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• Common factorizations
  \text{\textbullet}&&x^{n}-y^{n}&=(x-y)(x^{n-1}+x^{n-2}y+\cdots +xy^{n-2}+y^{n-1}) ...^2 \\\phantom{\text{\textbullet}}&&- b^4 + 2 b^2 d^2 - 4 b c^2 d + c^4 - d^4&=\det\begin{bmatrix}a&b&c&d\\d&a&b&c\\c&d&a&b\\b&c&d&a\end{bmatrix}\\&&&=(a
  2 KB (327 words) - 02:06, 28 April 2024
• Pi
  ...ost surprising places, such as in the sum <math>\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}</math>. Some common [[fraction]]al approximations for p ...approximates <math>\frac{\pi}{4}</math>. This can simply be multiplied by 4 to approximate <math>\pi</math>.
  8 KB (1,469 words) - 21:11, 16 September 2022
• Fibonacci sequence
  ...<math>F_1 = F_2 = 1</math> and <math>F_n=F_{n-1}+F_{n-2}</math> for <math>n \geq 3</math>. This is the simplest nontrivial example of a [[linear recur ...>n \geq 2</math>. In general, one can show that <math>F_n = (-1)^{n+1}F_{-n}</math>.
  6 KB (957 words) - 23:49, 7 March 2024
• Muirhead's Inequality
  ...equence <math>(5,1)</math> majorizes <math>(4,2)</math> (as <math>5>4, 5+1=4+2</math>), Muirhead's inequality states that for any positive <math>x,y</ma x^5y^1+y^5x^1&=\frac{3}{4}\left(x^5y^1+y^5x^1\right)+\frac{1}{4}\left(x^5y^1+y^5x^1\right)\\
  8 KB (1,346 words) - 12:53, 8 October 2023
• Cardinality
  ...t]]s in that set, i.e. the size of the set. The cardinality of <math>\{3, 4\}</math> is 2, the cardinality of <math>\{1, \{2, 3\}, \{1, 2, 3\}\}</math> ...4\}</math> is <math>|\{3, 4\}| = 2</math>. Sometimes, the notations <math>n(A)</math> and <math>\# (A)</math> are used.
  2 KB (263 words) - 00:54, 17 November 2019
• Fundamental Theorem of Calculus
  * Evaluate: <math>\int_2^5 x^3 dx</math> and <math>\int_{.2}^{.4} \cos(x) dx</math>. (The next few questions are meant as hints for how to ...ctly how far the object moved between times <math>t=.2</math> and <math>t=.4</math>. Interpret the distance that the object moved geometrically, as an
  11 KB (2,082 words) - 15:23, 2 January 2022
• Polygon
  ...et]] of [[vertex|vertices]], <math>\{A_1, A_2, \ldots, A_n\}</math>, <math>n \geq 3</math>, with [[edge]]s <math>\{\overline{A_1A_2}, \overline{A_2A_3} ...ewer -- it will have "degenerated" from an <math>n</math>-gon to an <math>(n - 1)</math>-gon. (In the case of triangles, this will result in either a l
  2 KB (372 words) - 19:04, 30 May 2015
• 2011 AMC 10B Problems/Problem 22
  <math> \textbf{(A)}\ 5\sqrt{2} - 7 \qquad\textbf{(B)}\ 7 - 4\sqrt{3} \qquad\textbf{(C)}\ \frac{2\sqrt{2}}{27} \qquad\textbf{(D)}\ \frac{ ...s$",(W--Z),E,red); label("$s$",(X--Y),W,red); label("$s\sqrt{2}$",(W--X),N,red);
  4 KB (691 words) - 18:38, 19 September 2021
• Vector
  ...rected line segment. In many situations, a vector is best considered as an n-tuple of numbers (often real or complex). Most generally, but also most abs ...sional vector can be described in this coordinate form as an ordered <math>n</math>-tuple of numbers within angle brackets or parentheses, <math>(x\,\,y
  7 KB (1,265 words) - 13:22, 14 July 2021
• Logarithm
  ...2.285669651531203956336043826\ldots=x</cmath> such that:<cmath>(^24)^x=4^{4^x}\approx(3^5)^6</cmath> # Evaluate <math>(\log_2 3)(\log_3 4)(\log_4 5)\cdots(\log_{2005} 2006)</math>.
