MHB Solve x,y,z: Real Solutions $(x,y,z)$

  • Thread starter Thread starter anemone
  • Start date Start date
AI Thread Summary
The discussion centers on finding real solutions for the equations xyz=8, x^2y+y^2z+z^2x=73, and x(y-z)^2+y(z-x)^2+z(x-y)^2=98. Participants share their approaches to solving the system, with one user acknowledging another's more efficient method. The conversation highlights the complexity of the problem, with one participant deeming their own solution too tedious to share. Overall, the focus remains on collaborative problem-solving and sharing insights on the mathematical challenge. The thread emphasizes the importance of efficiency in finding solutions to complex equations.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Find all the real solutions $(x, y, z)$ of the system:

$xyz=8$

$x^2y+y^2z+z^2x=73$

$x(y-z)^2+y(z-x)^2+z(x-y)^2=98$
 
Mathematics news on Phys.org
My solution
Expanding the third equation and using the first to eliminate $xyz$ and then using the second equation gives

$(x-y)(y-z)(z-x) = 0$

Clearly we cannot have $x = y= z$ as this violates the third equation. We haave three choices here. Let us pick $z = x$, the other follow

Thus

$x^2y = 8$ and $x^2y+xy^2+x^3=73$ or $xy^2 + x^3 = 65$ Now $y = \dfrac{8}{x^2}$ giving

$x^3 - 65 + \dfrac{64}{x^3}=0$ or $\dfrac{\left(x^3-1\right)\left(x^3-64\right)}{x^3} = 0$

Thus, $x = 1$ or $x = 4$ leading to $(1,8,1)$ and $(4,1/2,4)$.

The other solutions are obtained by interchanging these vlues.
 
Jester said:
My solution
Expanding the third equation and using the first to eliminate $xyz$ and then using the second equation gives

$(x-y)(y-z)(z-x) = 0$

Clearly we cannot have $x = y= z$ as this violates the third equation. We haave three choices here. Let us pick $z = x$, the other follow

Thus

$x^2y = 8$ and $x^2y+xy^2+x^3=73$ or $xy^2 + x^3 = 65$ Now $y = \dfrac{8}{x^2}$ giving

$x^3 - 65 + \dfrac{64}{x^3}=0$ or $\dfrac{\left(x^3-1\right)\left(x^3-64\right)}{x^3} = 0$

Thus, $x = 1$ or $x = 4$ leading to $(1,8,1)$ and $(4,1/2,4)$.

The other solutions are obtained by interchanging these vlues.

Hi Jester, thanks for participating and you got it right!

To be completely honest with you, my solution is tedious compared to yours and hence I don't think it's worth mentioning here... :o
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.

Similar threads

Back
Top