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• 1986 AIME Problems/Problem 11
  ...>C = \{1, (x+1), (x+1)^2, ..., (x+1)^{17} \}</math> be two bases for <math>V</math>. ...e polynomial given in the problem, and it is easy to see that <math>[ \vec{v} ]_B = \langle 1, -1, 1, -1, ... , 1, -1 \rangle.</math>
  6 KB (872 words) - 16:51, 9 June 2023
• 1988 AIME Problems/Problem 13
  ...(x)</math> are also roots of <math>f(x)</math>. Let these roots be <math>u,v</math>. We get the system If we multiply the first equation by <math>v^{16}</math> and the second by <math>u^{16}</math> we get <cmath>\begin{alig
  10 KB (1,585 words) - 03:58, 1 May 2023
• 1988 AIME Problems/Problem 10
  ...lies on exactly one vertex of a square/hexagon/octagon, we have that <math>V = 12 \cdot 4 = 8 \cdot 6 = 6 \cdot 8 = 48</math>. ...h of its endpoints, the number of edges <math>E</math> is <math>\frac{3}{2}V = 72</math>.
  5 KB (811 words) - 19:10, 25 January 2021
• 1990 AIME Problems/Problem 14
  Finally, we substitute <math>h</math> into the volume equation to find <math>V = 6\sqrt{133}\left(\frac{99}{\sqrt{133}}\right) = \boxed{594}</math>. ...ave the base area as <math>18\sqrt {133}</math>. Thus, the volume is <math>V = \frac {1}{3}\cdot18\sqrt {133}\cdot\frac {99}{\sqrt {133}} = 6\cdot99 = 5
  7 KB (1,086 words) - 08:16, 29 July 2023
• 1993 AIME Problems/Problem 12
  ...th>(u,v)</math> and <math>(p,q)</math>, then <math>u=2r-p</math> and <math>v=2s-q</math>. So we start with the point they gave us and work backwards. We
  4 KB (611 words) - 13:59, 15 July 2023
• 1993 AIME Problems/Problem 10
  ...<math>P</math> pentagonal faces meet. What is the value of <math>100P+10T+V</math>? ...ge). Thus, <math>E=60</math>. Finally, using Euler's formula we have <math>V=E-30=30</math>.
  4 KB (716 words) - 20:50, 17 April 2022
• 1994 AIME Problems/Problem 6
  ...(-20/sqrt(3),0)-2*u+i*u--(0,20)--(20/sqrt(3),0)+2*d-i*d;draw(shift(0,-2*i)*v);} ...(-20/sqrt(3),0)-2*u+i*u--(0,20)--(20/sqrt(3),0)+2*d-i*d);draw(shift(0,2*i)*v);}
  4 KB (721 words) - 16:14, 8 March 2021
• 1995 AIME Problems/Problem 12
  ...as <math>\vec{u}\cdot \vec{v} = \parallel \vec{u}\parallel \parallel \vec{v}\parallel \cos \theta</math>, we will be able to solve for <math>\cos \thet <cmath>\vec{v} = \overrightarrow{OB}\times \overrightarrow{OC} - \left|\begin{array}{ccc}
  8 KB (1,172 words) - 21:57, 22 September 2022
• 1997 AIME Problems/Problem 14
  ...c{m}{n}</math> be the [[probability]] that <math>\sqrt{2+\sqrt{3}}\le\left|v+w\right|</math>, where <math>m</math> and <math>n</math> are [[relatively p Now, let <math>v</math> be the root corresponding to <math>m\theta=2m\pi/1997</math>, and le
  5 KB (874 words) - 22:30, 1 April 2022
• 1999 AIME Problems/Problem 15
  ...coordinates of the vertex of the resulting pyramid. Call this point <math>V</math>. Clearly, the height of the pyramid is <math>z</math>. The desired v ...= QC</math>. We then use distance formula to find the distances from <math>V</math> to each of the vertices of the medial triangle. We thus arrive at a
