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• USAMO historical results
  Up to 2021, the top 12 scorers on the [[USAMO]] were designated as USAMO winners. In early years, this number was smaller. Starting 2022, the **In Sung Na
  10 KB (1,317 words) - 08:16, 23 April 2024
• Floor function
  ...as "rounding down." On a [[positive]] argument, this function is the same as "dropping everything after the decimal point," but this is ''not'' true for * [[Hermite's Identity]]: <cmath>\lfloor na\rfloor = \left\lfloor a\right\rfloor+\left\lfloor a+\frac{1}{n}\right\rfloo
  3 KB (508 words) - 21:05, 26 February 2024
• Modular arithmetic/Intermediate
  ...c structure called '''the integers modulo <math>n</math>''' (usually known as "the integers mod <math>n</math>," or <math>\mathbb{Z}_n</math> for short). ...h> the third. The symbol <math>\overline{3}</math> refers to the same set as <math>\overline{0}</math>, and so on.
  14 KB (2,317 words) - 19:01, 29 October 2021
• Proofs of AM-GM
  ...proofs of the weighted [[AM-GM Inequality]]. The inequality's statement is as follows: for all nonnegative reals <math>a_1, \dotsc, a_n</math> and nonneg ...we may disregard any <math>a_i</math> for which <math>\lambda_i= 0</math>, as they contribute to neither side of the desired inequality. We also note tha
  12 KB (2,171 words) - 07:55, 11 May 2023
• 2008 AIME II Problems
  ...(a)\sin(a) + \cos(4a)\sin(2a) + \cos(9a)\sin(3a) + \cdots + \cos(n^2a)\sin(na)] ...coordinate plane at <math>(5,0)</math>. Define a ''move'' for the particle as a counterclockwise rotation of <math>\pi/4</math> radians about the origin
  7 KB (1,167 words) - 21:33, 12 August 2020
• 2008 AIME II Problems/Problem 8
  ...(a)\sin(a) + \cos(4a)\sin(2a) + \cos(9a)\sin(3a) + \cdots + \cos(n^2a)\sin(na)] &= -\sin(0) + \sin(2a)- \sin(2a) + \sin(6a) - \cdots - \sin((n-1)na) + \sin(n(n+1)a)\\
  4 KB (621 words) - 18:26, 16 January 2023
• 2001 Iran NMO (Round 3) Problems/Problem 5
  Since <math>\widehat{MN}\equiv \widehat{NA}</math>, it follows that ...(\widehat{NA} + \widehat{AS}) \equiv \tfrac{1}{2} (\widehat{MN} + \widehat{AS}) \equiv \angle MIS \equiv \angle I_a IS . </cmath>
  3 KB (437 words) - 15:47, 27 April 2008
• Normal subgroup
  ...</math> but it does not imply that for every <math>n\in N</math>, <math>an=na</math>. Since any element of <math>{\rm H}</math> can be expressed as <math>xy^{-1}</math>, the statement "<math>{\rm H}</math> is normal in <mat
  15 KB (2,840 words) - 12:22, 9 April 2019
• 2009 AIME II Problems/Problem 15
  As <math>NV + MV =MW + NW = 1</math> we compute <math>NW = \frac{1}{1+\frac{3} ...es the quadratic inequality <math>d^2 - 14d + 1 \ge 0</math>, which solves as<cmath>d \in \left(-\infty, 7-4\sqrt3\right] \cup \left[7+4\sqrt3, \infty\ri
  11 KB (1,862 words) - 21:23, 23 May 2024
• 2010 AIME I Problems/Problem 13
  .../math>, and <math>UB = 168</math>. Then <math>DA</math> can be represented as <math>m\sqrt {n}</math>, where <math>m</math> and <math>n</math> are positi As in the above solutions we discover that <math>\angle AON = 60^\circ</math>,
  10 KB (1,418 words) - 23:05, 20 October 2021
• AoPS Wiki:FAQ
  ...have the ability to attach images, we do not generally recommend doing so, as we can not guarantee the images will be available through upgrades, restora ...of problems in the topic the Alcumus engine believes you will get correct. As you get more and more correct, the rating will go up slower and slower. How
  25 KB (4,184 words) - 12:22, 14 June 2024
• 1980 USAMO Problems/Problem 3
  ...n(C)=x^2\sin(2A)+y^2\sin(2B)+z^2\sin(2C)=0</math>, show that <math>x^n\sin(nA)+y^n \sin(nB) +z^n \sin(nC)=0</math> for any positive integer <math>n</math ...th>\frac{1}{2}\left((a+b+c)^2-(a^2+b^2+c^2)\right)=ab+bc+ac</math> is real as well.
