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• 2002 AIME I Problems/Problem 15
  ...nswer form asks for it, we get <math>128-16\sqrt{19}</math>, and <math>128+16+19=\fbox{163}</math>. ...ordinate of <math>E</math> is <math>12-\sqrt{(4\sqrt{10})^2-12^2}=12-\sqrt{16}=12-4=8</math>, so <math>E = (8,12,6)</math>.
  7 KB (1,181 words) - 15:56, 1 July 2024
• 2002 AIME I Problems/Problem 13
  ...int Theorem]] on <math>E</math>, we get <math>EF = \frac{12^2}{27} = \frac{16}{3}</math>. The [[Law of Cosines]] on <math>\triangle ACE</math> gives ...AEF] = 2 \cdot \frac 12 \cdot AE \cdot EF \sin \angle AEF = 12 \cdot \frac{16}{3} \cdot \frac{\sqrt{55}}{8} = 8\sqrt{55}</math>, and the answer is <math>
  7 KB (1,170 words) - 23:15, 11 July 2024
• 2003 AIME I Problems/Problem 2
  *For <math>8</math> circles, the ratio is <math>9/16</math>. {{AIME box|year=2003|n=I|num-b=1|num-a=3}}
  4 KB (523 words) - 15:49, 8 March 2021
• 2003 AIME II Problems/Problem 14
  ...{3}}{2} i\right) = -\frac{b}{2}-\sqrt{3}+\left(\frac{b\sqrt{3}}{2}-1\right)i</math>. We solve for <math>b</math> and <math>f</math> and find that <math ...rt{3x^2+12}</math>. Since AB=AF, we can also form the equation <math>4+x^2=16+z^2</math>.
  9 KB (1,463 words) - 18:08, 31 July 2024
• 2003 AIME II Problems/Problem 13
  <math>P_5=\frac{5}{16}</math>, ...riginal vertex, the net number of clockwise steps must be a multiple of 3, i.e., <math>\#CW - \#CCW \equiv 0 \pmod{3}</math>. Since <math>\#CW + \#CCW
  15 KB (2,406 words) - 23:56, 23 November 2023
• 2001 AIME II Problems/Problem 10
  ...ath>, where <math>i</math> and <math>j</math> are integers and <math>0\leq i < j \leq 99</math>? ...lies that <math>10^{6} - 1 | 10^{6k} - 1</math>, and so any <math>\boxed{j-i \equiv 0 \pmod{6}}</math> will work.
  4 KB (549 words) - 23:16, 19 January 2024
• 2000 AIME II Problems/Problem 14
  ...h>(f_1,f_2,f_3,\ldots,f_j)</math> is the factorial base expansion of <math>16!-32!+48!-64!+\cdots+1968!-1984!+2000!</math>, find the value of <math>f_1-f <math>(32m+16)!-(32m)! = \left(1+\sum_{k=1}^{32m+15} {k\cdot k!}\right)-\left(1+\sum_{k=1
  7 KB (1,131 words) - 14:49, 6 April 2023
• 2000 AIME II Problems/Problem 9
  ...we have <math>z = \frac{2\cos 3 \pm \sqrt{4\cos^2 3 - 4}}{2} = \cos 3 \pm i\sin 3 = \text{cis}\,3^{\circ}</math>. ...could let <math>z = a + bi</math> and solve to get <math>z=\cos 3^\circ + i\sin 3^\circ</math>.
