Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2005 OIM Problems/Problem 1: Difference between revisions

← Older edit
Creative raven (talk | contribs)
editor
1 edit
Solution
Eevee9406 (talk | contribs)
editor
662 edits
mNo edit summary
 
Line 37: Line 37:


~creativeRaven
~creativeRaven
{{solution}}


== See also ==
== See also ==
[[OIM Problems and Solutions]]
[[OIM Problems and Solutions]]

Latest revision as of 17:06, 25 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

See also

OIM Problems and Solutions

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_OIM_Problems/Problem_1&oldid=246061"