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Difference between revisions of "2005 OIM Problems/Problem 1"

(Created page with "== Problem == Determine all triples of real numbers <math>(x, y, z)</math> that satisfy the following system of equations: <cmath>xyz = 8</cmath> <cmath>x^2y+y^2z+z^2x=73</cm...")
 
(Solution)
Line 3: Line 3:
  
 
<cmath>xyz = 8</cmath>
 
<cmath>xyz = 8</cmath>
−
<cmath>x^2y+y^2z+z^2x=73</cmath>
+
<cmath>x^2y+y^2z+z^2x = 73</cmath>
−
<cmath>x(y-z)^2+y(z-x)^2+z(x-y)^2=98</cmath>
+
<cmath>x(y-z)^2+y(z-x)^2+z(x-y)^2 = 98</cmath>
  
 
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
 
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
  
 
== Solution ==
 
== Solution ==
 +
One can observe:
 +
<math>x(y-z)^2+y(z-x)^2+z(x-y)^2 = x(y^2+z^2)+y(z^2+x^2)+z(x^2+y^2)-6xyz</math>
 +
 +
A little more manipulation and we get:
 +
 +
<math>xy^2+yz^2+zx^2 = 73</math>
 +
 +
Subtract this from <math>x^2y+y^2z+z^2x = 73</math>
 +
 +
To get: <math>-(x-y)(y-z)(z-x) = 0</math>
 +
 +
Implying that <math>x = y, y = z</math> or <math>z = x</math>
 +
 +
WLOG, assume <math>x = z</math>,
 +
 +
So, <math>y = \frac{8}{x^2}</math>
 +
 +
<math>x^2y+y^2x+x^3 = 73</math>
 +
 +
Implying, <math>\frac{64}{x^3}+x^3 = 65</math>
 +
 +
Therefore, <math>x = 1</math> or <math>4</math>
 +
 +
So, <math>y = 8</math> or <math>1/2</math>
 +
 +
<math>(x,y,z) =</math> permutations of <math>(1,8,1)</math> and <math>(4,4,\frac{1}{2})</math>
 +
 +
~creativeRaven
 +
 
{{solution}}
 
{{solution}}
  
 
== See also ==
 
== See also ==
 
[[OIM Problems and Solutions]]
 
[[OIM Problems and Solutions]]

Revision as of 00:25, 24 March 2025

Problem

Determine all triples of real numbers $(x, y, z)$ that satisfy the following system of equations:

\[xyz = 8\] \[x^2y+y^2z+z^2x = 73\] \[x(y-z)^2+y(z-x)^2+z(x-y)^2 = 98\]

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

One can observe: $x(y-z)^2+y(z-x)^2+z(x-y)^2 = x(y^2+z^2)+y(z^2+x^2)+z(x^2+y^2)-6xyz$

A little more manipulation and we get:

$xy^2+yz^2+zx^2 = 73$

Subtract this from $x^2y+y^2z+z^2x = 73$

To get: $-(x-y)(y-z)(z-x) = 0$

Implying that $x = y, y = z$ or $z = x$

WLOG, assume $x = z$,

So, $y = \frac{8}{x^2}$

$x^2y+y^2x+x^3 = 73$

Implying, $\frac{64}{x^3}+x^3 = 65$

Therefore, $x = 1$ or $4$

So, $y = 8$ or $1/2$

$(x,y,z) =$ permutations of $(1,8,1)$ and $(4,4,\frac{1}{2})$

~creativeRaven

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See also

OIM Problems and Solutions

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