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• 2004 AIME I Problems/Problem 5
  == Problem == ...ss than Alpha's on that day. Alpha's two-day success ratio was 300/500 = 3/5. The largest possible two-day success ratio that Beta could achieve is <mat
  3 KB (436 words) - 18:31, 9 January 2024

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• Ptolemy's theorem
  ...lity, with equality holding when <math>A^*, B^*, C^*</math> are collinear, i.e. when <math>A,B,C</math> lie on a circle containing <math>D.</math> Addit ===2023 AIME I Problem 5===
  6 KB (922 words) - 17:34, 13 January 2025
• Constructive counting
  This is a problem where constructive counting is not the simplest way to proceed. This next e ...proceed with the construction. If we were to go like before and break the problem down by each box, we'd get a fairly messy solution.
  13 KB (2,018 words) - 15:31, 10 January 2025
• Recursion
  ...equence in terms of previous values: <math>F_0=1, F_1=1, F_2=2, F_3=3, F_4=5, F_5=8</math>, and so on. * [[Mock_AIME_2_2006-2007_Problems#Problem_8 | Mock AIME 2 2006-2007 Problem 8]] ([[number theory]])
  2 KB (316 words) - 16:03, 1 January 2024
• Fibonacci sequence
  ...he sum of the two preceding it. The first few terms are <math>1, 1, 2, 3, 5, 8, 13, 21, 34, 55,...</math>. ...ard, too. Then you get something like <math>\dots -8,5,-3,2,-1,1,0,1,1,2,3,5,8\dots</math> . The ratios between successive terms has you continue backwa
  7 KB (1,111 words) - 14:57, 24 June 2024
• Complex number
  ...ver, it is possible to define a number, <math> i </math>, such that <math> i = \sqrt{-1} </math>. If we add this new number to the reals, we will have ...rm <math> a + bi </math> where <math> a,b\in \mathbb{R} </math> and <math> i = \sqrt{-1} </math> is the [[imaginary unit]]. The set of complex numbers i
  5 KB (860 words) - 15:36, 10 December 2023
• Arithmetic sequence
  * [[2005_AMC_10A_Problems/Problem_17 | 2005 AMC 10A Problem 17]] * [[2006_AMC_10A_Problems/Problem_19 | 2006 AMC 10A Problem 19]]
  4 KB (736 words) - 02:00, 7 March 2024
• 2004 AIME I
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AIME I Problems]]
  1 KB (135 words) - 18:15, 19 April 2021
• 2004 AIME II
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AIME II Problems]]
  1 KB (135 words) - 12:24, 22 March 2011
• 2005 AIME I
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2005 AIME I Problems]]
  1 KB (154 words) - 12:30, 22 March 2011
• 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 ...amous theorems and formulas and see if there's any way you can make a good problem out of them.
  51 KB (6,175 words) - 21:41, 27 November 2024
• 2005 AIME I Problems
  {{AIME Problems|year=2005|n=I}} == Problem 1 ==
  6 KB (983 words) - 05:06, 20 February 2019
• 2004 AIME I Problems/Problem 15
  == Problem == We approach the problem by [[recursion]]. We [[partition]] the positive integers into the sets
  9 KB (1,491 words) - 01:23, 26 December 2022
• 2004 AIME I Problems/Problem 14
  == Problem == real x = 20 - ((750)^.5)/3, CE = 8*(6^.5) - 4*(5^.5), CD = 8*(6^.5), h = 4*CE/CD;
  4 KB (729 words) - 01:00, 27 November 2022
• 2004 AIME I Problems/Problem 12
  == Problem == <cmath>y \in \left(\frac{1}{5},1\right) \cup \left(\frac{1}{125},\frac{1}{25}\right) \cup \left(\frac{1}{
  2 KB (303 words) - 18:43, 16 October 2024
• 2004 AIME I Problems/Problem 11
  == Problem == ...ght]] of the cone. Using the [[Pythagorean Theorem]], we get <math>\ell = 5</math> and <math>A = 24\pi</math>.
  5 KB (839 words) - 22:12, 16 December 2015
• 2004 AIME I Problems/Problem 10
  == Problem == ...dot 36 /2}{(15+36+39)/2} = 6</math>. Thus <math>r_{A'B'C'} = r_{ABC} - 1 = 5</math>, and since the ratio of the areas of two similar figures is equal to
  5 KB (836 words) - 07:53, 15 October 2023
• 2004 AIME I Problems/Problem 9
  == Problem == Let <math> ABC </math> be a [[triangle]] with sides 3, 4, and 5, and <math> DEFG </math> be a 6-by-7 [[rectangle]]. A segment is drawn to d
  4 KB (647 words) - 17:43, 23 November 2024
• 2004 AIME I Problems/Problem 1
  == Problem == ...10(n + 1) + n = 3210 + 1111n</math>, for <math>n \in \lbrace0, 1, 2, 3, 4, 5, 6\rbrace</math>.
  2 KB (250 words) - 23:56, 2 December 2024
• 2004 AIME I Problems/Problem 2
  == Problem == ...ath> consecutive integers that sum to <math>2m</math>, the middle integer (i.e., the median) must be <math>2</math>. Therefore, the largest element in
  8 KB (1,431 words) - 17:50, 29 December 2024
• 2004 AIME I Problems/Problem 4
  == Problem == {{AIME box|year=2004|n=I|num-b=3|num-a=5}}
  4 KB (647 words) - 21:51, 12 January 2025

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