Question Number 28808 by ajfour last updated on 30/Jan/18
Answered by ajfour last updated on 30/Jan/18
![taking A as origin, and ∠AED=α =tan^(−1) 2 eq. of AC y=x eq. of DE y=a−2x F lies on both lines so x_F =y_F =(a/3) (r_1 /(cot 22.5°+cot ((α/2)))) = (a/2) let cot 22.5° =(1/m_1 ) ((2m_1 )/(1−m_1 ^2 ))=1 ⇒ m_1 ^2 +2m_1 =1 m_1 =((−2+2(√2))/2) = (√2)−1 (1/m_1 ) = (√2)+1 let tan (α/2)=m tan α==((2m)/(1−m^2 )) = 2 ⇒ 2m^2 +2m−2=0 or m=((−2+2(√5))/4) = (((√5)−1)/2) cot (α/2)= (1/m) = (((√5)+1)/2) As r_1 (cot 22.5°+cot (𝛂/2))=(a/2) , r_1 =((a/2)/( (√2)+1+(((√5)+1)/2))) =(a/(2(√2)+(√5)+3)) .... tan (45°−(α/2))=((1−((((√5)−1)/2)))/(1+(((√5)−1)/2))) =((3−(√5))/(1+(√5))) = cot (45°−(α/2))=((1+(√5))/(3−(√5))) =((3+5+2(√5))/4) = 2+((√5)/2) r_2 (cot 22.5°+cot (45°−(α/2))= a ⇒ r_2 =(a/( (√2)+3+((√5)/2))) ..... cot (45°+(α/2))=tan (45°−(α/2)) =((3−(√5))/( (√5)+1)) =((4(√5)−8)/4) =(√5)−2 r_3 [cot 22.5°+cot (45°+(α/2))]= a ⇒ r_3 =(a/( (√2)+(√5)−1)) .......](https://www.tinkutara.com/question/Q28810.png)
$${taking}\:{A}\:{as}\:{origin}, \\ $$$${and}\:\angle{AED}=\alpha\:=\mathrm{tan}^{−\mathrm{1}} \mathrm{2} \\ $$$${eq}.\:{of}\:{AC}\:\:\:\:\:\:\:\:\:\:\boldsymbol{{y}}=\boldsymbol{{x}} \\ $$$${eq}.\:{of}\:{DE}\:\:\:\:\:\:\:\:\boldsymbol{{y}}=\boldsymbol{{a}}−\mathrm{2}\boldsymbol{{x}} \\ $$$$\boldsymbol{{F}}\:{lies}\:{on}\:{both}\:{lines} \\ $$$${so}\:\:\boldsymbol{{x}}_{\boldsymbol{{F}}} \:=\boldsymbol{{y}}_{\boldsymbol{{F}}} =\frac{\boldsymbol{{a}}}{\mathrm{3}} \\ $$$$\frac{\boldsymbol{{r}}_{\mathrm{1}} }{\mathrm{cot}\:\mathrm{22}.\mathrm{5}°+\mathrm{cot}\:\left(\frac{\alpha}{\mathrm{2}}\right)}\:=\:\frac{{a}}{\mathrm{2}} \\ $$$${let}\:\mathrm{cot}\:\mathrm{22}.\mathrm{5}°\:=\frac{\mathrm{1}}{{m}_{\mathrm{1}} } \\ $$$$\frac{\mathrm{2}{m}_{\mathrm{1}} }{\mathrm{1}−{m}_{\mathrm{1}} ^{\mathrm{2}} }=\mathrm{1}\:\:\Rightarrow\:\:\:{m}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{2}{m}_{\mathrm{1}} =\mathrm{1} \\ $$$${m}_{\mathrm{1}} =\frac{−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}\:=\:\sqrt{\mathrm{2}}−\mathrm{1} \\ $$$$\frac{\mathrm{1}}{{m}_{\mathrm{1}} }\:=\:\sqrt{\mathrm{2}}+\mathrm{1} \\ $$$${let}\:\:\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}={m} \\ $$$$\mathrm{tan}\:\alpha==\frac{\mathrm{2}{m}}{\mathrm{1}−{m}^{\mathrm{2}} }\:=\:\mathrm{2} \\ $$$$\Rightarrow\:\:\:\:\mathrm{2}{m}^{\mathrm{2}} +\mathrm{2}{m}−\mathrm{2}=\mathrm{0} \\ $$$${or}\:\:\:{m}=\frac{−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}}\:=\:\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\mathrm{cot}\:\frac{\alpha}{\mathrm{2}}=\:\frac{\mathrm{1}}{{m}}\:=\:\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}} \\ $$$$\:{As}\:\:\:\:\boldsymbol{{r}}_{\mathrm{1}} \left(\mathrm{cot}\:\mathrm{22}.\mathrm{5}°+\mathrm{cot}\:\frac{\boldsymbol{\alpha}}{\mathrm{2}}\right)=\frac{{a}}{\mathrm{2}}\:\:\:, \\ $$$$\boldsymbol{{r}}_{\mathrm{1}} =\frac{{a}/\mathrm{2}}{\:\sqrt{\mathrm{2}}+\mathrm{1}+\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}}\:=\frac{\boldsymbol{{a}}}{\mathrm{2}\sqrt{\mathrm{2}}+\sqrt{\mathrm{5}}+\mathrm{3}} \\ $$$$…. \\ $$$$\:\:\:\:\:\mathrm{tan}\:\left(\mathrm{45}°−\frac{\alpha}{\mathrm{2}}\right)=\frac{\mathrm{1}−\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{1}+\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{1}+\sqrt{\mathrm{5}}}\:=\: \\ $$$$\mathrm{cot}\:\left(\mathrm{45}°−\frac{\alpha}{\mathrm{2}}\right)=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{3}−\sqrt{\mathrm{5}}}\:=\frac{\mathrm{3}+\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\mathrm{2}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\:\:\:\:\boldsymbol{{r}}_{\mathrm{2}} \left(\mathrm{cot}\:\mathrm{22}.