  2 KB (347 words) - 11:20, 6 May 2017
• 2013 AMC 12B Problems/Problem 24
  Let <math>BN=x</math> and <math>NA=y</math>. From the conditions, let's deduct some convenient conditions that ...des by <math>AM</math>. Doing so will make it so that <math>AM = 1</math>, as desired, and doing so will allow us to get the length of <math>BN</math>, w
  11 KB (1,876 words) - 00:08, 12 October 2023
• 1962 AHSME Problems/Problem 32
  ...g <math>a=1</math> and <math>r=\frac12</math>, we can rewrite the sequence as <math>a, a+r, \dots, a+(n-1)r</math>. ...e sum of the first <math>n</math> terms of an arithmetic sequence is <math>na+\binom{n}2r</math>.
  905 bytes (157 words) - 20:50, 31 May 2018
• 2015 AIME II Problems/Problem 11
  ...ath>r=BO</math>. Drawing perpendiculars, <math>BM=MC=2</math> and <math>BN=NA=2.5</math>. From there, <cmath>OM=\sqrt{r^2-4}</cmath> Thus, <cmath>OQ=\fra ...el lines. From here, <math>\angle P'PB = \angle PP'B = \angle A'P'Q</math> as <math>P, P', Q</math> collinear. From here, <math>A'P'QC</math> is cyclic,
  9 KB (1,249 words) - 21:59, 26 November 2023
• 2017 AIME II Problems/Problem 15
  ...+DX</math>. The least possible value of <math>f(X)</math> can be expressed as <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positiv ...e CDA\cong\triangle DCB</math>, and therefore <math>MC=MD</math> and <math>NA=NB</math>. It follows that <math>M</math> and <math>N</math> both lie on th
  6 KB (971 words) - 02:08, 22 January 2024
• 2019 USAJMO Problems/Problem 2
  ...that if <math>a=0</math> then <math>g(x)=g(x)+b</math> to <math>b=0</math> as well, so similarly if <math>b=0</math> then <math>a=0</math>, so now assume ...x-a</math>, so <math>|a|=|b|</math> does work and are the only solutions, as desired.
  4 KB (816 words) - 18:53, 7 April 2021
• 2019 AIME I Problems/Problem 11
  ...math> excenter. Denote the B excenter as <math>I_B</math> and the incenter as <math>I</math>. ...excenter lemma, <math>II_B=2IM</math> so <math>II_b=\frac{2b*IB}{a}</math> as desired.
  21 KB (3,915 words) - 19:55, 10 October 2023
• 2000 Pan African MO Problems/Problem 4
  ...b} = n</math> means that the ordered pair can be rewritten as <math>(a,b,c,na,nb,nc)</math>. ...he ordered pairs that are solutions are <math>\boxed{(0,0,0,x,y,z), (a,b,c,na,nb,nc)}</math>.
  2 KB (273 words) - 13:34, 17 December 2019
• 2001 Pan African MO Problems/Problem 4
  <cmath> \sum_{i=1}^{n}(x_i^2+(a-x_i)^2)= na^2 </cmath> \sum_{i=1}^{n} (x_i^2 + a^2 - 2ax_i + x_i^2) &= na^2 \\
  3 KB (629 words) - 00:26, 18 December 2019

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