solve for R^+ x^2 +y^2 −kxy=c^2 y^2 +z^2 −kyz=a^2 z^2 +x^2 −kzx=b^2 (k is constant)


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Question Number 211367 by mr W last updated on 07/Sep/24

$s o l v e f o r R^{+}$ $x^{2} + y^{2} - k x y = c^{2}$ $y^{2} + z^{2} - k y z = a^{2}$ $z^{2} + x^{2} - k z x = b^{2}$ $(k i s c o n s t a n t)$
Commented by ajfour last updated on 11/Sep/24
https://youtu.be/LgXNcnsOJtM?si=JMlLjooG3OzelvXT
	Answered by a.lgnaoui last updated on 09/Sep/24
	$k+2=(((x+y)^2 −c^2 )/(xy))=(((y+z)^2 −a^2 )/(yz))=(((x+z)^2 −b^2 )/(xz)) alors: xy=(((x+y)^2 )/(k+2)) −(c^2 /(k+2))(1) yz=(((y+z)^2 )/(k+2))−(a^2 /(k+2)) (2) xz=(((x+z)^2 )/(k+2))−(b^2 /(k+2))(3) 2(x^2 +y^2 +z^2 )−k(xy+yz+xz)=a^2 +b^2 +c^2 4(x^2 +y^2 +z^3 )−k[(x+y+z)^2 −(x^2 +y^2 +z^2 )]=2(a+b+c) (4+k)(x^2 +y^2 +z^2 )−k(x+y+z)^2 =2(a^2 +b^2 +c^2 ) or x^2 +y^2 +z^2 =(x+y+z)^2 −2(xy+yz+xz) ⇒ 3(x+y+z)^2 =2(a^2 +b^2 +c^2 ) x+y+z=(√((2/3)(a^2 +b^2 +c^2 ) )) (i) x+y=(√((2/3)(a^2 +b^2 +c^2 ))) −z (a^2 −b^2 )=(y^2 −x^2 )−kz(y−x) =(y−x)[(x+y)−kz] soit a^2 −b^2 =(y−x)[(√((2/3)(a^2 +b^2 +c^2 ))) −z−kz] y−x=((a^2 −b^2 )/( [(√((2/3)(a^2 +b^2 +c^3 ))) −(k+1)z])) { ((y=((a^2 −b^2 )/(2[(√((2/3)e))−(k+1)z])) −(z/2)+(1/2)(√((2/3)e)))),((e=(√(a^2 +b^2 +c^2 )))) :} a^2 =(y−(k/2)z)^2 +(((4−k^2 )/4))z^2 y=(k/2)z+(√(a^2 −(((4−k^2 )/4))z^2 )) donc on a lequation en z z^2 (√(a^2 −((4−k^2 )/4))) + z(((k+1)/2))−((a^2 −b^2 )/(2[(√((2/3)e)) −(k+1)z])) =(1/2)(√((2/3)e)) this equation is long othere methode: −−−−−−−−− ..(i)⇒ (√((2/3)e)) −z=(x+y) k+2=(((x+y)^2 −c^2 )/(xy))=(((y+z)^2 −a^2 )/(yz)) ⇒ ((x+y)−c)/x)=((y+z)−a)/z) ⇒a^2 + ((z[(x+y)−c^2 ])/x)=(y+z) ((x+y)−c)/x)=((y+z)−a)/z)=(((x+y)^2 −(y+z)^2 +a^2 −c^2 )/(x−z)) (((x−z)(x+z+2y)+)/(x−z))=(((x+z+2y))/1)+(((a−c)/(x−z))) (y+z)^2 =((z[(x+y)^2 −c^2 ])/x)+a^2 y+z=((k/2)+1)z+(√(a^2 −((4−k^2 )/4)z^2 )) (((y+z)^2 −a^2 )/(yz))=k+2 ⇒(y+z)^2 =(k+2)yz+a^2 =a^2 +(k+2)z((k/2)z+(√(a^2 −((4−k^2 )/4)z^2 )) ) donc (z+(k/2)z+(√(a^3 −((4−k^2 )/4)z^2 )) )^2 = a^2 +(((k+2)k)/2)z^2 +(k+2)z(√(a−((4−k^2 )/4)z^2 )) . soit apres calculs ((((k+2)^2 )/4)z^2 −((k(k+2))/2)z^2 )−((4−k)/4)z^2 =0 (((k+2)^2 −2(k+2))/4)=((4−k)/4) k^2 +3k−4=0 △=25 { ((k=+1)),((k=−4)) :} on choisit k>0 k=1 3=(((x+y)^2 −c^2 )/(xy)) =(((y+z)^2 −a^2 )/(yz)) (x+y)^2 −3xy−c^2 =0 (x−(3/2)y)^2 −((9/4)y^2 +c^2 )=0 (ii) (y−(3/2)z)^2 −((9/4)z^2 +c^2 )=0 (iii) y=(2/2)z+(√(a−(3/4)z^2 )) ⇒(a−(3/4)z^2 )=c^2 +(9/4)z^2 3z^2 =a^2 −c^2 z=(√((a^2 −c^2 )/3)) { (((x+y)=(√((2/3)(a^2 +b^2 +c^2 ))) −(√((a^2 −c^2 )/3)) )),((xy =(((x+y)^2 −c^2 )/3)=(((2/3)(a+b+c)+((a^2 −c^2 )/3)−(2/3)(√(2a^2 −c^2 )(a^2 +b^2 +c^2 )) −c^2 )/3))) :} { ((xy=((2/9)(√(a+b+c)) )[((√(a^2 +b^2 +c^2 )) −(√(a^2 −c^2 )) )]+((a^2 −4c^2 )/9))),((x+y =(√((2/3)(a^2 +b^2 +c^2 )) −(√((a^2 −c^2 )/3)))) :} { ((k^2 −sk+p=0)),((s=(x+y) p=xy)) :} k=((s±(√(s^2 −4p)))/2) s^(2 ) =3xy+c^2 p=xy ⇒ k=((3xy±(√(9xy−4xy)))/2) =((3xy+(√(5xy)))/2)=(1/2)(√(xy)) [3(√(xy)) +5] { (((√x)=(1/2)((√y) (3(√(xy)) +5) )),(((√y) =(1/2)(√x) (3(√(xy)) −5))) :} (√(xy)) =(1/4)(√(xy)) (3(√(xy)) +5)(3(√(xy)) −5) 1=(1/4)(3m+5)(3m−5) m=(√(xy)) 9m^2 −25=4 m=((√(29))/3) (√(xy)) =((√(29))/3) (x+y=(√(c^2 +3xy)) ⇒ x+y=(√(c^2 +((29)/3))) { ((x+y =(√(c^2 +((29)/3))))),((xy =((29)/9))) :} { ((△=((3c^2 +29)/3)−((116)/9))),( { ((x=(((1/3)((√(27c^2 −29)) −(√(9c^2 +29)) ))/6))),((y=(((1/3)((√(27c^2 −29)) +(√(9c^2 +29)) ))/6))) :}) :} Recap k=1 x=(((√(27c^2 −29)) −(√(9c^2 +29)))/(18)) y=(((√(27c^2 −29)) +(√(9c^2 +29)))/(18)) z=((√(a^2 −c^2 ))/3) pour ( k=−4): (meme procedure)$
	$k + 2 = \frac{{(x + y)}^{2} - c^{2}}{xy} = \frac{{(y + z)}^{2} - a^{2}}{yz} = \frac{{(x + z)}^{2} - b^{2}}{xz}$ $alors :$ $xy = \frac{{(x + y)}^{2}}{k + 2} - \frac{c^{2}}{k + 2} (1)$ $yz = \frac{{(y + z)}^{2}}{k + 2} - \frac{a^{2}}{k + 2} (2) xz = \frac{{(x + z)}^{2}}{k + 2} - \frac{b^{2}}{k + 2} (3)$ $2 (x^{2} + y^{2} + z^{2}) - k (xy + yz + xz) = a^{2} + b^{2} + c^{2}$ $4 (x^{2} + y^{2} + z^{3}) - k [{(x + y + z)}^{2} - (x^{2} + y^{2} + z^{2})] = 2 (a + b + c)$ $(4 + k) (x^{2} + y^{2} + z^{2}) - k {(x + y + z)}^{2} = 2 (a^{2} + b^{2} + c^{2})$ $or x^{2} + y^{2} + z^{2} = {(x + y + z)}^{2} - 2 (xy + yz + xz)$ $\Rightarrow$ $3 {(x + y + z)}^{2} = 2 (a^{2} + b^{2} + c^{2})$ $x + y + z = \sqrt{\frac{2}{3} (a^{2} + b^{2} + c^{2})} (i)$ $x + y = \sqrt{\frac{2}{3} (a^{2} + b^{2} + c^{2})} - z$ $(a^{2} - b^{2}) = (y^{2} - x^{2}) - kz (y - x)$ $= (y - x) [(x + y) - kz]$ $soit$ $a^{2} - b^{2} = (y - x) [\sqrt{\frac{2}{3} (a^{2} + b^{2} + c^{2})} - z - kz]$ $y - x = \frac{a^{2} - b^{2}}{[\sqrt{\frac{2}{3} (a^{2} + b^{2} + c^{3})} - (k + 1) z]}$ ${\begin{cases} y = \frac{a^{2} - b^{2}}{2 [\sqrt{\frac{2}{3} e} - (k + 1) z]} - \frac{z}{2} + \frac{1}{2} \sqrt{\frac{2}{3} e} \\ e = \sqrt{a^{2} + b^{2} + c^{2}} \end{cases}$ $a^{2} = {(y - \frac{k}{2} z)}^{2} + (\frac{4 - k^{2}}{4}) z^{2}$ $y = \frac{k}{2} z + \sqrt{a^{2} - (\frac{4 - k^{2}}{4}) z^{2}}$ $donc on a lequati o n en z$ $z^{2} \sqrt{a^{2} - \frac{4 - k^{2}}{4}} + z (\frac{k + 1}{2}) - \frac{a^{2} - b^{2}}{2 [\sqrt{\frac{2}{3} e} - (k + 1) z]}$ $= \frac{1}{2} \sqrt{\frac{2}{3} e}$ $this equation is long$ $othere methode :$ $- - - - - - - - -$ $. . (i) \Rightarrow \sqrt{\frac{2}{3} e} - z = (x + y)$ $k + 2 = \frac{{(x + y)}^{2} - c^{2}}{xy} = \frac{{(y + z)}^{2} - a^{2}}{yz}$ $\Rightarrow \frac{x + y) - c}{x} = \frac{y + z) - a}{z}$ $\Rightarrow a^{2} + \frac{z [(x + y) - c^{2}]}{x} = (y + z)$ $\frac{x + y) - c}{x} = \frac{y + z) - a}{z} = \frac{{(x + y)}^{2} - {(y + z)}^{2} + a^{2} - c^{2}}{x - z}$ $\frac{(x - z) (x + z + 2 y) +}{x - z} = \frac{(x + z + 2 y)}{1} + (\frac{a - c}{x - z})$ ${(y + z)}^{2} = \frac{z [{(x + y)}^{2} - c^{2}]}{x} + a^{2}$ $y + z = (\frac{k}{2} + 1) z + \sqrt{a^{2} - \frac{4 - k^{2}}{4} z^{2}}$ $\frac{{(y + z)}^{2} - a^{2}}{yz} = k + 2$ $\Rightarrow {(y + z)}^{2} = (k + 2) yz + a^{2}$ $= a^{2} + (k + 2) z (\frac{k}{2} z + \sqrt{a^{2} - \frac{4 - k^{2}}{4} z^{2}})$ $donc$ ${(z + \frac{k}{2} z + \sqrt{a^{3} - \frac{4 - k^{2}}{4} z^{2}})}^{2} =$ $a^{2} + \frac{(k + 2) k}{2} z^{2} + (k + 2) z \sqrt{a - \frac{4 - k^{2}}{4} z^{2}} .$ $soit apres calculs$ $(\frac{{(k + 2)}^{2}}{4} z^{2} - \frac{k (k + 2)}{2} z^{2}) - \frac{4 - k}{4} z^{2} = 0$ $\frac{{(k + 2)}^{2} - 2 (k + 2)}{4} = \frac{4 - k}{4}$ $k^{2} + 3 k - 4 = 0$ $△ = 25 {\begin{cases} k = + 1 \\ k = - 4 \end{cases}$ $on choisit k > 0 k = 1$ $3 = \frac{{(x + y)}^{2} - c^{2}}{xy} = \frac{{(y + z)}^{2} - a^{2}}{yz}$ ${(x + y)}^{2} - 3 xy - c^{2} = 0$ ${(x - \frac{3}{2} y)}^{2} - (\frac{9}{4} y^{2} + c^{2}) = 0 (ii)$ ${(y - \frac{3}{2} z)}^{2} - (\frac{9}{4} z^{2} + c^{2}) = 0 (iii)$ $y = \frac{2}{2} z + \sqrt{a - \frac{3}{4} z^{2}}$ $\Rightarrow (a - \frac{3}{4} z^{2}) = c^{2} + \frac{9}{4} z^{2}$ $3 z^{2} = a^{2} - c^{2} z = \sqrt{\frac{a^{2} - c^{2}}{3}}$ ${\begin{cases} (x + y) = \sqrt{\frac{2}{3} (a^{2} + b^{2} + c^{2})} - \sqrt{\frac{a^{2} - c^{2}}{3}} \\ xy = \frac{{(x + y)}^{2} - c^{2}}{3} = \frac{\frac{2}{3} (a + b + c) + \frac{a^{2} - c^{2}}{3} - \frac{2}{3} \sqrt{2 a^{2} - c^{2}) (a^{2} + b^{2} + c^{2}} - c^{2}}{3} \end{cases}$ ${\begin{cases} xy = (\frac{2}{9} \sqrt{a + b + c}) [(\sqrt{a^{2} + b^{2} + c^{2}} - \sqrt{a^{2} - c^{2}})] + \frac{a^{2} - 4 c^{2}}{9} \\ x + y = \sqrt{\frac{2}{3} (a^{2} + b^{2} + c^{2}} - \sqrt{\frac{a^{2} - c^{2}}{3}} \end{cases}$ ${\begin{cases} k^{2} - sk + p = 0 \\ s = (x + y) p = xy \end{cases}$ $k = \frac{s \pm \sqrt{s^{2} - 4 p}}{2}$ $s^{2} = 3 xy + c^{2} p = xy$ $\Rightarrow k = \frac{3 xy \pm \sqrt{9 xy - 4 xy}}{2}$ $= \frac{3 xy + \sqrt{5 xy}}{2} = \frac{1}{2} \sqrt{xy} [3 \sqrt{xy} + 5]$ ${\begin{cases} \sqrt{x} = \frac{1}{2} (\sqrt{y} (3 \sqrt{xy} + 5) \\ \sqrt{y} = \frac{1}{2} \sqrt{x} (3 \sqrt{xy} - 5) \end{cases}$ $\sqrt{xy} = \frac{1}{4} \sqrt{xy} (3 \sqrt{xy} + 5) (3 \sqrt{xy} - 5)$ $1 = \frac{1}{4} (3 m + 5) (3 m - 5) m = \sqrt{xy}$ $9 m^{2} - 25 = 4 m = \frac{\sqrt{29}}{3}$ $\sqrt{xy} = \frac{\sqrt{29}}{3} (x + y = \sqrt{c^{2} + 3 xy}$ $\Rightarrow x + y = \sqrt{c^{2} + \frac{29}{3}}$ ${\begin{cases} x + y = \sqrt{c^{2} + \frac{29}{3}} \\ xy = \frac{29}{9} \end{cases}$ ${\begin{cases} △ = \frac{3 c^{2} + 29}{3} - \frac{116}{9} \\ {\begin{cases} x = \frac{\frac{1}{3} (\sqrt{27 c^{2} - 29} - \sqrt{9 c^{2} + 29})}{6} \\ y = \frac{\frac{1}{3} (\sqrt{27 c^{2} - 29} + \sqrt{9 c^{2} + 29})}{6} \end{cases} \end{cases}$ $R e c a p k = 1$ $x = \frac{\sqrt{27 c^{2} - 29} - \sqrt{9 c^{2} + 29}}{18}$ $y = \frac{\sqrt{27 c^{2} - 29} + \sqrt{9 c^{2} + 29}}{18}$ $z = \frac{\sqrt{a^{2} - c^{2}}}{3}$ $pour (k = - 4) : (meme procedure)$
	Commented by mr W last updated on 09/Sep/24

	$t h a n k s f o r t r y i n g!$ $b u t y o u c a n n o t d e t e r m i n e t h e v a l u e$ $o f k b y y o u r s e l f . k i s a g i v e n c o n s t a n t,$ $f o r e x a m p l e k = 1.5 .$
	Commented by a.lgnaoui last updated on 09/Sep/24

	$thank you$
	Answered by ajfour last updated on 08/Oct/24

	$x^{2} + y^{2} + z^{2} = s$ $k x y + z^{2} = s - c^{2}$ $k y z + x^{2} = s - a^{2}$ $k z x + y^{2} = s - b^{2}$ $I f x = p y, z = q y$ $\Rightarrow (k p + q^{2}) y^{2} = s - c^{2}$ $(k q + p^{2}) y^{2} = s - a^{2}$ $& (k p q + 1) y^{2} = s - b^{2}$ $s = (1 + p^{2} + q^{2}) y^{2}$ $\frac{1 + k p - p^{2}}{1 + k q - q^{2}} = \frac{c^{2}}{a^{2}} . . . (i)$ $& \frac{2 + k (p + q) - (p^{2} + q^{2})}{p^{2} + q^{2} - k p q} = \frac{a^{2} + c^{2}}{b^{2}} . . (i i)$ $. . .$
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