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Difference between revisions of "2000 JBMO Problems/Problem 1"

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Let <math>x</math> and <math>y</math> be positive reals such that <cmath> x^3 + y^3 + (x + y)^3 + 30xy = 2000. </cmath> Show that <math>x + y = 10</math>.
 
Let <math>x</math> and <math>y</math> be positive reals such that <cmath> x^3 + y^3 + (x + y)^3 + 30xy = 2000. </cmath> Show that <math>x + y = 10</math>.
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== Solution 1 ==
 
== Solution 1 ==
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<math>Kris17</math>
 
<math>Kris17</math>
  
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(This is wrong :|)
  
  
 
== Solution 2 ==
 
== Solution 2 ==
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By expanding the sum of cubes and substituting <math>x+y=10</math>, we get <math>(x+y)(x^2-xy+y^2)+1000+30xy=2000</math> dividing by <math>x+y=10</math> from both sides yields <math>x^2-xy+y^2+100+3xy=200</math>. Simplifying this we get <math>x^2+2xy+y^2+100=200</math> to <math>(x+y)^2+100=200</math> to <math>200=200</math> showing that <math>x+y=10</math>.
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krakkurz
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(dunno if this is a valid proof, my first time writing a solution)
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== Solution 3 ==
  
 
Rearranging the equation yields
 
Rearranging the equation yields

Latest revision as of 10:20, 21 July 2025

Problem

Let $x$ and $y$ be positive reals such that \[x^3 + y^3 + (x + y)^3 + 30xy = 2000.\] Show that $x + y = 10$.


Solution 1

After some manipulation we get: $3xy(10 - (x+y)) = 2((x+y)^3 - 10^3)$

Case 1: $x+y > 10:$ $RHS > 0$, So $LHS$ has to be $> 0$, so $10 - (x+y) > 0$ or $x+y < 10 => contradiction!$

Case 2: $x+y < 10:$ $RHS < 0$, So $LHS$ has to be $< 0$, so $10 - (x+y) < 0$ or $x+y > 10 => contradiction!$

Thus $x + y = 10$


$Kris17$

(This is wrong :|)


Solution 2

By expanding the sum of cubes and substituting $x+y=10$, we get $(x+y)(x^2-xy+y^2)+1000+30xy=2000$ dividing by $x+y=10$ from both sides yields $x^2-xy+y^2+100+3xy=200$. Simplifying this we get $x^2+2xy+y^2+100=200$ to $(x+y)^2+100=200$ to $200=200$ showing that $x+y=10$.


krakkurz (dunno if this is a valid proof, my first time writing a solution)

Solution 3

Rearranging the equation yields \[x^3 + y^3 + (x + y)^3 + 30xy - 2000 = 0.\] If $x+y=10$ in the large equation, then $x+y-10$ must be a factor of the large equation. Note that we can rewrite the large equation as \begin{align*} 0 &= (x+y)^3 - 1000 + x^3 + 3x^2y - 3x^2y + 3xy^2 - 3xy^2 + y^3 - 1000 + 30xy \\ &= 2[(x+y)^3 - 1000] - 3x^2y - 3xy^2 + 30xy. \end{align*} We can factor the difference of cubes in the first part and factor $3xy$ in the second part, resulting in \[0 = 2(x+y-10)((x+y)^2 + 10x + 10y + 100) - 3xy(x+y-10)\] Finally, we can factor by grouping, which results in \begin{align*} 0 &= (x+y-10)(2(x+y)^2 + 20x + 20y + 200 - 3xy) \\ &= (x+y-10)(2x^2 + xy + 2y^2 + 20x + 20y + 200). \end{align*} By the Zero Product Property, either $x+y=10$ or $2x^2 + xy + 2y^2 + 20x + 20y + 200 = 0.$ However, since $x$ and $y$ are both positive, $2x^2 + xy + 2y^2 + 20x + 20y + 200$ can not equal zero, so we have proved that $x+y = 10.$

See Also

2000 JBMO (Problems • Resources)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4
All JBMO Problems and Solutions
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