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Question Number 161742 by mr W last updated on 21/Dec/21

Commented by mr W last updated on 21/Dec/21

$$findtheradiiofthecirclesinside$$$$thebigquatercircle.$$

Answered by FongXD last updated on 22/Dec/21

$$\overline{)\begin{array}{c}\mathrm{Find}\mathrm{the}\mathrm{radius}\left(\mathrm{r}\right)\mathrm{of}\mathrm{the}\mathrm{small}\mathrm{circle}\end{array}}:$$$$\bullet \tan 2\alpha =\frac{8}{4}=2=\frac{2\tan \alpha }{1-{\tan }^2\alpha },\mathrm{but}\tan \alpha =\frac{\mathrm{r}}{\mathrm{FE}}$$$$\iff \frac{2\mathrm{r}\left(\mathrm{FE}\right)}{{\left(\mathrm{FE}\right)}^2-{\mathrm{r}}^2}=2,\iff {\left(\mathrm{FE}\right)}^2-\mathrm{r}\left(\mathrm{FE}\right)-{\mathrm{r}}^2=0$$$$\Rightarrow \mathrm{FE}=\mathrm{r}\left(\frac{1+\sqrt{5}}{2}\right)$$$$▴\mathrm{in}\mathrm{the}\mathrm{right}\mathrm{triangle}\mathrm{OFD}:$$$$\bullet {\mathrm{FD}}^2={\mathrm{OD}}^2-{\mathrm{OF}}^2$$$$\iff \mathrm{FD}=\sqrt{{(8-\mathrm{r})}^2-{\mathrm{r}}^2}$$$$\iff 4+\mathrm{FE}=4+\mathrm{r}\left(\frac{1+\sqrt{5}}{2}\right)=\sqrt{64-16\mathrm{r}}$$$$\iff 64+16\mathrm{r}(1+\sqrt{5})+{\mathrm{r}}^2(6+2\sqrt{5})=256-64\mathrm{r}$$$$\iff (3+\sqrt{5}){\mathrm{r}}^2+8\mathrm{r}(5+\sqrt{5})-96=0$$$$\Rightarrow \mathrm{r}=\frac{-4(5+\sqrt{5})+\sqrt{16(30+10\sqrt{5})+96(3+\sqrt{5})}}{3+\sqrt{5}}$$$$\Rightarrow \mathrm{r}=\frac{-4(5+\sqrt{5})+4\sqrt{48+16\sqrt{5}}}{9-5}\times (3-\sqrt{5})$$$$\Rightarrow \mathrm{r}=(5+\sqrt{5})(\sqrt{5}-3)+4(3-\sqrt{5})\sqrt{3+\sqrt{5}}$$$$\Rightarrow \mathrm{r}=2\sqrt{5}-10+(12-4\sqrt{5})\frac{\sqrt{5}+1}{\sqrt{2}}$$$$\Rightarrow \mathrm{r}=2\sqrt{5}-10+\frac{8\sqrt{5}-8}{\sqrt{2}}$$$$\mathrm{therefore},\overline{)\begin{array}{c}\mathrm{r}=4\sqrt{10}+2\sqrt{5}-4\sqrt{2}-10\end{array}}$$$$\overline{)\begin{array}{c}\mathrm{Find}\mathrm{the}\mathrm{radius}\left(\mathrm{R}\right)\mathrm{of}\mathrm{the}\mathrm{big}\mathrm{circle}\end{array}}:$$$$▴\mathrm{in}\mathrm{the}\mathrm{right}\mathrm{triangle}{\mathrm{O}}^{\prime }\mathrm{DG}:$$$$\bullet {\mathrm{DG}}^2={\mathrm{O}}^{\prime }{\mathrm{D}}^2-{\mathrm{O}}^{\prime }{\mathrm{R}}^2$$$$\Rightarrow \mathrm{DG}=\sqrt{{(8-\mathrm{R})}^2-{\mathrm{R}}^2}=4\sqrt{4-\mathrm{R}}$$$$\bullet {\mathrm{O}}^{\prime }\mathrm{H}=\mathrm{CG}=8-\mathrm{DG}=8-4\sqrt{4-\mathrm{R}}$$$$▴\mathrm{in}\mathrm{the}\mathrm{right}\mathrm{triangle}{\mathrm{O}}^{\prime }\mathrm{JE}:$$$$\bullet {\mathrm{O}}^{\prime }{\mathrm{E}}^2={\mathrm{O}}^{\prime }{\mathrm{J}}^2+{\mathrm{JE}}^2$$$$\iff {\mathrm{O}}^{\prime }{\mathrm{E}}^2={\left(4\sqrt{4-\mathrm{R}}\right)}^2+{(4-\mathrm{R})}^2$$$$\Rightarrow {\mathrm{O}}^{\prime }{\mathrm{E}}^2={\mathrm{R}}^2-24\mathrm{R}+80$$$$▴\mathrm{in}\mathrm{the}\mathrm{right}\mathrm{triangle}{\mathrm{O}}^{\prime }\mathrm{IE}:$$$$\bullet {\mathrm{EI}}^2={\mathrm{O}}^{\prime }{\mathrm{E}}^2-{\mathrm{O}}^{\prime }{\mathrm{I}}^2$$$$\Rightarrow \mathrm{EI}=\sqrt{({\mathrm{R}}^2-24\mathrm{R}+80)-{\mathrm{R}}^2}=\sqrt{80-24\mathrm{R}}$$$$▴\mathrm{in}\mathrm{the}\mathrm{right}\mathrm{triangle}\mathrm{ABE}:$$$$\bullet {\mathrm{BE}}^2={\mathrm{AB}}^2+{\mathrm{AE}}^2$$$$\Rightarrow \mathrm{BE}=\sqrt{8^2+4^2}=4\sqrt{5}$$$$\bullet \mathrm{IB}=\mathrm{BE}-\mathrm{EI}=4\sqrt{5}-\sqrt{80-24\mathrm{R}}$$$$▴\mathrm{in}\mathrm{the}\mathrm{right}\mathrm{triangle}{\mathrm{O}}^{\prime }\mathrm{BI}\mathrm{and}{\mathrm{O}}^{\prime }\mathrm{BH}$$$$\bullet {\mathrm{O}}^{\prime }{\mathrm{I}}^2+{\mathrm{IB}}^2={\mathrm{O}}^{\prime }{\mathrm{H}}^2+{\mathrm{HB}}^2$$$$\iff {\mathrm{R}}^2+{(4\sqrt{5}-\sqrt{80-24\mathrm{R}})}^2={(8-4\sqrt{4-\mathrm{R}})}^2+{(8-\mathrm{R})}^2$$$$\iff {\mathrm{R}}^2+(160-24\mathrm{R}-16\sqrt{100-30\mathrm{R}})=(128-16\mathrm{R}-64\sqrt{4-\mathrm{R}})+({\mathrm{R}}^2-16\mathrm{R}+64)$$$$\iff \mathrm{R}-4=2\sqrt{100-30\mathrm{R}}-8\sqrt{4-\mathrm{R}}$$$$\iff {\mathrm{R}}^2-8\mathrm{R}+16=656-184\mathrm{R}-32\sqrt{{30\mathrm{R}}^2-220\mathrm{R}+400}$$$$\iff {\mathrm{R}}^2+176\mathrm{R}-640=-32\sqrt{{30\mathrm{R}}^2-220\mathrm{R}+400}$$$$\iff {\mathrm{R}}^4+{30976\mathrm{R}}^2+409600+{352\mathrm{R}}^3-225280\mathrm{R}-{1280\mathrm{R}}^2={30720\mathrm{R}}^2-225280\mathrm{R}+409600$$$$\iff {\mathrm{R}}^4+{352\mathrm{R}}^3-{1024\mathrm{R}}^2=0$$$$\iff {\mathrm{R}}^2+352\mathrm{R}-1024=0$$$$\mathrm{therefore},\overline{)\begin{array}{c}\mathrm{R}=80\sqrt{5}-176\end{array}}$$

Commented by Tawa11 last updated on 22/Dec/21

$$\mathrm{Great}\mathrm{sir}$$

Commented by mr W last updated on 27/Dec/21

$$allcorrect!thanks!$$

Answered by FongXD last updated on 22/Dec/21

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