55rd Ukrainian National Mathematical Olympiad - Fourth Round

Question

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)

Accepted Answer

Let's rewrite the condition of harmonious average: b = 2aca+c, and now use the fact c = a+b2: 2a a+b2 = b(a + a+b2) a^2 + ab = ba + ab+b^22 2a^2 = ba + b^2 (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.