2021/22 AHSMC Part 1 Problem 12

$a = -(b+c)$, $b = -(a+c)$, and $c = -(a+b)$. $$\begin{align*} \frac{a}{b} + \frac{b}{c} + \frac{c}{a} &= \frac{a}{c} + \frac{c}{b} + \frac{b}{a} + 2021 \\ &= \frac{-(b+c)}{c} + \frac{-(a+b)}{b} + \frac{-(c+a)}{a} + 2021 \\ &= -\frac{b}{c} - 1 - \frac{a}{b} - 1 - \frac{c}{a} - 1 + 2021 \\ &= 2018 - \left( \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \right), \end{align*}$$ so $$2\left( \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \right) = 2018, \qquad \frac{a}{b} + \frac{b}{c} + \frac{c}{a} = \frac{2018}{2} = 1009.$$ The answer is (A).

Remark: There exist real numbers $a$, $b$, $c$ which satisfy the condition of the problem. Indeed, if we take $a = 1$ then $b$, $c$ need to be non-zero real numbers such that $$1 + b + c = 0, \qquad \frac{1}{b} + \frac{b}{c} + c = \frac{1}{c} + \frac{c}{b} + b + 2021.$$ Since $c = -b - 1$, by using the second equation from above one obtains the cubic equation $$b^3 + 1012b^2 + 1009b - 1 = 0,$$ which has at least one real non-zero solution.