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Purple Comet 2018 HS problem 15

Source:

March 19, 2020
algebra

Problem Statement

Let aa and bb be real numbers such that 1a2+3b2=2018a\frac{1}{a^2} +\frac{3}{b^2} = 2018a and 3a2+1b2=290b\frac{3}{a^2} +\frac{1}{b^2} = 290b. Then abba=mn\frac{ab}{b-a }= \frac{m}{n} , where mm and nn are relatively prime positive integers. Find m+nm + n.