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• 2005 AIME II Problems/Problem 12
  ...BG</math>). Then <math>\tan \angle EOG = \frac{x}{450}</math>, and <math>\tan \angle FOG = \frac{y}{450}</math>. ...frac{y}{450}}{1 - \frac{x}{450} \cdot \frac{y}{450}}.</cmath> Since <math>\tan 45 = 1</math>, this simplifies to <math>1 - \frac{xy}{450^2} = \frac{x + y}
  13 KB (2,080 words) - 21:20, 11 December 2022
• 1991 AIME Problems/Problem 11
  for (int i=0; i<12; ++i) dot((cos(i*pi/6), sin(i*pi/6)));
  4 KB (740 words) - 19:33, 28 December 2022
• 1999 AIME Problems/Problem 11
  ...</math> are relatively prime positive integers that satisfy <math>\frac mn<90,</math> find <math>m+n.</math> ...manipulate this sum by wrapping the terms around (since the first half is equal to the second half), but it quickly becomes apparent that this way is diffi
  4 KB (614 words) - 04:38, 8 December 2023
• Mock AIME 1 2006-2007/Problems
  ...angle BAC</math> hits <math>BC</math> at <math>D</math>. If <math>\angle C=90^\circ</math>, and the maximum value of <math>\frac{[ABD]}{[ACD]}=\frac{m}{n .../math>. <math>\lfloor x\rfloor</math> is the greatest integer less than or equal to <math>x</math>. If for fixed <math>n</math>, there exists an integer <ma
  8 KB (1,355 words) - 14:54, 21 August 2020
• Euler line
  <i><b>Acute triangle</b></i> ...vec BD}{\vec DA} = n = \frac {\tan \alpha – \tan \gamma}{\tan \beta – \tan \gamma} > 0.</math>
  59 KB (10,203 words) - 04:47, 30 August 2023
• 1985 IMO Problems/Problem 1
  ...</math> is clearly a cyclic quadrilateral, the angle <math>COE </math> is equal to half the angle <math>GAO </math>. Then \begin{matrix} {CE} & = & r \tan(COE) \\
  4 KB (684 words) - 07:28, 3 October 2021
• 2007 AIME I Problems/Problem 9
  ...n the exterior of the triangle and are the centers of two [[circle]]s with equal [[radius|radii]]. The circle with center <math>O_1</math> is tangent to th [[Image:AIME I 2007-9.png]]
  11 KB (1,851 words) - 12:31, 21 December 2021
• 2007 AIME I Problems/Problem 12
  ...their [[complement]]s are <math>15^{\circ}</math>, and <math>\angle DAB' = 90 - 2(15) = 60^{\circ}</math>. We know that <math>\angle DB'A = 75^{\circ}</m ...h>30-60-90</math> triangle is <math>10</math>, the base of the <math>45-45-90</math> is <math>10\sqrt{3}</math>, and their common height is <math>10\sqrt
  10 KB (1,458 words) - 20:50, 3 November 2023
• 2007 AIME II Problems/Problem 15
  Q=r*(rotate(-90)*A/length(A)); X=extension(A,P,Q,Q+A); Y=extension(B,P,Q,Q+A); R=rotate(90,C)*(C+r*(A-C)/length(A-C)); Z=extension(C,P,R,R+A-C);
  11 KB (2,099 words) - 17:51, 4 January 2024
• LaTeX:Commands
  | <math>\textstyle 2^{23}</math>||2^{23}||<math>\textstyle n_{i-1}</math>||n_{i-1} | <math>a^{i+1}_3</math>||a^{i+1}_3||<math>x^{3^2}</math>||x^{3^2}
  12 KB (1,898 words) - 15:31, 22 February 2024
• 1999 AHSME Problems
  I. The digit is 1. <math> \mathrm{(A) \ I\ is\ true} \qquad \mathrm{(B) \ I\ is\ false} \qquad \mathrm{(C) \ II\ is\ true} \qquad \mathrm{(D) \ III\ is
  13 KB (1,945 words) - 18:28, 19 June 2023
• 2008 AMC 12A Problems
  Which of the following is equal to the [[product]] ...rs <math>15\%</math> off the sticker price followed by a <math>\textdollar 90</math> rebate, and store <math>B</math> offers <math>25\%</math> off the sa
  13 KB (2,025 words) - 13:56, 2 February 2021
• 2008 AMC 12B Problems/Problem 24
  ...he resulting heights <math>h_n</math> and <math>h_{n+1}</math>. From 30-60-90 rules, the distance between the points where these altitudes meet the x-axi ...th> coordinates are the squares of the heights). Setting these expressions equal and dividing throughout by <math>h_{n+1}+h_n</math> leaves <math>h_{n+1}-h_
  9 KB (1,482 words) - 13:52, 4 April 2024
• 2008 AMC 10B Problems/Problem 24
  ...rm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95</math> ...</math> equals <math>55</math> degrees, thus making angle <math>AYB</math> equal to <math>60</math> degrees. We can also find out that angle <math>CYB</math
  12 KB (1,944 words) - 17:15, 20 January 2024
• 2008 iTest Problems
  for(int i=0;i<=5;i=i+1) draw(circle(7/2+d*i,3/2));
  71 KB (11,749 words) - 01:31, 2 November 2023
• 2009 AIME I Problems/Problem 12
  ...> be the circle having <math>\overline{CD}</math> as a diameter. Let <math>I</math> be a point outside <math>\triangle ABC</math> such that <math>\overl ...length <math>x</math>. Then the perimeter of <math>\triangle ABI</math> is equal to <cmath>2(x+AD+DB)=2(x+37).</cmath> It remains to compute <math>\dfrac{2(
  12 KB (1,970 words) - 22:53, 22 January 2024
• 2009 AIME I Problems/Problem 15
  ...t{3})=14-7\sqrt3</math>. Then, the area of [<math>\triangle BPC</math>] is equal to <math>7 \cdot OP=98-49\sqrt{3}</math>, and so the answer is <math>98+49+ {{AIME box|year=2009|n=I|num-b=14|after=Last Question}}
  6 KB (1,048 words) - 19:35, 2 January 2023
• Mock AIME 2 2010 Answer Key
  Notice that removing as many lemmings as possible at each step (i.e., using the greedy algorithm) is optimal. This can be easily proved throu ...{10})</math>, with <math>0 \le b_i \le 2, 3|a_i-b_i</math> for <math>1 \le i \le 10</math>. Given that <math>f</math> can take on <math>K</math> distinc
  36 KB (6,214 words) - 20:22, 13 July 2023
• 2011 AIME I Problems/Problem 14
  ...A_6}</math>, and <math>\overline{A_7 A_8}</math>, respectively. For <math>i = 1, 3, 5, 7</math>, ray <math>R_i</math> is constructed from <math>M_i</ma label("$M_1$",M,dir(90));
  8 KB (1,344 words) - 18:39, 9 February 2023
• 2011 USAJMO Problems/Problem 3
  ...th> is on the circumcircle of <math>\Delta XYZ</math>, and similarly <math>I</math> is on the circumcircle of <math>\Delta XYZ</math>. Therefore <math>\ ...th>HM=HB/2=HZ/2</math>. Therefore <math>\Delta HMZ</math> is a <math>30-60-90</math> right triangle and so <math>\angle ZHB=60^{\circ}</math>. So <math>\
  15 KB (2,593 words) - 13:37, 29 January 2021

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