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Romanian Mathematical Olympiad

Algebra Difficulty 5.0 AIME Prove it Romania

Consider positive real numbers m,n,a,b,cm, n, a, b, c, such that m>nm > n and ∣ma−nb∣≤c(m−n)|ma - nb| \le c(m - n), ∣mb−nc∣≤a(m−n)|mb - nc| \le a(m - n), ∣mc−na∣≤b(m−n)|mc - na| \le b(m - n). Prove that a=b=ca = b = c.

Lucian Petrescu

Solution

If x≥0x \ge 0, y≥0y \ge 0, and z≥0z \ge 0, then ma−nb=c(m−n)ma - nb = c(m - n) and analogously, we have
m(a−c)=n(b−c),m(b−a)=n(c−a),m(c−b)=n(a−b).(∗) m(a - c) = n(b - c), \quad m(b - a) = n(c - a), \quad m(c - b) = n(a - b). \quad (*)
If all parentheses in (∗)(*) are nonzero, then we obtain m3(a−c)(b−a)(c−b)=−n3(b−c)(c−a)(a−b)m^3(a-c)(b-a)(c-b) = -n^3(b-c)(c-a)(a-b), and therefore m=−nm = -n, false. Thus at least one difference is zero, say a−c=0a - c = 0. It follows that b=cb = c, hence a=b=ca = b = c.

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