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• 2006 AMC 12B Problems/Problem 22
  The power of <math>10</math> for any factorial is given by the well-known algorithm ...of <math>2</math> that divides <math>n!</math> is larger or equal than the power of <math>5</math> which divides
  5 KB (881 words) - 15:52, 23 June 2021
• 2006 AMC 12A Problems/Problem 17
  ...> is an isosceles right triangle. Thus <math>DG = r\sqrt{2}</math>. By the Power of a Point Theorem,
  6 KB (958 words) - 23:29, 28 September 2023
• 2006 AMC 12A Problems/Problem 24
  ...rent expression is irreducible as each term has a different <math>x</math> power. Thus, when we write <math>a</math> and <math>b</math> back to their origin
  8 KB (1,332 words) - 17:37, 17 September 2023
• 2005 AMC 12B Problems/Problem 23
  We can then put <math>x+y</math> to the third power or <math>(x+y)^{3}=10^{3z}</math>. Basic polynomial multiplication shows us
  5 KB (786 words) - 11:36, 19 May 2024
• Lassiter Invitational
  ...consists of a multiple-choice test, ten ciphering questions, and a pair of power questions, i.e., more in-depth questions on which members of teams collabor
  1 KB (161 words) - 18:35, 25 November 2007
• Power of a Point Theorem/Introductory Problem 1
  Applying the [[Power of a Point Theorem]], we get <math> 3\cdot(3+5) = x (x+10) \rightarrow x^2 ''Back to the [[Power of a Point Theorem]].''
  448 bytes (67 words) - 15:15, 23 March 2020
• Power of a Point Theorem/Introductory Problem 2
  Applying the Power of a Point Theorem gives <math> 6\cdot x = 4\cdot 1 </math>, so <math> x = ''Back to the [[Power of a Point Theorem]].''
  289 bytes (45 words) - 13:14, 16 July 2017
• Power of a Point Theorem/Introductory Problem 3
  From the Power of a Point Theorem, we have that ''Back to the [[Power of a Point Theorem]].''
  969 bytes (154 words) - 14:40, 3 July 2006
• Power of a Point Theorem/Introductory Problem 4
  ...3 </math> (or by just knowing your [[Pythagorean Triple]]s). Applying the Power of a Point Theorem gives <math> AE\cdot BE = CE\cdot DE </math>, or <math> ''Back to the [[Power of a Point Theorem]].''
  1 KB (177 words) - 02:14, 26 November 2020
• Power of a Point Introductory Problem 3
  #REDIRECT [[Power of a Point Theorem/Introductory Problem 3]]
  61 bytes (8 words) - 12:07, 10 July 2006
• Power of a Point Introductory Problem 4
  #REDIRECT [[Power of a Point Theorem/Introductory Problem 4]]
  61 bytes (8 words) - 18:49, 10 July 2006
• Set
  ==Power Sets== {{main|power set}}
  11 KB (2,021 words) - 00:00, 17 July 2011
• 2006 USAMO Problems
  ...> or to <math>n+2^{m_n+1}</math> where <math>2^{m_n}</math> is the largest power of 2 that is a factor of <math>n</math>. Show that if <math>k\ge2</math> is
  3 KB (520 words) - 09:24, 14 May 2021
• 2005 AIME II Problems/Problem 4
  If <math>n</math> is the power of a single prime, then there are 11 possibilities (<math>2^1</math> to <ma
  3 KB (377 words) - 18:36, 1 January 2024
• 2005 AIME II Problems/Problem 5
  ...> and <math>3^6</math>. However, the ordered pairs where b is to the sixth power are distinct, so they are not redundant. (For example, the pairs (4, 64) an
  3 KB (547 words) - 19:15, 4 April 2024
• 2005 AIME II Problems/Problem 8
  ...until it reaches the circle on both sides; call them <math>P,Q</math>. By Power of a Point,
  4 KB (693 words) - 13:03, 28 December 2021
• Problem solving
  ...es, students’ formula sheets were the source of knowledge, the source of power that fueled the top students and the top schools. They were studied, memori
  6 KB (1,039 words) - 17:43, 30 July 2018
• Euler's identity
  *[[Power series]]
  3 KB (452 words) - 23:17, 4 January 2021
• 2005 AIME I Problems/Problem 6
  ...1)^4</math>. Thus, we add 1 to each side in order to complete the fourth power and get If we don't see the fourth power, we can always factor the LHS to try to create a quadratic substitution. Ch
  4 KB (686 words) - 01:55, 5 December 2022
• 2005 AIME I Problems/Problem 15
  ...ength of the median be <math>3m</math>. Then by two applications of the [[Power of a Point Theorem]], <math>DE^2 = 2m \cdot m = AF^2</math>, so <math>DE = Our earlier result from Power of a Point was that <math>2m^2 = (10 - c)^2</math>, so we combine these two
  5 KB (906 words) - 23:15, 6 January 2024

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