2000 JBMO Problems/Problem 1

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 $x + y = 10$

$Kris17$

Solution 2

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 $x+y = 10.$

2000 JBMO (Problems • Resources)
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2000 JBMO Problems/Problem 1

Contents

Problem

Solution 1

Solution 2

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