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• Hölder's Inequality
  ...bda_a = \lambda_b = 1/2</math>, this is the elementary form of the [[Cauchy-Schwarz Inequality]]. ...he left-hand side of the desired inequality but may contribute to the right-hand side.
  4 KB (762 words) - 00:11, 18 June 2023
• Mobius function
  ...>b</math>. Via a combinatorial argument using the [[Principle of Inclusion-Exclusion]] (PIE), it can be seen that the number of integers less than or e ...u{p_i}+\sum_{i,j}\mu(p_ip_j)+\ldots=1-k+\binom{k}{2}-\binom{k}{3}-\ldots=(1-1)^k=0^k=0</math>.</center>
  5 KB (910 words) - 02:58, 1 March 2022
• 2012 IMO Problems/Problem 1
  Let <math>R</math> be the radius of the A-excircle. Because <math>CM = CL</math>, we have <math>CML</math> isosceles a ...w of Sines on triangle <math>BFM</math> and <math>ABC</math> and the double-angle formula for sine, we have
  7 KB (1,189 words) - 01:22, 19 November 2023
• Energy
  636 bytes (99 words) - 21:44, 12 April 2021
• 2002 AMC 8 Problems
  A birdbath is designed to overflow so that it will be self-cleaning. Water flows in at the rate of 20 milliliters per minute and drains Right isosceles triangles are constructed on the sides of a 3-4-5 right triangle, as shown. A capital letter represents the area of each tri
  15 KB (2,102 words) - 09:58, 5 May 2024
• 1994 USAMO Problems/Problem 4
  ...timal equation is <math>a_n = \sqrt{n} - \sqrt{n-1} = 1/(\sqrt{n} + \sqrt{n-1})</math>. Note that this is strictly greater than <math>1/(2\sqrt{n})</mat {{USAMO box|year=1994|num-b=3|num-a=5}}
  1 KB (210 words) - 13:30, 4 July 2013
• 2011 AIME I Problems/Problem 11
  ...ed by <math>d = \text{ord}_{125}(2).</math> To begin with, we'll use a well-known property of the order to get a bound on <math>d.</math> <cmath>2^b\left(2^{a-b} - 1\right) \equiv 0 \mod 1000</cmath>
  11 KB (1,668 words) - 22:10, 24 February 2023
• 2011 AIME I Problems/Problem 7
  ...h> to be anything we want (including 0), all we care about is that <math>(m-1)\text{ }|\text{ } 2010</math> which happens <math>\boxed{016}</math> times ...<math>(m-1)Q(m) = 2010</math>. Thus, if <math>P(m) = 0</math>, then <math>m-1 | 2010</math> .
  10 KB (1,677 words) - 19:59, 29 May 2024
• 2011 AIME II Problems/Problem 8
  <center>[[File:2011_AIME_II_-8.png‎]]</center> <cmath>(z^6-2^{18})(z^6+2^{18})=0</cmath>
  5 KB (805 words) - 18:46, 27 January 2024
• 2011 AIME II Problems
  The degree measures of the angles in a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Fi
  8 KB (1,301 words) - 08:43, 11 October 2020
• 2011 AIME II Problems/Problem 14
  Each POSITION in the 30-position permutation is uniquely defined by an ordered triple <math>(i, j, k ...or the number 2, there are (2-1) choices for i, (3-1) choices for j, and (5-1) choices for k. Thus, there are 1*2*4=8 possible placements for the number
  10 KB (1,581 words) - 16:55, 27 May 2024
• 2011 AIME II Problems/Problem 9
  Let <math>x_1, x_2, ... , x_6</math> be non-negative real numbers such that <math>x_1 +x_2 +x_3 +x_4 +x_5 +x_6 =1</math> ...n is the product of three non-negative terms whose sum is equal to 1. By AM-GM this product is at most <math>\frac1{27}</math>. Since we have added at l
  3 KB (466 words) - 15:06, 16 January 2023
• 1998 USAMO Problems
  ...> (<math>1\leq i\leq 999</math>) so that for all <math>i</math>, <math>|a_i-b_i|</math> equals <math>1</math> or <math>6</math>. Prove that the sum <cmath> |a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}| </cmath>
  3 KB (486 words) - 06:11, 24 November 2020
• 1999 AMC 8 Problems
  ...is at milepost 160. There is a service center on the highway located three-fourths of the way from the third exit to the tenth exit. At what milepost w Walter can buy eggs by the half-dozen. How many half-dozens should he buy to make enough cookies? (Some eggs and some cookies may
  17 KB (2,394 words) - 19:51, 8 May 2023
• KGS math club/solution 11 5
  ...<math>\frac{1}{y}, \frac{2}{y}, \ldots, \frac{y}{y}=1</math>, i.e. <math>y-x</math> needs to divide <math>y</math> perfectly. This condition is also en ...now that for every pair <math>i</math> and <math>j</math> we know <math>s_j-s_i</math> divides <math>s_j</math> exactly an integer number of times. We w
  4 KB (712 words) - 10:51, 11 May 2011
• 2011 USAJMO Problems/Problem 3
  ..._1), l(P_2), l(P_3)</math> are <cmath>y=2a_1x-a_1^2, y=2a_2x-a_2^2, y=2a_3x-a_3^2,</cmath> respectively. It is easy to deduce that the three points of i &= \frac{\Im(1+6a_ni-12a_n^2-8a_n^3i)}{\Re(1+6a_ni-12a_n^2-8a_n^3i)}\\
  15 KB (2,593 words) - 13:37, 29 January 2021
• 2011 USAMO Problems/Problem 6
  ...ach intersection of three of the 11 sets. Since each pair of sets is in 9 3-way intersections&mdash;one with each of the 9 remaining sets&mdash;any two ...he number of <math>b_k>0</math>, where <math>1\le k\le225</math>. By the QM-AM inequality, we know <math>\frac{\sum_{k=1}^{225}b_k^2}n\ge\Bigg(\frac{\su
  7 KB (1,209 words) - 12:50, 25 August 2023
• 2011 AMC 10B Problems/Problem 16
  ...}} = \frac{1}{2+2\sqrt{2}} \times \frac{2-2\sqrt{2}}{2-2\sqrt{2}} = \frac{2-2\sqrt{2}}{-4} = \boxed{\textbf{(A)} \frac{\sqrt{2}-1}{2}}</cmath> ...he denominator reduces to <math>-4</math> and the fraction is <math>\frac{2-2\sqrt{2}}{-4}=\frac{\sqrt{2}-1}{2}</math> which is <math>\boxed{\textbf{(A)
  4 KB (606 words) - 06:14, 25 June 2022
• 2001 AMC 8 Problems/Problem 11
  Using the diagram above, the figure can be divided along the x-axis into two familiar regions that do not overlap: a right triangle and a r {{AMC8 box|year=2001|num-b=10|num-a=12}}
  2 KB (383 words) - 19:36, 24 December 2023
• 2002 AMC 10B Problems/Problem 23
  ...ore, <math>a_m = a_{m-1} + m, a_{m-1}=a_{m-2}+(m-1), a_{m-2} = a_{m-3} + (m-2)</math>, and so on until <math>a_2 = a_1 + 2</math>. ...Sides of all of these equations gives <math>a_m + a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_2</math>.
  5 KB (781 words) - 06:38, 23 October 2023

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