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Question Number 28808 by ajfour last updated on 30/Jan/18

Answered by ajfour last updated on 30/Jan/18

$$takingAasorigin,$$$$and\mathrm{\angle }AED=\alpha ={\tan }^{-1}2$$$$eq.ofAC\mathit{y}=\mathit{x}$$$$eq.ofDE\mathit{y}=\mathit{a}-2\mathit{x}$$$$\mathit{F}liesonbothlines$$$$so{\mathit{x}}_{\mathit{F}}={\mathit{y}}_{\mathit{F}}=\frac{\mathit{a}}{3}$$$$\frac{{\mathit{r}}_1}{\cot 22.5°+\cot \left(\frac{\alpha }{2}\right)}=\frac{a}{2}$$$$let\cot 22.5°=\frac{1}{m_1}$$$$\frac{2m_1}{1-m_1^2}=1\Rightarrow m_1^2+2m_1=1$$$$m_1=\frac{-2+2\sqrt{2}}{2}=\sqrt{2}-1$$$$\frac{1}{m_1}=\sqrt{2}+1$$$$let\tan \frac{\alpha }{2}=m$$$$\tan \alpha ==\frac{2m}{1-m^2}=2$$$$\Rightarrow 2m^2+2m-2=0$$$$orm=\frac{-2+2\sqrt{5}}{4}=\frac{\sqrt{5}-1}{2}$$$$\cot \frac{\alpha }{2}=\frac{1}{m}=\frac{\sqrt{5}+1}{2}$$$$As{\mathit{r}}_1(\cot 22.5°+\cot \frac{\mathit{\alpha }}{2})=\frac{a}{2},$$$${\mathit{r}}_1=\frac{a/2}{\sqrt{2}+1+\frac{\sqrt{5}+1}{2}}=\frac{\mathit{a}}{2\sqrt{2}+\sqrt{5}+3}$$$$....$$$$\tan (45°-\frac{\alpha }{2})=\frac{1-\left(\frac{\sqrt{5}-1}{2}\right)}{1+\frac{\sqrt{5}-1}{2}}$$$$=\frac{3-\sqrt{5}}{1+\sqrt{5}}=$$$$\cot (45°-\frac{\alpha }{2})=\frac{1+\sqrt{5}}{3-\sqrt{5}}=\frac{3+5+2\sqrt{5}}{4}$$$$=2+\frac{\sqrt{5}}{2}$$$${\mathit{r}}_2(\cot 22.5°+\cot (45°-\frac{\alpha }{2})=\mathit{a}$$$$\Rightarrow {\mathit{r}}_2=\frac{\mathit{a}}{\sqrt{2}+3+\frac{\sqrt{5}}{2}}$$$$.....$$$$\cot (45°+\frac{\alpha }{2})=\tan (45°-\frac{\alpha }{2})$$$$=\frac{3-\sqrt{5}}{\sqrt{5}+1}=\frac{4\sqrt{5}-8}{4}=\sqrt{5}-2$$$${\mathit{r}}_3[\cot 22.5°+\cot (45°+\frac{\alpha }{2})]=\mathit{a}$$$$\Rightarrow {\mathit{r}}_3=\frac{\mathit{a}}{\sqrt{2}+\sqrt{5}-1}$$$$.......$$

