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  • == Problem == {{AIME box|year=2006|n=I|num-b=13|num-a=15}}
    6 KB (980 words) - 21:45, 31 March 2020
  • == Problem == [[Image:2005_I_AIME-14.png|center]]
    3 KB (561 words) - 14:11, 18 February 2018
  • == Problem == In [[triangle]] <math> ABC, AB=13, BC=15, </math> and <math>CA = 14. </math> Point <math> D </math> is on <math> \overline{BC} </math> with <ma
    14 KB (2,340 words) - 16:38, 21 August 2024
  • == Problem == {{AIME box|year=2004|n=I|num-b=13|num-a=15}}
    4 KB (729 words) - 01:00, 27 November 2022
  • == Problem == ...n't need to be nearly as rigorous). A more natural manner of attacking the problem is to think of the process in reverse, namely seeing that <math>n \equiv 1
    11 KB (1,857 words) - 12:57, 18 July 2024
  • == Problem == draw(circumcircle((14,0),p,r));
    14 KB (2,351 words) - 21:06, 8 December 2024
  • == Problem == ...s 190 or lower. We use the fact that sufficiently high multiples of 6, 10, 14, 22, etc. can be represented as <math>n+n</math>. We bash each case until w
    8 KB (1,365 words) - 15:38, 10 December 2024
  • == Problem == ...alf of their points by playing the <math>n-10</math> stronger players. The problem also tells us that the <math>n-10</math> people who aren't part of the lose
    5 KB (772 words) - 22:14, 18 June 2020
  • == Problem == {{AIME box|year=1986|num-b=13|num-a=15}}
    2 KB (346 words) - 13:13, 22 July 2020
  • == Problem == ~ pi_is_3.14
    7 KB (965 words) - 23:39, 11 September 2024
  • == Problem == {{AIME box|year=1988|num-b=13|num-a=15}}
    4 KB (700 words) - 17:21, 3 May 2021
  • == Problem == {{AIME box|year=1989|num-b=13|num-a=15}}
    2 KB (410 words) - 01:37, 25 August 2024
  • == Problem == Alternatively, since you can use rulers on the AIME, you can measure out a <math>12\sqrt{3}</math> by <math>13\sqrt{3}</math> r
    8 KB (1,148 words) - 22:35, 4 February 2025
  • == Problem == </asy></center><!-- asy replaced Image:AIME 1991 Solution 14.png by minsoens -->
    4 KB (547 words) - 04:46, 1 December 2024
  • == Problem == {{AIME box|year=1992|num-b=13|num-a=15}}
    4 KB (669 words) - 18:17, 20 January 2025
  • == Problem == [[Image:1993 AIME 14 Diagram.png|center]]
    3 KB (601 words) - 09:25, 19 November 2023
  • == Problem == {{AIME box|year=1994|num-b=13|num-a=15}}
    2 KB (303 words) - 00:03, 28 December 2017
  • == Problem == {{AIME box|year=1995|num-b=13|num-a=15}}
    3 KB (484 words) - 13:11, 14 January 2023
  • == Problem == {{AIME box|year=1996|num-b=13|num-a=15}}
    5 KB (923 words) - 21:21, 22 September 2023
  • == Problem == ...<math>166</math> of them <math>< m</math> (<math>m \neq n</math> since the problem says that <math>m</math> and <math>n</math> are distinct). Since <math>m</
    6 KB (932 words) - 15:55, 16 November 2024

Page text matches

  • * <math>14! = 87178291200</math> ...9277696409600000000000000</math> (Note: this number is 82 digits long with 14 terminal zeroes!)
    10 KB (809 words) - 16:40, 17 March 2024
  • ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AIME II Problems]]
    1 KB (133 words) - 12:32, 22 March 2011
  • ...uality]]. Ptolemy's theorem frequently shows up as an intermediate step in problems involving inscribed figures. == Problems ==
    6 KB (922 words) - 17:34, 13 January 2025
  • == Problems == * [[1991 AIME Problems/Problem 12]]
    1 KB (179 words) - 19:41, 3 January 2025
  • 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
  • ...0</math>, dividing both sides by <math>\gcd(93,42) = 3</math> gives <math>14 \cdot 93 - 31 \cdot 42 = 0</math>. We can add <math>k</math> times this eq * [[2020 AMC 10A Problems/Problem 24]]
    6 KB (923 words) - 17:39, 30 September 2024
  • ==Problems== *The ellipse with axis lengths <math>14</math> and <math>16</math> has the general equation of <math>\frac{x^2}{a^2
    5 KB (892 words) - 21:52, 1 May 2021
  • ==Problems== ([[University of South Carolina High School Math Contest/1993 Exam/Problem 29|Source]])
    6 KB (1,003 words) - 00:02, 20 May 2024
  • == Problems == *[[2007 AMC 12A Problems/Problem 18]]
    5 KB (860 words) - 15:36, 10 December 2023
  • ...c sequence with common difference <math>-8</math>; however, <math>7, 0, 7, 14</math> and <math>4, 12, 36, 108, \ldots</math> are not arithmetic sequences == Problems ==
    4 KB (736 words) - 02:00, 7 March 2024
  • == Practice Problem == ...ath>. If the sides of triangle <math>ABC</math> are <math>13,</math> <math>14,</math> and <math>15,</math> the radius of <math>\omega</math> can be repre
    3 KB (533 words) - 13:51, 2 September 2024
  • ...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
  • ...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
  • ...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
  • ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AIME I Problems]]
    1 KB (135 words) - 12:31, 22 March 2011
  • ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2005 AIME II Problems]]
    1 KB (135 words) - 12:30, 22 March 2011
  • {{AIME Problems|year=2006|n=I}} == Problem 1 ==
    7 KB (1,173 words) - 03:31, 4 January 2023
  • == Problem == At first, this problem looks kind of daunting, but we can easily solve this problem by finding patterns, recursion and algebraic manipulations.
    10 KB (1,702 words) - 22:23, 25 July 2024
  • == Problem == ...\overline{CD}</math>, <math>AB=18</math>, <math>BC=21</math>, and <math>CD=14</math>. Find the [[perimeter]] of <math>ABCD</math>.
    1 KB (174 words) - 16:10, 3 February 2025
  • == Problem == ...olutions so <math>P(x)</math> is at least quadratic. Let us first try this problem out as if <math>P(x)</math> is a quadratic polynomial. Thus <math>P(n)-(n+3
    4 KB (642 words) - 02:14, 1 June 2024

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