  4 KB (680 words) - 12:54, 16 October 2023
• Law of Cosines
  draw(D--(30,4)--(34,4)--(34,0)--D); 1. If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?
  6 KB (1,003 words) - 09:11, 7 June 2023
• Angle bisector
  A=(4,2); D=(3,4);
  3 KB (575 words) - 15:27, 19 March 2023
• Law of Sines
  pair A=(-1,5), B=(-4,-1), C=(4,-1), D, O; *If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?
  4 KB (658 words) - 16:19, 28 April 2024
• Sequence
  ...ath>\mathbb{N}</math> corresponds to the sequence <math>X = (x_n) = (0, 1, 4, 9, 16, \ldots)</math>. ...> is called the [[limit]] of <math>(x_n)</math> and is written <math>\lim_{n \to \infty} x_n</math>. The statement that <math>(x_n)</math> converges to
  2 KB (413 words) - 21:18, 13 November 2022
• Geometric sequence
  For example, <math>1, 2, 4, 8</math> is a geometric sequence with common ratio <math>2</math> and <mat ...progression if and only if <math>a_2 / a_1 = a_3 / a_2 = \cdots = a_n / a_{n-1}</math>. A similar definition holds for infinite geometric sequences. It
  4 KB (644 words) - 12:55, 7 March 2022
• Arithmetic sequence
  ...on difference <math>-8</math>; however, <math>7, 0, 7, 14</math> and <math>4, 12, 36, 108, \ldots</math> are not arithmetic sequences, as the difference ...progression if and only if <math>a_2 - a_1 = a_3 - a_2 = \cdots = a_n - a_{n-1}</math>. A similar definition holds for infinite arithmetic sequences. It
  4 KB (736 words) - 02:00, 7 March 2024
• Fermat's Last Theorem
  ...\geq 3</math>, there are no solutions to the equation <math>a^n + b^n = c^n</math>. ...he never published it, though he did publish a proof for the case <math>n=4</math>. It seems unlikely that he would have circulated a proof for the sp
  3 KB (453 words) - 11:13, 9 June 2023
• 2005 USAMO Problems/Problem 4
  ...ece of length <math>k_i</math> from the end of leg <math>L_i \; (i = 1,2,3,4)</math> and still have a stable table? For <math>0 \le x \le n</math>, it is easy to see that the number of stable tables is <math>(x+1)^2
  7 KB (1,276 words) - 20:51, 6 January 2024
• Diophantine equation
  If <math>n>1</math>, <math>2n, n^2 - 1, n^2 + 1</math> is a Pythagorean triple. ...ny <math>m,n</math>(<math>m>n</math>), we have <math>m^2 - n^2, 2mn, m^2 + n^2</math> is a Pythagorean triple.
  9 KB (1,434 words) - 13:10, 20 February 2024
• Area
  ...s to the [[circumcenter]]. This creates a triangle that is <math>\frac{1}{n},</math> of the total area (consider the regular [[octagon]] below as an ex ...ound using [[trigonometry]] to be of length <math>\frac s2 \cot \frac{180}{n}^{\circ}</math>.
