Question Number 205535 by cortano12 last updated on 24/Mar/24
Answered by A5T last updated on 24/Mar/24
Commented by A5T last updated on 24/Mar/24
$$\frac{{sin}\mathrm{90}=\mathrm{1}}{{BC}}=\frac{{sin}\left(\mathrm{90}−\theta\right)={cos}\theta}{{AB}}\Rightarrow{AB}={BCcos}\theta \\ $$$$\frac{{sin}\mathrm{90}=\mathrm{1}}{{BC}}=\frac{{sin}\theta}{{AC}}\Rightarrow{AC}={BCsin}\theta \\ $$$$\angle{DBA}=\mathrm{180}−\theta;\angle{ACE}=\mathrm{90}+\theta \\ $$$$\Rightarrow{AD}^{\mathrm{2}} ={x}^{\mathrm{2}} {BC}^{\mathrm{2}} +{BC}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta+\mathrm{2}{xBC}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta \\ $$$$\Rightarrow{AE}^{\mathrm{2}} ={BC}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta+{x}^{\mathrm{2}} {BC}^{\mathrm{2}} +\mathrm{2}{xBC}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta \\ $$$${DE}^{\mathrm{2}} =\left(\mathrm{2}{xBC}+{BC}\right)^{\mathrm{2}} ={BC}^{\mathrm{2}} \left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\frac{{AD}^{\mathrm{2}} +{DE}^{\mathrm{2}} +{EA}^{\mathrm{2}} }{{BC}^{\mathrm{2}} }=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}+\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{6}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{2}\Rightarrow{f}\left(\frac{\sqrt{\mathrm{1347}}−\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2021} \\ $$
Answered by mr W last updated on 24/Mar/24
Commented by mr W last updated on 24/Mar/24
$${AD}^{\mathrm{2}} =\left({x}+\mathrm{1}\right)^{\mathrm{2}} {a}^{\mathrm{2}} +{q}^{\mathrm{2}} −\mathrm{2}{a}\left({x}+\mathrm{1}\right){q}×\frac{{q}}{{a}} \\ $$$$\:\:\:\:\:=\left({x}+\mathrm{1}\right)^{\mathrm{2}} {a}^{\mathrm{2}} −{q}^{\mathrm{2}} \left(\mathrm{2}{x}+\mathrm{1}\right) \\ $$$${similarly} \\ $$$${AE}^{\mathrm{2}} =\left({x}+\mathrm{1}\right)^{\mathrm{2}} {a}^{\mathrm{2}} −{p}^{\mathrm{2}} \left(\mathrm{2}{x}+\mathrm{1}\right) \\ $$$${AD}^{\mathrm{2}} +{AE}^{\mathrm{2}} =\mathrm{2}\left({x}+\mathrm{1}\right)^{\mathrm{2}} {a}^{\mathrm{2}} −\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)\left(\mathrm{2}{x}+\mathrm{1}\right) \\ $$$$\:\:\:\:=\mathrm{2}\left({x}+\mathrm{1}\right)^{\mathrm{2}} {a}^{\mathrm{2}} −{a}^{\mathrm{2}} \left(\mathrm{2}{x}+\mathrm{1}\right) \\ $$$$\:\:\:\:=\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right){a}^{\mathrm{2}} \\ $$$${DE}^{\mathrm{2}} =\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} {a}^{\mathrm{2}} =\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}\right){a}^{\mathrm{2}} \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right){a}^{\mathrm{2}} +\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}\right){a}^{\mathrm{2}} }{{a}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:=\mathrm{6}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{2}=\mathrm{6}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${f}\left(\frac{\sqrt{\mathrm{1349}}−\mathrm{1}}{\mathrm{2}}\right)=\mathrm{6}\left(\frac{\sqrt{\mathrm{1349}}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{3}×\mathrm{1349}+\mathrm{1}}{\mathrm{2}}=\mathrm{2024}\checkmark \\ $$
Commented by cortano12 last updated on 24/Mar/24