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• Mock AIME 3 Pre 2005 Problems/Problem 4
  ==Problem== <cmath>\zeta_1^2+\zeta_2^2+\zeta_3^2=3</cmath>
  2 KB (221 words) - 02:49, 19 March 2015

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• American Invitational Mathematics Examination
  ...ed States at the [[International Mathematics Olympiad]] (IMO). While most AIME participants are high school students, some bright middle school students a High scoring AIME students are invited to take the prestigious [[United States of America Mat
  8 KB (1,057 words) - 12:02, 25 February 2024
• United States of America Mathematical Olympiad
  ...administered by the [[American Mathematics Competitions]] (AMC). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC and of the recent expansion ...7-9|breakdown=<u>Problem 1/4</u>: 7<br><u>Problem 2/5</u>: 8<br><u>Problem 3/6</u>: 9}}
  6 KB (869 words) - 12:52, 20 February 2024
• Brahmagupta's Formula
  <cmath>4[ABCD]^2=\sin^2 B(ab+cd)^2</cmath> <cmath>4[ABCD]^2=(1-\cos^2B)(ab+cd)^2=(ab+cd)^2-\cos^2B(ab+cd)^2</cmath>
  3 KB (465 words) - 18:31, 3 July 2023
• Mock AMC
  ...A number of '''Mock AMC''' competitions have been hosted on the [[Art of Problem Solving]] message boards. They are generally made by one community member ...AMC]] competition. There is no guarantee that community members will make Mock AMCs in any given year, but there probably will be one.
  51 KB (6,175 words) - 20:58, 6 December 2023
• Mock AIME 2 Pre 2005
  The '''Mock AIME 2 Pre 2005''' was written by [[Art of Problem Solving]] community member Mildorf. * [[Mock AIME 2 Pre 2005 Problems|Entire Exam]]
  2 KB (181 words) - 10:58, 18 March 2015
• Mock AIME 7 Pre 2005
  The '''Mock AIME 7 Pre 2005''' was written by [[Art of Problem Solving]] community member Mildorf. * [[Mock AIME 7 Pre 2005 Problems|Entire Exam]]
  1 KB (146 words) - 16:33, 14 October 2022
• Mock AIME 1 2005-2006
  The '''Mock AIME 1 2005-2006''' was written by [[Art of Problem Solving]] community member paladin8. * [[Mock AIME 1 2005-2006/Answer Key|Answer Key]]
  1 KB (135 words) - 17:41, 21 January 2017
• Mock AIME 1 Pre 2005 Problems
  == Problem 1 == [[Mock AIME 1 Pre 2005 Problems/Problem 1|Solution]]
  6 KB (1,100 words) - 22:35, 9 January 2016
• Mock AIME 3 Pre 2005 Problems
  ==Problem 1== ...rcles are mutually externally tangent. Two of the circles have radii <math>3</math> and <math>7</math>. If the area of the triangle formed by connecting
  7 KB (1,135 words) - 23:53, 24 March 2019
• Mock AIME 3 Pre 2005 Problems/Problem 1
  ==Problem== ...rcles are mutually externally tangent. Two of the circles have radii <math>3</math> and <math>7</math>. If the area of the triangle formed by connecting
  795 bytes (129 words) - 10:22, 4 April 2012
• Mock AIME 3 Pre 2005 Problems/Problem 3
  ==Problem== <cmath>2f\left(x\right) + f\left(\frac{1}{x}\right) = 5x + 4</cmath>
  1 KB (191 words) - 10:22, 4 April 2012
• Mock AIME 3 Pre 2005 Problems/Problem 5
  ==Problem== ...VC</tt> - the only other combination, two vowels, is impossible due to the problem statement). Then, note that:
  5 KB (795 words) - 16:03, 17 October 2021
• Mock AIME 3 Pre 2005 Problems/Problem 7
  ==Problem== ...ect at <math>P</math>. If <math>AB = 1, CD = 4,</math> and <math>BP : DP = 3 : 8,</math> then the area of the inscribed circle of <math>ABCD</math> can
  2 KB (330 words) - 10:23, 4 April 2012
• Mock AIME 3 Pre 2005 Problems/Problem 8
  ==Problem== Here are some thoughts on the problem:
  3 KB (520 words) - 12:55, 11 January 2019
• Mock AIME 3 Pre 2005 Problems/Problem 10
  ==Problem== Therefore we have <math>a_n \equiv 6\cdot 16 - 4^2 - 4\cdot 4 - 6 = \boxed{058} \pmod{1000}</math>.
  2 KB (306 words) - 10:36, 4 April 2012
• Mock AIME 3 Pre 2005 Problems/Problem 14
  ==Problem== ...th>C</math> and <math>D</math> respectively. If <math>AD = 3, AP = 6, DP = 4,</math> and <math>PQ = 32</math>, then the area of triangle <math>PBC</math
  3 KB (563 words) - 02:05, 25 November 2023
• Mock AIME 3 Pre 2005 Problems/Problem 15
  ==Problem== <math>\sum_{k=1}^{40} \cos^{-1}\left(\frac{k^2 + k + 1}{\sqrt{k^4 + 2k^3 + 3k^2 + 2k + 2}}\right)</math>
  2 KB (312 words) - 10:38, 4 April 2012
• Mock AIME 5 2005-2006 Problems/Problem 1
  == Problem == ...divisors of <math>n</math> less than <math>50</math> (e.g. <math>f(12) = 2+3 = 5</math> and <math>f(101) = 0</math>). Evaluate the remainder when <math>
  2 KB (209 words) - 12:43, 10 August 2019
• Mock AIME 1 Pre 2005 Problems/Problem 1
  == Problem == ...e tens and units digits. Thus the sum of the hundreds places is <math>(1+2+3+\cdots+9)(72) \times 100 = 45 \cdot 72 \cdot 100 = 324000</math>.
  1 KB (194 words) - 13:44, 5 September 2012
• Mock AIME 1 Pre 2005 Problems/Problem 7
  == Problem == ...numbers in the middle (those mentioned in condition [2]). There are <math>4-k</math> <tt>A</tt>s amongst the last six numbers then. Also, there are <ma
  1 KB (221 words) - 17:27, 23 February 2013

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