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• 2020 CIME II Problems/Problem 10
  Case II) <math>a+b=2c\Rightarrow 2+c=ab\Rightarrow4+a+b=2ab\Rightarrow(2a-1)(2b-1)=9</math> Case V) <math>a+b=5c\Rightarrow (5a-1)(5b-1)=126</math> for which there are 2 solu
  2 KB (332 words) - 08:37, 30 December 2021
• Cauchy-Schwarz Inequality
  ...ly if there exists a scalar <math>t</math> such that <math>\overrightarrow{v} = t \overrightarrow{w}</math>, or if one of the vectors is zero. This form ...is equal to <math> \left( ||\mathbf{a}|| * ||\mathbf{b}|| \right)^2 = a^2b^2</math>. The inequality then follows from <math> |\cos\theta | \le 1 </ma
  13 KB (2,048 words) - 14:28, 22 February 2024
• 1988 AIME Problems/Problem 13
  3a+2b &= 0. 3a+2b &= -1, \\
  10 KB (1,595 words) - 15:30, 24 August 2024
• 2002 IMO Shortlist Problems/N3
  ..., then the [[greatest common factor]] of <math>2^u + 1 </math> and <math>2^v + 1 </math> is 3. ...ath>t </math>, contradicting our assumption that <math>u </math> and <math>v </math> are relatively prime.
  10 KB (1,739 words) - 05:38, 12 November 2019
• 2007 AIME II Problems/Problem 15
  ...ale the triangle with the circumradius by a [[line]]ar scale factor, <math>v</math>. :<math>\frac{65}{8}v=8u</math>
  11 KB (2,099 words) - 22:44, 6 October 2024
• 2005 USAMO Problems/Problem 3
  ...math> Substituting in <math>P</math> yields <cmath>-a^2p(1-p)+vp=0\implies v = a^2(1-p).</cmath> Substituting in <math>B_1</math> yields ...c{k-\sqrt{k^2+4b^2c^2q(1-q)}}{2c^2}, z'=\frac{-k-\sqrt{k^2+4b^2c^2q(1-q)}}{2b^2}</cmath>
  6 KB (1,117 words) - 00:17, 11 October 2021
• Characteristic polynomial
  ...non-zero vector that satisfies the relation <math>A\bold{v} = \lambda\bold{v}</math>, for some scalar <math>\lambda \in K</math>. In other words, applyi ...bold{v} = \lambda \bold{v}</math>, then <math>\lambda I \bold{v} - A \bold{v} = \bold{O}</math>. But then, the column vectors of <math>\lambda I - A</ma
  19 KB (3,412 words) - 13:57, 21 September 2022
• 2011 AIME II Problems/Problem 15
  ...math>P(v) = 4</math>, which will give us the working range <math>5 \le x < v</math>. v^2 - 3v - 9 &= 4 \\
  8 KB (1,307 words) - 16:05, 20 November 2024
• 2011 USAMO Problems/Problem 3
  ...> <cmath>\vec{w} = \vec{CD} + \vec{FA}</cmath> Clearly, <math>\vec{u}+\vec{v}+\vec{w}=\textbf{0}</math>. ...th>. Thus, <math>|\vec{u}|=2p \cos \gamma</math>. Similarly, <math>|\vec{v}|=2q \cos \alpha</math> and <math>|\vec{w}| = 2r \cos \beta</math>.
  11 KB (1,925 words) - 11:07, 31 August 2023
• 1989 AHSME Problems
  ...>, <math>c</math>, and <math>d</math> be positive integers with <math> a < 2b</math>, <math>b < 3c </math>, and <math>c<4d</math>. If <math>d<100</math>, ...{v} \rfloor</math> means the greatest integer less than or equal to <math>v</math>.)
