Competition problem — Algebra — Olympiad Maths Prep

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Algebra Difficulty 5.3 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 geometric average number of *a*, *c* is equal to the number *b*, so $b = \sqrt{ac}$ . Is it necessary that numbers *a*, *b*, *c* are equal?

Solution

Let's rewrite the equality $b^2 = ac$ using $c = \frac{a+b}{2}$ :

$b^2 = a \cdot \frac{a+b}{2} \Leftrightarrow 2b^2 = ba + a^2 \Leftrightarrow (b-a)(2b+a) = 0.$
Now let's denote, for example, $b=2$ , which means $a=-4$ and $c=-1$ , hence we receive three different numbers satisfying the conditions.

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