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ChallengeSuperchallengeAdvanced Number Theory

Advanced Number Theory · Superchallenge problem

In triangle ABCABC, cevians ADAD, BEBE, CFCF are concurrent, with FF on ABAB satisfying AF:FB=1:2AF:FB = 1:2 and DD on BCBC satisfying BD:DC=3:4BD:DC = 3:4. Then CE:EA=mnCE:EA = \tfrac{m}{n} for relatively prime positive integers mm and nn. Find m+nm + n.