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Question Number 224018 by hardmath last updated on 14/Aug/25
x ≠ y λ ≥ 1 { ((x + λ^2 = (y − λ)^2 )),((y + λ^2 = (x − λ)^2 )) :} Find: (((x^2 + y^2 )/(4λ^2 − 1)))^(2025) = ?
$$\mathrm{x}\:\neq\:\mathrm{y} \\ $$$$\lambda\:\geqslant\:\mathrm{1} \\ $$$$\begin{cases}{\mathrm{x}\:+\:\lambda^{\mathrm{2}} \:=\:\left(\mathrm{y}\:−\:\lambda\right)^{\mathrm{2}} }\\{\mathrm{y}\:+\:\lambda^{\mathrm{2}} \:=\:\left(\mathrm{x}\:−\:\lambda\right)^{\mathrm{2}} }\end{cases} \\ $$$$\mathrm{Find}:\:\:\:\left(\frac{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} }{\mathrm{4}\lambda^{\mathrm{2}} \:−\:\mathrm{1}}\right)^{\mathrm{2025}} =\:\:? \\ $$
Answered by mahdipoor last updated on 14/Aug/25
x+r^2 =y^2 +r^2 −2ry eq 1 y+r^2 =x^2 +r^2 −2rx eq 2 ⇒1−2⇒ x−y=y^2 −x^2 −2ry+2rx=(y−x)(y+x)−2r(y−x) x−y≠0⇒−1=(y+x)−2r ⇒2r−1=y+x ⇒1+2⇒ x+yx^2 +y^2 −2rx−2ry⇒x^2 +y^2 =(x+y)(1+2r) x^2 +y^2 =(2r−1)(1+2r)=4r^2 −1 ⇒((x^2 +y^2 )/(4r^2 −1))=1
$${x}+{r}^{\mathrm{2}} ={y}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{ry}\:\:\:\:\:\:\:\:{eq}\:\mathrm{1} \\ $$$${y}+{r}^{\mathrm{2}} ={x}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{rx}\:\:\:\:\:\:\:\:{eq}\:\mathrm{2} \\ $$$$\Rightarrow\mathrm{1}−\mathrm{2}\Rightarrow \\ $$$${x}−{y}={y}^{\mathrm{2}} −{x}^{\mathrm{2}} −\mathrm{2}{ry}+\mathrm{2}{rx}=\left({y}−{x}\right)\left({y}+{x}\right)−\mathrm{2}{r}\left({y}−{x}\right) \\ $$$${x}−{y}\neq\mathrm{0}\Rightarrow−\mathrm{1}=\left({y}+{x}\right)−\mathrm{2}{r} \\ $$$$\Rightarrow\mathrm{2}{r}−\mathrm{1}={y}+{x} \\ $$$$\Rightarrow\mathrm{1}+\mathrm{2}\Rightarrow \\ $$$${x}+{yx}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{rx}−\mathrm{2}{ry}\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\left({x}+{y}\right)\left(\mathrm{1}+\mathrm{2}{r}\right) \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\left(\mathrm{2}{r}−\mathrm{1}\right)\left(\mathrm{1}+\mathrm{2}{r}\right)=\mathrm{4}{r}^{\mathrm{2}} −\mathrm{1} \\ $$$$\Rightarrow\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{4}{r}^{\mathrm{2}} −\mathrm{1}}=\mathrm{1} \\ $$$$ \\ $$
Answered by A5T last updated on 14/Aug/25
(i)−(ii)⇒x−y=(y−x)(y−2λ+x) y≠x⇒−1=y−2λ+x⇒x+y=2λ−1 (i)+(ii)⇒x+y=x^2 +y^2 −2λ(x+y) ⇒x^2 +y^2 =4λ^2 −1 ⇒(((x^2 +y^2 )/(4λ^2 −1)))^(2025) =1
$$\left(\mathrm{i}\right)−\left(\mathrm{ii}\right)\Rightarrow\mathrm{x}−\mathrm{y}=\left(\mathrm{y}−\mathrm{x}\right)\left(\mathrm{y}−\mathrm{2}\lambda+\mathrm{x}\right) \\ $$$$\mathrm{y}\neq\mathrm{x}\Rightarrow−\mathrm{1}=\mathrm{y}−\mathrm{2}\lambda+\mathrm{x}\Rightarrow\mathrm{x}+\mathrm{y}=\mathrm{2}\lambda−\mathrm{1} \\ $$$$\left(\mathrm{i}\right)+\left(\mathrm{ii}\right)\Rightarrow\mathrm{x}+\mathrm{y}=\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{2}\lambda\left(\mathrm{x}+\mathrm{y}\right) \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{4}\lambda^{\mathrm{2}} −\mathrm{1} \\ $$$$\Rightarrow\left(\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }{\mathrm{4}\lambda^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2025}} =\mathrm{1} \\ $$

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