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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=12(a+b)c = \frac{1}{2}(a+b), and that the harmonious average number of a, c is equal to the number b, so b=21a+1cb = \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=2aca+cb = \frac{2ac}{a+c}, and now use the fact c=a+b2c = \frac{a+b}{2}:
2a⋅a+b2=b(a+a+b2)⇔a2+ab=ba+ab+b22⇔2a2=ba+b2⇔(a−b)(2a+b)=0. 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=2a=2, which means b=−4b=-4 and c=−1c=-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.