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Question Number 77504 by Maclaurin Stickker last updated on 07/Jan/20

Answered by mr W last updated on 07/Jan/20

$$saytheradiusoflargeoutercircle=R$$

Commented by mr W last updated on 07/Jan/20

Commented by mr W last updated on 07/Jan/20

$$\frac{a}{R-a}=\sin 30°=\frac{1}{2}$$$$\Rightarrow a=\frac{R}{3}$$$$\frac{x}{R-2a-x}=\sin 30°=\frac{1}{2}$$$$3x=R-2a=R-\frac{2R}{3}=\frac{R}{3}$$$$\Rightarrow x=\frac{R}{9}$$

Commented by mr W last updated on 07/Jan/20

Commented by mr W last updated on 07/Jan/20

$$OT=R\cos 30°=\frac{\sqrt{3}R}{2}$$$$2b=R-\frac{\sqrt{3}R}{2}=\frac{(2-\sqrt{3})R}{2}$$$$\Rightarrow b=\frac{(2-\sqrt{3})R}{4}$$$$\cos \beta =\frac{{OP}^2+{OQ}^2-{PQ}^2}{2\times OP\times OQ}=\frac{{(R-r)}^2+{(R-b)}^2-{(r+b)}^2}{2(R-r)(R-b)}$$$$\cos \beta =\frac{R^2-(R+b)r-Rb}{(R-r)(R-b)}$$$$\cos \beta =\frac{R-(15-8\sqrt{3})r}{(R-r)}$$$${ST}^2={(b+r)}^2-{(b-r)}^2=4br$$$$\Rightarrow ST=2\sqrt{br}=\sqrt{(2-\sqrt{3})Rr}$$$$ST=PS\tan \beta +OT\tan \beta =(r+\frac{\sqrt{3}R}{2})\tan \beta$$$$ST=(r+\frac{\sqrt{3}R}{2})\frac{\sqrt{{(R-r)}^2-{[R-(15-8\sqrt{3})r]}^2}}{R-(15-8\sqrt{3})r}$$$$ST=(\sqrt{3}R+2r)(2-\sqrt{3})\frac{\sqrt{[R-4(2-\sqrt{3})r]r}}{R-(15-8\sqrt{3})r}$$$$\sqrt{(2-\sqrt{3})Rr}=(\sqrt{3}R+2r)(2-\sqrt{3})\frac{\sqrt{[R-4(2-\sqrt{3})r]r}}{R-(15-8\sqrt{3})r}$$$$\sqrt{(2+\sqrt{3})R}=(\sqrt{3}R+2r)\frac{\sqrt{R-4(2-\sqrt{3})r}}{R-(15-8\sqrt{3})r}$$$$\frac{1-(15-8\sqrt{3})\frac{r}{R}}{\sqrt{3}+2\frac{r}{R}}=\sqrt{(2-\sqrt{3})[1-4(2-\sqrt{3})\frac{r}{R}]}$$$$\Rightarrow \frac{1-(15-8\sqrt{3})\lambda }{\sqrt{3}+2\lambda }=\sqrt{(2-\sqrt{3})[1-4(2-\sqrt{3})\lambda ]}$$$$\Rightarrow \lambda =\frac{1}{16}$$$$\Rightarrow r=\frac{R}{16}$$$$\Rightarrow \frac{x}{r}=\frac{\frac{R}{9}}{\frac{R}{16}}=\frac{16}{9}$$

Commented by Maclaurin Stickker last updated on 07/Jan/20

$$Amazing$$

Commented by TawaTawa last updated on 07/Jan/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{Thanks}\mathrm{for}\mathrm{your}\mathrm{time}.$$

Commented by mr W last updated on 08/Jan/20

$$alternative:$$$$ST=\sqrt{(2-\sqrt{3})Rr}=OP\times \sin \beta$$$$(2-\sqrt{3})Rr={OP}^2\times (1-{\cos }^2\beta )$$$$(2-\sqrt{3})Rr={(R-r)}^2(1-{\left[\frac{R-(15-8\sqrt{3})r}{(R-r)}\right]}^2)$$$$(2-\sqrt{3})Rr={(R-r)}^2-{[R-(15-8\sqrt{3})r]}^2$$$$(2-\sqrt{3})Rr=4[R-4(2-\sqrt{3})r]{(2-\sqrt{3})}^{2}r$$$$R=4[R-4(2-\sqrt{3})r](2-\sqrt{3})$$$$\Rightarrow r=\frac{R}{4(2-\sqrt{3})}[1-\frac{1}{4(2-\sqrt{3})}]$$$$\Rightarrow r=\frac{R}{16}\left[\frac{7-4\sqrt{3}}{{(2-\sqrt{3})}^2}\right]$$$$\Rightarrow r=\frac{R}{16}\left[\frac{{(2-\sqrt{3})}^2}{{(2-\sqrt{3})}^2}\right]$$$$\Rightarrow r=\frac{R}{16}$$

Commented by TawaTawa last updated on 08/Jan/20

$$\mathrm{Wow},\mathrm{great}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by Tawa11 last updated on 29/Dec/21

$$\mathrm{This}\mathrm{too}.\mathrm{Hahaha}$$

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