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• 2011 AIME I Problems/Problem 15
  == Problem == ...ath> and if <math>a < 0</math>, then <math>b,c > 0</math>. We have <cmath>-2011=ab+bc+ac = a(b+c)+bc = -a^2+bc</cmath>
  8 KB (1,302 words) - 04:07, 24 July 2023
• 2011 AIME II Problems/Problem 15
  ==Problem== ...ber <math>x</math> is chosen at random from the interval <math>5 \le x \le 15</math>. The probability that <math>\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfl
  8 KB (1,273 words) - 14:03, 7 January 2023

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• 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
• Joining an ARML team
  ...The three top teams usually all place in the top 20, often even in the top 15 or 10. * Performance on problems at practice sessions.
  21 KB (3,500 words) - 18:41, 23 April 2024
• Principle of Inclusion-Exclusion
  ...d only once. In particular, memorizing a formula for PIE is a bad idea for problem solving. ...thrice! To account for the elements that are in two of the three sets, we first subtract out <math>|A_1\cap A_2|+|A_2\cap A_3| + |A_3\cap A_1|</math>. Now
  9 KB (1,703 words) - 07:25, 24 March 2024
• Mock AMC
  ...ake. Sometimes, the administrator may ask other people to sign up to write problems for the contest. * Look at past [[AMC]]/[[AHSME]] tests to get a feel for what kind of problems you should write and what difficulty level they should be.
  51 KB (6,175 words) - 20:58, 6 December 2023
• Mock MathCounts
  ...istered to the AoPS community, while others may be sourced from a group of problem writers. Different users may have a different way of participating; some ma ...ms from one subject. Having a group is also good so they can discuss which problems are good or need improvement, and fix errors. More than one person working
  26 KB (3,265 words) - 21:34, 20 March 2024
• Mass points
  ...eorems concerning [[polygon]]s, and is helpful in solving complex geometry problems involving lengths. In essence, it involves using a local [[coordinate syst ...school students made it popular. The technique greatly simplifies certain problems.
  5 KB (815 words) - 22:31, 1 April 2024
• 2010 AIME II
  ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2010 AIME II Problems|Entire Test]]
  1 KB (160 words) - 18:08, 20 June 2020
• 2010 AIME II Problems
  {{AIME Problems|year=2010|n=II}} == Problem 1 ==
  8 KB (1,246 words) - 21:58, 10 August 2020
• 2011 AIME I Problems/Problem 4
  == Problem == ...th>. Our formula now becomes <math>MN^2 = \frac{393120}{181} + \frac{2534}{15 \cdot 117}</math>. After simplifying, we get <math>MN^2 = 3136 \implies MN
  6 KB (1,068 words) - 18:52, 2 August 2023
• 2011 AIME I
  ...contains the full set of test problems. The rest contain each individual problem and its solution. *[[2011 AIME I Problems]]
  1 KB (157 words) - 20:49, 29 June 2020
• 2011 AIME I Problems
  {{AIME Problems|year=2011|n=I}} == Problem 1 ==
  10 KB (1,634 words) - 22:21, 28 December 2023
• 2011 AIME I Problems/Problem 11
  ==Problem== ...25 (proof below). Thus, <math>i = 103</math> and <math>j =3</math> are the first two integers such that <math>2^i \equiv 2^j \pmod{1000}</math>. All that is
  11 KB (1,668 words) - 22:10, 24 February 2023
• 2011 AIME I Problems/Problem 14
  == Problem == {{AIME box|year=2011|n=I|num-b=13|num-a=15}}
  8 KB (1,344 words) - 18:39, 9 February 2023
• 2011 AIME II
  ...k contains the full set of test problems. The rest contain each individual problem and its solution. *[[2011 AIME II Problems]]
  1 KB (156 words) - 19:47, 15 July 2020
• 2011 AIME II Problems/Problem 6
  ==Problem 6== * two parts equal to one, <math>\binom62 = 15</math> ways,
  9 KB (1,535 words) - 01:28, 16 January 2023
• 2011 AIME II Problems/Problem 7
  ==Problem== ...nted here) and 5 "bars," the green marbles. So the answer is <math>\binom{15}{5} = 3003</math>, take the remainder when divided by 1000 to get the answe
  3 KB (605 words) - 10:30, 12 January 2024
• 2011 AIME II Problems
  {{AIME Problems|year=2011|n=II}} == Problem 1 ==
  8 KB (1,301 words) - 08:43, 11 October 2020
• 2011 AIME II Problems/Problem 12
  == Problem 12 == ...<cmath> \frac mn = \frac{30\cdot 56 - 18\cdot 25}{30\cdot 56} = \frac{56 - 15}{56} = \frac{41}{56}.</cmath> Thus <math>m + n = 56 + 41 = \fbox{097}.</mat
  5 KB (848 words) - 19:15, 30 April 2023
• 2011 AIME II Problems/Problem 14
  ==Problem 14== ...e of these triples are repeated and all are used. By the conditions of the problem, if <math>i</math> is the same in two different triples, then the two numbe
  10 KB (1,581 words) - 22:09, 27 August 2023
• 2011 AIME II Problems/Problem 10
  == Problem 10 == Since <math>E</math> and <math>F</math> are midpoints, <math>BE = 15</math> and <math>CF = 7</math>.
  11 KB (1,720 words) - 03:12, 18 December 2023

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