R}−{r}_{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{2}\left({s}+{r}_{\mathrm{2}} \right){d}} \\ $$$$\frac{{s}^{\mathrm{2}} +{sr}_{\mathrm{1}} +\left({s}−{r}_{\mathrm{1}} \right){r}_{\mathrm{2}} }{{s}+{r}_{\mathrm{1}} }=\frac{{s}^{\mathrm{2}} +\mathrm{2}\left({R}+{s}\right){r}_{\mathrm{2}} +{d}^{\mathrm{2}} −{R}^{\mathrm{2}} }{\mathrm{2}{d}} \\ $$$$\frac{\left(\frac{\mathrm{45}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{45}}{\mathrm{2}}×\frac{\mathrm{51}}{\mathrm{2}}+\left(\frac{\mathrm{45}}{\mathrm{2}}−\frac{\mathrm{51}}{\mathrm{2}}\right){r}_{\mathrm{2}} }{\frac{\mathrm{45}}{\mathrm{2}}+\frac{\mathrm{51}}{\mathrm{2}}}=\frac{\left(\frac{\mathrm{45}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{68}+\frac{\mathrm{45}}{\mathrm{2}}\right){r}_{\mathrm{2}} +\left(\frac{\mathrm{11}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{68}^{\mathrm{2}} }{\mathrm{2}×\frac{\mathrm{11}}{\mathrm{2}}} \\ $$$$\Rightarrow{r}_{\mathrm{2}} =\frac{\mathrm{1360}}{\mathrm{57}}\approx\mathrm{23}.\mathrm{859649} \\ $$$${similarly} \\ $$$$\frac{\mathrm{2}{s}^{\mathrm{2}} +\mathrm{2}{sr}_{\mathrm{4}} +\mathrm{2}\left({s}−{r}_{\mathrm{4}} \right){r}_{\mathrm{3}} }{{s}+{r}_{\mathrm{4}} }=\frac{{s}^{\mathrm{2}} +\mathrm{2}\left({R}+{s}\right){r}_{\mathrm{3}} +{d}^{\mathrm{2}} −{R}^{\mathrm{2}} }{−{d}} \\ $$$$\frac{\left(\frac{\mathrm{45}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{45}}{\mathrm{2}}×\mathrm{20}+\left(\frac{\mathrm{45}}{\mathrm{2}}−\mathrm{20}\right){r}_{\mathrm{3}} }{\frac{\mathrm{45}}{\mathrm{2}}+\mathrm{20}}=\frac{\left(\frac{\mathrm{45}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{68}+\frac{\mathrm{45}}{\mathrm{2}}\right){r}_{\mathrm{3}} +\left(\frac{\mathrm{11}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{68}^{\mathrm{2}} }{−\mathrm{2}×\frac{\mathrm{11}}{\mathrm{2}}} \\ $$$$\Rightarrow{r}_{\mathrm{3}} =\frac{\mathrm{4080}}{\mathrm{193}}\approx\mathrm{21}.\mathrm{139}\:\mathrm{896} \\ $$$${c}={r}_{\mathrm{2}} +{r}_{\mathrm{3}} =\frac{\mathrm{1360}}{\mathrm{57}}+\frac{\mathrm{4080}}{\mathrm{193}}=\frac{\mathrm{495040}}{\mathrm{11001}}\approx\mathrm{44}.\mathrm{999}\:\mathrm{545}\:\mathrm{496} \\ $$
Commented by mr W last updated on 23/Mar/24
