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Difference between revisions of "2006 Indonesia MO Problems/Problem 1"

(Solution to Problem 1 — nonlinear system)
 
m (Solution edits)
 
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'''Case 1: <math>x=y</math> and <math>x=-y</math>'''
 
'''Case 1: <math>x=y</math> and <math>x=-y</math>'''
  
−
Solving this system results in <math>(0,0)</math>.
+
Solving this system results in <math>(0,0)</math>.  Plugging the x and y values back in satisfies the original equation.
  
 
'''Case 2: <math>x=y</math> and <math>x^2 - xy + y^2 = 2</math>'''
 
'''Case 2: <math>x=y</math> and <math>x^2 - xy + y^2 = 2</math>'''
  
−
Substitution results in <math>x^2 = 2</math>, so solving this system results in <math>(\sqrt{2},\sqrt{2})</math> and <math>(-\sqrt{2},-\sqrt{2})</math>.
+
Substitution results in <math>x^2 = 2</math>, so solving this system results in <math>(\sqrt{2},\sqrt{2})</math> and <math>(-\sqrt{2},-\sqrt{2})</math>.  Plugging the x and y values back in for each ordered pair satisfies the original equation.
  
 
'''Case 3: <math>x^2 + xy + y^2 = 4</math> and <math>x=-y</math>'''
 
'''Case 3: <math>x^2 + xy + y^2 = 4</math> and <math>x=-y</math>'''
  
−
Substitution results in <math>y^2 = 4</math>, so solving this system results in <math>(2,-2)</math> and <math>(-2,2)</math>.
+
Substitution results in <math>y^2 = 4</math>, so solving this system results in <math>(2,-2)</math> and <math>(-2,2)</math>.  Plugging the x and y values back in for each ordered pair satisfies the original equation.
  
 
'''Case 4: <math>x^2 + xy + y^2 = 4</math> and <math>x^2 - xy + y^2 = 2</math>'''
 
'''Case 4: <math>x^2 + xy + y^2 = 4</math> and <math>x^2 - xy + y^2 = 2</math>'''
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<br>
 
<br>
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Therefore, there are four ordered pairs — <math>(\sqrt{\frac{3 + \sqrt{5}}{2}},\sqrt{\frac{3 - \sqrt{5}}{2}}), (-\sqrt{\frac{3 + \sqrt{5}}{2}}, -\sqrt{\frac{3 - \sqrt{5}}{2}}), </math> <math>(\sqrt{\frac{3 - \sqrt{5}}{2}}, \sqrt{\frac{3 + \sqrt{5}}{2}}), (-\sqrt{\frac{3 - \sqrt{5}}{2}}, -\sqrt{\frac{3 + \sqrt{5}}{2}}).</math>
+
Therefore, there are four ordered pairs — <math>(\sqrt{\frac{3 + \sqrt{5}}{2}},\sqrt{\frac{3 - \sqrt{5}}{2}}), (-\sqrt{\frac{3 + \sqrt{5}}{2}}, -\sqrt{\frac{3 - \sqrt{5}}{2}}), </math> <math>(\sqrt{\frac{3 - \sqrt{5}}{2}}, \sqrt{\frac{3 + \sqrt{5}}{2}}), (-\sqrt{\frac{3 - \sqrt{5}}{2}}, -\sqrt{\frac{3 + \sqrt{5}}{2}}).</math> Plugging the x and y values back in for each ordered pair satisfies the original equation.
  
 
<br>
 
<br>
−
In total, there are nine solutions, which are listed in each case.
+
In total, there are nine ordered pairs of real numbers that satisfy the original equation, which are listed in each case.
  
 
==See Also==
 
==See Also==

Latest revision as of 11:29, 17 March 2020

Problem

Find all pairs $(x,y)$ of real numbers which satisfy $x^3-y^3=4(x-y)$ and $x^3+y^3=2(x+y)$.

Solution

Factoring and rearranging terms for both equations results in $(x-y)(x^2 + xy + y^2 - 4) = 0$ and $(x+y)(x^2 - xy + y^2 - 2) = 0$. By the Zero Product Property, $x = y$ or $x^2 + xy + y^2 = 4$ in the first equation, and $x = -y$ and $x^2 - xy + y^2 = 2$ in the second equation. In order for the solution to satisfy both equations, it must satisfy at least one of the conditions in each equation. Now there are four cases to consider.

Case 1: $x=y$ and $x=-y$

Solving this system results in $(0,0)$. Plugging the x and y values back in satisfies the original equation.

Case 2: $x=y$ and $x^2 - xy + y^2 = 2$

Substitution results in $x^2 = 2$, so solving this system results in $(\sqrt{2},\sqrt{2})$ and $(-\sqrt{2},-\sqrt{2})$. Plugging the x and y values back in for each ordered pair satisfies the original equation.

Case 3: $x^2 + xy + y^2 = 4$ and $x=-y$

Substitution results in $y^2 = 4$, so solving this system results in $(2,-2)$ and $(-2,2)$. Plugging the x and y values back in for each ordered pair satisfies the original equation.

Case 4: $x^2 + xy + y^2 = 4$ and $x^2 - xy + y^2 = 2$

Subtracting the second equation from the first equation results in $2xy = 2$, so $xy = 1$. Thus, $x^2 + y^2 = 3$ and $y = \tfrac{1}{x}$, so $x^2 + \tfrac{1}{x^2} = 3$.


Multiplying both sides by $x^2$ and bringing all terms to one side results in $x^4 - 3x^2 + 1 = 0$. By using the Quadratic Formula, $x^2 = \tfrac{3 \pm \sqrt{5}}{2}$, and both quantities are positive. If $x^2 = \tfrac{3 + \sqrt{5}}{2}$, then $y^2 = \tfrac{3 - \sqrt{5}}{2}$ (and vice versa). Since $xy = 1$, $x$ and $y$ must have the same sign.


Therefore, there are four ordered pairs — $(\sqrt{\frac{3 + \sqrt{5}}{2}},\sqrt{\frac{3 - \sqrt{5}}{2}}), (-\sqrt{\frac{3 + \sqrt{5}}{2}}, -\sqrt{\frac{3 - \sqrt{5}}{2}}),$ $(\sqrt{\frac{3 - \sqrt{5}}{2}}, \sqrt{\frac{3 + \sqrt{5}}{2}}), (-\sqrt{\frac{3 - \sqrt{5}}{2}}, -\sqrt{\frac{3 + \sqrt{5}}{2}}).$ Plugging the x and y values back in for each ordered pair satisfies the original equation.


In total, there are nine ordered pairs of real numbers that satisfy the original equation, which are listed in each case.

See Also

2006 Indonesia MO (Problems)
Preceded by
First Problem 		1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 Followed by
Problem 2
All Indonesia MO Problems and Solutions
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