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• 1996 AIME Problems/Problem 10
  ...h> <math>\dfrac{2\sin{141^{\circ}}\cos{45^{\circ}}}{2\cos{141^{\circ}}\sin{45^{\circ}}} = \tan{141^{\circ}}</math>. ...]] function is <math>180^\circ</math>, and the tangent function is [[one-to-one]] over each period of its domain.
  4 KB (503 words) - 15:46, 3 August 2022
• 1996 AIME Problems/Problem 3
  ...itive [[integer]] <math>n</math> for which the expansion of <math>(xy-3x+7y-21)^n</math>, after like terms have been collected, has at least 1996 terms. ...fter <math>1996</math> is <math>2025 = 45^2</math>, so our answer is <math>45 - 1 = \boxed{044}</math>.
  3 KB (515 words) - 04:29, 27 November 2023
• 1997 AIME Problems/Problem 11
  ...la <math>\cos x + \cos y = 2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)</math> ...+\cos(\frac{41}{2})+\dots+\cos(\frac{1}{2})]} \Rightarrow \frac{\cos(\frac{45}{2})}{\cos(\frac{135}{2})}</cmath>
  10 KB (1,512 words) - 17:16, 18 June 2024
• 1997 AIME Problems/Problem 4
  [[Image:1997_AIME-4.png]] \sqrt{10r+r^2}&=& 4-r\\
  2 KB (354 words) - 22:33, 2 February 2021
• 1998 AIME Problems/Problem 11
  ...font-size:100%">"For non-asymptote version of image, see [[:Image:1998_AIME-11.png]]"</span> ...[hypotenuse]]s are <math>5\sqrt{5}</math>. The other two are of <math>45-45-90 \triangle</math>s with legs of length 15, so their hypotenuses are <math>
  7 KB (1,084 words) - 11:48, 13 August 2023
• 1998 AIME Problems/Problem 10
  [[Image:1998_AIME-10a.png|450px]] [[Image:1998_AIME-10b.png|450px]]
  3 KB (496 words) - 13:02, 5 August 2019
• 1998 AIME Problems/Problem 9
  ...inutes past 9 a.m.). The two mathematicians meet each other when <math>|M_1-M_2| \leq m</math>. Also because the mathematicians arrive between 9 and 10, real m=60-12*sqrt(15);
  4 KB (624 words) - 19:00, 19 June 2024
• 1999 AIME Problems/Problem 10
  ...lso be picked. Since the triangle accounts for 3 segments, there are <math>45 - 3 = 42</math> segments remaining. ...se3} \cdot 42}{{45\choose4}} = \frac{10 \cdot 9 \cdot 8 \cdot 42 \cdot 4!}{45 \cdot 44 \cdot 43 \cdot 42 \cdot 3!} = \frac{16}{473}</math>. The solution
  3 KB (524 words) - 17:25, 17 July 2023
• 1999 AIME Problems/Problem 4
  pair W=dir(225), X=dir(315), Y=dir(45), Z=dir(135), O=origin; {{AIME box|year=1999|num-b=3|num-a=5}}
  3 KB (398 words) - 13:27, 12 December 2020
• 1999 AIME Problems/Problem 2
  Consider the [[parallelogram]] with [[vertex|vertices]] <math>(10,45)</math>, <math>(10,114)</math>, <math>(28,153)</math>, and <math>(28,84)</m ...The slope of the line (since it passes through the origin) is <math>\frac{45 + \frac{135}{19}}{10} = \frac{99}{19}</math>, and the solution is <math>m +
  3 KB (423 words) - 11:06, 27 April 2023
• 1999 AIME Problems/Problem 8
  [[Image:1999_AIME-8.png]] [[Image:1999_AIME-8a.png]]
  3 KB (445 words) - 19:40, 4 July 2013
• 2000 AIME I Problems/Problem 4
  The diagram shows a [[rectangle]] that has been dissected into nine non-overlapping [[square]]s. Given that the width and the height of the rectangl draw((34,36)--(34,45)--(25,45));
  3 KB (485 words) - 00:31, 19 January 2024
• 2001 AIME I Problems/Problem 9
  ...=(0,0),B=(13,0),C=IP(CR(A,17),CR(B,15)), D=A+p*(B-A), E=B+q*(C-B), F=C+r*(A-C); ...ot \sin \angle CAB}{\frac 12 \cdot AB \cdot AC \cdot \sin \angle CAB} = p(1-r)
  4 KB (673 words) - 20:52, 28 June 2024
• 2001 AIME I Problems/Problem 6
  Recast the problem entirely as a block-walking problem. Call the respective dice <math>a, b, c, d</math>. In the ...combination]] of four numbers, there is only one way to order them in a non-decreasing order. It suffices now to find the number of combinations for fou
  11 KB (1,729 words) - 20:50, 28 November 2023
• 2001 AIME I Problems/Problem 4
  ...math>A</math> and <math>B</math> measure <math>60</math> degrees and <math>45</math> degrees, respectively. The bisector of angle <math>A</math> intersec <math>\angle ABC=45^{\circ}</math> and <math>\angle BAC=60^{\circ}</math>, so <math>BC = 12\sqr
  3 KB (534 words) - 03:22, 23 January 2023
• 2001 AIME I Problems/Problem 1
  Find the sum of all positive two-digit integers that are divisible by each of their digits. ...r expand on solution 2, it would be tedious to test all <math>90</math> two-digit numbers. We can reduce the amount to look at by focusing on the tens d
  4 KB (687 words) - 18:37, 27 November 2022
• 2002 AIME I Problems/Problem 14
  ...>. After any element <math>x</math> is removed, we are given that <math>n|N-x</math>, so <math>x\equiv N\pmod{n}</math>. Since <math>1\in\mathcal{S}</ma ...02</math>, so <math>n \leq 44</math>. The largest factor of 2001 less than 45 is 29, so <math>n=29</math> and <math>n+1</math> <math>\Rightarrow{\fbox{30
  2 KB (267 words) - 19:18, 21 June 2021
• 2002 AIME I Problems/Problem 1
  ...ree-digit arrangement that reads the same left-to-right as it does right-to-left) is <math>\dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> ...larly, there is a <math>\frac 1{26}</math> probability of picking the three-letter palindrome.
  3 KB (369 words) - 23:36, 6 January 2024
• 2003 AIME I Problems/Problem 11
  ...ility is symmetric around <math>45^\circ</math>. Thus, take <math>0 < x < 45</math> so that <math>\sin x < \cos x</math>. Then <math>\cos^2 x</math> is ...math>\cos 2x > \frac 12 \sin 2x</math>. Since we've chosen <math>x \in (0, 45)</math>, <math>\cos 2x > 0</math> so
  5 KB (895 words) - 15:24, 21 June 2024
• 2003 AIME II Problems/Problem 15
  ...that <math> \left( x^{23} + x^{22} + \cdots + x^2 + x + 1 \right) \cdot (x-1) = x^{24} - 1 </math>. The five-element sum is just <math>\sin 30^\circ + \sin 60^\circ + \sin 90^\circ + \s
  4 KB (675 words) - 17:23, 30 July 2022

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