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Belarus — Selection and Training Session

Algebra Difficulty 5.7 AIME, harder Prove it Belarus

Given nonzero real numbers $a, b, c$ , with
$a + b + c = a^2 + b^2 + c^2 = a^3 + b^3 + c^3. \quad (*)$
a) Find $\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) (a + b + c - 2)$ .
b) Do there exist pairwise different nonzero $a, b, c$ satisfying $(*)$ ?
(D. Bazylev)

Solution

a) (Solution of M.Mankevich, E.Dovgialo.) Let $\sigma_1 = a + b + c$ , $\sigma_2 = ab + bc + ca$ , $\sigma_3 = abc$ . From the given equalities we have the system
$\sigma_1 = \sigma_1^2 - 2\sigma_2 \quad (1), \quad \sigma_1^2 = \sigma_1^3 - 3\sigma_1\sigma_2 + 3\sigma_3. \quad (2)$
From (1) we have $\sigma_1^2 = \sigma_1^3 - 2\sigma_1\sigma_2$ . Then from (2) it follows that
$-2\sigma_1\sigma_2 = -\sigma_1\sigma_2 + 3\sigma_3. \quad (3)$
Finally, the expression we search is
$\frac{\sigma_2}{\sigma_3}(\sigma_1 - 2) = \frac{\sigma_1\sigma_2 - 2\sigma_2}{\sigma_3} = [\text{in view of (3)}] = \frac{3\sigma_3}{\sigma_3} = 3.$

b) It is easy to verify, that $a = \frac{1}{2}$ , $b = \frac{2+\sqrt{6}}{4}$ , $c = \frac{2-\sqrt{6}}{4}$ satisfy the given system.

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