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• 1983 AIME Problems
  pair O=(0,0), A=r*dir(45),B=(A.x,A.y-r); path P=circle(O,r);
  7 KB (1,104 words) - 03:13, 27 May 2024
• 1992 AIME Problems
  ..., <math>BB'</math>, and <math>CC'</math> are concurrent at the point <math>O^{}_{}</math>, and that <math>\frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{
  8 KB (1,117 words) - 05:32, 11 November 2023
• 1996 AIME Problems
  In parallelogram <math>ABCD,</math> let <math>O</math> be the intersection of diagonals <math>\overline{AC}</math> and <mat
  6 KB (931 words) - 17:49, 21 December 2018
• 2000 AIME II Problems
  ..., <math>BC=14</math>, <math>CA=15</math>, and that the distance from <math>O</math> to triangle <math>ABC</math> is <math>\frac{m\sqrt{n}}k</math>, wher
  6 KB (947 words) - 21:11, 19 February 2019
• 2002 AIME II Problems
  ...th> is a right angle. A circle of radius <math>19</math> with center <math>O</math> on <math>\overline{AP}</math> is drawn so that it is tangent to <mat
  7 KB (1,177 words) - 15:42, 11 August 2023
• 1983 AIME Problems/Problem 4
  pair O=(0,0), path P=circle(O,r);
  11 KB (1,741 words) - 22:40, 23 November 2023
• 1983 AIME Problems/Problem 12
  ...distance from their intersection point <math>H</math> to the center <math>O</math> is a positive rational number. Determine the length of <math>AB</mat label("O",(0,0),NE);</asy>
  2 KB (412 words) - 18:23, 1 January 2024
• 1983 AIME Problems/Problem 15
  If we take <math>O</math> to be the center of the given circle, then this means that <math>OD< pair O = (0,0), D = (0, 5), B = (-3, 4), C = (3, 4), A = (-4, 3), EE = (4, 3);
  20 KB (3,497 words) - 15:37, 27 May 2024
• 1984 AIME Problems/Problem 11
  ...om{8}{5}</math>. But then, we can rearrange the <math>M</math>'s and <math>O</math>'s in <math>7!/(3!4!)=\binom{7}{3}</math> ways. So then there are <ma
  7 KB (1,115 words) - 00:52, 7 September 2023
• 1984 AIME Problems/Problem 13
  On the coordinate plane, let <math>O=(0,0)</math>, <math>A_1=(3,0)</math>, <math>A_2=(3,1)</math>, <math>B_1=(21
  3 KB (473 words) - 12:06, 18 December 2018
• 1985 AIME Problems/Problem 12
  ...\frac{1}{3}O_{k-2}</cmath>. Substituting this into our equation for <math>O</math>, we have that <cmath>O_n = \frac{1}{3}O_{n-2} + \frac{2}{3}O_{n-1}</
  19 KB (3,128 words) - 00:53, 22 June 2024
• 1985 AIME Problems/Problem 9
  pair O = (0,0), A = r*expi(pi/3); D(CR(O,r));
  5 KB (763 words) - 16:20, 28 September 2019
• 1989 AIME Problems/Problem 10
  ...c</math> are legs of right triangle <math>abc</math> with <math>\beta = 90^o</math> and <math>c=1</math>
  8 KB (1,401 words) - 21:41, 20 January 2024
• 1990 AIME Problems/Problem 14
  triple O=(0,0,0); triple O=(0,0,0);
  7 KB (1,086 words) - 08:16, 29 July 2023
• 1991 AIME Problems/Problem 12
  ...)",S,W); label("\(15\)",B/2+P/2,N);label("\(20\)",B/2+Q/2,E);label("\(O\)",O,SW); </asy></center> ...h>\triangle APS \cong \triangle CRQ</math>). Quickly we realize that <math>O</math> is also the center of the rectangle.
  8 KB (1,270 words) - 23:36, 27 August 2023
• 1991 AIME Problems/Problem 11
  ...the center of the dodecagon, which we denote <math>A, M,</math> and <math>O</math> respectively. Notice that <math>OM=1</math>, and that <math>\triangl
  4 KB (740 words) - 17:46, 24 May 2024
• 1991 AIME Problems/Problem 9
  ...ot \beta = \frac{(o+h)(a+h)}{oa} = \frac{oa +oh +ha +h^2}{oa} = 1+ \frac{h(o+a+h)}{oa} = 1+ \alpha + \beta</math>. From the information provided in the
  10 KB (1,590 words) - 14:04, 20 January 2023
• 1992 AIME Problems/Problem 14
  ..., <math>BB'</math>, and <math>CC'</math> are concurrent at the point <math>O^{}_{}</math>, and that <math>\frac{AO^{}_{}}{OA'}+\frac{BO}{OB'}+\frac{CO}{ ...bove solutions, find <math>\sum_{cyc} \frac{y+z}{x}=92</math> (where <math>O=(x:y:z)</math> in barycentric coordinates). Now letting <math>y=z=1</math>
  4 KB (667 words) - 01:26, 16 August 2023
• 2006 AMC 10B Problems/Problem 19
  A circle of radius <math>2</math> is centered at <math>O</math>. Square <math>OABC</math> has side length <math>1</math>. Sides <mat pair O=origin, A=(1,0), C=(0,1), B=(1,1), D=(1, sqrt(3)), E=(sqrt(3), 1), point=B;
  5 KB (873 words) - 15:39, 29 May 2023
• 2006 AMC 10B Problems/Problem 24
  .... Points <math>A</math> and <math>B</math> on the circle with center <math>O</math> and points <math>C</math> and <math>D</math> on the circle with cent pair X=(-6,0), O=origin, P=(6,0), B=tangent(X, O, 2, 1), A=tangent(X, O, 2, 2), C=tangent(X, P, 4, 1), D=tangent(X, P, 4, 2);
  4 KB (558 words) - 14:38, 6 April 2024

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