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• 2011 AIME II Problems/Problem 13
  ...onals are perpendicular, both triangle <math>DMP</math> and triangle <math>BMP</math> are 30-60-90 right triangles.
  13 KB (2,055 words) - 05:25, 9 September 2022
• 2011 USAJMO Problems/Problem 5
  ...t, so <math>\angle BOP = \angle BOD/2 = \angle BED</math>, so <math>\angle BMP = \angle BED</math>. Because <math>AC</math> and <math>DE</math> are parall
  4 KB (717 words) - 17:00, 14 April 2024
• 2013 AIME II Problems/Problem 4
  ...rt{3}\right)</math>. At this point, it is useful to note that <math>\Delta BMP</math> is a 30-60-90 triangle with <math>\overline{MB}</math> measuring <ma
  8 KB (1,268 words) - 14:10, 31 January 2024
• 2013 USAMO Problems/Problem 1
  ...cmath><cmath>= \angle BRM - \angle PRM</cmath><cmath>= \angle BRP = \angle BMP.</cmath> ...e PBM - \angle BMP</cmath><cmath>= \left(180^{\circ} - \angle PBM - \angle BMP\right) - \angle MCP</cmath><cmath>= \angle BPM - \angle MCP</cmath><cmath>=
  14 KB (1,830 words) - 18:22, 10 May 2023
• 2014 USAMO Problems/Problem 5
  ...iral similarity, <math> BPM \sim XPY </math> and <math>\angle XYP = \angle BMP = \angle BCA</math>. Hence, <math>\angle XYP = \angle BCA</math>, <math>XY
  3 KB (530 words) - 12:08, 27 March 2022
• 2018 AIME II Problems/Problem 12
  ...endicular from <math>D</math> is <math>N</math>, then right triangle <math>BMP</math> is congruent to right triangle <math>DNP</math>.
  18 KB (2,912 words) - 13:12, 24 January 2024
• 1979 IMO Problems/Problem 3
  ...since <math>P</math> is the antipode of <math>B</math>, then <math>\angle BMP = 90^{\circ} = \angle EMP = \angle FMP</math> so <math>EP=FP</math> for all
  8 KB (1,485 words) - 22:55, 29 January 2021
• 2023 USAMO Problems/Problem 1
  .... Thus, <math>\triangle AXM \sim \triangle MPC</math>, and <math>\triangle BMP \sim \triangle AMQ</math>.
  4 KB (684 words) - 01:55, 6 February 2024