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• 2005 AIME II Problems/Problem 12
  ...ying using the fact that <math>e^{i \frac{\pi}{4}} = \dfrac{\sqrt{2}}{2} + i \dfrac{\sqrt{2}}{2}</math>, clearing the denominator, and setting the imagi I hope you like expanding
  13 KB (2,080 words) - 21:20, 11 December 2022
• Euler's identity
  '''Euler's Formula''' is <math>e^{i\theta}=\cos \theta+ i\sin\theta</math>. It is named after the 18th-century mathematician [[Leon ...eta) + i\sin(\theta))^n = (e^{i\theta})^n = e^{in\theta} = \cos(n\theta) + i\sin(n\theta)</math>.
  3 KB (452 words) - 23:17, 4 January 2021
• 2005 AIME I Problems/Problem 1
  [[Image:2005 AIME I Problem 1.png]] ...ath>. <math>K = 30^2\pi - 6(10^2\pi) = 300\pi</math>, so <math>\lfloor 300\pi \rfloor = \boxed{942}</math>.
  1 KB (213 words) - 13:17, 22 July 2017
• 2005 AIME I Problems/Problem 6
  ...urth root of 2006. Then the roots of the above equation are <math>x = 1 + i^n r</math> for <math>n = 0, 1, 2, 3</math>. The two non-real members of th Starting like before,
  4 KB (686 words) - 01:55, 5 December 2022
• 2004 AIME I Problems/Problem 14
  pair Cxy = 8*expi((3*pi)/2-CE/8); pair L = 8*expi(pi+0.05), R = 8*expi(-0.22); /* left and right cylinder lines, numbers from tr
  4 KB (729 words) - 01:00, 27 November 2022
• 2004 AIME I Problems/Problem 13
  ...4</math> complex roots of the form <math> z_k = r_k[\cos(2\pi a_k)+i\sin(2\pi a_k)], k=1, 2, 3,\ldots, 34, </math> with <math> 0 < a_1 \le a_2 \le a_3 \l {{AIME box|year=2004|n=I|num-b=12|num-a=14}}
  2 KB (298 words) - 20:02, 4 July 2013
• 2004 AIME I Problems/Problem 11
  ...the [[Pythagorean Theorem]], we get <math>\ell = 5</math> and <math>A = 24\pi</math>. ...ace area of the original solid, we find that <math>A_f=24\pi - \frac{5}{3}\pi x^2</math>.
  5 KB (839 words) - 22:12, 16 December 2015
• 2004 AIME I Problems/Problem 4
  ...ed by all of the midpoints is <math>4-4\cdot \left(\frac{\pi}{4}\right)=4-\pi \approx .86</math> to the nearest hundredth. Thus <math>100\cdot k=\boxed{8 ...rea of the square minus the area of the quarter circles, which is <math>4-\pi \approx 0.86</math>, so <math>100k = \boxed{086}</math>. ~Extremelysupercoo
  3 KB (532 words) - 09:22, 11 July 2023
• 2004 AIME I Problems
  {{AIME Problems|year=2004|n=I}} [[2004 AIME I Problems/Problem 1|Solution]]
  9 KB (1,434 words) - 13:34, 29 December 2021
• 2004 AIME II Problems
  ...to the smaller can be expressed in the form <math> \frac{a\pi+b\sqrt{c}}{d\pi-e\sqrt{f}}, </math> where <math> a, b, c, d, e, </math> and <math> f </math {{AIME box|year = 2004|n=II|before=[[2004 AIME I Problems]]|after=[[2005 AIME I Problems]]}}
  9 KB (1,410 words) - 05:05, 20 February 2019
• 1989 AIME Problems
  ...up heads in exactly <math>3</math> out of <math>5</math> flips. Find <math>i+j^{}_{}</math>. pair A=(0,0),B=(10,0),C=6*expi(pi/3);
  7 KB (1,045 words) - 20:47, 14 December 2023
• 1991 AIME Problems
  ...th>x^{}_{}</math> satisfy the equation <math>\frac{1}{5}\log_2 x = \sin (5\pi x)</math>? ...The sum of the areas of the twelve disks can be written in the form <math>\pi(a-b\sqrt{c})</math>, where <math>a,b,c^{}_{}</math> are positive integers a
  7 KB (1,106 words) - 22:05, 7 June 2021
• 1995 AIME Problems
  ...> are the perpendicular bisectors of two adjacent sides of square <math>S_{i+2}.</math> The total area enclosed by at least one of <math>S_{1}, S_{2}, ...math> and the sum of the other two roots is <math>3+4i,</math> where <math>i=\sqrt{-1}.</math> Find <math>b.</math>
  6 KB (1,000 words) - 00:25, 27 March 2024
• 1998 AIME Problems
  Given that <math>A_k = \frac {k(k - 1)}2\cos\frac {k(k - 1)\pi}2,</math> find <math>|A_{19} + A_{20} + \cdots + A_{98}|.</math> ...,x_2,x_3,x_4)</math> of positive odd [[integer]]s that satisfy <math>\sum_{i = 1}^4 x_i = 98.</math> Find <math>\frac n{100}.</math>
  7 KB (1,084 words) - 02:01, 28 November 2023
• 2003 AIME I Problems
  {{AIME Problems|year=2003|n=I}} [[2003 AIME I Problems/Problem 1|Solution]]
  6 KB (965 words) - 16:36, 8 September 2019
• 2002 AIME II Problems
  ...math> radians are <math>\frac{m\pi}{n-\pi}</math> and <math>\frac{p\pi}{q+\pi}</math>, where <math>m</math>, <math>n</math>, <math>p</math>, and <math>q< {{AIME box|year = 2002|n=II|before=[[2002 AIME I Problems]]|after=[[2003 AIME I Problems]]}}
  7 KB (1,177 words) - 15:42, 11 August 2023
• 2003 AIME II Problems
  ...e log. The number of cubic inches in the wedge can be expressed as <math>n\pi</math>, where n is a positive integer. Find <math>n</math>. ...istinct zeros of <math>P(x),</math> and let <math>z_{k}^{2} = a_{k} + b_{k}i</math> for <math>k = 1,2,\ldots,r,</math> where <math>a_{k}</math> and <mat
  7 KB (1,127 words) - 09:02, 11 July 2023
• 1983 AIME Problems/Problem 15
  ...the minor arc <math>AD</math>, we see that <math>\alpha \in \left(-\frac{\pi}{2}-\alpha_0,\alpha_0\right)</math>. However, since the midpoint of <math>A Let I be the intersection of AD and BC.
  19 KB (3,221 words) - 01:05, 7 February 2023
• 1984 AIME Problems/Problem 13
  Recall that <math>\cot^{-1}\theta = \frac{\pi}{2} - \tan^{-1}\theta</math> and that <math>\arg(a + bi) = \tan^{-1}\frac{b ...{2} - \arg x + \frac{\pi}{2} - \arg y + \frac{\pi}{2} - \arg z) = 10\cot(2\pi - \arg wxyz)</cmath>
  3 KB (473 words) - 12:06, 18 December 2018
• 1985 AIME Problems/Problem 12
  ...can generalize the following relationships for all <i><b>nonnegative</b></i> integers <math>k:</math> Let <math>\omega = e^{i\pi / 2}</math>. We have that if <math>G(x) = (x+x^2+x^3)^7</math>, then <cmath
  17 KB (2,837 words) - 13:34, 4 April 2024

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