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• United States of America Mathematical Olympiad
  The '''United States of America Mathematical Olympiad''' ('''USAMO''') is the third test in a series of exams used to challenge bright students on th ...rican Mathematics Competitions]] (AMC). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC and of the recent expansion of USAMO participant
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• Fermat's Little Theorem
  ...if they have a hard time following the rest of this article). This theorem is credited to [[Pierre de Fermat]]. ...n [[integer]], <math>{p}</math> is a [[prime number]] and <math>{a}</math> is not [[divisibility|divisible]] by <math>{p}</math>, then <math>a^{p-1}\equi
  16 KB (2,658 words) - 16:02, 8 May 2024
• 1986 AIME Problems/Problem 3
  If <math>\tan x+\tan y=25</math> and <math>\cot x + \cot y=30</math>, what is <math>\tan(x+y)</math>? Since <math>\cot</math> is the reciprocal function of <math>\tan</math>:
  3 KB (545 words) - 23:44, 12 October 2023
• 1988 AIME Problems/Problem 14
  ...icular]] to <math>y=2x</math>, so the slope of <math>\overline{PP'}</math> is <math>\frac{-1}{2}</math>. Thus <math>\frac{y' - y}{x' - x} = \frac{-1}{2} ...math>, which is unchanged by the reflection, into the expression. But this is not necessary. We see that <math>b=-7</math>, <math>c=-12</math>, so <math>
  4 KB (700 words) - 17:21, 3 May 2021
• 1988 AIME Problems/Problem 9
  ...<math>20k + 8 \equiv 88 \pmod{100}</math>. This is true if the tens digit is either <math>4</math> or <math>9</math>. Casework: ...le value for the hundreds digit is <math>4</math>, and so <math>442</math> is a valid solution.
  6 KB (893 words) - 08:15, 2 February 2023
• 1989 AIME Problems/Problem 11
  ...r D\rfloor</math>? (For real <math>x</math>, <math>\lfloor x\rfloor</math> is the [[floor function|greatest integer]] less than or equal to <math>x</math ...> (in other words, <math>x \in [1,1000]</math>). Indeed, <math>D(x)</math> is symmetric about <math>x = 500.5</math>; consider replacing all of numbers <
  5 KB (851 words) - 18:01, 28 December 2022
• 1996 AIME Problems/Problem 6
  The probability that one team wins all games is <math>5\cdot \left(\frac{1}{2}\right)^4=\frac{5}{16}</math>. Similarity, the probability that one team loses all games is <math>\frac{5}{16}</math>.
  3 KB (461 words) - 00:33, 16 May 2024
• 1998 AIME Problems/Problem 4
  ...values. The [[probability]] that all three players obtain an [[odd]] sum is <math>m/n,</math> where <math>m</math> and <math>n</math> are [[relatively ...that it matters in what order the people pick the tiles; the final answer is the same if we assume the opposite, that order doesn't matter.)
  5 KB (917 words) - 02:37, 12 December 2022
• 2003 AIME I Problems/Problem 3
  Therefore the desired sum is <math>34\cdot7+21\cdot6+13\cdot5+8\cdot4+5\cdot3+3 \cdot2+2\cdot1+1\cdot0=\ Finding <math>7x</math> isn't hard, and we see that it is equal to <math>609</math>:
  2 KB (317 words) - 00:09, 9 January 2024
• 2001 AIME II Problems/Problem 7
  ...h> and <math>C_{3}</math> can be written as <math>\sqrt {10n}</math>. What is <math>n</math>? ...> and <math>b</math> are the legs of the right triangle and <math>c</math> is the hypotenuse. (This formula should be used ''only for right triangles''.)
  7 KB (1,112 words) - 02:15, 26 December 2022
• 2000 AIME II Problems/Problem 10
  A [[circle]] is [[inscribe]]d in [[quadrilateral]] <math>ABCD</math>, [[tangent]] to <math> which is nowhere near as hard of a calculation. In fact, this is basically the same exact calculation done at the end of solution 1, just wi
  2 KB (399 words) - 17:37, 2 January 2024
• 2000 AIME II Problems/Problem 7
  ...N{1!18!}</math></center> find the [[floor function|greatest integer]] that is less than <math>\frac N{100}</math>. ...ing these up we have <math>13796</math> therefore <math>\boxed{137}</math> is the desired answer.
