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Question Number 201604 by ajfour last updated on 09/Dec/23

Commented by ajfour last updated on 09/Dec/23

$$Bothcirclesinbluehaveequal$$$$radii.Find.$$

Answered by mr W last updated on 09/Dec/23

Commented by mr W last updated on 09/Dec/23

$$\lambda =\frac{r}{R}$$$$\sin \beta =\frac{R-2r}{R}=1-2\lambda$$$$\tan \frac{\beta }{2}=\frac{\sin \beta }{1+\cos \beta }=\frac{1-2\lambda }{1+2\sqrt{\lambda (1-\lambda )}}$$$$\alpha =\frac{\frac{\pi }{2}-\beta }{2}=\frac{\pi }{4}-\frac{\beta }{2}$$$$DC=DG=\frac{r}{\tan \alpha }=\frac{r(1+\tan \frac{\beta }{2})}{1-\tan \frac{\beta }{2}}$$$$=\frac{r(1+\frac{1-2\lambda }{1+2\sqrt{\lambda (1-\lambda )}})}{1-\frac{1-2\lambda }{1+2\sqrt{\lambda (1-\lambda )}}}=\frac{r(1-\lambda +\sqrt{\lambda (1-\lambda )})}{\sqrt{\lambda (1-\lambda )}+\lambda }$$$$EF=\sqrt{{(R+r)}^2-{(R-r)}^2}=2\sqrt{Rr}$$$$DH=\frac{r(1-\lambda +\sqrt{\lambda (1-\lambda )})}{\sqrt{\lambda (1-\lambda )}+\lambda }+2\sqrt{Rr}$$$$DH=2R\tan \beta =\frac{2R(1-2\lambda )}{2\sqrt{\lambda (1-\lambda )}}=\frac{R(1-2\lambda )}{\sqrt{\lambda (1-\lambda )}}$$$$\frac{r(1-\lambda +\sqrt{\lambda (1-\lambda )})}{\sqrt{\lambda (1-\lambda )}+\lambda }+2\sqrt{Rr}=\frac{R(1-2\lambda )}{\sqrt{\lambda (1-\lambda )}}$$$$\frac{\lambda (1-\lambda +\sqrt{\lambda (1-\lambda )})}{\sqrt{\lambda (1-\lambda )}+\lambda }+2\sqrt{\lambda }=\frac{1-2\lambda }{\sqrt{\lambda (1-\lambda )}}$$$$\Rightarrow \lambda \approx 0.21995157$$

Commented by ajfour last updated on 09/Dec/23

$$Thankyousir.Igotsame.Iwill$$$$analyseyourwayonlynow.$$

Answered by ajfour last updated on 09/Dec/23

Commented by ajfour last updated on 09/Dec/23

$$Eq.ofline$$$$y+x\tan 2\theta =1$$$$\cos 2\theta =1-2r$$$$\Rightarrow r=\sin {}^2\theta$$$$2\sqrt{r}+\frac{r\cos \theta }{\sin \theta }=\frac{2}{\tan 2\theta }$$$$2\sin \theta +\sin \theta \cos \theta =\frac{2\cos {}^2\theta -1}{\sin \theta \cos \theta }$$$$say\cos \theta =t$$$$(1-t^2)t(2+t)=2t^2-1$$$$\Rightarrow t\approx 0.8832$$$$r=\sin {}^2\theta =1-t^2\approx 0.21995$$

Commented by mr W last updated on 09/Dec/23

$$nicesolution!$$

Commented by Frix last updated on 10/Dec/23

$$\mathrm{We}\mathrm{can}\mathrm{give}\mathrm{the}\mathrm{exact}\mathrm{solution}$$$$(1-t^2)t(2+t)=2t^2-1$$$$\iff$$$$t^4+2t^3+t^2-2t-1=0$$$$t=\frac{-1+\sqrt{2}\pm \sqrt{-1+2\sqrt{2}}}{2}$$$$t=\frac{-1-\sqrt{2}\pm \mathrm{i}\sqrt{1+2\sqrt{2}}}{2}$$

Commented by ajfour last updated on 28/Oct/24

$$Thanksfortheexact!$$

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