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• Pi
  ...mber]], as proved by Lindemann in 1882) denoted by the Greek letter <math>\pi </math>. ...^2}=\frac{\pi^2}{6}</math>. Some common [[fraction]]al approximations for pi are <math>\frac{22}{7} \approx 3.14285</math> and <math>\frac{355}{113} \ap
  8 KB (1,469 words) - 21:11, 16 September 2022
• 1998 AIME Problems
  Given that <math>A_k = \frac {k(k - 1)}2\cos\frac {k(k - 1)\pi}2,</math> find <math>|A_{19} + A_{20} + \cdots + A_{98}|.</math> ...,x_2,x_3,x_4)</math> of positive odd [[integer]]s that satisfy <math>\sum_{i = 1}^4 x_i = 98.</math> Find <math>\frac n{100}.</math>
  7 KB (1,084 words) - 02:01, 28 November 2023
• 2006 Romanian NMO Problems
  ...thocenter of the triangle. We consider the points <math>D</math> and <math>E</math> on the segments <math>AA_1</math> and <math>BC</math> such that <mat ...<math>a_i \in \{ -1, 1 \}</math>, such that <center><math>n = \sum_{1\leq i < j \leq k } a_ia_j</math>.</center>
  11 KB (1,779 words) - 14:57, 7 May 2012
• 2007 iTest Problems
  Star flips a quarter four times. Find the probability that the quarter lands heads exactly twice. ...\,2097151\quad\mathrm{(C)}\,2097153\quad\mathrm{(D)}\,2097157\quad\mathrm{(E)}\,2097161</math>
  33 KB (5,177 words) - 21:05, 4 February 2023
• 2008 AIME I Problems
  {{AIME Problems|year=2008|n=I}} [[2008 AIME I Problems/Problem 1|Solution]]
  9 KB (1,536 words) - 00:46, 26 August 2023
• 2014 AIME I Problems/Problem 15
  ...Circle <math>\omega</math> intersects <math>\overline{AB}</math> at <math>E</math> and <math>B</math>, <math>\overline{BC}</math> at <math>B</math> and pair E = (0, 1.21);
  10 KB (1,643 words) - 22:30, 28 January 2024
• 1992 AHSME Problems
  If <math>3(4x+5\pi)=P</math> then <math>6(8x+10\pi)=</math> \text{(E) } 18P</math>
  16 KB (2,548 words) - 13:40, 19 February 2020
• Gmaas
  ...eorem. You wouldn't waste that much money to solve one problem. Therefore, I proved that you cannot use the Games theorem to solve problems. But using t ...dreadful. EDIT: He now eats gnats again. He thinks that they taste like pi(e). EDIT: He ate 3,141,592,653,589,793,238,462,643,383 so far.
  69 KB (11,805 words) - 20:49, 18 December 2019
• 2008 iTest Problems/Problem 56
  ...e people will have reasonably inexpensive options for switching to cleaner power sources. Even now most families could switch to biomass for between <math>\ ...Alexis begins walking forward. As she walks, the tricorder displays at all times her distance from her starting point at the origin. When Alexis is <math>24
  7 KB (1,092 words) - 19:05, 17 December 2021
• What is the definition of Pure Mathematics?
  ...row \frac{p^2}{q^2}=n \Rightarrow (q^2)n=p^2</math>. But no perfect square times a nonperfect square positive integer is a perfect square. Therefore <math>\ ===== pi and e =====
  35 KB (5,882 words) - 18:08, 28 June 2021
• 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 (13,954 words) - 17:25, 22 March 2024
• 2020 AMC 12B Problems/Problem 25
  <cmath>\sin^2{(\pi x)} + \sin^2{(\pi y)} > 1</cmath> ...rac{1+\sqrt{2}}{4} \qquad\textbf{(D)}\ \frac{\sqrt{5}-1}{2} \qquad\textbf{(E)}\ \frac{5}{8}</math>
  8 KB (1,412 words) - 06:17, 30 December 2023
• 2022 AIME I Problems/Problem 11
  label("$6\sqrt{3}$", (30,7.5), E); label("$E$", EE, N);
  16 KB (2,517 words) - 20:22, 31 January 2024
• 2022 AMC 12B Problems/Problem 24
  for (int i = 0; i < 7; ++i) { for (int j = 0; j < i; ++j) {
  21 KB (3,265 words) - 17:06, 15 November 2023
• 2023 AIME I Problems/Problem 8
  We apply the Power of a Point Theorem to <math>R</math> and <math>T:</math> ...orem on right <math>\triangle PRX.</math> Together, we conclude that <math>E=X.</math> Therefore, points <math>S,P,</math> and <math>X</math> must be co
  17 KB (2,612 words) - 14:54, 3 July 2023
• 2023 AIME II Problems/Problem 8
  ...mega = \cos\frac{2\pi}{7} + i \cdot \sin\frac{2\pi}{7},</math> where <math>i = \sqrt{-1}.</math> Find the value of the product<cmath>\prod_{k=0}^6 \left ...>z_n = \left(\textrm{cis }\frac{2n\pi}{7}\right)^3 + \textrm{cis }\frac{2n\pi}{7} + 1</math>.
  9 KB (1,284 words) - 23:37, 31 January 2024
• Know the Facts: Gmaas
  ...ng). He can be anything, but his form currently is a cat, though his godly power can turn him into anything, he just prefers to be a cat. As you read on, yo ...Pre-algebra class. He enters every class by posting a meme and announcing "I have arrived!" Sseraj is the human servant of Gmaas and is one of only 5 en
  88 KB (14,927 words) - 01:36, 16 April 2024