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- == Problem == *Person 2: <math>\frac{6 \cdot 4 \cdot 2}{6 \cdot 5 \cdot 4} = \frac 25</math>4 KB (628 words) - 11:28, 14 April 2024
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- ...d only once. In particular, memorizing a formula for PIE is a bad idea for problem solving. ==== Problem ====9 KB (1,703 words) - 01:20, 7 December 2024
- * <math>\sum_{x=0}^{n} \binom{n}{x} = 2^n</math> ...ath>2^n</math>, and a and b disappear because they are 1, the sum is <math>2^n</math>.4 KB (638 words) - 21:55, 5 January 2025
- ...llipse is the origin this simplifies to <math>\frac{x^2}{a^2}+\frac{y^2}{b^2}=1</math>. ...<math>P</math>. Then it is true that the bisector of angle <math>F_{1}PF_{2}</math> is perpendicular to <math>\ell</math>.5 KB (892 words) - 21:52, 1 May 2021
- ...mplex numbers''' arise when we try to solve [[equation]]s such as <math> x^2 = -1 </math>. ...from this addition, we are not only able to find the solutions of <math> x^2 = -1 </math> but we can now find ''all'' solutions to ''every'' polynomial.5 KB (860 words) - 15:36, 10 December 2023
- ..., -50, 25, -25/2</math> is a geometric sequence with common ratio <math>-1/2</math>; however, <math>1, 3, 9, -27</math> and <math>-3, 1, 5, 9, \ldots</m ...onsecutive terms <math>a_{n-1}, a_n, a_{n+1}</math>. In symbols, <math>a_n^2 = a_{n-1}a_{n+1}</math>. This is mostly used to perform substitutions, thou4 KB (649 words) - 21:09, 19 July 2024
- ...ree [[integer]]s that satisfy the [[Pythagorean Theorem]], <math>a^2+b^2=c^2</math>. There are three main methods of finding Pythagorean triples: ...th> is an [[odd]] number, then <math>m, \frac {m^2 -1}{2}, \frac {m^2 + 1}{2}</math> is a Pythagorean triple.9 KB (1,434 words) - 01:15, 4 July 2024
- ...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
- ...in the AoPSWiki, please consider adding them. Also, if you notice that a problem in the Wiki differs from the original wording, feel free to correct it. Fi ! Year || Test I || Test II3 KB (400 words) - 20:21, 13 February 2025
- {{AIME Problems|year=2006|n=I}} == Problem 1 ==7 KB (1,173 words) - 03:31, 4 January 2023
- == Problem == ...he [[binomial coefficient]] <math>{n \choose 6} = \frac{n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot(n-4)\cdot(n-5)}{6\cdot5\cdot4\cdot3\cdot2\cdot1}</math>.1 KB (239 words) - 11:54, 31 July 2023
- == Problem == *Person 2: <math>\frac{6 \cdot 4 \cdot 2}{6 \cdot 5 \cdot 4} = \frac 25</math>4 KB (628 words) - 11:28, 14 April 2024
- == Problem == An [[infinite]] [[geometric series]] has sum 2005. A new series, obtained by squaring each term of the original series, has 13 KB (581 words) - 21:19, 22 September 2024
- == Problem == <math>10^{10} = 2^{10}\cdot 5^{10}</math> so <math>10^{10}</math> has <math>11\cdot11 = 121</3 KB (377 words) - 18:36, 1 January 2024
- == Problem == ==Solution 2==4 KB (642 words) - 02:14, 1 June 2024
- == Problem == ...og_a b + 6\log_b a=5, 2 \leq a \leq 2005, </math> and <math> 2 \leq b \leq 2005. </math>3 KB (547 words) - 19:15, 4 April 2024
- == Problem == ...th>n = 131 + 65 = 196</math>; the total number of cards is <math>196 \cdot 2 = \boxed{392}</math>.2 KB (384 words) - 00:31, 26 July 2018
- {{AIME Problems|year=2005|n=II}} == Problem 1 ==7 KB (1,119 words) - 21:12, 28 February 2020
