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• 1996 AIME Problems/Problem 7
  for(int i=0;i<7;++i) D(shift(i,j)*unitsquare);
  4 KB (551 words) - 11:44, 26 June 2020
• 1998 AIME Problems/Problem 13
  ...ed to be <math>a_1i + a_2i^2+ a_3i^3 + \cdots + a_ni^n,</math> where <math>i^2 = - 1.</math> Let <math>S_n</math> be the sum of the complex power sums ...2(-176 -64i) + 144i = -352 + 16i</math>. Thus, <math>|p| + |q| = |-352| + |16| = 368</math>.
  2 KB (384 words) - 14:47, 14 June 2024
• 1998 AIME Problems/Problem 2
  <div style="text-align:center;"><math>A = I + \frac B2 - 1</math></div> ...yle="text-align:center;"><math>450 = I + \frac {60}2 - 1</math><br /><math>I = 421</math></div>
  6 KB (913 words) - 16:34, 6 August 2020
• 1999 AIME Problems/Problem 12
  pair A=(0,0),B=(50,0),C=IP(circle(A,23+245/2),circle(B,27+245/2)), I=incenter(A,B,C); Let the incenter be denoted <math>I</math>. It is commonly known that the incenter is the intersection of the
  3 KB (472 words) - 19:03, 21 June 2024
• 1999 AIME Problems/Problem 9
  ...finition of <math>f(z)</math>, <cmath>f(1+i) = (a+bi)(1+i) = (a-b) + (a+b)i,</cmath> this image must be equidistant to <math>(1,1)</math> and <math>(0, ...55}}{16}</math>. Finally, <math>b = |a+bi|\sin\theta = 8\frac{\sqrt{255}}{16} = \frac{\sqrt{255}}{2}</math>, and <math>b^2 = \frac{255}{4}</math>, so th
  6 KB (1,010 words) - 19:01, 24 May 2023
• 2000 AIME I Problems/Problem 15
  Round 7: <math>b</math> to <math>b</math>, <math>15</math> to right, <math>16</math> left in deck, <math>n = -2 + 8k</math>, because <math>n = 2 + 4k</ma ...index, categorizing them further as <math>I\equiv 2\pmod 4</math> or <math>I\equiv 4\pmod 4</math> in that order (The card with the smaller remainder wi
  15 KB (2,673 words) - 19:16, 6 January 2024
• 2000 AIME I Problems/Problem 8
  ...the container minus the volume of the liquid, which is <math>\frac{925\pi}{16}</math>, which is <math>\frac{37}{64}</math> of the volume of the container The volume of the cone with height 9 is <math>\frac{675}{16}\pi</math>.
  4 KB (677 words) - 16:33, 30 December 2023
• 2000 AIME I Problems/Problem 6
  <cmath>((x^2 + 2xy + y^2) + 16 - 2(4)(x+y)) = 4xy</cmath> <cmath>x^2 - 2xy + y^2 + 16 - 8x - 8y = 0</cmath>
  6 KB (966 words) - 11:31, 24 June 2024
• 2000 AIME I Problems/Problem 5
  ...18 = 7</math> white marbles total, the denominator becomes <math>(9 + w_1)(16 - w_1) = 150</math>. Testing <math>w_1 = 1</math> gives a solution, and th {{AIME box|year=2000|n=I|num-b=4|num-a=6}}
  7 KB (1,011 words) - 20:09, 4 January 2024
• 2001 AIME I Problems/Problem 15
  ...H</math>, and thus to an octagonal circuit of these vertices. The cube has 16 diagonal segments that join nonadjacent vertices. In effect, the problem as ...for any vertex, it can be linked to at most one different-colored vertex, i.e. its opposite vertex. Thus, there are only two possible types of circuits
  11 KB (1,837 words) - 18:53, 22 January 2024
• 2001 AIME I Problems/Problem 14
  n&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18&19\\\hline a_n&1&1&1&2&2&3&4&5&7&9&12&16&21&28&37&49&65&86\\\hline
  13 KB (2,298 words) - 19:46, 9 July 2020
• 2001 AIME I Problems/Problem 12
  triple I = (2/3,2/3,2/3); draw(L--F--N--E--M--G--L--I--M--I--N--I--J);
  6 KB (1,050 words) - 18:44, 27 September 2023
• 2001 AIME I Problems/Problem 9
  /* -- arbitrary values, I couldn't find nice values for pqr please replace if possible -- */ /* -- arbitrary values, I couldn't find nice values for pqr please replace if possible -- */
  4 KB (673 words) - 20:52, 28 June 2024
• 2001 AIME I Problems/Problem 5
  ...e side length of the equilateral triangle is twice of this, so <math>\frac{16\sqrt{3}}{13}</math>. ...n=\dfrac{4\sqrt{3}}{13}</math> which means the side length is <math>\dfrac{16\sqrt{3}}{13}=\sqrt{\dfrac{768}{169}}.</math>
  6 KB (1,043 words) - 10:09, 15 January 2024
• 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 (1,461 words) - 15:09, 18 August 2023
• 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

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