  2 KB (281 words) - 12:09, 5 April 2024
• 2007 AIME I Problems/Problem 14
  A [[sequence]] is defined over [[non-negative]] integral indexes in the following way: <math> ...<math>a_1/a_0 = 1</math>, this means that the sequence <math>(a_n)</math> is increasing. Since <math>a_3=672</math> already, we must have <math>a_{2006}
  13 KB (2,185 words) - 23:30, 5 December 2022
• 2007 AIME I Problems/Problem 15
  ..., where <math>p</math> and <math>q</math> are rational, and <math>r</math> is an [[integer]] not divisible by the [[square]] of a [[prime]]. Find <math> [[Image:AIME I 2007-15.png]]
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• How should I prepare?
  ...ld also try to strengthen in the areas you are not as good at. This guide is intended to help you get started. * Art of Problem Solving Volume 2 - [[AMC 12]], [[AIME]], [[USAMO]], [[MOP]]
  13 KB (1,926 words) - 11:22, 30 November 2023
• 2009 AIME I Problems/Problem 12
  ...<math>AC = 12</math>, <math>BC = 35</math>, and <math>\overline{CD}</math> is the altitude to <math>\overline{AB}</math>. Let <math>\omega</math> be the ...ve length <math>x</math>. Then the perimeter of <math>\triangle ABI</math> is equal to <cmath>2(x+AD+DB)=2(x+37).</cmath> It remains to compute <math>\df
  12 KB (1,970 words) - 22:53, 22 January 2024
• 2012 AIME II Problems/Problem 15
  Triangle <math>ABC</math> is inscribed in circle <math>\omega</math> with <math>AB=5</math>, <math>BC=7< ...lemy's we see <math>AF=\frac{30}{\sqrt{19}}</math>. We conclude the answer is <math>900+19=\boxed{919}</math>.
  13 KB (2,298 words) - 12:56, 10 September 2023
• 2013 AIME I Problems/Problem 7
  ...h>, the altitude of the triangle from the base with length <math>10</math> is <math>6</math>. ...r <math>x</math> gives us <math>x=\frac{36}{5}</math>. Since this fraction is simplified:
  6 KB (934 words) - 20:06, 24 January 2021
• 2013 AIME I Problems/Problem 13
  ...and smaller units. Note that the area of the triangle, by Heron's formula, is 90. ...if I say some area has ratio <math>\frac{1}{2}</math>, that means its area is 45.
  7 KB (1,085 words) - 22:48, 17 July 2023
• 2014 AIME II Problems/Problem 14
  ...> is the midpoint of the segment <math>HM</math>, and point <math>P</math> is on ray <math>AD</math> such that <math>PN\perp{BC}</math>. Then <math>AP^2= <math>M</math> is the midpoint of <math>BC</math> and <math>N</math> is the midpoint of <math>HM</math>
  11 KB (1,442 words) - 19:28, 21 October 2023
• 2015 AMC 10B Problems/Problem 22
  ...elow, <math>ABCDE</math> is a regular pentagon and <math>AG=1</math>. What is <math>FG + JH + CD</math>? //(0,0) is a convenient point
  7 KB (1,146 words) - 14:09, 13 April 2024
• 2015 AIME I Problems/Problem 5
  ...e probability that Wednesday is the first day Sandy selects matching socks is <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relat ...th sock be arbitrary; the probability that the sixth sock matches in color is <math>\dfrac{1}{9}</math>.
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• 2015 AIME II Problems/Problem 11
  ...O</math> is the circumcenter of triangle <math>ABC</math>, <math>OM</math> is a perpendicular bisector by the definition of a circumcenter. Hence, <math> ...A</math> so <math>\angle{BOC}=2A</math>. Since <math>\triangle{BOC}</math> is isosceles (<math>BO</math> and <math>OC</math> are radii), <math>\angle{CBO
  9 KB (1,249 words) - 21:59, 26 November 2023
• 2015 AIME II Problems/Problem 15
  ...ngle DBA</math> and <math>\triangle ACE</math> are equal. This common area is <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relat ...re completely lost and you have a lot of time left in your mocking of this AIME, go ahead and use analytic geometry.
  31 KB (5,086 words) - 19:15, 20 December 2023
• 2016 AIME II Problems/Problem 3
  ...now <math>xyz</math> has to be a power of <math>5</math>, and so it is not hard to test values in the equation <math>3xyz+\log_5{xyz} = 378</math> in order {{AIME box|year=2016|n=II|num-b=2|num-a=4}}
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• 2016 AIME II Problems/Problem 14
  ...math> whose distance from each of <math>A,B,C,P,</math> and <math>Q</math> is <math>d</math>. Find <math>d</math>. ...=900</math>. Since we want <math>\dfrac{x+y}{2}</math>, the desired answer is <math>\boxed{450}</math>.
