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• Work
  1 KB (192 words) - 00:46, 16 July 2018
• Celsius
  4 KB (357 words) - 22:40, 17 April 2024
• 1999 USAMO Problems/Problem 4
  <cmath> \sum_{i=k+1}^n -a_i \le 2k-n . </cmath> <cmath> \sum_{i=k+1}^n a_i^2 \le \left( \sum_{i=k+1}^n-a_i \right)^2 \le (2k-n)^2 . </cmath>
  3 KB (499 words) - 09:47, 20 July 2016
• 2001 IMO Shortlist Problems/A3
  By the [[Cauchy-Schwarz Inequality]], By the [[Cauchy-Schwarz Inequality]]
  3 KB (441 words) - 00:54, 24 October 2023
• 1999 IMO Problems/Problem 2
  ...ath> and the right-hand side becomes <math>(x_a + x_b)^4/8</math>. By [[AM-GM]], ...ase the left-hand side of the desired inequality without changing the right-hand side; and then to use the inductive hypothesis.
  4 KB (838 words) - 01:04, 17 November 2023
• Proofs of AM-GM
  This pages lists some proofs of the weighted [[AM-GM Inequality]]. The inequality's statement is as follows: for all nonnegati ...and <math>\lambda_i \neq 0</math>, for some <math>i</math>, then the right-hand side of the inequality is zero and the left hand of the inequality is g
  12 KB (2,171 words) - 07:55, 11 May 2023
• Carleman's Inequality
  Define <math>c_n = n \left( 1 + \frac{1}{n} \right)^n = \frac{(n+1)^n}{n^{n-1}}</math>. Then for all positive integers <math>k</math>, Now, by [[AM-GM]],
  2 KB (278 words) - 16:39, 29 December 2021
• 2008 AMC 12B Problems/Problem 23
  ...square. When these are multiplied, they equal <math>2^{a+n-a} \times 5^{b+n-b} = 10^n</math>. <math>\log 10^n=n</math> so the number of factors divided .../math> satisfies <math>ab = 10^n</math> -- ex. <math>(1, 10^n), (2, 5*10^{n-1})</math>, etc. Then the sum of the base-<math>10</math> logarithms is <mat
  5 KB (814 words) - 18:02, 17 January 2023
• Vieta's formulas
  Let <math>P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0</math> be any polynomial with [[Complex number | c ....</cmath> This can be compactly summarized as <math>s_j = (-1)^j \frac{a_{n-j}}{a_n}</math> for some <math>j</math> such that <math>1 \leq j \leq n</mat
  3 KB (521 words) - 15:58, 13 July 2024
• 2000 AMC 12 Problems/Problem 16
  <math>f(i, j) = 17(i-1) + j = 17i + j - 17</math> {{AMC12 box|year=2000|num-b=15|num-a=17}}
  2 KB (310 words) - 11:28, 3 August 2021
• 2007 AMC 10A Problems
  draw((-h-1,-h-1)--(-h-1,h+1)--(h+1,h+1)--(h+1,-h-1)--cycle); ...at random from 0 to 2007, inclusive. What is the probability that <math>ad-bc</math> is even?
  13 KB (2,058 words) - 17:54, 29 March 2024
• 1990 USAMO Problems/Problem 4
  '''Lemma.''' Let <math>a</math> be an integer. Then there are <math>2^{k-1}</math> <math>k</math>-good sequences starting on <math>a</math>, and furt ...M</math>. Then the only possibilities for <math>a_{k+1}</math> are <math>m-1</math> and <math>M+1</math>; either way, <math>\{ a_i \}_{i=1}^{k+1}</math
  3 KB (529 words) - 19:15, 18 July 2016
• 2004 AMC 10A Problems/Problem 16
  ...math>1</math> - <math>5\times5</math> square), therefore there are <math>55-36 = 19</math> squares that contain the black square, which is <math>\boxed{ {{AMC10 box|year=2004|ab=A|num-b=15|num-a=17}}
  2 KB (377 words) - 01:37, 14 August 2024
• 2007 AMC 10A Problems/Problem 18
  ...\cdot 12 = \frac {36}5 </math>. (An alternate way is by seeing that the set-up AHGCM is similar to the 2 pole problem (http://www.artofproblemsolving.co {{AMC10 box|year=2007|ab=A|num-b=17|num-a=19}}
  6 KB (867 words) - 00:17, 20 May 2023
• 1974 USAMO Problems/Problem 1
  <cmath> a-b \mid P(a) - P(b) = b-c \mid P(b)-P(c) = c-a \mid P(c) - P(a) = a-b , </cmath> ...absolute value. In fact, two of them, say <math>(a-b)</math> and <math>(b-c)</math>, must be equal. Then
  7 KB (1,291 words) - 20:30, 27 April 2020
• 1994 USAMO Problems
  The sides of a 99-gon are initially colored so that consecutive sides are red, blue, red, blue ...sual, let <math>\, \binom{n}{k} \,</math> denote <math>\, n! \over k! \, (n-k)!</math>. Prove that <cmath> \sum_{U \subseteq S} (-1)^{|U|} \binom{m - \s
  2 KB (391 words) - 07:58, 19 July 2016
• Isogonal conjugate
  Let <math>I'</math> and <math>J'</math> be the A-excenters of triangles <math>\triangle ABC</math> and <math>\triangle ADC,</ ...oordinates <cmath>(p' : q' : r'), p' = \frac {|B - C|^2}{p}, q' = \frac {|A-C|^2}{q}, r' = \frac {|A - B|^2}{r}.</cmath>
  54 KB (9,416 words) - 08:40, 18 April 2024
• 1976 IMO Problems/Problem 2
  ...inct real solutions, where <math>2^{n-1}</math> are positive and <math>2^{n-1}</math> are negative. Also, for ever root <math>r</math>, <math>|r|<2</mat ...values less than 2, where <math>2^{k-1}</math> are positive and <math>2^{k-1}</math> are negative.
  3 KB (596 words) - 16:19, 28 July 2015
• 1976 IMO Problems/Problem 5
  {{IMO box|year=1976|num-b=4|num-a=6}}
  2 KB (377 words) - 16:28, 29 January 2021
• Partition (combinatorics)
  ...iven by <math>F(x)= \sum_{n \geq 0}P(n) x^n = \prod_{n=1}^\infty \frac{1}{1-x^n}</math>. Partitions can also be studied by using the [[Jacobi theta func ...epresents a different addend in the partition. The rows are ordered in non-increasing order so that that the row with the most dots is on the top and t
  10 KB (1,508 words) - 14:24, 17 September 2017

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