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- ==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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- ...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 (1,057 words) - 12:02, 25 February 2024
- ...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
- <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
- ...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
- 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
- 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
- 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
- == Problem 1 == [[Mock AIME 1 Pre 2005 Problems/Problem 1|Solution]]6 KB (1,100 words) - 22:35, 9 January 2016
- ==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 connecting7 KB (1,135 words) - 23:53, 24 March 2019
- ==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 connecting795 bytes (129 words) - 10:22, 4 April 2012
- ==Problem== <cmath>2f\left(x\right) + f\left(\frac{1}{x}\right) = 5x + 4</cmath>1 KB (191 words) - 10:22, 4 April 2012
- ==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
- ==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> can2 KB (330 words) - 10:23, 4 April 2012
- ==Problem== Here are some thoughts on the problem:3 KB (520 words) - 12:55, 11 January 2019
- ==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
- ==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</math3 KB (563 words) - 02:05, 25 November 2023
- ==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
- == 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
- == 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
- == 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) - 17:27, 23 February 2013
- == Problem 1 == [[Mock AIME 5 Pre 2005 Problems/Problem 1|Solution]]6 KB (909 words) - 07:27, 12 October 2022
- == Problem == Let <math>m = 101^4 + 256</math>. Find the sum of the digits of <math>m</math>.517 bytes (55 words) - 20:01, 23 March 2017
- ...<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 (775 words) - 23:53, 13 April 2024
- == Problem == ..., we have that <math>\cos B = \frac{7}{16}</math> and <math>\sin B = \frac{3\sqrt{23}}{16}</math>. Since <math>\sin B = \frac{b}{2R}</math>, we have tha2 KB (340 words) - 01:44, 3 March 2020
- == Problem == : <math>P_1(x) = 1+x+x^3+x^4+\cdots+x^{96}+x^{97}+x^{99}+x^{100}</math>522 bytes (77 words) - 21:17, 8 October 2014
- ==Problem 1== [[Mock AIME 4 Pre 2005/Problems/Problem 1 | Solution]]7 KB (1,094 words) - 15:39, 24 March 2019
- == Problem 1 == [[Mock AIME 2 Pre 2005 Problems/Problem 1|Solution]]6 KB (1,052 words) - 13:52, 9 June 2020
- == Problem == <math>x</math> is a real number with the property that <math>x+\tfrac1x = 3</math>. Let <math>S_m = x^m + \tfrac{1}{x^m}</math>. Determine the value of883 bytes (128 words) - 16:14, 4 August 2019
- == Problem == ...7 \cdot 3^3 \cdot 37,</cmath> the number <math>10^{12} -1</math> has <math>4 \cdot 2^6 = 256</math> divisors and our answer is <math>256 - 1 = \boxed{251 KB (171 words) - 17:38, 4 August 2019
- ...are positive integers. What is <math>a+b</math>? ([[2022 AMC 10B Problems/Problem 9|Source]]) ...math>(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?</math> (Hint: difference of squares!)3 KB (500 words) - 14:11, 21 November 2023
- == Problem == ...left(1+2x^{3^2}\right)\cdots \left(1+kx^{3^k}\right) \cdots \left(1+1997x^{3^{1997}}\right) = 1+a_1 x^{k_1} + a_2 x^{k_2} + \cdots + a_m x^{k_m}</cmath>2 KB (232 words) - 00:22, 1 January 2021
