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• 2003 AIME II Problems/Problem 8
  If you multiply the corresponding terms of two arithmetic sequences, you get t
  5 KB (793 words) - 15:18, 14 July 2023
• 2003 AIME II Problems/Problem 4
  ...on <math>ABD</math>. Clearly, the two regular tetrahedrons are similar, so if we can find the ratio of the sides, we can find the ratio of the volumes. L
  3 KB (563 words) - 17:36, 30 July 2022
• 2002 AIME II Problems/Problem 15
  .... Let <math>\tan{\frac{\theta}{2}}=m_1</math>, for convenience. Therefore if <math>(x,y)</math> is on the angle bisector, then <math>x=\frac{y}{m_1}</ma
  7 KB (1,182 words) - 09:56, 7 February 2022
• 2002 AIME II Problems/Problem 12
  ...iangle type method below, computing the number of paths to each point that only move right and up. ...ed. The first two shots must fail, and the last shot must succeed. Thus, only slots 3-9 need to be counted, and can be abbreviated as follows:
  7 KB (1,127 words) - 13:34, 19 June 2022
• 2002 AIME II Problems/Problem 9
  Let <math>A</math> and <math>B</math> be the disjoint subsets. If <math>A</math> has <math>n</math> elements, then the number of elements of
  3 KB (404 words) - 23:07, 4 May 2024
• 2002 AIME II Problems/Problem 8
  Note that if <math>\frac{2002}n - \frac{2002}{n+1}\leq 1</math>, then either <math>\left ...e, <math>k=49</math> is the smallest such case. (If unsure, we could check if the result holds for <math>k=48,</math> and as it turns out, it doesn't.) T
  6 KB (908 words) - 14:22, 14 July 2023
• 2002 AIME II Problems/Problem 7
  <math>\frac{k(k+1)(2k+1)}{6}</math> is a multiple of <math>200</math> if <math>k(k+1)(2k+1)</math> is a multiple of <math>1200 = 2^4 \cdot 3 \cdot 5 Since <math>2k+1</math> is always odd, and only one of <math>k</math> and <math>k+1</math> is even, either <math>k, k+1 \eq
  3 KB (403 words) - 12:10, 9 September 2023
• 2002 AIME II Problems/Problem 4
  If <math>n=202</math>, then the area of the garden enclosed by the path, not i
  2 KB (268 words) - 07:28, 13 September 2020
• 2001 AIME II Problems/Problem 14
  ...n figure this out by adding 360 repeatedly to the number 60 to try and see if it will satisfy what we need. We realize that it does not work in the integ ...345) ^\circ</math> which is equal to <math>\boxed {840}</math> degrees. We only want the sum of a certain number of theta, not all of it.
  2 KB (380 words) - 15:03, 22 July 2018
• 2001 AIME II Problems/Problem 13
  Since <math>\angle{BAD}=\angle{ADM}</math>, if we extend AB and DC, they must meet at one point to form a isosceles triang
  4 KB (743 words) - 03:32, 23 January 2023
• 2001 AIME II Problems/Problem 12
  ...ng the midpoint triangle provide the other <math>3</math>. This is because if you read this question carefully, it asks to add new tetrahedra to each fac
  2 KB (380 words) - 00:28, 5 June 2020
• 2001 AIME II Problems/Problem 10
  ...solutions of <math>10^6 - 10^0, 10^7 - 10^1, \dots\implies 94</math> ways. If <math>j - i = 12</math>, we can have the solutions of <math>10^{12} - 10^{0 ...\dbinom{17}{2} = 544</math>. However, if <math>n = 4, 5</math>, there are only <math>16</math> choices, giving us <math>2 \cdot \dbinom{16}{2} = 240</math
  4 KB (549 words) - 23:16, 19 January 2024
• 2001 AIME II Problems/Problem 9
  ...of the squares, so 32 ways to do that. For the second case, there will be only two ways to pick two squares, and <math>2^2</math> ways to color the other :Choosing three such squares leaves only one square left, with four places to place it. This is <math>2 \cdot 4 = 8<
  8 KB (1,207 words) - 20:04, 5 September 2023
• 2001 AIME II Problems/Problem 6
  ...e{CD}</math> and vertices <math>G</math> and <math>H</math> on the circle. If the area of square <math>ABCD</math> is <math>1</math>, then the area of sq
  4 KB (772 words) - 19:31, 6 December 2023
• 2001 AIME II Problems/Problem 5
  A [[set]] of positive numbers has the ''triangle property'' if it has three distinct elements that are the lengths of the sides of a [[tri If we wanted to find this for a much larger number (say 2001), we could have n
  2 KB (286 words) - 22:32, 5 January 2024
• 2001 AIME II Problems/Problem 3
  The recursive formula suggests telescoping. Indeed, if we add <math>x_n</math> and <math>x_{n-1}</math>, we have <math>x_n + x_{n-
  2 KB (300 words) - 01:28, 12 November 2022
• 2000 AIME II Problems/Problem 15
  Recall that the approximation of <math>\sin(x)</math> in radians is x if x is close to zero. In this case x is close to zero. Converting to radians ...+...+\cot(133)-\cot(134)</math>. Observe that this "almost telescopes," if only we had some extra terms. Consider adding the sequence <math>\frac{1}{\sin(4
  3 KB (469 words) - 21:14, 7 July 2022
• 2000 AIME II Problems/Problem 14
  Therefore we have <math>f_{16} = 1</math>, <math>f_k=k</math> if <math>32m\le k \le 32m+15</math> for some <math>m=1,2,\ldots,62</math>, and ...{10}</math>) the result will be an integer in base <math>n</math> composed only of the digits <math>n - 1</math> and <math>0</math> (in this example, <math
  7 KB (1,131 words) - 14:49, 6 April 2023
• 2000 AIME II Problems/Problem 13
  It would be really nice if the coefficients were symmetrical. What if we make the substitution, <math>x = -\frac{i}{\sqrt{10}}y</math>. The the p Now, if we let <math>z = y + \frac{1}{y}</math>, we can get the equations
  6 KB (1,060 words) - 17:36, 26 April 2024
• 2000 AIME II Problems/Problem 11
  ...des, and <math>\overline{AB}</math> and <math>\overline{CD}</math> are the only [[parallel]] sides. The sum of the absolute values of all possible slopes f (Note: This Solution is a lot faster if you rule out <math>(Y, Z) = (1, 7)</math> due to degeneracy.)
  4 KB (750 words) - 22:55, 5 February 2024

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