Question Number 201322 by cherokeesay last updated on 04/Dec/23
Answered by AST last updated on 05/Dec/23
![Let ∠EDC=β⇒∠ABD=2β−90 EC^2 =2−2cosβ...(i) BC^2 =BD^2 +1−2BDDCcos(2β)...(ii) ((sin(2β−90))/1)=((sin90)/(BD))=(1/(BD))=−cos(2β)...(iii) cos(2β)=2cos^2 β−1 (iii)&(ii)⇒BC^2 =BD^2 +3=(1/(cos^2 (2β)))+3 ⇒BD^2 +3=(1/(cos^2 (2β)))+3=(1/((2cos^2 β−1)^2 ))+2 ⇒(4/([EC^4 −4EC^2 +2]^2 ))+3=BD^2 +3 ⇒(4/([x^4 −4x^2 +2]^2 ))=y^2 +1](https://www.tinkutara.com/question/Q201340.png)
$${Let}\:\angle{EDC}=\beta\Rightarrow\angle{ABD}=\mathrm{2}\beta−\mathrm{90} \\ $$$${EC}^{\mathrm{2}} =\mathrm{2}−\mathrm{2}{cos}\beta…\left({i}\right) \\ $$$${BC}^{\mathrm{2}} ={BD}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{BD}\boldsymbol{{D}}{Ccos}\left(\mathrm{2}\beta\right)…\left({ii}\right) \\ $$$$\frac{{sin}\left(\mathrm{2}\beta−\mathrm{90}\right)}{\mathrm{1}}=\frac{{sin}\mathrm{90}}{{BD}}=\frac{\mathrm{1}}{{BD}}=−{cos}\left(\mathrm{2}\beta\right)…\left({iii}\right) \\ $$$${cos}\left(\mathrm{2}\beta\right)=\mathrm{2}{cos}^{\mathrm{2}} \beta−\mathrm{1} \\ $$$$\left({iii}\right)\&\left({ii}\right)\Rightarrow{BC}^{\mathrm{2}} ={BD}^{\mathrm{2}} +\mathrm{3}=\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left(\mathrm{2}\beta\right)}+\mathrm{3} \\ $$$$\Rightarrow{BD}^{\mathrm{2}} +\mathrm{3}=\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left(\mathrm{2}\beta\right)}+\mathrm{3}=\frac{\mathrm{1}}{\left(\mathrm{2}{cos}^{\mathrm{2}} \beta−\mathrm{1}\right)^{\mathrm{2}} }+\mathrm{2} \\ $$$$\Rightarrow\frac{\mathrm{4}}{\left[{EC}^{\mathrm{4}} −\mathrm{4}{EC}^{\mathrm{2}} +\mathrm{2}\right]^{\mathrm{2}} }+\mathrm{3}={BD}^{\mathrm{2}} +\mathrm{3} \\ $$$$\Rightarrow\frac{\mathrm{4}}{\left[{x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}\right]^{\mathrm{2}} }={y}^{\mathrm{2}} +\mathrm{1} \\ $$
Answered by mr W last updated on 05/Dec/23
$${BD}={x} \\ $$$${BA}^{\mathrm{2}} ={x}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} \\ $$$${BC}=\sqrt{\mathrm{2}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} }=\sqrt{{x}^{\mathrm{2}} +\mathrm{3}} \\ $$$$\frac{{BE}}{{BD}}=\frac{{EC}}{{DC}}\Rightarrow\frac{{BE}}{{x}}=\frac{{EC}}{\mathrm{1}}\Rightarrow{BE}={xEC} \\ $$$${BE}+{EC}=\left({x}+\mathrm{1}\right){EC}=\sqrt{{x}^{\mathrm{2}} +\mathrm{3}} \\ $$$$\Rightarrow{EC}=\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}}{{x}+\mathrm{1}} \\ $$$${EC}=\mathrm{2}×\mathrm{1}\:\mathrm{cos}\:{C}=\mathrm{2}×\frac{\mathrm{1}+\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}}=\frac{\mathrm{4}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}} \\ $$$$\frac{\mathrm{4}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}}=\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}}{{x}+\mathrm{1}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +\mathrm{3}=\mathrm{4}\left({x}+\mathrm{1}\right) \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{2}+\sqrt{\mathrm{5}}\:\checkmark \\ $$
Commented by cherokeesay last updated on 05/Dec/23
$${thank}\:{you}\:{sir}\:! \\ $$