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• Mock AIME 2 2006-2007 Problems
  A positive integer is called a ''dragon'' if it can be written as the sum of four positive intege ...ow many elements <math>n</math> in <math>S</math> is <math>f(n) = \frac{2n^3+n^2-n-2}{n^2-1}</math> an integer?
  5 KB (848 words) - 23:49, 25 February 2017
• Mock AIME 2 2006-2007 Problems/Problem 13
  ...ce appears. If the expected (average) number of cards Richard will turn up is <math>m/n,</math> where <math>m</math> and <math>n</math> are relatively pr ...b+c+3</math>, and thus the expected value of its position is <math>E[a+b+c+3]</math>.
  3 KB (553 words) - 10:13, 19 May 2024
• Expected value
  ...erage outcome if the event were to be repeated many times. Note that this is ''not'' the same as the "most likely outcome." ...h> where the sum is over all outcomes <math>z</math> and <math>P(z)</math> is the probability of that particular outcome. If the event <math>Z</math> ha
  5 KB (789 words) - 20:56, 10 May 2024
• Base numbers/Conversion
  The next natural question is: how do we convert a number from another base into base 10? For example, w <center><math>4201_5 = (4\cdot 5^3 + 2\cdot 5^2 + 0\cdot 5^1 + 1\cdot 5^0)_{10}</math></center>
  7 KB (1,177 words) - 15:56, 18 April 2020
• 2005 AMC 10A Problems/Problem 9
  ...marked <math>O</math>. The five tiles are randomly arranged in a row. What is the probability that the arrangement reads <math>XOXOX</math>? ...} \frac{1}{6}\qquad \textbf{(D) } \frac{1}{4}\qquad \textbf{(E) } \frac{1}{3} </math>
  764 bytes (112 words) - 12:01, 13 December 2021
• Arithmetico-geometric series
  An arithmetico-geometric series is the sum of consecutive terms in an arithmetico-geometric sequence defined a ...Or, <math>\frac{a_ng_{n+1}-x_1-drS_g}{r-1}</math>, where <math>S_g</math> is the sum of the first <math>n</math> terms of <math>g_n</math>.
  2 KB (477 words) - 19:39, 17 August 2020
• IMO Problems and Solutions
  * [[1959 IMO Problems/Problem 3 | Problem 3]] proposed by Hungary * [[1960 IMO Problems/Problem 3 | Problem 3]] proposed by Gheorghe D. Simionescu, Romania
  35 KB (4,009 words) - 20:25, 21 February 2024
• 1959 IMO Problems
  Prove that <math>\frac{21n+4}{14n+3}</math> is irreducible for every natural number <math>n</math>. For what real values of <math>x</math> is
  3 KB (480 words) - 11:57, 17 September 2012
• 1959 IMO Problems/Problem 1
  Prove that the fraction <math>\frac{21n+4}{14n+3}</math> is irreducible for every natural number <math>n</math>. <cmath>(21n+4, 14n+3) = (7n+1, 14n+3) = (7n+1, 1) = 1</cmath>
  5 KB (767 words) - 10:59, 23 July 2023
• 1977 Canadian MO Problems
  ...h> on the circumference of the circle such that the angle <math>OPA</math> is a maximum. == Problem 3 ==
  3 KB (560 words) - 19:23, 10 March 2015
• 1971 Canadian MO Problems/Problem 1
  <math>DEB</math> is a chord of a circle such that <math>DE=3</math> and <math>EB=5 .</math> Let <math>O</math> be the center of the circ
  680 bytes (114 words) - 21:38, 9 July 2019
• 1959 IMO Problems/Problem 4
  ...ven hypotenuse <math>c</math> such that the median drawn to the hypotenuse is the [[geometric mean]] of the two legs of the triangle. ...o a segment from any point on the circle to the midpoint of the hypotenuse is a radius.)
  6 KB (939 words) - 17:31, 15 July 2023
• 1971 Canadian MO Problems/Problem 4
  It is easy to see that this value works for the second polynomial as well. ...)(x-t) </math> and <math> x^2+x+a = (x-s)(x-u)</math> where <math>s</math> is the common root. From Vieta's Formulas, we have: <math>-(s+t) = a,...
  2 KB (346 words) - 23:20, 18 January 2023
• 1970 IMO Problems/Problem 4
  ...gers <math>n</math> with the property that the set <math>\{ n, n+1, n+2, n+3, n+4, n+5 \} </math> can be partitioned into two sets such that the product ...4, 5, 6 \}</math>, but that does not work because only one of the numbers is a multiple of 5. So there are no such sets.
  1 KB (250 words) - 03:31, 2 January 2023
• 2006 Romanian NMO Problems
  ...math> and <math>NP</math> are perpendicular if and only if <math>PN</math> is the interior angle bisector of <math>\angle MPC</math>. ...uch that for each coloring, there exists a line and a column with at least 3 unit squares of the same color (on the same line or column).
  11 KB (1,779 words) - 14:57, 7 May 2012
• 2006 Romanian NMO Problems/Grade 9/Problem 2
  ...ath>\angle BDE = \angle ADP = \angle CDF</math>. Prove that <math>P</math> is the midpoint of <math>EF</math> and <math>DP \perp EF</math>. .../math>. Since P and P' are on the same ray (<math>DP</math>), P = P' and P is the midpoint of <math>EF</math>.
  3 KB (488 words) - 14:05, 15 December 2022
• 2006 Romanian NMO Problems/Grade 9/Problem 3
  ...[inscribe]]d in a [[circle]] of [[radius]] <math>r</math>, for which there is a [[point]] <math>P</math> on <math>CD</math> such that <math>CB=BP=PA=AB</ ...elow; <math>P=(0,0)</math>, <math>A=(-1,\sqrt{3})</math>, <math>B=(1,\sqrt{3})</math>, <math>C=(2,0)</math>, and <math>D=(-2,0)</math>.
  6 KB (1,080 words) - 19:28, 21 September 2014
• 1969 Canadian MO Problems/Problem 2
  ...numbers <math>\sqrt{c+1}-\sqrt{c}</math>, <math>\sqrt{c}-\sqrt{c-1}</math> is greater for any <math>c\ge 1</math>. Thus <math>\boxed{\sqrt{c}-\sqrt{c-1}}</math> is greater.
  1 KB (220 words) - 21:05, 13 August 2023
• 1969 Canadian MO Problems/Problem 4
  ...<math>P</math> is chosen, <math>\frac{PD+PE+PF}{AB+BC+CA}=\frac{1}{2\sqrt{3}}</math>. ...<math>P</math>) Because the sum of the sides is <math>3s</math>, the ratio is always <math>\cfrac{s\frac{\sqrt3}{2}}{3s}=\frac{1}{2\sqrt3}.</math>
  1 KB (215 words) - 00:23, 9 October 2020
• 2005 Canadian MO Problems
  ...r row to a higher row) or revisits a triangle. An example of one such path is illustrated below for <math>n=5</math>. Determine the value of <math>f(2005 ==Problem 3==
  2 KB (436 words) - 11:45, 26 December 2018

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