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• 2008 AMC 12B Problems/Problem 25
  ...d <math>r+s=19-11=8</math>. By the Pythagorean Theorem on <math>\triangle ADR</math> and <math>\triangle BCS</math>, we get <math>r^2+(2x)^2=7^2</math> a
  13 KB (2,248 words) - 14:44, 5 January 2025
• 2011 USAMO Problems/Problem 5
  ...</math> meet at <math>R</math>. Notice that with respect to triangle <math>ADR</math>, <math>P</math> and <math>Q_2</math> are isogonal conjugates (this c
  5 KB (846 words) - 18:22, 15 February 2025
• 2012 IMO Problems/Problem 5
  So <math>\angle ADR = \angle ALR = 90^{\circ}</math>, and in the same way <math>\angle BKR = 90 ...th>RK = RL</math>. Since <math>RM</math> is in the middle and <math>\angle ADR = \angle BKR = 90^{\circ}</math>,
  2 KB (437 words) - 08:30, 20 November 2023
• 2021 AIME I Problems/Problem 9
  ...math>ARQD</math> are both cyclic,it is clear that <math>\angle AQR =\angle ADR</math> and <math>\angle APR =\angle ABR</math>,which shows <math>\triangle
  22 KB (3,581 words) - 05:30, 13 January 2025