\mathrm{5}°+\mathrm{cot}\:\left(\mathrm{45}°−\frac{\alpha}{\mathrm{2}}\right)=\:\boldsymbol{{a}}\right. \\ $$$$\Rightarrow\:\:\boldsymbol{{r}}_{\mathrm{2}} =\frac{\boldsymbol{{a}}}{\:\sqrt{\mathrm{2}}+\mathrm{3}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}} \\ $$$$….. \\ $$$$\:\:\:\:\mathrm{cot}\:\left(\mathrm{45}°+\frac{\alpha}{\mathrm{2}}\right)=\mathrm{tan}\:\left(\mathrm{45}°−\frac{\alpha}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{5}}+\mathrm{1}}\:=\frac{\mathrm{4}\sqrt{\mathrm{5}}−\mathrm{8}}{\mathrm{4}}\:=\sqrt{\mathrm{5}}−\mathrm{2} \\ $$$$\:\:\boldsymbol{{r}}_{\mathrm{3}} \left[\mathrm{cot}\:\mathrm{22}.\mathrm{5}°+\mathrm{cot}\:\left(\mathrm{45}°+\frac{\alpha}{\mathrm{2}}\right)\right]=\:\boldsymbol{{a}} \\ $$$$\Rightarrow\:\:\boldsymbol{{r}}_{\mathrm{3}} =\frac{\boldsymbol{{a}}}{\:\sqrt{\mathrm{2}}+\sqrt{\mathrm{5}}−\mathrm{1}} \\ $$$$……. \\ $$
Answered by mrW2 last updated on 30/Jan/18
$${AF}=\frac{{AC}}{\mathrm{3}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}{a} \\ $$$${EF}=\frac{{ED}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{3}}×\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}{a}=\frac{\sqrt{\mathrm{5}}}{\mathrm{6}}{a} \\ $$$${FC}=\frac{\mathrm{2}}{\mathrm{3}}{AC}=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}{a} \\ $$$${FD}=\frac{\mathrm{2}}{\mathrm{3}}{ED}=\frac{\sqrt{\mathrm{5}}}{\mathrm{3}}{a} \\ $$$$ \\ $$$${Using}\:{general}\:{formula}: \\ $$$${r}=\sqrt{\frac{\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)}{{s}}} \\ $$$$ \\ $$$${Circle}\:\mathrm{1}\:{in}\:\Delta{AEF}: \\ $$$${s}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}{a}+\frac{\sqrt{\mathrm{5}}}{\mathrm{6}}{a}+\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}{a}\right)=\frac{\mathrm{3}+\sqrt{\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{12}}{a} \\ $$$${r}_{\mathrm{1}} ={a}\sqrt{\frac{\left(\frac{\mathrm{3}+\sqrt{\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\frac{\mathrm{3}+\sqrt{\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{12}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{6}}\right)\left(\frac{\mathrm{3}+\sqrt{\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{12}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}\right)}{\frac{\mathrm{3}+\sqrt{\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{12}}}} \\ $$$${r}_{\mathrm{1}} =\frac{{a}}{\mathrm{12}}\sqrt{\frac{\left(\sqrt{\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{3}\right)\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}−\sqrt{\mathrm{5}}\right)\left(\mathrm{3}+\sqrt{\mathrm{5}}−\mathrm{2}\sqrt{\mathrm{2}}\right)}{\mathrm{3}+\sqrt{\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{2}}}} \\ $$$$\Rightarrow{r}_{\mathrm{1}} =\frac{{a}}{\mathrm{12}}\sqrt{\mathrm{2}\left(\mathrm{9}−\mathrm{4}\sqrt{\mathrm{2}}−\sqrt{\mathrm{5}}\right)}\approx\mathrm{0}.