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• Pick's Theorem
  A = ab, \; I = (a-1)(b-1), \; B = 2a+2b, ...L_1</math> is <math>\frac{1}{\sqrt{x_1^2 + y_1^2}}</math> or <math>\frac{d-1}{\sqrt{x_1^2 + y_1^2}}</math>, this point must be different from the vert
  7 KB (1,253 words) - 14:29, 26 June 2024
• Fermat point
  We shall present a standard triangle inequality proof as well as a less-known vector proof: ...\vec{0}</math> and adding, we see that <math>|a|+|b|+|c|\leq |a-x|+|b-x|+|c-x|</math>, or <math>AP+BP+CP\leq AX+BX+CX</math>. Thus, the origin or point
  4 KB (769 words) - 16:07, 29 December 2019
• Chinese Remainder Theorem
  ...or instance, [[Fermat's Little Theorem]] may be generalized to the [[Fermat-Euler Theorem]] in this manner. ...: the proof of the general case follows by induction to the above result (k-1) times.
  6 KB (1,022 words) - 14:57, 6 May 2023
• Matrix
  4 KB (856 words) - 15:29, 30 March 2013
• 1985 IMO Problems/Problem 3
  Now assume that the problem holds for <math>k-1</math>. We now have two cases: <math> i_1 \ge k</math>, and <math>i_1 < k ..._{j=0}^{k-1}a_j x^j + (1+x)^k \sum_{j=0}^{k-1} b_j x^j \equiv \sum_{j=0}^{k-1}\left[ (a_j + b_j)x^j + b_j x^{j+k} \right] \pmod{2}
  2 KB (354 words) - 04:56, 11 March 2023
• 2001 AMC 8 Problems
  Granny Smith has \$63. Elberta has \$2 more than Anjou and Anjou has one-third as much as Granny Smith. How many dollars does Elberta have? ...3, 4 and 9 are each used once to form the smallest possible '''even''' five-digit number. The digit in the tens place is
  13 KB (1,990 words) - 09:51, 19 June 2024
• 2003 AMC 12A Problems/Problem 12
  {{AMC10 box|year=2003|ab=A|num-b=23|num-a=25}} {{AMC12 box|year=2003|ab=A|num-b=11|num-a=13}}
  2 KB (368 words) - 18:04, 28 December 2020
• Minkowski Inequality
  1 KB (240 words) - 16:49, 29 December 2021
• 2001 IMO Shortlist Problems/A2
  ...>b_0, b_1, \ldots</math> of positive integers for which <math>1+b_n\le b_{n-1}\sqrt[n]{2}</math>, it is clear that there will not exist any sequence <ma ...exist such a sequence. Then, define <math>x_0=b_0</math> and <math>x_n=x_{n-1}\sqrt[n]{2} -1</math>. It is clear that <math>x_n\ge b_n</math> for all <m
  2 KB (411 words) - 04:38, 17 June 2019
• Quaternion
  1,007 bytes (155 words) - 20:47, 14 October 2013
• 2020 AMC 10A Problems
  <cmath>\frac{a-3}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{c-5}{4-b}</cmath> What is the sum of all real numbers <math>x</math> for which <math>|x^2-12x+34|=2?</math>
  13 KB (1,968 words) - 18:32, 29 February 2024
• Asymptote: Logical Operators and Loops
  if (floor((i-j)/2)==((i-j)/2)) if (floor((i-j)/2)==((i-j)/2))
  2 KB (324 words) - 16:50, 2 October 2016
• Pigeonhole Principle/Solutions
  ...riends, then the remaining friends must have from <math>1</math> to <math>n-2</math> friends for the remaining friends not to also have no friends. By p ...same remainder when divided by 5. Since there are 5 possible remainders (0-4), by the pigeonhole principle, at least two of the integers must share the
  10 KB (1,617 words) - 01:34, 26 October 2021
• 2002 USAMO Problems/Problem 3
  ...it is necessary and sufficient that <math> P(x) = Q(x) + \prod_{i=1}^{n}(x-x_i) </math>. ...s. But a polynomial with real coefficients must have an even number of non-real roots, so <math>P(x) </math> must have <math>n </math> real roots. Sim
  4 KB (688 words) - 13:38, 4 July 2013
• 2007 AIME I Problems/Problem 10
  [[Image:AIME I 2007-10.png]] ...ath> (<math>j < k</math>) here. We can now use the [[Principle of Inclusion-Exclusion]] based on the stipulation that <math>j\ne k</math> to solve the p
  13 KB (2,328 words) - 00:12, 29 November 2023
• 2004 USAMO Problems
  ...egers <math> x, y \in S </math>, if <math> x+y \in S </math>, then <math> x-y \in S </math>. * [http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2004-ua/04usamo-test.shtml 2004 USAMO Problems]
  3 KB (474 words) - 09:20, 14 May 2021
• 2007 AIME II Problems/Problem 3
  ...ince <math>\triangle ABE \cong \triangle CDF</math> and are both <math>5-12-13</math> [[right triangle]]s, we can come to the conclusion that the two ne A slightly more analytic/brute-force approach:
  6 KB (933 words) - 00:05, 8 July 2023
• 2007 AIME II Problems/Problem 13
  {{AIME box|year=2007|n=II|num-b=12|num-a=14}}
  3 KB (600 words) - 11:10, 22 January 2023
• 2007 AIME II Problems
  <div style="text-align:center;">[[Image:2007 AIME II-3.png]]</div> ..._{i+1}</math> if <math>a_{i}</math> is [[even]]. How many four-digit parity-monotonic integers are there?
  9 KB (1,435 words) - 01:45, 6 December 2021
• 2003 USAMO Problems/Problem 1
  ...-1} + a </math> is divisible by 5, which is true when <math> k \equiv -3^{n-1}a \pmod{5} </math>. Since there is an odd digit in each of the residue cl ...s after it, and all multiples of 5 end in 5. Therefore, <math> a*10^x*5^{n-x}</math> always contains a 5 as its <math> (x+1)^{st}</math> digit, and we
  4 KB (736 words) - 22:17, 3 March 2023

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