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Philippine Mathematical Olympiad

Algebra Difficulty 5.4 AIME, harder Prove it Philippines

Problem:
Let $a$ , $b$ , $c$ be real numbers such that

$3 a b + 2 = 6 b, \quad 3 b c + 2 = 5 c, \quad 3 c a + 2 = 4 a$

Suppose the only possible values for the product $a b c$ are $r / s$ and $t / u$ , where $r / s$ and $t / u$ are both fractions in lowest terms. Find $r+s+t+u$ .

Solution

Solution:
The three given equations can be written as

$3 a + \frac{2}{b} = 12, \quad 3 b + \frac{2}{c} = 10, \quad 3 c + \frac{2}{a} = 8$

The product of all the three equations gives us
$27 a b c + 6\left(3 a + \frac{2}{b}\right) + 6\left(3 b + \frac{2}{c}\right) + 6\left(3 c + \frac{2}{a}\right) + \frac{8}{a b c} = 120$

Plugging in the values and simplifying the equation gives us
$27(a b c)^2 - 30 a b c + 8 = 0$
This gives $a b c$ as either $4 / 9$ or $2 / 3$ , so $r+s+t+u=4+9+2+3=18$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.