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• 2006 AMC 12A Problems/Problem 22
  ...that <math>\triangle APO</math> is an [[isosceles triangle]]: <math>\angle AOP = 15^{\circ}</math> and <math>\angle OAP = OAB - 60^{\circ} = \frac{180^{\c
  2 KB (343 words) - 15:39, 14 June 2023
• 2000 AIME II Problems/Problem 10
  Thus, <math>\angle{AOP}+\angle{POB}+\angle{COQ}+\angle{QOD}=180</math>, or <math>(\arctan(\tfrac{1
  2 KB (399 words) - 17:37, 2 January 2024
• Mock AIME 1 Pre 2005 Problems
  ..., and <math>P</math> is a point on <math>l</math> such that triangle <math>AOP</math> is isosceles. Let <math>p</math> denote the value of <math>AP \cdot
  6 KB (1,100 words) - 22:35, 9 January 2016
• Mock AIME 1 Pre 2005 Problems/Problem 10
  ..., and <math>P</math> is a point on <math>l</math> such that triangle <math>AOP</math> is isosceles. Let <math>p</math> denote the value of <math>AP \cdot
  540 bytes (96 words) - 00:28, 23 December 2023
• 2011 AMC 12A Problems/Problem 15
  ...angent. Let <math>O</math> be the centre of the hemisphere. Triangle <math>AOP</math> is a right triangle, since <math>P</math> is ninety degrees. Apply t
  2 KB (376 words) - 23:14, 5 January 2024
• 2011 AIME II Problems/Problem 13
  ...^{\circ}</math>. Then <math>\angle O_2PA=75^{\circ}-\theta \implies \angle AOP = 180^{\circ}-2\angle O_2PA = 2\theta+30^{\circ} \implies \angle ABP=\theta
  13 KB (2,055 words) - 05:25, 9 September 2022
• 2016 USAJMO Problems/Problem 5
  ...AS similarity. Thus, <math>\angle AOP = \angle AQH,</math> so <math>\angle AOP</math> is a right angle. Since <math>\angle AOP</math> and <math>\angle AOQ</math> are both right angles, we get <math>\ang
  10 KB (1,733 words) - 19:15, 14 June 2020
• 2017 AIME I Problems/Problem 4
  ...\overline {OP} = h</math>. Using the Pythagorean Theorem on triangle <math>AOP</math>, <math>BOP</math>, or <math>COP</math> (all three are congruent by S Then continue as before to use the Pythagorean Theorem on <math>\triangle AOP</math>, find <math>h</math>, and find the volume of the pyramid.
  7 KB (1,177 words) - 15:55, 5 January 2024
• 2017 USAJMO Problems/Problem 5
  In addition, <math>\angle AON=\angle AOP=\angle AOB+\angle BOP=2\angle C+\angle A</math>. Combining these two equati
  5 KB (805 words) - 18:31, 21 September 2021
• 1978 IMO Problems/Problem 2
  Let <math>\alpha=\angle AOP,\;\beta=\angle BPD,\;\theta=\angle APE</math>
  6 KB (1,039 words) - 13:00, 13 March 2024
• 2021 Fall AMC 10B Problems/Problem 17
  Because of how reflections work, we have that <math>\angle AOP' = \angle POA</math> and <math>\angle P'OB = \angle BOP''</math>; adding th
  8 KB (1,331 words) - 22:44, 16 December 2023
• 2023 AIME I Problems/Problem 5
  ...and center of the circumcircle respectively, and let <math>\theta = \angle AOP</math>. .... We can use the [[Law of Cosines]] on isosceles triangles <math>\triangle AOP, \, \triangle COP, \, \triangle BOP, \, \triangle DOP</math> to get
  19 KB (3,107 words) - 23:31, 17 January 2024
• 2023 AMC 10B Problems/Problem 7
  ...</math> and <math>H</math> point <math>P</math>. We know that <math>\angle{AOP} = 20</math> and that <math>\angle{A} = 90</math>. Subtracting <math>20</ma
  4 KB (577 words) - 02:09, 27 November 2023