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• 2015 AIME II Problems/Problem 15
  <cmath>[AEC] = \frac{1}{2}AE*CE*\sin{\angle{ACE}}=\frac{1}{2}AE*8\sin{\angle{CYE}}*\frac{1}{\sqrt{5}}</cmath> ...ath>\angle{CAY} = 90 - \angle{AYC}</math>, <math>\angle{XYC} = 90 - \angle{CYE}</math>, <math>XY = AE</math>, we have
  31 KB (5,086 words) - 19:15, 20 December 2023