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• 2005 AIME II Problems/Problem 12
  ...s on the perpendicular bisector of <math>EF</math>, which means that <math>EJ=JF=200</math>, <math>EG=GF=200\sqrt{2}</math> and <math>GH=250</math>. Now,
  13 KB (2,080 words) - 21:20, 11 December 2022
• 2005 AIME II Problems/Problem 14
  ...13}</math>. Now we see that in <math>\triangle{EJA}</math>, assuming <math>EJ=6x;AJ=13x</math> since <math>\tan\angle{JAE}=\tan\angle{DAN}</math>. Now we
  13 KB (2,129 words) - 18:56, 1 January 2024
• 2001 AIME II Problems
  ...}</math>, and <math>\overline{EC}</math>, respectively, so that <math>EI = EJ = EK = 2</math>. A solid <math>S</math> is obtained by drilling a tunnel th
  8 KB (1,282 words) - 21:12, 19 February 2019
• 2001 AIME II Problems/Problem 15
  ...}</math>, and <math>\overline{EC}</math>, respectively, so that <math>EI = EJ = EK = 2</math>. A solid <math>S</math> is obtained by drilling a tunnel th
  4 KB (518 words) - 15:01, 31 December 2021
• 2002 AMC 10A Problems/Problem 20
  ...Therefore, <math>\frac{CH}{EJ} = \left(\frac{CH}{AG}\right)\div\left(\frac{EJ}{AG}\right) = \left(\frac{1}{3}\right)\div\left(\frac{1}{5}\right) = \boxed
  3 KB (439 words) - 22:15, 9 June 2023
• 2014 AMC 12B Problems/Problem 21
  ...3} \cdot KE=\frac{2\sqrt{3}-3}{2}</math>, which gives <math>BE=\frac{1}{2}-EJ=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}</math>, so the answer is <math>\textbf{(C) <cmath>EJ = x\sqrt{1-y^2}</cmath>
  8 KB (1,307 words) - 02:07, 1 January 2023
• 2018 AMC 8 Problems/Problem 24
  Notice that <math>CI=CJ=EI=EJ=\sqrt{5}</math> so <math>EJCI</math> is a rhombus. Furthermore, by the dist
  10 KB (1,684 words) - 13:25, 14 January 2024
• 2019 AIME I Problems/Problem 13
  ...ngle{EFJ}</math>, getting that <math>\frac{CB}{EF}=\frac{BJ}{FJ}=\frac{CJ}{EJ}=\frac{5}{7}</math> Using Power of Point, we can get that <math>BJ * EJ=CJ*FJ; DJ * AJ=CJ * FJ</math> Assume that <math>DJ=5k,CJ=15k</math>, gettin
  10 KB (1,620 words) - 20:44, 20 December 2023
• 2023 USAMO Problems/Problem 6
  ...</math> is a parallelogram with center <math>J</math>, so <math>\tfrac{EJ}{EJ'}=\tfrac{1}{2}</math>. Similarly, we get that if line <math>\overline{E'K}<
  14 KB (2,600 words) - 23:37, 10 March 2024
• 2015 OIM Problems/Problem 4
  <cmath>\frac{FK}{KD}=\frac{EJ}{JD}</cmath>
  694 bytes (121 words) - 15:05, 14 December 2023