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  • ...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 (1,062 words) - 18:04, 17 January 2025
  • ...th>, <math>d</math> are the four side lengths and <math>s = \frac{a+b+c+d}{2}</math>. .../math>. Hence, <math>[ABCD]=\frac{\sin B(ab+cd)}{2}</math>. Multiplying by 2 and squaring, we get:
    3 KB (543 words) - 18:35, 29 October 2024
  • ...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) - 20:41, 27 November 2024
  • 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) - 09:58, 18 March 2015
  • 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) - 15:33, 14 October 2022
  • 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) - 16:41, 21 January 2017
  • == Problem 1 == [[Mock AIME 1 Pre 2005 Problems/Problem 1|Solution]]
    6 KB (1,100 words) - 21:35, 9 January 2016
  • ==Problem 1== [[Mock AIME 3 Pre 2005/Problem 1|Solution]]
    7 KB (1,135 words) - 22:53, 24 March 2019
  • ==Problem== ...alent to <math>8</math> dividers, and there are <math>{8 + 7 \choose 7} = {15 \choose 7} = 6435 \equiv \boxed{435} \pmod{1000}</math>.
    950 bytes (137 words) - 09:16, 29 November 2019
  • ==Problem== <math>\left|a_1^{2} - a_2^{2}\right| = 10</math>
    3 KB (520 words) - 11:55, 11 January 2019
  • ==Problem== .../math> such that <math>\angle{BAE}</math> is right. If <math>BD = 15, DE = 2,</math> and <math>BC = 16</math>, then <math>CD</math> can be expressed as
    2 KB (278 words) - 15:32, 27 December 2019
  • ==Problem== ...second equation and rearranging variables gives <math>45x=8\cdot 15+bx=15^2\implies x=5</math>. Back-substitution yields <math>AC=21</math> and consequ
    2 KB (379 words) - 00:27, 6 December 2024
  • ==Problem== ...ence then repeats itself. We hence find that there are a total of <math>11*15 - 1</math> or <math>\boxed{164}</math> numbers that satisfy the inequality.
    714 bytes (105 words) - 22:59, 24 April 2013
  • ==Problem== ...ath>QP</math>. Thus, <math>O_3R \bot PQ</math> and <math>|PR|=\frac{|PQ|}{2}=16</math>
    3 KB (563 words) - 01:05, 25 November 2023
  • == Problem == Let <math>N</math> denote the number of permutations of the <math>15</math>-character string <math>AAAABBBBBCCCCCC</math> such that
    1 KB (221 words) - 16:27, 23 February 2013
  • == Problem 1 == [[Mock AIME 5 Pre 2005 Problems/Problem 1|Solution]]
    6 KB (909 words) - 06:27, 12 October 2022
  • ...e number of ways to do such is <math>{4+3-1 \choose 3-1} = {6 \choose 2} = 15</math>. == Reasoning 2 ==
    5 KB (795 words) - 16:39, 31 December 2024
  • == Problem == ...inradius of <math>5</math> and a circumradius of <math>16</math>. If <math>2\cos{B} = \cos{A} + \cos{C}</math>, then the area of triangle <math>ABC</mat
    2 KB (340 words) - 00:44, 3 March 2020
  • == Problem == ...a triangle with <math>AB = 13</math>, <math>BC = 14</math>, and <math>AC = 15</math>. Let <math>D</math> be the foot of the altitude from <math>A</math>
    2 KB (294 words) - 15:24, 24 August 2022
  • == Problem == <math>CosB=\frac{68^2+100^2-112^2}{2.68.100}=\frac{13}{85}</math>
    2 KB (282 words) - 09:06, 9 August 2022

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