Maths Olympiad Prep

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, 2009

Algebra Difficulty 3.0 AMC 10/12 Prove it Canada

If a>0a > 0 and b>0b > 0, a new operation \nabla is defined as follows: ab=a+b1+aba\nabla b = \dfrac{a+b}{1+ab}.
For example, 36=3+61+3×6=9193\nabla 6 = \dfrac{3+6}{1+3\times 6} = \dfrac{9}{19}.
(a) Calculate 252\nabla 5.
(b) Calculate (12)3(1\nabla 2)\nabla 3.
(c) If 2x=572\nabla x = \dfrac{5}{7}, what is the value of xx?
(d) For some values of xx and yy, the value of xyx\nabla y is equal to x+y17\dfrac{x+y}{17}. Determine all possible ordered pairs of positive integers xx and yy for which this is true.

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project. Solutions are the publisher's, linked not copied.