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Ukrainian National Mathematical Olympiad - Fourth Round

Algebra Difficulty 5.2 AIME, harder Prove it Ukraine

It is known that the arithmetic average of the numbers a, b is equal to the number c, so $c = \frac{1}{2}(a+b)$ , and that the harmonious average number of a, c is equal to the number b, so $b = \frac{2}{\frac{1}{a}+\frac{1}{c}}$ . Is it necessary that numbers a, b, c are equal?

(Bogdan Rublyov)

Solution

Let's rewrite the condition of harmonious average: $b = \frac{2ac}{a+c}$ , and now use the fact $c = \frac{a+b}{2}$ :
$2a \cdot \frac{a+b}{2} = b\left(a + \frac{a+b}{2}\right) \Leftrightarrow a^2 + ab = ba + \frac{ab+b^2}{2} \Leftrightarrow 2a^2 = ba + b^2 \Leftrightarrow (a-b)(2a+b) = 0.$
Let's denote, for example, $a=2$ , which means $b=-4$ and $c=-1$ , hence we receive three different numbers satisfying the conditions.

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