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• Gmaas
  gmaaas is our leader. To prove yourself worthy of GMAAS, you don't have to memorize f -1. GMAAS looks like this when he is mad: https://cdn.artofproblemsolving.com/images/7/4/b/74b21fec95225e1621c71
  69 KB (11,805 words) - 20:49, 18 December 2019
• 2018 AIME II Problems/Problem 5
  ...both, simply multiples of a root of unity. It turns out that root of unity is <math>e^{\frac{3\pi i}{2}}</math>. Anyway this results in getting that <mat ...d <math>z=3-3i</math>. More division is needed to find what <math>x</math> is. <cmath>x=-20-12i</cmath> <cmath>x+y+z=-7-5i</cmath> <cmath>(-7)^2+(-5)^2=\
  11 KB (2,077 words) - 20:15, 12 January 2024
• 2018 AIME II Problems/Problem 13
  ...e rolls. The probability that she will roll the die an odd number of times is <math>\dfrac{m}{n}</math> where <math>m</math> and <math>n</math> are relat The ratio of <math>\frac{P_{odd}}{P_{even}}</math> is just <math>\frac{P_{odd}}{1-P_{odd}}</math>.
  11 KB (1,860 words) - 13:12, 24 January 2024
• 2019 AIME I Problems/Problem 11
  ...two sides. Suppose that the excircle tangent to <math>\overline{BC}</math> is internally tangent to <math>\omega</math>, and the other two excircles are ...e sum of the radius of <math>\omega</math> and <math>\frac{rs}{s-b}</math> is equivalent to the distance between the incenter and the the <math>B/C</math
  21 KB (3,915 words) - 19:55, 10 October 2023
• 2020 AIME I Problems/Problem 5
  ...the <math>2</math>. Therefore, any cases with two adjacent numbers swapped is overcounted, so we subtract <math>5</math> cases (namely, <math>213456, 132 ...above, a <math>1-1</math> correspondence between ascending and descending is established by subtracting each number from <math>7</math>.
  14 KB (2,102 words) - 19:05, 11 July 2023
• 2020 AIME I Problems/Problem 13
  ...<math>p</math> are relatively prime positive integers, and <math>n</math> is a positive integer not divisible by the square of any prime. Find <math>m+n ...ABC]</math>. From here, we get that the area of <math>\triangle ABC</math> is <math>\frac{15\sqrt{7}}{14} \implies \boxed{036}</math>, by similarity. ~aw
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• GMAAS
  4. The Infinity Numeral, PI, is the second deity of the Almighty Gmaas's heaven. 5. The Infinite Logarithm, E, is the third deity of the Almighty Gmaas's heaven.
  85 KB (13,971 words) - 18:02, 16 May 2024
• 2020 AIME II Problems/Problem 9
  ...e number of working permutations with <math>D-F</math> as the first person is the same as if it were <math>A-C</math>, hence, after doing such a bash, we ...to each other after the break. Then the required number of seating orders is given by the Inclusion-Exclusion Principle as
  9 KB (1,460 words) - 21:17, 28 December 2023
• 2020 AIME II Problems/Problem 14
  ...for <math>0\leq x\leq 2020</math>. Find the remainder when <math>N</math> is divided by <math>1000</math>. ..., by observing the function composition from right to left, <math>N</math> is (note that the summation starts from the right to the left):
  7 KB (1,186 words) - 15:31, 5 January 2024
• Know the Facts: Gmaas
  This book is dedicated to GMAAS, the supreme leader. ...er of Gmaas(trademark pending). He can be anything, but his form currently is a cat, though his godly power can turn him into anything, he just prefers t
  88 KB (14,928 words) - 13:54, 29 April 2024
• Problems Collection
  This is a page where you can share the problems you made (try not to use past exams ==AIME styled==
  43 KB (7,006 words) - 14:24, 19 February 2024
• 2024 AIME II Problems/Problem 12
  ...t quadrant with \(P\) on the \(x\)-axis and \(Q\) on the \(y\)-axis. There is a unique point \(C\) on \(\overline{AB}\), distinct from \(A\) and \(B\), ...h>. However, a critical condition that must be satisfied by our parameters is that <math>a^2+b^2=1</math>, since the length of <math>\overline{PQ}=1</mat
  13 KB (2,178 words) - 11:34, 14 May 2024