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- ...administered by the [[American Mathematics Competitions]] (AMC). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC and of the recent expansion ...ficulty=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 (874 words) - 22:02, 10 November 2024
- <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 (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== ...VC</tt> - the only other combination, two vowels, is impossible due to the problem statement). Then, note that:5 KB (795 words) - 15:03, 17 October 2021
- ==Problem== ...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 be e2 KB (330 words) - 09:23, 4 April 2012
- ==Problem== Let <math>N</math> denote the number of <math>8</math>-tuples <math>(a_1, a_2, \dots, a_8)</math> of real numbers such that3 KB (520 words) - 11:55, 11 January 2019
- ==Problem== <cmath> F = \sum_{k=0}^{n-1} k(k-1)2^k = (n^2-5n+8)2^n - 8 </cmath>2 KB (306 words) - 09:36, 4 April 2012
- == Problem == ...8,9\}</math> may appear as the hundreds digit, and there are <math>9 \cdot 8 = 72</math> choices for the tens and units digits. Thus the sum of the hund1 KB (194 words) - 12:44, 5 September 2012
- == 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 <ma1 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
- ...h> objects in <math>3</math> bins. The number of ways to do such is <math>{4+3-1 \choose 3-1} = {6 \choose 2} = 15</math>. ...ach urn, then there would be <math>{n \choose k}</math> possibilities; the problem is that you can repeat urns, so this does not work.<math>n</math> and then5 KB (795 words) - 16:39, 31 December 2024
- == Problem == ...}}{2}</math>. Combining this with the fact that <math>a^2+c^2 - \frac{7ac}{8} = 36 \cdot 23</math>, we have that: <math>\frac{(a+c)(-2ac \cdot \frac{7}{2 KB (340 words) - 00:44, 3 March 2020
- == Problem == ...assigned to these houses such that there is at least one house with <math>4</math> residents?461 bytes (62 words) - 20:18, 8 October 2014
- ==Problem 1== [[Mock AIME 4 Pre 2005/Problems/Problem 1 | Solution]]7 KB (1,094 words) - 14:39, 24 March 2019
- == Problem 1 == [[Mock AIME 2 Pre 2005 Problems/Problem 1|Solution]]6 KB (1,052 words) - 12:52, 9 June 2020