  4 KB (675 words) - 13:42, 4 April 2024
• User talk:JBL
  Yeah.. I didn't see that the article's sequence was defined starting with <math>F_1< I viewed the thread, and it seems that theorems would be capitalized, which i
  4 KB (519 words) - 15:22, 26 April 2016
• Quadratic reciprocity
  ...other [[ring of integers|rings of integers]]. (I'll put that here later if I remember.) <cmath> P(x) = x^q - 1 = (x-1) \sum_{i=0}^{q-1} x^i, </cmath>
  7 KB (1,182 words) - 16:46, 28 April 2016
• 2000 AMC 12 Problems/Problem 8
  <cmath>4,8,12,16,...</cmath> for (int i = 0; i <= 2*fig; ++i) {
  8 KB (1,226 words) - 23:15, 25 July 2024
• Sigma notation
  * <math>\sum_{i = 33}^{56} 1 = 1 + 1 + \ldots + 1 + 1 = 24</math> ...fty} \left(\dfrac{1}{2}\right)^i = \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{16} + \ldots = 1</math>
  2 KB (335 words) - 17:17, 8 February 2024
• University of South Carolina High School Math Contest/1993 Exam/Problems
  <cmath> \mathrm{(A) \ }\sqrt{3}-\frac{\pi}2 \qquad \mathrm{(B) \ } \frac 16 \qquad \mathrm{(C) \ }\frac 13 \qquad \mathrm{(D) \ } \frac{\sqrt{3}}2 - \f If <math>(1 + i)^{100}</math> is expanded and written in the form <math>a + bi</math> where
  14 KB (2,102 words) - 22:03, 26 October 2018
• User:Rowechen
  ...^{}_{}</math> are integers, can be uniquely expressed in the base <math>-n+i^{}_{}</math> using the integers <math>1,2^{}_{},\ldots,n^2</math> as digits <center><math>r+si=a_m(-n+i)^m+a_{m-1}(-n+i)^{m-1}+\cdots +a_1(-n+i)+a_0</math></center>
  22 KB (3,700 words) - 21:57, 8 August 2024
• Mock AIME 1 2006-2007/Problems
  ...et circles <math>A''</math>, <math>B''</math>, <math>C''</math>, and <math>I</math> have radii <math>a</math>, <math>b</math>, <math>c</math>, and <math ...}=3</math>, <math>d_{4}=-7</math>, <math>d_{5}=13</math>, and <math>d_{6}=-16</math>, find <math>d_{7}</math>.
  8 KB (1,355 words) - 14:54, 21 August 2020
• Mock AIME 2 2006-2007 Problems
  ...3}{z^2}+\frac{4}{z ^3}+\frac{5}{z^4}+\cdots</math> and <math>z=n\pm \sqrt{-i},</math> find <math> \lfloor 100n \rfloor.</math> ...ons, in degrees, to the equation <math>\sin^{10}x + \cos^{10}x = \frac{29}{16}\cos^4 2x,</math> where <math>0^\circ \le x^\circ \le 2007^\circ.</math>
  5 KB (848 words) - 23:49, 25 February 2017
• Alternating sum
  ...ac12 + \frac13 - \frac 14 + \ldots = \sum_{i = 1}^\infty \frac{(-1)^{i+1}}{i}</math>. Given an [[infinite]] alternating sum, <math>\sum_{i = 0}^\infty (-1)^i a_i</math>, with <math>a_i \geq 0</math>, if corresponding sequence <math>a
  2 KB (301 words) - 22:13, 19 February 2022
• 2007 iTest Problems
  <math>\mathrm{(A)}\,\frac{1}{8}\quad\mathrm{(B)}\,\frac{3}{16}\quad\mathrm{(C)}\,\frac{3}{8}\quad\mathrm{(D)}\,\frac{1}{2}</math> <cmath>\sum_{i=1}^{100}(i!)^{2}</cmath>
  30 KB (4,794 words) - 23:00, 8 May 2024
• 2005 AMC 10A Problems
  <math> \mathrm{(A) \ } -16\qquad \mathrm{(B) \ } -8\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } 8\qq == Problem 16 ==
  14 KB (2,026 words) - 11:45, 12 July 2021
• 2003 AMC 10A Problems/Problem 22
  <math> \mathrm{(A) \ } 16\qquad \mathrm{(B) \ } 20\qquad \mathrm{(C) \ } 24\qquad \mathrm{(D) \ } 28\ ...lifying this equation, we get that <math> y=2x+8 </math>. Let point <math> I </math> be the point on line <math> FG </math> so that lines <math> CI </ma
  9 KB (1,446 words) - 12:43, 18 September 2024

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