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  • 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
  • ...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
  • ...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 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
  • ...(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
  • ...(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
  • 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
  • ...</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
  • 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 (1,849 words) - 19:43, 2 January 2023
  • .../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
  • ...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
  • 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
  • ...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
  • ...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
  • ...+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
  • ...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
  • ...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
  • ...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
  • <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
  • ...math> for which the sum <cmath>U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor</cmath> is an integer strictly between <math>-1000</math> U & \leq \sum_{n=1}^{2023} \frac{n^2 - na}{5} \\
    10 KB (1,578 words) - 09:48, 24 April 2024

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