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• 2006 AMC 12A Problems/Problem 20
  The bug then has only <math>1</math> way to go to a point <math>E</math> on the opposite face, therefore having a <math>\f
  5 KB (908 words) - 19:23, 22 September 2022
• 2006 AMC 12A Problems/Problem 24
  ...nt of <math>y</math> is <math>3</math>, the exponent of <math>z</math> can go up to <math>2002</math>, so there are <math>1002</math> terms.
  8 KB (1,332 words) - 17:37, 17 September 2023
• 2000 AIME I Problems/Problem 6
  ...y. Thus we get that <math>y</math> can go up to 999 and <math>x</math> can go down to <math>3</math>, leaving <math>997</math> possibilities for <math>x,
  6 KB (966 words) - 21:48, 29 January 2024
• 2001 AIME II Problems/Problem 6
  ...</math>. Let <math>DF</math> = <math>b</math> and diameter <math>HI</math> go through <math>J</math> the midpoint of <math>EF</math>. Since a diameter al
  4 KB (772 words) - 19:31, 6 December 2023
• System of equations
  ...is to graph them in the [[plane]] and observe where they intersect. We'll go back to our same example to illustrate this.
  5 KB (784 words) - 23:27, 30 July 2020
• Euler line
  <cmath>GO^2 = R^2 - \frac {a^2 + b^2 + c^2}{9}.</cmath> <cmath>HO^2 = 9R^2 - a^2 - b^2 - c^2, GO^2 = \frac {HO^2}{9} = R^2 - \frac {a^2 + b^2 + c^2}{9}.</cmath>
  59 KB (10,203 words) - 04:47, 30 August 2023
• 2003 AMC 12A Problems/Problem 25
  ...Complex numbers are a bit harder to address than real numbers, so I won't go into depth on complex domain/range (You don't need to, as this is a 2003 qu
  9 KB (1,606 words) - 11:34, 10 July 2020
• 2007 AIME I Problems/Problem 10
  ...ces, and another fixed empty square. The remaining shaded squares can only go in 4 places, so we get <math>{4\choose2} = 6</math>. We get <math>180 \cdot ...y need to choose 2, since 1 of the shaded squares in the third column must go to the unoccupied 2-row).
  13 KB (2,328 words) - 00:12, 29 November 2023
• 2007 AIME II Problems/Problem 12
  ...math>, from which we simplify to <math>2a+7n=77</math>. From here, we just go brute force using the second statement (that <math>3^{56}\leq 3^{a}+...+3^{
  5 KB (829 words) - 12:22, 8 January 2024
• 2001 USAMO Problems/Problem 2
  ...and <math>\mathbf{H}_2(B) = M_2</math>. Under the dilation, parallel lines go to parallel lines and the intersection of two lines goes to the intersectio
  11 KB (2,091 words) - 08:35, 16 November 2017
• De Longchamps point
  <math>2\vec {HG} = \vec GL_o, 2\vec {GO} = \vec HG \implies L_o = L</math> as desired. <cmath>OK = R, HK = 2R, HG = 2GO \implies GK^2 = 2(R^2 – GO^2), GO^2 = \frac {HO^2}{9} \implies</cmath>
  10 KB (1,780 words) - 09:23, 17 November 2022
• 2008 AIME II Problems/Problem 13
  ...artesian plane over the x-axis)! If you're familiar with inversion you can go plot the inverted figure's Cartesian Plane Equivalent. Then simply continue
  6 KB (894 words) - 18:56, 25 December 2022
• 2021 GMC 10B
  Ary wants to go to the park at afternoon. He walked to halfway, and he's pretty hungry. The
  11 KB (1,695 words) - 14:33, 7 March 2022
• 1988 AJHSME Problems/Problem 4
  ...the number of black squares in each row increases by 1 continuously as we go down the pyramid.
  2 KB (354 words) - 15:22, 17 January 2023
• 2010 AIME II Problems/Problem 11
  ...h>'s, then we have two <math>1</math>'s left to place. Clearly they cannot go in the same row, because then the other row is filled with <math>0</math>'s
  6 KB (1,057 words) - 01:58, 8 January 2023
• Barycentric coordinates
  <cmath>\frac{HO}{BO} = \frac {|a-c|}{b}; \frac{GO}{BO} = \frac {|b-c|}{a}.</cmath>
  25 KB (5,067 words) - 22:15, 31 March 2024
• 1953 AHSME Problems
  ...e diameter of <math>6</math> feet, required to cause a point on the rim to go one mile is:
  21 KB (3,123 words) - 14:24, 20 February 2020
• 1955 AHSME Problems
  The length of the shortest wire that will go around them is:
  22 KB (3,509 words) - 21:29, 31 December 2023
• Mock Geometry AIME 2011 Problems/Problem 12
  ...ath>4 \cos^2{X} - 1</math> since the solution, <math>60^\circ</math> would go out of range for the triangle sum of angles. ...for <math>\cos{X}</math>, and neglecting the negative root since it would go out of bounds too, we find <math>\cos{X} = \frac{1 + \sqrt{33}}{8}</math>.
  5 KB (831 words) - 17:55, 21 July 2018
• 2013 AIME II Problems/Problem 15
  Let's draw the triangle. Since the problem only deals with angles, we can go ahead and set one of the sides to a convenient value. Let <math>BC = \sin{A
  8 KB (1,411 words) - 23:48, 11 January 2023

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