  6 KB (1,181 words) - 22:37, 22 January 2023
• Floor function
  ...r+\left\lfloor a+\frac{1}{n}\right\rfloor+\ldots+\left\lfloor a+\frac{n-1}{n}\right\rfloor</cmath> *<math>\lfloor -3.2 \rfloor = -4</math>
  3 KB (508 words) - 21:05, 26 February 2024
• Pascal's triangle
  ...he sum of the values on row <math>n</math> of Pascal's Triangle is <math>2^n</math>. ...ved from the combinatorics identity <math>{n \choose k}+{n \choose k+1} = {n+1 \choose k+1}</math>. Thus, any number in the interior of Pascal's Triang
  5 KB (838 words) - 17:20, 3 January 2023
• Newton's Sums
  Consider a polynomial <math>P(x)</math> of degree <math>n</math>, <center><math> P(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0</math></center>
  4 KB (690 words) - 13:11, 20 February 2024
• 1985 IMO Problems/Problem 5
  ...ath> and <math>BC</math> again at distinct points <math>K</math> and <math>N</math> respectively. Let <math>M</math> be the point of intersection of the {{IMO box|year=1985|num-b=4|num-a=6}}
  3 KB (496 words) - 13:35, 18 January 2023
• Trigonometry
  ...=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} \cdots</cmath>
  8 KB (1,217 words) - 20:15, 7 September 2023
• Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality
  ...</cmath> where <math>n_1>1,~~0<n_2<1,~~-1<n_3<0,~~n_4<-1</math>, and <math>n</math> is the root mean power. ...1}{a}}</math> and the harmonic mean's root mean power is -1 as <math>\frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}}=\sqrt[-1]{\frac{x_1^{-1}+\cdots+x_a^{-
  5 KB (912 words) - 20:06, 14 March 2023
• Goldbach Conjecture
  <math>4 = 2 + 2</math> ...might expect the total number of ways to write a large even integer <math>n</math> as the sum of two odd primes to be roughly
  7 KB (1,201 words) - 16:59, 19 February 2024
• Riemann Hypothesis
  ...function, it is easy to see that <math>\zeta(s)=0</math> when <math>s=-2,-4,-6,\ldots</math>. These are called the trivial zeros. This hypothesis is o ...uld hold. The Riemann Hypothesis would also follow if <math>M(n)\le C\sqrt{n}</math> for any constant <math>C</math>.
  2 KB (425 words) - 12:01, 20 October 2016
• Quadratic residues
  ...ulo]] <math>m</math> if there is some integer <math>n</math> so that <math>n^2-a</math> is [[divisibility | divisible]] by <math>m</math>. ...p-1}{2}}</math>, so <math>\left(\frac{-1}{p}\right)=1 \iff p \equiv 1 \mod 4</math>
  5 KB (778 words) - 13:10, 29 November 2017
• Power of a Point Theorem
  Let <math>P</math> be a point, and let <math>S</math> be an <math>n</math>-sphere. Let two arbitrary lines passing through <math>P</math> inter ...th> intersect at <math>R</math>. If <math>AR:BR=1:4</math> and <math>CR:DR=4:9</math>, find the ratio <math>AB:CD</math>.
  5 KB (827 words) - 17:30, 21 February 2024
• 2004 AIME I
  * [[2004 AIME I Problems/Problem 4]] {{AIME box|year=2004|n=I|before=[[2003 AIME I]], [[2003 AIME II|II]]|after=[[2004 AIME II]]}}
  1 KB (135 words) - 18:15, 19 April 2021
• Collatz Problem
  ...sitive integer]] <math>n</math>, the sequence <math>\{n,f(n),f(f(n)),f(f(f(n))),\ldots\}</math> contains 1. This conjecture is still open. Some people h ==Properties of <math>f(n)</math> ==
  1 KB (231 words) - 19:45, 24 February 2020
• 2004 AIME II
  * [[2004 AIME II Problems/Problem 4]] {{AIME box|year=2004|n=II|before=[[2004 AIME I]]|after=[[2005 AIME I]], [[2005 AIME II|II]]}}
  1 KB (135 words) - 12:24, 22 March 2011
• 2005 AIME I
  * [[2005 AIME I Problems/Problem 4 | Problem 4]] {{AIME box|year=2005|n=I|before=[[2004 AIME I|2004 AIME I]], [[2004 AIME II|II]]|after=[[2005 AIME
