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• 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 1 Pre 2005 Problems/Problem 3
  == Problem == ...> and <math>E</math> are collinear in that order such that <math>AB = BC = 1, CD = 2,</math> and <math>DE = 9</math>. If <math>P</math> can be any point
  1 KB (217 words) - 06:18, 2 July 2015

Page text matches

• 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
  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 (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 2
  ==Problem== ...sets of <math>7</math> digits, consider <math>9</math> urns labeled <math>1,2,\cdots,9</math> (note that <math>0</math> is not a permissible digit); th
  950 bytes (137 words) - 10:16, 29 November 2019
• 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 4
  ==Problem== <cmath>\zeta_1+\zeta_2+\zeta_3=1</cmath>
  2 KB (221 words) - 02:49, 19 March 2015
• Mock AIME 3 Pre 2005 Problems/Problem 5
  ==Problem== === Solution 1 (recursive) ===
  5 KB (795 words) - 16:03, 17 October 2021
• Mock AIME 3 Pre 2005 Problems/Problem 6
  ==Problem== <cmath>\sum_{n = 1}^{9800} \frac{1}{\sqrt{n + \sqrt{n^2 - 1}}}</cmath>
  3 KB (501 words) - 14:48, 29 November 2019
• 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 9
  ==Problem== ==Solution 1==
  2 KB (278 words) - 16:32, 27 December 2019
• Mock AIME 3 Pre 2005 Problems/Problem 10
  ==Problem== <math>\{A_n\}_{n \ge 1}</math> is a sequence of positive integers such that
  2 KB (306 words) - 10:36, 4 April 2012
• Mock AIME 3 Pre 2005 Problems/Problem 12
  ==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) - 23:59, 24 April 2013
• Mock AIME 3 Pre 2005 Problems/Problem 13
  ==Problem== <math>\left(\frac{2}{3}\right)^{2005} \cdot \sum_{k=1}^{2005} \frac{k^2}{2^k} \cdot {2005 \choose k}</math>
  3 KB (502 words) - 14:53, 19 July 2020
• Mock AIME 3 Pre 2005 Problems/Problem 14
  ==Problem== ..._1</math> at <math>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
  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 == ...5</math> and <math>f(101) = 0</math>). Evaluate the remainder when <math>f(1)+f(2)+\cdots+f(99)</math> is divided by <math>1000</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 4
  == Problem == When <math>1 + 7 + 7^2 + \cdots + 7^{2004}</math> is divided by <math>1000</math>, a rem
  685 bytes (81 words) - 10:51, 11 June 2013
• Mock AIME 1 Pre 2005 Problems/Problem 7
  == Problem == ...</tt>s amongst the middle five numbers, and so there are <math>6-(5-k) = k+1</math> <tt>C</tt>s amongst the first four numbers.
  1 KB (221 words) - 17:27, 23 February 2013
• Mock AIME 1 Pre 2005 Problems/Problem 8
  == Problem == Thus, the height of <math>P</math> is <math>\sqrt [3]{8} = 2</math> times the height of <math>P'</math>, and thus the height of
  3 KB (446 words) - 00:18, 10 February 2020
• Mock AIME 5 Pre 2005 Problems
  == Problem 1 == <cmath>6g(1 + (1/y)) + 12g(y + 1) = \log_{10} y</cmath>
  6 KB (909 words) - 07:27, 12 October 2022
• Mock AIME 5 Pre 2005 Problems/Problem 12
  == Problem == ...56 = 104060657, \end{aligned}</math> and so the sum of the digits is <math>1+4+6+6+5+7 = \boxed{29}.</math>
  517 bytes (55 words) - 20:01, 23 March 2017
• Mock AIME 1 Pre 2005 Problems/Problem 11
  == Problem == <cmath>\sum_{n=0}^{668} (-1)^{n} {2004 \choose 3n}</cmath>
  2 KB (272 words) - 10:51, 2 July 2015
• Ball-and-urn
  ...ath>3</math> bins. The 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 k-1}</math> or <math>\dbinom{n+k-1}n</math>.
  5 KB (775 words) - 23:53, 13 April 2024
• Mock AIME 1 Pre 2005 Problems/Problem 15
  == 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 tha
  2 KB (340 words) - 01:44, 3 March 2020
• Mock AIME 5 2005-2006 Problems/Problem 2
  == Problem == {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=1|num-a=3}}
  752 bytes (117 words) - 21:16, 8 October 2014
• Mock AIME 5 2005-2006 Problems/Problem 4
  == Problem == Let <math>m</math> and <math>n</math> be integers such that <math>1 < m \le 10</math> and <math>m < n \le 100</math>. Given that <math>x = \log
  645 bytes (109 words) - 20:41, 22 March 2016
• Mock AIME 5 2005-2006 Problems/Problem 6
  == 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
• Mock AIME 5 2005-2006 Problems/Problem 7
  == Problem == ...that <math>P_n > P</math>. Find the value of <math>\left\lfloor \frac{n_0}{3} \right\rfloor</math>.
  853 bytes (134 words) - 21:18, 8 October 2014
• Mock AIME 5 2005-2006 Problems/Problem 14
  == 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) - 10:06, 9 August 2022
• Mock AIME 4 Pre 2005/Problems
  ==Problem 1== For how many integers <math>n>1</math> is it possible to express <math>2005</math> as the sum of <math>n</m
  7 KB (1,094 words) - 15:39, 24 March 2019
• Mock AIME 2 Pre 2005 Problems
  == Problem 1 == [[Mock AIME 2 Pre 2005 Problems/Problem 1|Solution]]
  6 KB (1,052 words) - 13:52, 9 June 2020
• Mock AIME 2 Pre 2005 Problems/Problem 2
  == Problem == ...he property that <math>x+\tfrac1x = 3</math>. Let <math>S_m = x^m + \tfrac{1}{x^m}</math>. Determine the value of <math>S_7</math>.
  883 bytes (128 words) - 16:14, 4 August 2019
• Mock AIME 2 Pre 2005 Problems/Problem 4
  == Problem == ...th> digits, how many elements of <math>S</math> begin with the digit <math>1</math>?
  817 bytes (114 words) - 17:16, 4 August 2019
• Mock AIME 2 Pre 2005 Problems/Problem 5
  == Problem == Let <math>S</math> be the set of integers <math>n > 1</math> for which <math>\tfrac1n = 0.d_1d_2d_3d_4\ldots</math>, an infinite
  1 KB (171 words) - 17:38, 4 August 2019
• Telescoping series
  ...one by using a form of <math>\sum_{k=1}^{n} f(k)=\sum_{k=1}^{n}(a_{k}-a_{k-1})=a_{k}-a_0</math> for some expression <math>a_{k}</math>. ==Example 1==
  3 KB (500 words) - 14:11, 21 November 2023
• Mock AIME 2 Pre 2005 Problems/Problem 9
  == Problem == ...ht)\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> where <math>a_i \n
  2 KB (232 words) - 00:22, 1 January 2021