Question-205535

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

((sin90=1)/(BC))=((sin(90−θ)=cosθ)/(AB))⇒AB=BCcosθ ((sin90=1)/(BC))=((sinθ)/(AC))⇒AC=BCsinθ ∠DBA=180−θ;∠ACE=90+θ ⇒AD^2 =x^2 BC^2 +BC^2 cos^2 θ+2xBC^2 cos^2 θ ⇒AE^2 =BC^2 sin^2 θ+x^2 BC^2 +2xBC^2 sin^2 θ DE^2 =(2xBC+BC)^2 =BC^2 (2x+1)^2 ⇒((AD^2 +DE^2 +EA^2 )/(BC^2 ))=2x^2 +2x+1+(2x+1)^2 ⇒f(x)=6x^2 +6x+2⇒f((((√(1347))−1)/2))=2021

$$\frac{{sin}\mathrm{90}=\mathrm{1}}{{BC}}=\frac{{sin}\left(\mathrm{90}−\theta\right)={cos}\theta}{{AB}}\Rightarrow{AB}={BCcos}\theta \\ $$$$\frac{{sin}\mathrm{90}=\mathrm{1}}{{BC}}=\frac{{sin}\theta}{{AC}}\Rightarrow{AC}={BCsin}\theta \\ $$$$\angle{DBA}=\mathrm{180}−\theta;\angle{ACE}=\mathrm{90}+\theta \\ $$$$\Rightarrow{AD}^{\mathrm{2}} ={x}^{\mathrm{2}} {BC}^{\mathrm{2}} +{BC}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta+\mathrm{2}{xBC}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta \\ $$$$\Rightarrow{AE}^{\mathrm{2}} ={BC}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta+{x}^{\mathrm{2}} {BC}^{\mathrm{2}} +\mathrm{2}{xBC}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta \\ $$$${DE}^{\mathrm{2}} =\left(\mathrm{2}{xBC}+{BC}\right)^{\mathrm{2}} ={BC}^{\mathrm{2}} \left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\frac{{AD}^{\mathrm{2}} +{DE}^{\mathrm{2}} +{EA}^{\mathrm{2}} }{{BC}^{\mathrm{2}} }=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}+\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{6}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{2}\Rightarrow{f}\left(\frac{\sqrt{\mathrm{1347}}−\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2021} \\ $$

Answered by mr W last updated on 24/Mar/24

Commented by mr W last updated on 24/Mar/24

AD^2 =(x+1)^2 a^2 +q^2 −2a(x+1)q×(q/a) =(x+1)^2 a^2 −q^2 (2x+1) similarly AE^2 =(x+1)^2 a^2 −p^2 (2x+1) AD^2 +AE^2 =2(x+1)^2 a^2 −(p^2 +q^2 )(2x+1) =2(x+1)^2 a^2 −a^2 (2x+1) =(2x^2 +2x+1)a^2 DE^2 =(2x+1)^2 a^2 =(4x^2 +4x+1)a^2 ⇒f(x)=(((2x^2 +2x+1)a^2 +(4x^2 +4x+1)a^2 )/a^2 ) =6x^2 +6x+2=6(x+(1/2))^2 +(1/2) f((((√(1349))−1)/2))=6(((√(1349))/2))^2 +(1/2) =((3×1349+1)/2)=2024✓

$${AD}^{\mathrm{2}} =\left({x}+\mathrm{1}\right)^{\mathrm{2}} {a}^{\mathrm{2}} +{q}^{\mathrm{2}} −\mathrm{2}{a}\left({x}+\mathrm{1}\right){q}×\frac{{q}}{{a}} \\ $$$$\:\:\:\:\:=\left({x}+\mathrm{1}\right)^{\mathrm{2}} {a}^{\mathrm{2}} −{q}^{\mathrm{2}} \left(\mathrm{2}{x}+\mathrm{1}\right) \\ $$$${similarly} \\ $$$${AE}^{\mathrm{2}} =\left({x}+\mathrm{1}\right)^{\mathrm{2}} {a}^{\mathrm{2}} −{p}^{\mathrm{2}} \left(\mathrm{2}{x}+\mathrm{1}\right) \\ $$$${AD}^{\mathrm{2}} +{AE}^{\mathrm{2}} =\mathrm{2}\left({x}+\mathrm{1}\right)^{\mathrm{2}} {a}^{\mathrm{2}} −\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)\left(\mathrm{2}{x}+\mathrm{1}\right) \\ $$$$\:\:\:\:=\mathrm{2}\left({x}+\mathrm{1}\right)^{\mathrm{2}} {a}^{\mathrm{2}} −{a}^{\mathrm{2}} \left(\mathrm{2}{x}+\mathrm{1}\right) \\ $$$$\:\:\:\:=\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right){a}^{\mathrm{2}} \\ $$$${DE}^{\mathrm{2}} =\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} {a}^{\mathrm{2}} =\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}\right){a}^{\mathrm{2}} \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right){a}^{\mathrm{2}} +\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}\right){a}^{\mathrm{2}} }{{a}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:=\mathrm{6}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{2}=\mathrm{6}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${f}\left(\frac{\sqrt{\mathrm{1349}}−\mathrm{1}}{\mathrm{2}}\right)=\mathrm{6}\left(\frac{\sqrt{\mathrm{1349}}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{3}×\mathrm{1349}+\mathrm{1}}{\mathrm{2}}=\mathrm{2024}\checkmark \\ $$

Commented by cortano12 last updated on 24/Mar/24

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