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Page title matches
- == Problem == ...extbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10</math>2 KB (395 words) - 22:29, 3 December 2024
- == Problem ==4 KB (651 words) - 17:27, 22 May 2021
- == Problem==3 KB (409 words) - 16:10, 30 April 2024
- #REDIRECT [[2006 AMC 12A Problems/Problem 9]]45 bytes (5 words) - 13:42, 14 January 2016
- {{duplicate|[[2005 AMC 12B Problems|2005 AMC 12B #9]] and [[2005 AMC 10B Problems|2005 AMC 10B #19]]}} == Problem ==2 KB (280 words) - 14:35, 16 December 2021
- == Problem == This problem begs us to use the familiar identity <math>e^{it} = \cos(t) + i \sin(t)</ma6 KB (1,154 words) - 02:30, 11 January 2024
- #REDIRECT [[2006 AMC 12A Problems/Problem 9]]45 bytes (5 words) - 10:02, 20 February 2016
- == Problem == [[Image:2005_I_AIME-9.png]]4 KB (600 words) - 20:44, 20 November 2023
- == Problem == draw((9*4/14,0)--(9*4/14,5*3/14),dashed);4 KB (647 words) - 16:43, 23 November 2024
- == Problem == We can find the value of <math>a_{9}</math> by its bounds using three conditions:3 KB (547 words) - 10:41, 7 August 2024
- == Problem == <math>f'(y)</math> = <math>9 - 4y^{-2}</math>5 KB (824 words) - 18:34, 20 July 2024
- == Problem == ...B \cdot h_{ABD}</math>, we find that <math>h_{ABD} = 8</math>. Because the problem does not specify, we may assume both <math>ABC</math> and <math>ABD</math>6 KB (947 words) - 19:44, 26 November 2021
- == Problem == == Solution 1== <!-- Images obsoleted Image:1985_AIME-9.png, Image:1985_AIME-9a.png by asymptote -->5 KB (784 words) - 20:05, 8 December 2024
- == Problem == </asy></center> <!-- Asymptote replacement for Image:1986_AIME-9.png by azjps -->11 KB (1,879 words) - 20:04, 8 December 2024
- == Problem ==1 KB (200 words) - 17:44, 5 February 2024
- == Problem == ...}</math>. This is true if the tens digit is either <math>4</math> or <math>9</math>. Casework:6 KB (893 words) - 07:15, 2 February 2023
- == Problem ==6 KB (874 words) - 14:50, 20 January 2024
- == Problem == ...<math>\frac{144}{1024} = \frac{9}{64}</math>. Thus, our solution is <math>9 + 64 = \boxed{073}</math>.3 KB (427 words) - 14:03, 24 June 2024
- == Problem ==1 KB (140 words) - 23:54, 1 April 2023
- == Problem == Note: The problem is much easier computed if we consider what <math>\sec (x)</math> is, then10 KB (1,590 words) - 13:04, 20 January 2023
Page text matches
- <cmath>9-40-41</cmath> * [[2006_AIME_I_Problems/Problem_1 | 2006 AIME I Problem 1]]5 KB (885 words) - 22:03, 5 October 2024
- == Problem == pair O0=(9,9), O1=(1,1), O2=(3,1), O3=(1,3);2 KB (307 words) - 23:58, 17 November 2024
- == Problem == {{AMC10 box|year=2016|ab=A|num-b=7|num-a=9}}1 KB (169 words) - 13:59, 8 August 2021
- == Problem == ...extbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10</math>2 KB (395 words) - 22:29, 3 December 2024
- == Problem == The problem states that <math>x, 2x+6, 2x+14</math> is an arithmetic progression, meani2 KB (337 words) - 13:56, 25 June 2023
- == Problem == label("$7$",(9/2,5),dir(90));8 KB (1,016 words) - 23:17, 30 December 2023
- == Problem == {{AMC12 box|year=2016|ab=A|num-b=9|num-a=11}}2 KB (402 words) - 13:54, 25 June 2023
- *[[Doane triMATHlon Math Contest]]. Takes place at Doane University for 9-12 graders, includes material up to Algebra II. [https://www.doane.edu/trim ...t]]. Consists of two problem solving rounds and a speed round - for grades 9-12.2 KB (200 words) - 09:11, 29 December 2020
- These '''math books''' are recommended by [[Art of Problem Solving]] administrators and members of the [http://aops.com/community AoPS * [http://www.amazon.com/exec/obidos/ASIN/0387982191/artofproblems-20 Problem Solving Strategies] by Arthur Engel contains significant material on inequa24 KB (3,198 words) - 19:44, 4 December 2024
- ...ks''' page is for compiling a list of [[textbook]]s for mathematics -- not problem books, contest books, or general interest books. See [[math books]] for mo * Getting Started is recommended for students grades 6 to 9.7 KB (901 words) - 13:11, 6 January 2022
- ...A/B, 1/2</u>: 7<br><u>Problem A/B, 3/4</u>: 8<br><u>Problem A/B, 5/6</u>: 9}} ...chool olympiads are, although they include more advanced mathematics. Each problem is graded on a scale of 0 to 10. The top five scorers (or more if there are4 KB (623 words) - 12:11, 20 February 2024
- ==Problem== =\frac{1}{2}+\frac{4}{4}+\frac{9}{8}+\frac{16}{16}+\cdots\1 KB (193 words) - 20:13, 18 May 2021
- == Problem 46== pair A=(-3*sqrt(3)/32,9/32), B=(3*sqrt(3)/32, 9/32), C=(0,9/16);3 KB (415 words) - 17:01, 24 May 2020
- ...administered by the [[American Mathematics Competitions]] (AMC). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC and of the recent expansion ...down=<u>Problem 1/4</u>: 7<br><u>Problem 2/5</u>: 8<br><u>Problem 3/6</u>: 9}}6 KB (874 words) - 22:02, 10 November 2024
- '''Valentin Vornicu''' is a former [[Art of Problem Solving]] administrator and webmaster. In 2002, he founded [[MathLinks]], Vornicu also has 9 World Series of Poker gold rings and is currently the 486th ranked poker pl740 bytes (108 words) - 10:22, 27 September 2024
- In team rounds, teams of 4 people work fast to answer a problem in 4 minutes. However, the faster a team turns in an answer, the more point A=<insert some kind of math problem here>4 KB (632 words) - 16:09, 11 October 2020
- ...heavily on developing deep understanding of the methods of [[mathematical problem solving]]. [https://artofproblemsolving.com/school/handbook/prospective/ab * [[Math Jams]] are free classes that include information sessions, problem solving lessons, and competition solution discussion immediately after majo8 KB (965 words) - 02:41, 17 September 2020
- == Problem == \textbf{(B)}\ 9\text{ inches} \qquad878 bytes (143 words) - 19:56, 1 April 2017
- ...hat is the largest area that this triangle can have? ([[1992 AIME Problems/Problem 13|AIME 1992]]) ...ntegers. Find <math>r+s</math>. (Solution [[Problems Collection|here]] see problem 3 solution 1)3 KB (583 words) - 20:20, 2 August 2024
- ...ded to this, then we would have a [[perfect square]], <math>(x-3)^2=x^2-6x+9</math>. To do this, add <math>7</math> to each side of the equation to get <math>x^2-6x+9=7\Rightarrow(x-3)^2=7\Rightarrow x-3=\pm{\sqrt7}\Rightarrow x=3\pm\sqrt{7}<2 KB (422 words) - 15:20, 5 March 2023