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Mathematical Olympiad in Ukraine

Algebra Difficulty 5.3 AIME, harder Prove it Ukraine

Distinct real numbers a,b,ca, b, c satisfy the condition a+1b=b+1c=c+1aa + \frac{1}{b} = b + \frac{1}{c} = c + \frac{1}{a}. Find all possible values of the product abcabc:
1) for all real a,b,ca, b, c;
2) for positive real a,b,ca, b, c?

Solution

a) See problem 9.5

b) We will show that there are no such positive numbers that satisfy the condition of the problem. Without loss of generality we suppose that a>ba > b, then from the equality (a−b)=b−ab−c(a - b) = \frac{b - a}{b - c} we conclude that b>cb > c. From the condition b−c=c−aa−cb - c = \frac{c - a}{a - c}, we can get a contradiction: c>a>b>cc > a > b > c.

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