\mathrm{12}{a} \\ $$$$ \\ $$$${Circle}\:\mathrm{2}\:{in}\:\Delta{AFD}: \\ $$$${s}=\frac{\mathrm{1}}{\mathrm{2}}\left({a}+\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}{a}+\frac{\sqrt{\mathrm{5}}}{\mathrm{3}}{a}\right)=\frac{\mathrm{3}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{5}}}{\mathrm{6}}{a} \\ $$$${r}_{\mathrm{2}} ={a}\sqrt{\frac{\left(\frac{\mathrm{3}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{5}}}{\mathrm{6}}−\mathrm{1}\right)\left(\frac{\mathrm{3}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{5}}}{\mathrm{6}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}\right)\left(\frac{\mathrm{3}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{5}}}{\mathrm{6}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{3}}\right)}{\frac{\mathrm{3}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{5}}}{\mathrm{6}}}} \\ $$$${r}_{\mathrm{2}} =\frac{{a}}{\mathrm{6}}\sqrt{\frac{\left(−\mathrm{3}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{5}}\right)\left(\mathrm{3}−\sqrt{\mathrm{2}}+\sqrt{\mathrm{5}}\right)\left(\mathrm{3}+\sqrt{\mathrm{2}}−\sqrt{\mathrm{5}}\right)}{\mathrm{3}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{5}}}} \\ $$$$\Rightarrow{r}_{\mathrm{2}} =\frac{{a}}{\mathrm{6}}\sqrt{\mathrm{2}\left(\mathrm{12}−\mathrm{7}\sqrt{\mathrm{2}}−\mathrm{5}\sqrt{\mathrm{5}}+\mathrm{3}\sqrt{\mathrm{10}}\right)}\approx\mathrm{0}.\mathrm{15}{a} \\ $$$$ \\ $$$${Circle}\:\mathrm{3}\:{in}\:\Delta{DFC}: \\ $$$${s}=\frac{\mathrm{1}}{\mathrm{2}}\left({a}+\frac{\sqrt{\mathrm{5}}}{\mathrm{3}}{a}+\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}{a}\right)=\frac{\mathrm{3}+\sqrt{\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{6}}{a} \\ $$$${r}_{\mathrm{3}} ={a}\sqrt{\frac{\left(\frac{\mathrm{3}+\sqrt{\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{6}}−\mathrm{1}\right)\left(\frac{\mathrm{3}+\sqrt{\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{6}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{3}}\right)\left(\frac{\mathrm{3}+\sqrt{\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{6}}−\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}\right)}{\frac{\mathrm{3}+\sqrt{\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{6}}}} \\ $$$${r}_{\mathrm{3}} =\frac{{a}}{\mathrm{6}}\sqrt{\frac{\left(−\mathrm{3}+\sqrt{\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{2}}\right)\left(\mathrm{3}−\sqrt{\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{2}}\right)\left(\mathrm{3}+\sqrt{\mathrm{5}}−\mathrm{2}\sqrt{\mathrm{2}}\right)}{\mathrm{3}+\sqrt{\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{2}}}} \\ $$$$\Rightarrow{r}_{\mathrm{3}} =\frac{{a}}{\mathrm{6}}\sqrt{\mathrm{2}\left(\mathrm{9}−\mathrm{4}\sqrt{\mathrm{2}}−\sqrt{\mathrm{5}}\right)}=\mathrm{2}{r}_{\mathrm{1}} \approx\mathrm{0}.\mathrm{25}{a} \\ $$$$ \\ $$$${Circle}\:\mathrm{4}\:{in}\:\Delta{ABC}: \\ $$$${s}=\frac{\mathrm{1}}{\mathrm{2}}\left({a}+{a}+\sqrt{\mathrm{2}}{a}\right)=\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}}{a} \\ $$$${r}_{\mathrm{4}} ={a}\sqrt{\frac{\left(\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}}−\mathrm{1}\right)\left(\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}}−\mathrm{1}\right)\left(\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}}−\sqrt{\mathrm{2}}\right)}{\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}}}} \\ $$$${r}_{\mathrm{4}} =\frac{{a}}{\mathrm{2}}\sqrt{\mathrm{2}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)}\approx\mathrm{0}.\mathrm{29}{a} \\ $$
Commented by ajfour last updated on 30/Jan/18
$$\mathcal{S}{plendid}!\:{Sir}\:. \\ $$