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Question Number 53168 by ajfour last updated on 18/Jan/19

Commented by ajfour last updated on 19/Jan/19

$$Equilateral△ABCcontains$$$$twocirclesofradiiaandbinthe$$$$mannershown.Findtheradius$$$$Rofacirclethattouchesthesetwo$$$$circlesexternallyandpassesthrough$$$$vertexAintermsof\mathit{a}and\mathit{b}.$$

Commented by behi83417@gmail.com last updated on 18/Jan/19

$$R=b.\frac{a+b}{a-b}?$$

Commented by ajfour last updated on 18/Jan/19

$$soquick!howdoyougetit?$$

Commented by behi83417@gmail.com last updated on 18/Jan/19

$$sidesismissinginyourcommonet.$$

Commented by ajfour last updated on 18/Jan/19

$$sisnotevenrequired,solutionsir?$$

Commented by MJS last updated on 20/Jan/19

$$s=1\wedge \frac{\sqrt{3}}{18}<a<\frac{\sqrt{3}}{6}:b=\frac{1}{3}(\sqrt{3}-a-2\sqrt{\sqrt{3}a-2a^2})$$$$\mathrm{for}\mathrm{these}\mathrm{values}\mathrm{both}\mathrm{circles}\mathrm{stay}\mathrm{within}\mathrm{the}$$$$\mathrm{triangle}\mathrm{and}\mathrm{the}\mathrm{big}\mathrm{circle}\mathrm{exists}$$$$\mathrm{I}{\mathrm{haven}}^{\prime }\mathrm{t}\mathrm{been}\mathrm{able}\mathrm{to}\mathrm{get}\mathrm{the}\mathrm{coordinates}$$$$\mathrm{of}P\mathrm{or}\mathrm{the}\mathrm{value}\mathrm{of}r$$

Answered by mr W last updated on 19/Jan/19

Commented by mr W last updated on 19/Jan/19

$$s=\sqrt{3}(a+b)+\sqrt{{(a+b)}^2-{(a-b)}^2}=\sqrt{3}(a+b)+2\sqrt{ab}$$$$AM=\sqrt{{(s-\sqrt{3}a)}^2+a^2}=\sqrt{{(\sqrt{3}b+2\sqrt{ab})}^2+a^2}=\sqrt{a^2+3b^2+4ab+4b\sqrt{3ab}}$$$$AN=\sqrt{(s-\sqrt{3}b)+b^2}=\sqrt{3a^2+b^2+4ab+4a\sqrt{ab}}$$$$MN=a+b$$$$PM=a+R$$$$PN=b+R$$$$AP=R$$$$let\mathrm{\angle }APM=\alpha ,\mathrm{\angle }MPN=\beta$$$$\cos \alpha =\frac{R^2+{(R+a)}^2-(a^2+3b^2+4ab+4b\sqrt{3ab})}{2R(R+a)}$$$$\Rightarrow \cos \alpha =\frac{2R^2+2aR-3b^2-4ab-4b\sqrt{3ab}}{2R(R+a)}$$$$\cos \beta =\frac{{(R+a)}^2+{(R+b)}^2-{(a+b)}^2}{2(R+a)(R+b)}$$$$\Rightarrow \cos \beta =\frac{R^2+(a+b)R-ab}{(R+a)(R+b)}$$$$\cos (\alpha +\beta )=\frac{R^2+{(R+b)}^2-(3a^2+b^2+4ab+4a\sqrt{ab})}{2R(R+b)}$$$$\Rightarrow \cos (\alpha +\beta )=\frac{2R^2+2bR-3a^2-4ab-4a\sqrt{ab}}{2R(R+b)}$$$$$$$$\Rightarrow {\cos }^{-1}\frac{2R^2+2bR-3a^2-4ab-4a\sqrt{ab}}{2R(R+b)}={\cos }^{-1}\frac{2R^2+2aR-3b^2-4ab-4b\sqrt{3ab}}{2R(R+a)}+{\cos }^{-1}\frac{R^2+(a+b)R-ab}{(R+a)(R+b)}$$$$let\mu =\frac{b}{a},\lambda =\frac{R}{a}$$$$\Rightarrow {\cos }^{-1}\{1-\frac{3+4\mu +4\sqrt{\mu }}{2\lambda (\lambda +\mu )}\}={\cos }^{-1}\{1-\frac{3{\mu }^2+4\mu +4\mu \sqrt{3\mu }}{2\lambda (\lambda +1)}\}+{\cos }^{-1}\{1-\frac{2\mu }{(\lambda +1)(\lambda +\mu )}\}$$$$\Rightarrow \lambda =\lambda \left(\mu \right)....$$

Commented by ajfour last updated on 19/Jan/19

$$Thankssir,verybetterway,ihad$$$$resortedtoacoordinatelikemethod.$$

Answered by ajfour last updated on 19/Jan/19

Commented by ajfour last updated on 19/Jan/19

$$letmidpointofBCbeoriginO,$$$$withx-axisalongBC.$$$$M(-\frac{s}{2}+a\sqrt{3},a);N(\frac{s}{2}-b\sqrt{3},b)$$$$A(0,\frac{s\sqrt{3}}{2});s=\sqrt{3}(a+b)+2\sqrt{ab}$$$$letP(h,k).AsMP=a+R$$$${(h+\frac{s}{2}-a\sqrt{3})}^2+{(k-a)}^2={(a+R)}^2$$$$AsNP=b+R$$$${(h-\frac{s}{2}+b\sqrt{3})}^2+{(k-b)}^2={(b+R)}^2$$$$asAP=R$$$$h^2+{(\frac{s\sqrt{3}}{2}-k)}^2=R^2.$$

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