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• 2005 AIME I Problems/Problem 15
  ...ength of the median be <math>3m</math>. Then by two applications of the [[Power of a Point Theorem]], <math>DE^2 = 2m \cdot m = AF^2</math>, so <math>DE = Our earlier result from Power of a Point was that <math>2m^2 = (10 - c)^2</math>, so we combine these two
  5 KB (906 words) - 23:15, 6 January 2024
• 2004 AIME I Problems/Problem 14
  Note that by Power of a Point, the point the unicorn is at has power <math>4 \cdot 20 = 80</math> which implies that the tangent from that point
  4 KB (729 words) - 01:00, 27 November 2022
• 2004 AIME II Problems/Problem 10
  ...so if we raise <math>2^{3}</math>, which we know already works, to an odd power, we will also satisfy the congruence. Thus, <math>2^{3}, 2^{9}, 2^{15},</ma
  8 KB (1,283 words) - 19:19, 8 May 2024
• 2004 AIME II Problems/Problem 8
  For a single power of 2004, we have three choices (2, 3, and 167) to give a power of 2003 to. ...501, there are three choices to give a power of 500 to and the rest get a power of 1.
  2 KB (353 words) - 18:08, 25 November 2023
• 1998 AIME Problems
  ...re <math>i^2 = - 1.</math> Let <math>S_n</math> be the sum of the complex power sums of all nonempty [[subset]]s of <math>\{1,2,\ldots,n\}.</math> Given t
  7 KB (1,084 words) - 02:01, 28 November 2023
• 1983 AIME Problems/Problem 1
  ...w</math>, and <math>(xyz)^{12}=w</math>. If we now convert everything to a power of <math>120</math>, it will be easy to isolate <math>z</math> and <math>w< ...ng both sides of <math>y^{40}=w</math> to the <math>\frac{12}{40}</math>th power gives <math>y^{12}=w^{\frac{3}{10}}</math>.
  4 KB (642 words) - 03:14, 17 August 2022
• 1983 AIME Problems/Problem 9
  The derivative of <math>f(y)</math>, using the Power Rule, is
  4 KB (722 words) - 20:25, 14 January 2023
• 1983 AIME Problems/Problem 12
  Since H is the midpoint of <math>CD</math>, by [[Power of a Point]], <math>CH^2=(AH)(BH)</math>. Because <math>AH=r-OH</math> and
  2 KB (412 words) - 18:23, 1 January 2024
• 1983 AIME Problems/Problem 14
  Thus by Power of a Point in the circle passing through <math>Q</math>, <math>R</math>, an
  13 KB (2,151 words) - 17:48, 27 May 2024
• 1983 AIME Problems/Problem 15
  Suppose that <math>\Delta > 0</math>, which would mean that there could be two real roots of <math>f(x)</math>, one lying in the i ...which does not make sense in the original problem statement. (For it would mean that the point <math>A</math> lies in the half-plane above the line <math>3
  20 KB (3,497 words) - 15:37, 27 May 2024
• 1995 IMO Problems/Problem 2
  By the [[Power Mean Inequality]],
  6 KB (1,121 words) - 05:32, 1 June 2024
• 1985 AIME Problems/Problem 13
  ...ath>n</math>, we can multiply the left integer, <math>100+n^2</math>, by a power of two without affecting the greatest common divisor. Since the <math>n^2</
  4 KB (671 words) - 20:04, 6 March 2024
• 1985 AIME Problems/Problem 7
  ...[perfect power | perfect fourth power]], <math>b</math> is a perfect fifth power, <math>c</math> is a [[perfect square]] and <math>d</math> is a [[perfect c
  1 KB (222 words) - 11:04, 4 November 2022
• 1986 AIME Problems/Problem 11
  We want the coefficient of the <math>y^2</math> term of each power of each binomial, which by the binomial theorem is <math>{2\choose 2} + {3\ ...c{d^2f}{dx^2} = 2\cdot 1 - 3\cdot 2x+\cdots-17\cdot 16x^{15}</math> by the power rule.
  6 KB (872 words) - 16:51, 9 June 2023
• 1986 AIME Problems/Problem 8
  ...question asks for proper divisors, we exclude <math>2^65^6</math>, so each power is actually <math>141</math> times. The answer is thus <math>S = \log 2^{14
  3 KB (487 words) - 20:52, 16 September 2020
• 1986 AIME Problems/Problem 7
  ...math>k</math>th term after the <math>n</math>th power of 3 is equal to the power plus the <math>k</math>th term in the entire sequence. Thus, the <math>100< ...= <math>2187</math>. Writing out more terms of the sequence until the next power of 3 again (81) we can see that the (<math>2^n</math>+<math>2^{n+1}</math>)
  5 KB (866 words) - 00:00, 22 December 2022
• 1987 AIME Problems/Problem 11
  ...of the two factors will be a power of three, and the other will be twice a power of three. <math>(2n + m + 1)</math> will represent the greater factor while
  3 KB (418 words) - 18:30, 20 January 2024
• 1989 AIME Problems/Problem 9
  ...>'s units digit is <math>0, 2, 4, 6,</math> or <math>8.</math> When to the power of <math>5,</math> they each give <math>0, 2, 4, 6,</math> and <math>8</mat
  6 KB (874 words) - 15:50, 20 January 2024
• 1990 AIME Problems/Problem 13
  '''Lemma''': For all positive integers n, there's exactly one n-digit power of 9 that does not have a left-most digit 9 ...ove by contradiction that there must be at least either one or two n-digit power of 9 for all n.
  5 KB (762 words) - 01:18, 10 February 2023
• 1990 AIME Problems/Problem 5
  ...<math>5</math>, the greatest of all the factors, to be raised to the least power. Therefore, <math>n = 2^43^45^2</math> and <math>\frac{n}{75} = \frac{2^43^
  1 KB (175 words) - 03:45, 21 January 2023

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