2024 SSMO Team Round Problems/Problem 7

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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2024 SSMO Team Round Problems/Problem 7

Problem

Solution