Answered by mrW2 last updated on 30/Jan/18

$$AF=\frac{AC}{3}=\frac{\sqrt{2}}{3}a$$$$EF=\frac{ED}{3}=\frac{1}{3}\times \frac{\sqrt{5}}{2}a=\frac{\sqrt{5}}{6}a$$$$FC=\frac{2}{3}AC=\frac{2\sqrt{2}}{3}a$$$$FD=\frac{2}{3}ED=\frac{\sqrt{5}}{3}a$$$$$$$$Usinggeneralformula:$$$$r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$$$$$$$$Circle1in\mathrm{\Delta }AEF:$$$$s=\frac{1}{2}(\frac{1}{2}a+\frac{\sqrt{5}}{6}a+\frac{\sqrt{2}}{3}a)=\frac{3+\sqrt{5}+2\sqrt{2}}{12}a$$$$r_1=a\sqrt{\frac{(\frac{3+\sqrt{5}+2\sqrt{2}}{12}-\frac{1}{2})(\frac{3+\sqrt{5}+2\sqrt{2}}{12}-\frac{\sqrt{5}}{6})(\frac{3+\sqrt{5}+2\sqrt{2}}{12}-\frac{\sqrt{2}}{3})}{\frac{3+\sqrt{5}+2\sqrt{2}}{12}}}$$$$r_1=\frac{a}{12}\sqrt{\frac{(\sqrt{5}+2\sqrt{2}-3)(3+2\sqrt{2}-\sqrt{5})(3+\sqrt{5}-2\sqrt{2})}{3+\sqrt{5}+2\sqrt{2}}}$$$$\Rightarrow r_1=\frac{a}{12}\sqrt{2(9-4\sqrt{2}-\sqrt{5})}\approx 0.12a$$$$$$$$Circle2in\mathrm{\Delta }AFD:$$$$s=\frac{1}{2}(a+\frac{\sqrt{2}}{3}a+\frac{\sqrt{5}}{3}a)=\frac{3+\sqrt{2}+\sqrt{5}}{6}a$$$$r_2=a\sqrt{\frac{(\frac{3+\sqrt{2}+\sqrt{5}}{6}-1)(\frac{3+\sqrt{2}+\sqrt{5}}{6}-\frac{\sqrt{2}}{3})(\frac{3+\sqrt{2}+\sqrt{5}}{6}-\frac{\sqrt{5}}{3})}{\frac{3+\sqrt{2}+\sqrt{5}}{6}}}$$$$r_2=\frac{a}{6}\sqrt{\frac{(-3+\sqrt{2}+\sqrt{5})(3-\sqrt{2}+\sqrt{5})(3+\sqrt{2}-\sqrt{5})}{3+\sqrt{2}+\sqrt{5}}}$$$$\Rightarrow r_2=\frac{a}{6}\sqrt{2(12-7\sqrt{2}-5\sqrt{5}+3\sqrt{10})}\approx 0.15a$$$$$$$$Circle3in\mathrm{\Delta }DFC:$$$$s=\frac{1}{2}(a+\frac{\sqrt{5}}{3}a+\frac{2\sqrt{2}}{3}a)=\frac{3+\sqrt{5}+2\sqrt{2}}{6}a$$$$r_3=a\sqrt{\frac{(\frac{3+\sqrt{5}+2\sqrt{2}}{6}-1)(\frac{3+\sqrt{5}+2\sqrt{2}}{6}-\frac{\sqrt{5}}{3})(\frac{3+\sqrt{5}+2\sqrt{2}}{6}-\frac{2\sqrt{2}}{3})}{\frac{3+\sqrt{5}+2\sqrt{2}}{6}}}$$$$r_3=\frac{a}{6}\sqrt{\frac{(-3+\sqrt{5}+2\sqrt{2})(3-\sqrt{5}+2\sqrt{2})(3+\sqrt{5}-2\sqrt{2})}{3+\sqrt{5}+2\sqrt{2}}}$$$$\Rightarrow r_3=\frac{a}{6}\sqrt{2(9-4\sqrt{2}-\sqrt{5})}=2r_1\approx 0.25a$$$$$$$$Circle4in\mathrm{\Delta }ABC:$$$$s=\frac{1}{2}(a+a+\sqrt{2}a)=\frac{2+\sqrt{2}}{2}a$$$$r_4=a\sqrt{\frac{(\frac{2+\sqrt{2}}{2}-1)(\frac{2+\sqrt{2}}{2}-1)(\frac{2+\sqrt{2}}{2}-\sqrt{2})}{\frac{2+\sqrt{2}}{2}}}$$$$r_4=\frac{a}{2}\sqrt{2(3-2\sqrt{2})}\approx 0.29a$$

Commented by ajfour last updated on 30/Jan/18

$$\mathcal{S}plendid!Sir.$$

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