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Question Number 205507 by mr W last updated on 23/Mar/24

Commented by mr W last updated on 23/Mar/24

$$anunsolvedoldquestion$$

Answered by mr W last updated on 25/Mar/24

Commented by mr W last updated on 23/Mar/24

$$R=\frac{85+51}{2}=68$$$$r_1=\frac{51}{2}=25.5$$$$R+d=s+2r_1$$$$\Rightarrow d=s-17$$$$\lambda =\frac{R^2+s^2-d^2}{2Rs}=\frac{{68}^2+s^2-d^2}{136s}=\frac{{68}^2+s^2-{(s-17)}^2}{136s}=\frac{34s+4335}{136s}$$$$for6circlesbetweentheouter$$$$andtheinnercirclestheinversive$$$$distancefromthemshouldbe$$$$\delta ={\cosh }^{-1}\lambda =2\ln \left[\tan (\frac{\pi }{4}+\frac{\pi }{2n})\right]$$$$withn=6$$$${\cosh }^{-1}\lambda =2\ln \left[\tan (\frac{\pi }{4}+\frac{\pi }{2\times 6})\right]$$$${\cosh }^{-1}\lambda =2\ln \left(\tan \frac{\pi }{3}\right)$$$${\cosh }^{-1}\lambda =\ln 3$$$$\ln (\lambda +\sqrt{{\lambda }^2-1})=\ln 3$$$$\lambda +\sqrt{{\lambda }^2-1}=3$$$$\Rightarrow \lambda =\frac{5}{3}=\frac{34s+4335}{136s}$$$$\Rightarrow s=\frac{45}{2}=22.5$$$$\Rightarrow d=22.5-17=5.5$$$$\Rightarrow r_4=R-s-r_1=68-\frac{51}{2}-\frac{45}{2}=20$$$$\frac{{(s+r_2)}^2+{(s+r_1)}^2-{(r_1+r_2)}^2}{2(s+r_2)(s+r_1)}=\frac{{(s+r_2)}^2+d^2-{(R-r_2)}^2}{2(s+r_2)d}$$$$\frac{s^2+{sr}_1+(s-r_1)r_2}{s+r_1}=\frac{s^2+2(R+s)r_2+d^2-R^2}{2d}$$$$\frac{{\left(\frac{45}{2}\right)}^2+\frac{45}{2}\times \frac{51}{2}+(\frac{45}{2}-\frac{51}{2})r_2}{\frac{45}{2}+\frac{51}{2}}=\frac{{\left(\frac{45}{2}\right)}^2+2(68+\frac{45}{2})r_2+{\left(\frac{11}{2}\right)}^2-{68}^2}{2\times \frac{11}{2}}$$$$\Rightarrow r_2=\frac{1360}{57}\approx 23.859649$$$$similarly$$$$\frac{2s^2+2{sr}_4+2(s-r_4)r_3}{s+r_4}=\frac{s^2+2(R+s)r_3+d^2-R^2}{-d}$$$$\frac{{\left(\frac{45}{2}\right)}^2+\frac{45}{2}\times 20+(\frac{45}{2}-20)r_3}{\frac{45}{2}+20}=\frac{{\left(\frac{45}{2}\right)}^2+2(68+\frac{45}{2})r_3+{\left(\frac{11}{2}\right)}^2-{68}^2}{-2\times \frac{11}{2}}$$$$\Rightarrow r_3=\frac{4080}{193}\approx 21.139896$$$$c=r_2+r_3=\frac{1360}{57}+\frac{4080}{193}=\frac{495040}{11001}\approx 44.999545496$$

Commented by mr W last updated on 23/Mar/24

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