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Difference between revisions of "2024 SSMO Team Round Problems/Problem 7"

(Created page with "==Problem== Let <math>a</math> and <math>b</math> be real numbers that satisfy <math>a^3+8ab^2=8b^3+4a^2b=375</math>. Find <math>\lfloor ab \rfloor</math>. ==Solution==")
 
 
Line 4: Line 4:
  
 
==Solution==
 
==Solution==
 +
We can factor the first equation as such:
 +
\begin{align*}
 +
a^3+8ab^2&=8b^3+4a^2b\implies\\
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a^3-8b^3&=4a^2b-8ab^2\implies\\
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(a-2b)(a^2+2ab+4b^2)&=4ab(a-2b)\implies\\
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(a-2b)(a^2-2ab+4b^2)&=0.
 +
\end{align*}
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Consider the case where <math>a-2b</math> is nonzero. Then, the discriminant of the quadratic factor (in <math>a</math>) is <math>(2b)^2-4(4b^2) = -12b^2.</math> Since <math>a</math> and <math>b</math> are reals, this means <math>b=0</math> which is clearly not possible. So, <math>a = 2b.</math> Substituting, we have <cmath>8b^3+16b^3 = 375\implies b^3 = \frac{125}{8}\implies b = \frac{5}{2}.</cmath> So, <cmath>\left\lfloor ab\right\rfloor = \left\lfloor (5)\left(\frac{5}{2}\right)\right\rfloor = \left\lfloor\frac{25}{2}\right\rfloor = \boxed{12}.</cmath>
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 +
~SMO_Team

Latest revision as of 14:38, 10 September 2025

Problem

Let $a$ and $b$ be real numbers that satisfy $a^3+8ab^2=8b^3+4a^2b=375$. Find $\lfloor ab \rfloor$.

Solution

We can factor the first equation as such: \begin{align*} a^3+8ab^2&=8b^3+4a^2b\implies\\ a^3-8b^3&=4a^2b-8ab^2\implies\\ (a-2b)(a^2+2ab+4b^2)&=4ab(a-2b)\implies\\ (a-2b)(a^2-2ab+4b^2)&=0. \end{align*} Consider the case where $a-2b$ is nonzero. Then, the discriminant of the quadratic factor (in $a$) is $(2b)^2-4(4b^2) = -12b^2.$ Since $a$ and $b$ are reals, this means $b=0$ which is clearly not possible. So, $a = 2b.$ Substituting, we have \[8b^3+16b^3 = 375\implies b^3 = \frac{125}{8}\implies b = \frac{5}{2}.\] So, \[\left\lfloor ab\right\rfloor = \left\lfloor (5)\left(\frac{5}{2}\right)\right\rfloor = \left\lfloor\frac{25}{2}\right\rfloor = \boxed{12}.\]

~SMO_Team

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