  7 KB (1,107 words) - 20:34, 27 January 2023
• 2000 AIME I Problems/Problem 8
  (Computational) The volume of a cone can be found by <math>V = \frac{\pi}{3}r^2h</math>. In the second container, if we let <math>h',r'< From the formula <math>V=\frac{\pi r^2h}{3}</math>, we can find that the volume of the container is
  4 KB (677 words) - 16:33, 30 December 2023
• 2000 AIME I Problems/Problem 2
  ...rea of [[pentagon]] <math>ABCDE</math> is <math>451</math>. Find <math>u + v</math>. D(D(MP("A\ (u,v)",A,(1,0)))--D(MP("B",B,N))--D(MP("C",C,N))--D(MP("D",D))--D(MP("E",E))--cy
  3 KB (434 words) - 22:43, 16 May 2021
• 2001 AIME I Problems/Problem 12
  ...ath>P</math> perpendicular to plane <math>ABC</math> can be found as <math>V=(A-C)\times(B-C)=\langle 8, 12, 24 \rangle</math> ...r each pyramid(base times height divided by 3) we have <math>\dfrac{rF}{3}=V</math>. The surface area of the pyramid is <math>\dfrac{6\cdot{4}+6\cdot{2}
  6 KB (1,050 words) - 18:44, 27 September 2023
• 2014 USAJMO Problems
  ...line{CA}</math> and <math>\overline{AB}</math>, respectively. Let <math>U,V</math> be the intersections of line <math>EF</math> with line <math>MN</mat
  3 KB (600 words) - 16:42, 5 August 2023
• 2003 AIME II Problems/Problem 5
  ...another identical wedge and sticking it to the existing one). Thus, <math>V=\dfrac{6^2\cdot 12\pi}{2}=216\pi</math>, so <math>n=\boxed{216}</math>. Part 1: https://www.youtube.com/watch?v=EzE53aPGbrQ&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=5
  1 KB (204 words) - 17:41, 30 July 2022
• 2001 AIME II Problems/Problem 15
  triple S=(1,0,0), T=(2,0,2), U=(8,6,8), V=(8,8,6), W=(2,2,0), X=(6,8,8); ...-U--V--W--cycle); draw((0,0,1)--T--U--X--(0,2,2)--cycle); draw((0,1,0)--W--V--X--(0,2,2)--cycle);
  4 KB (518 words) - 15:01, 31 December 2021
• 2001 AIME II Problems/Problem 7
  ...ine{UV}</math> with <math>U</math> on <math>\overline{PQ}</math> and <math>V</math> on <math>\overline{QR}</math> such that <math>\overline{UV}</math> i pair P = (0,0), Q = (90, 0), R = (0, 120), S=(0, 60), T=(45, 60), U = (60,0), V=(60, 40), O1 = (30,30), O2 = (15, 75), O3 = (70, 10);
  7 KB (1,112 words) - 02:15, 26 December 2022
• Integration by parts
  <math>\int u\, dv=uv-\int v\,du</math> ...math>u</math> will show up as <math>du</math> and <math>dv</math> as <math>v</math> in the integral on the RHS, <math>u</math> should be chosen such tha
  1 KB (235 words) - 17:01, 11 March 2022
• Holomorphic function
  Specifically, let <math>u, v : \mathbb{R \times R \to R}</math> be definted <cmath> u(x,y) = \text{Re}\,f(x+iy), \qquad v(x,y) = \text{Im}\,f(x+iy) . </cmath>
  9 KB (1,537 words) - 21:04, 26 July 2017
• Expected value
  https://www.youtube.com/watch?v=BBD66Q3KXuI ...enter connecting the midpoints of the two sides of the small triangle with V as an endpoint. Find, with proof, the expected value of the number of full
  4 KB (719 words) - 19:41, 25 November 2020

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