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Question Number 212949 by Spillover last updated on 27/Oct/24

Answered by som(math1967) last updated on 27/Oct/24

Commented by som(math1967) last updated on 27/Oct/24

$$AB=BC=CD=AD=x$$$$DE=r\sqrt{2}\Rightarrow OD=r\sqrt{2}+r+R$$$$BD=2(r\sqrt{2}+r+R)$$$$\therefore x=\frac{BD}{\sqrt{2}}=\sqrt{2}(r\sqrt{2}+r+R)$$$$\frac{FG}{BG}=\tan 22\frac{1}{2}$$$$BG=R\times cot22\frac{1}{2}=R(\sqrt{2}+1)$$$$AB=x=AG+GB=R+R(\sqrt{2}+1)$$$$\therefore \sqrt{2}(r\sqrt{2}+r+R)=R(2+\sqrt{2})$$$$\Rightarrow r(2+\sqrt{2})=R(2+\sqrt{2}-\sqrt{2})$$$$\frac{r}{R}=\frac{2}{(2+\sqrt{2})}=\frac{2-\sqrt{2}}{1}$$

Answered by ajfour last updated on 27/Oct/24

$$r(1+\sqrt{2})+R=R(1+\sqrt{2})=\frac{a}{\sqrt{2}}$$$$\frac{r}{R}=\frac{\sqrt{2}}{1+\sqrt{2}}=2-\sqrt{2}\approx 0.586$$$$R=a(1-\frac{1}{\sqrt{2}})\approx 0.393a$$$$r=(3-2\sqrt{2})a\approx 0.172a$$

Answered by A5T last updated on 27/Oct/24

$$Letsidebes\Rightarrow diagonal=s\sqrt{2}$$$$\frac{s^2}{2}=\frac{R}{2}(2s+s\sqrt{2})\Rightarrow R=\frac{s}{2+\sqrt{2}}\Rightarrow \frac{s}{R}=2+\sqrt{2}$$$$r\sqrt{2}+r+R=\frac{s\sqrt{2}}{2}\stackrel{/R}{\Rightarrow }\frac{r}{R}(1+\sqrt{2})=\frac{\sqrt{2}s}{2R}-1$$$$\Rightarrow \frac{r}{R}=\frac{\sqrt{2}}{1+\sqrt{2}}=\frac{\left(\sqrt{2}\right)(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}=2-\sqrt{2}$$

Answered by Spillover last updated on 27/Oct/24

Answered by Spillover last updated on 27/Oct/24

$$$$R+√2R=R+r+√2r √2R=r+√2r r/R=√2/(1+√2) =√2(√2-1)/(2-1) =2-√2

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