  15 KB (2,247 words) - 12:44, 19 February 2020
• 1954 AHSME Problems
  ...an upright V.}\\ \textbf{(D)}\ \text{Two line segments forming an inverted V.}\\ \textbf{(E)}\ \text{None of these.} </math> ...le by <math>8</math>. If <math>a>b</math>, and <math>2a+1</math> and <math>2b+1</math> are the odd numbers,
  23 KB (3,534 words) - 18:16, 8 November 2024
• 1960 AHSME Problems
  \text{(B) if a }\neq\text{2b}\qquad The volume <math>V = \pi R^2H</math> is to be increased by the same fixed positive amount when
  21 KB (3,242 words) - 20:27, 30 December 2020
• 1988 AHSME Problems
  <math>\frac{bx(a^2x^2 + 2a^2y^2 + b^2y^2) + ay(a^2x^2 + 2b^2x^2 + b^2y^2)}{bx + ay}</math> ...),-sqrt(5)), W=(4+2/sqrt(5),sqrt(5)), T=(4,0), U=(4+2/sqrt(5),-4/sqrt(5)), V=(4+2/sqrt(5),1/sqrt(5));
  17 KB (2,535 words) - 12:45, 19 February 2020
• 1964 AHSME Problems
  .../math> and the remainder is <math>v</math>, where <math>u</math> and <math>v</math> are integers. \textbf{(D)}\ v \qquad
  19 KB (2,907 words) - 13:16, 20 February 2020
• 2015 AMC 10A Problems/Problem 21
  ...dron and <math>TD</math> as the height. Thus, the desired volume is <cmath>V = \dfrac{1}{3} bh = \dfrac{1}{3}\cdot[ABC]\cdot TD = \dfrac{1}{3} \cdot 6 \ <cmath>16A^2 = 2a^2 b^2 + 2b^2 c^2 + 2c^2 a^2 - a^4 - b^4 - c^4 = -(a^2 + b^2 - c^2)^2 + 4a^2 b^2.</cmat
  4 KB (689 words) - 19:16, 18 June 2024
• 1969 IMO Problems/Problem 6
  <math>H</math> is positive definite if for any 3x1 vector <math>v</math> we have <math>v^t H v \ge 0</math>. (<math>v^t</math> is the transpose of <math>v</math>, a matrix which is 1x3.)
  11 KB (2,086 words) - 13:50, 1 August 2024
• 2018 AIME I Problems/Problem 12
  <math>5a,2b</math> ...</math> be the number of elements <math>2 \pmod{3}</math>. Then, <math>a + 2b \equiv 0 \pmod{3} \implies a - b \equiv 0 \pmod{3}</math>. Since <math>0 \l
  26 KB (4,044 words) - 12:58, 24 January 2024
• 2019 AIME II Problems/Problem 4
  'v' .../math> multiplicity using <cmath>\frac{f(1, b, c)+f(-1, b, c)}{2}=\frac{(3+2b+c)^4+(c+1)^4}{2}=g(b,c).</cmath> Similarly, now filter out the coefficients
  20 KB (3,220 words) - 02:24, 14 August 2024
• 2020 AMC 12B Problems/Problem 7
  ...equation <math>\frac{6b}{a}=\frac{a+b}{a-b}\implies a^2-5ab+6b^2=(a-3b)(a-2b)=0</math> and this gives our answer to be <math>\mathbf{(C)} \frac{3}{2}</m ...the origin and select a point on each line to define vectors <math>\mathbf{v}_{i}=(x_{i},y_{i})</math>.
  5 KB (895 words) - 22:54, 1 October 2024
• 2022 AIME I Problems/Problem 15
  and solve for <math>m/n = (abc)^2 = a^2b^2c^2</math> Since <math>a^2=b^2+c^2-2bc=\frac{1}{2}</math>, <math>m/n = a^2b^2c^2 = a^2(bc)^2 = \frac{1}{2}\left(\frac{1}{4}\right)^2=\frac{1}{32}</math
  15 KB (2,208 words) - 00:25, 1 February 2024

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