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• 2000 AIME II Problems/Problem 9
  There are other ways we can come to this conclusion. Note that if <math>z</math> is on the [[unit circle]] in the complex plane, then <math>z <math>r</math> must be <math>1</math> (or else if you take the magnitude would not be the same). Therefore, <math>z=e^{i\frac
  4 KB (675 words) - 13:42, 4 April 2024
• 2000 AIME II Problems/Problem 7
  ..., \frac{15}{5}, \frac{14}{6},</math> etc. until <math>\frac{11}{9}</math>. If you were taking the test under normal time conditions, it shouldn't be too
  2 KB (281 words) - 12:09, 5 April 2024
• 2000 AIME II Problems/Problem 4
  ...dots p_k^{e_k}</math> is <math>(e_1 + 1)(e_2 + 1) \cdots (e_k + 1)</math>. If a number has <math>18 = 2 \cdot 3 \cdot 3</math> factors, then it can have ...he number <math>n</math> has two even factors for every odd factor. Thus, if <math>x</math> is an odd factor of <math>n</math>, then <math>2x</math> and
  2 KB (397 words) - 15:55, 11 May 2022
• 2000 AIME II Problems/Problem 3
  ...not removed, which can happen in <math>9{4 \choose 2} = 54</math> ways, or if the two cards are the remaining two cards of the number that was removed, w
  1 KB (191 words) - 04:27, 4 November 2022
• 2005 AMC 10B Problems/Problem 14
  Note: Even if you don't know the value of <math>\sin 120^\circ</math>, you can use the fa
  5 KB (882 words) - 22:12, 30 April 2024
• 2005 AMC 10B Problems/Problem 20
  We first look at how many times each number will appear in each slot. If we fix a number in a slot, then there are <math> 4! = 24 </math> ways to ar
  2 KB (356 words) - 19:15, 17 September 2023
• 2005 AMC 10B Problems/Problem 24
  ...h>33^2(a-b) = m^2</math>, so <math>a-b</math> is a perfect square, but the only perfect squares that are within our bound on <math>a-b</math> are <math>1</ ...of the equation is a perfect square because <math>m</math> is an integer. If we factor <math>99</math> into its prime factors, we get <math>3^2\cdot 11<
  5 KB (845 words) - 19:23, 17 September 2023
• 2005 AMC 10B Problems/Problem 25
  ...nce the numbers <math>1</math> to <math>24</math> are in the set there are only the numbers <math>25</math> to <math>100</math> to consider. As <math>62+63 ==Solution 2 (If you have no time)==
  3 KB (517 words) - 19:15, 15 October 2023
• 2006 USAMO Problems/Problem 1
  if and only if <math>s </math> is not a divisor of <math>p-1 </math>.
  3 KB (506 words) - 17:54, 22 June 2023
• 2006 USAMO Problems/Problem 3
  <math>f</math> by multiplying by <math>-1</math>, if necessary, to ensure that the A theorem of T.&nbsp;Nagell implies that if <math>d(g_i) \ge 2</math> the ratio
  9 KB (1,699 words) - 13:48, 11 April 2020
• 2006 USAMO Problems/Problem 5
  ...h> is the largest power of 2 that is a factor of <math>n</math>. Show that if <math>k\ge 2</math> is a positive integer and <math>i</math> is a nonnegati ...longest jump made in generating <math>J_{i_0,k_0}</math>. Such a jump can only be made from a number that is divisible by <math>2^M</math> (and by no high
  7 KB (1,280 words) - 17:23, 26 March 2016
• 2006 USAMO Problems/Problem 4
  ...a sum and product of <math>n</math>. For <math>p_1+p_2=n</math>, which is only possible in one case, <math>n=4</math>, we consider <math>p_1=p_2=2</math>. ...a_3=\frac{1}{2}</math> and let the rest of the <math>a_k</math> be 1. The only numbers we now need to check are those such that <math>\frac{n}{2}+4+\frac{
  3 KB (486 words) - 22:43, 5 August 2014
• 2006 USAMO Problems/Problem 6
  ...ath> and <math>z_4</math> in the complex plane lie on a circle if and only if the cross-ratio
  5 KB (986 words) - 22:46, 18 May 2015
• 2006 AMC 12A Problems/Problem 8
  ...average (which could be a fraction) must be a divisor of <math>15</math>. If the number of integers in the list is odd, then the average must be either ...e list is even, then the average will have a <math>\frac{1}{2}</math>. The only possibility is <math>\frac{15}{2}</math>, from which we get:
  3 KB (450 words) - 02:00, 13 January 2024
• 2006 AMC 12A Problems/Problem 9
  ...ath>e = 16</math>. This contradicts the pencil being more expensive. The only remaining value for <math>p</math> is 7; then the 13 pencils cost <math>7 \ ...</math> pencils, we can choose numbers to be the value for the pencil, and if the remainder to <math>1.00</math> is divisible by 3, we know our answer is
  3 KB (429 words) - 18:14, 26 September 2020
• Incircle
  ...ath>(\dfrac{aA_x+bB_x+cC_x}{a+b+c}, \dfrac{aA_y+bB_y+cC_y}{a+b+c})</math>, if the coordinates of each vertex are <math>A(A_x, A_y)</math>, <math>B(B_x, B
  2 KB (384 words) - 18:38, 9 March 2023
• Hypotenuse
  ...of the [[triangle]] is the center of the circle. The converse also holds: if the length of the median of <math>\triangle ABC</math> from <math>C</math>
  810 bytes (133 words) - 19:02, 15 October 2018
• Negative number
  ...integer]], [[rational number]] or [[irrational number]]) is '''negative''' if it is less than [[zero (constant) | zero]].
  635 bytes (93 words) - 14:07, 16 January 2023
• Inradius
  ...ABC</math> is <math>rs</math>. This formula holds true for other polygons if the incircle exists.
  2 KB (336 words) - 00:44, 23 April 2024
• Centroid
  ...'''centroid''' of a plane figure is, roughly speaking, its center of mass. If the plane figure is cut out from uniform cardboard, say, and you connected
  3 KB (612 words) - 22:32, 25 February 2024

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