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- == Problem == Using the identity <math>\cos A + \cos B + \cos C = 1+\frac{r}{R}</math>, we have that <math>\cos A + \cos B + \cos C = \frac{21}2 KB (340 words) - 00:44, 3 March 2020
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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 Mat8 KB (1,062 words) - 18:04, 17 January 2025
- Substituting <math>\sin^2B=1-\cos^2B</math> results in <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== ...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== Here are some thoughts on the problem:3 KB (520 words) - 11:55, 11 January 2019
- ==Problem== ...<math>D</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 expre2 KB (278 words) - 15:32, 27 December 2019
- ==Problem== ...<math>n</math> such that <math>1 \le n \le 1000</math> and <math>n^{12} - 1</math> is divisible by <math>73</math>.714 bytes (105 words) - 22:59, 24 April 2013
- ==Problem== <math>A_1=\frac{\sqrt{\left(13\right)\left(7\right)\left(1\right)\left(5\right)}}{4}=\frac{\sqrt{455}}{4}</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 that1 KB (221 words) - 16:27, 23 February 2013
- == Problem 1 == <cmath>6g(1 + (1/y)) + 12g(y + 1) = \log_{10} y</cmath>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>. ...ath>n</math> bins is <math>{n+k-1 \choose n-1}</math> or <math>\dbinom{n+k-1}k</math>.5 KB (795 words) - 16:39, 31 December 2024
- == Problem == Note: The problem is meant to be interpreted so that if you cannot produce one arrangement fr692 bytes (107 words) - 12:01, 24 November 2021
- == Problem == Using the identity <math>\cos A + \cos B + \cos C = 1+\frac{r}{R}</math>, we have that <math>\cos A + \cos B + \cos C = \frac{21}2 KB (340 words) - 00:44, 3 March 2020
- == Problem == So <math>Sin^2B=1-Cos^2B=\frac{85^2-13^2}{85^2}=\frac{84^2}{85^2}</math>2 KB (282 words) - 09:06, 9 August 2022
- ==Problem 1== For how many integers <math>n>1</math> is it possible to express <math>2005</math> as the sum of <math>n</m7 KB (1,094 words) - 14:39, 24 March 2019