  1 KB (154 words) - 12:30, 22 March 2011
• 2006 AIME I
  * [[2006 AIME I Problems/Problem 4]] {{AIME box|year=2006|n=I|before=[[2005 AIME I]], [[2005 AIME II|II]]|after=[[2006 AIME II]]}}
  1 KB (135 words) - 12:31, 22 March 2011
• 2005 AIME II
  * [[2005 AIME II Problems/Problem 4]] {{AIME box|year=2005|n=II|before=[[2005 AIME I]]|after=[[2006 AIME I]], [[2006 AIME II|II]]}}
  1 KB (135 words) - 12:30, 22 March 2011
• Mock AMC
  | n/a | n/a
  51 KB (6,175 words) - 20:58, 6 December 2023
• Bertrand's Postulate
  ...em. A more widely known version states that there is a prime between <math>n</math> and <math>2n</math>. ...closer look at the [[combinations|binomial coefficient]] <math>\binom{2n}{n}</math>. Assuming that the reader is familiar with that proof, the Bertrand
  2 KB (309 words) - 21:43, 11 January 2010
• Riemann zeta function
  <cmath>\zeta (s)=\sum_{n=1}^{\infty}\frac{1}{n^s}= 1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\cdots</cmath>
  9 KB (1,547 words) - 03:04, 13 January 2021
• Modular arithmetic/Introduction
  draw((0,0), linewidth(4)); <math>1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, \ldots </math>
  15 KB (2,396 words) - 20:24, 21 February 2024
• Perfect square
  ...so that <math>m^2=n</math>. The first few perfect squares are <math>0, 1, 4, 9, 16, 25, 36</math>. ...th>n</math> square numbers (starting with <math>1</math>) is <math>\frac{n(n+1)(2n+1)}{6}</math>
  954 bytes (155 words) - 01:14, 29 November 2023
• Modular arithmetic/Intermediate
  ...th>n</math>), if the difference <math>{a - b}</math> is divisible by <math>n</math>. ...ntegers modulo <math>n</math>''' (usually known as "the integers mod <math>n</math>," or <math>\mathbb{Z}_n</math> for short). This structure gives us
  14 KB (2,317 words) - 19:01, 29 October 2021
• 2006 AIME I Problems
  {{AIME Problems|year=2006|n=I}} == Problem 4 ==
  7 KB (1,173 words) - 03:31, 4 January 2023
• 2006 AIME I Problems/Problem 15
  === Solution 4 === {{AIME box|year=2006|n=I|num-b=14|after=Last Question}}
  6 KB (910 words) - 19:31, 24 October 2023
• 2006 AIME I Problems/Problem 14
  ...> is not divisible by the square of any prime. Find <math> \lfloor m+\sqrt{n}\rfloor. </math> (The notation <math> \lfloor x\rfloor </math> denotes the triple M=(B+C)/2,S=(4*A+T)/5;
  6 KB (980 words) - 21:45, 31 March 2020
• 2006 AIME I Problems/Problem 13
  ...h> S_n=\sum_{k=1}^{2^{n-1}}g(2k). </math> Find the greatest integer <math> n </math> less than 1000 such that <math> S_n </math> is a [[perfect square]] .../math> but not by <math>2^n, \ldots,</math> and <math>2^{n-1}-2^{n-2} = 2^{n-2}</math> elements of <math>S</math> that are divisible by <math>2^1</math>
  10 KB (1,702 words) - 00:45, 16 November 2023
• 2006 AIME I Problems/Problem 10
  ...ough the tangent point of the other two circles. This clearly will cut the 4 circles into two regions of equal area. Using super advanced linear algebra {{AIME box|year=2006|n=I|num-b=9|num-a=11}}
  4 KB (731 words) - 17:59, 4 January 2022
• 2006 AIME I Problems/Problem 8
  pair A=(0,0), B=(4.2,0), C=(5.85,-1.6), D=(4.2,-3.2), EE=(0,-3.2), F=(-1.65,-1.6), G=(0.45,-1.6), H=(3.75,-1.6), I=(2.1, label("$\mathcal{Q}$",(4.2,-1),NW);
  5 KB (730 words) - 15:05, 15 January 2024
• 2006 AIME I Problems/Problem 7
  for(int i=0; i<4; i=i+1) { pair A=(1/3,4), B=A+7.5*dir(-17), C=A+7*dir(10);
  4 KB (709 words) - 01:50, 10 January 2022
• 2006 AIME I Problems/Problem 5
  <math> b = 4 </math> <math>b=4</math>
  3 KB (439 words) - 18:24, 10 March 2015

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