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Question Number 124821 by ajfour last updated on 06/Dec/20

Commented by ajfour last updated on 06/Dec/20

$$Ifthecirclesinredandbluehave$$$$equalradii,findAB.$$

Commented by ajfour last updated on 06/Dec/20

Answered by mr W last updated on 06/Dec/20

Commented by mr W last updated on 06/Dec/20

$$R=1=radiusofsemicircle$$$$let\lambda =\frac{r}{R}$$$$\cos \theta =\frac{R-2r}{R}=1-2\lambda$$$$\tan \alpha =\frac{r}{R+\sqrt{{(R-r)}^2-r^2}}=\frac{\lambda }{1+\sqrt{1-2\lambda }}$$$$2\alpha =\frac{\pi }{2}-\theta$$$$\tan 2\alpha =\frac{1}{\tan \theta }$$$$\frac{2\times \frac{\lambda }{1+\sqrt{1-2\lambda }}}{1-{\left(\frac{\lambda }{1+\sqrt{1-2\lambda }}\right)}^2}=\frac{1-2\lambda }{\sqrt{1-{(1-2\lambda )}^2}}$$$$\frac{2\lambda (1+\sqrt{1-2\lambda })}{{(1+\sqrt{1-2\lambda })}^2-{\lambda }^2}=\frac{1-2\lambda }{\sqrt{1-{(1-2\lambda )}^2}}$$$$\Rightarrow \lambda \approx 0.3129$$$$\sin \beta =\frac{r}{R-r}=\frac{\lambda }{1-\lambda }$$$$AB=2(R-r)\sin \frac{\pi -\theta -\beta }{2}$$$$\frac{AB}{R}=2(1-\lambda )\cos \frac{\theta +\beta }{2}$$$$=2(1-\lambda )\cos \frac{1}{2}[{\cos }^{-1}(1-2\lambda )+{\sin }^{-1}\left(\frac{\lambda }{1-\lambda }\right)]$$$$\approx 0.9274$$

Commented by MJS_new last updated on 06/Dec/20

$$\mathrm{btw}.\lambda \mathrm{is}\mathrm{the}\mathrm{real}\mathrm{solution}\mathrm{of}$$$${\lambda }^3-5{\lambda }^2+\frac{57}{4}\lambda -4=0$$$$\lambda =\frac{5}{3}+\frac{1}{6}\sqrt[3]{-1133+120\sqrt{114}}-\frac{1}{6}\sqrt[3]{1133+120\sqrt{114}}$$$$\mathrm{but}\mathrm{this}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{make}\mathrm{anything}\mathrm{better}...$$

Commented by mr W last updated on 06/Dec/20

$$wow!istopedtoconsiderifthe$$$$equationcanbesimplifiedfurther.$$$$youshowedanexactsolutionexists.$$$$thanks!$$

Commented by MJS_new last updated on 06/Dec/20

$$\frac{2\lambda (1+\sqrt{1-2\lambda })}{{(1+\sqrt{1-2\lambda })}^2-{\lambda }^2}=\frac{1-2\lambda }{\sqrt{1-{(1-2\lambda )}^2}}$$$$\mathrm{let}t=\sqrt{1-2x}\wedge t⩾0\iff x=\frac{1-t^2}{2}$$$$\frac{4(t-1)}{(t-3)(t+1)}=\frac{t^2}{\sqrt{1-t^4}}$$$$\mathrm{squaring}\mathrm{and}\mathrm{transforming}$$$$(t+1){(t^2-t+2)}^2(t^3-3t^2+8t-4)=0$$$$\mathrm{as}t⩾0$$$$t^3-3t^2+8t-4=0$$$$\mathrm{and}\mathrm{from}\mathrm{here}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}$$

Commented by ajfour last updated on 06/Dec/20

$$ThanksmrWSirforsolving,$$$$NMjSSirforinterfering!$$

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