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• 2005 AIME II Problems/Problem 14
  Substituting our knowns, we have <math>\frac{CE}{BE} = \frac{3 \cdot 14^2}{2 \cdot 13^2} = \frac{BC - BE}{BE} = \frac{15}{BE} - 1 \Longrig ==Solution 3 (LoC and LoS bash)==
  13 KB (2,129 words) - 18:56, 1 January 2024