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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 (1,351 words) - 22:12, 12 October 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) - 20:05, 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) - 18:01, 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) - 13:45, 11 October 2024
• 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 (1,166 words) - 20:56, 28 June 2024
• 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) - 17:26, 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) - 14:47, 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) - 11:22, 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) - 20:23, 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) - 22:05, 20 October 2021
• 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) - 10:20, 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) - 21:49, 20 October 2024
• 2013 AIME I Problems/Problem 9
  ...ngth of the line segment along which the triangle is folded can be written as <math>\frac{m\sqrt{p}}{n}</math>, where <math>m</math>, <math>n</math>, and Proceed with the same labeling as in Solution 1.
  11 KB (1,786 words) - 21:56, 24 November 2024
• 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) - 19:50, 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) - 20:59, 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) - 01:08, 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) - 17:53, 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) - 18:55, 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) - 12:34, 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) - 23:26, 17 December 2019

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