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• Isogonal conjugates
  #REDIRECT[[Isogonal conjugate]]
  31 bytes (3 words) - 16:18, 8 May 2021
• Isogonal conjugate
  '''Isogonal conjugates''' are pairs of [[point]]s in the [[plane]] with respect to a ce ==The isogonal theorem==
  54 KB (9,416 words) - 07:40, 18 April 2024

Page text matches

• 2005 AIME II Problems/Problem 14
  ==Solution 5 (Isogonal lines with respect to A angle bisesector)== ...x}{AC\sin y}</cmath> and multiplying these together proves the formula for isogonal lines. Hence, we have <cmath>\frac{BE}{15-BE}\cdot \frac{9}{6}=\frac{169}{1
  14 KB (2,340 words) - 15:38, 21 August 2024
• Geometry/Olympiad
  * [[Isogonal conjugates]] and [[Isotomic conjugates]]
  2 KB (242 words) - 09:16, 18 June 2023
• Orthocenter
  * The orthocenter and the circumcenter of a triangle are [[isogonal conjugate]]s.
  5 KB (828 words) - 13:09, 18 October 2024
• Euler line
  *[[Isogonal conjugate]]
  59 KB (10,203 words) - 03:47, 30 August 2023
• Isogonal conjugates
  #REDIRECT[[Isogonal conjugate]]
  31 bytes (3 words) - 16:18, 8 May 2021
• Isogonal conjugate
  '''Isogonal conjugates''' are pairs of [[point]]s in the [[plane]] with respect to a ce ==The isogonal theorem==
  54 KB (9,416 words) - 07:40, 18 April 2024
• 2008 USAMO Problems/Problem 2
  ''Proof.'' This is fairly easy to prove (as H, O are isogonal conjugates, plus using SAS similarity), but the author lacks time to write === Solution 6 (isogonal conjugates) ===
  20 KB (3,565 words) - 10:54, 1 May 2024
• 1995 USAMO Problems/Problem 3
  *[[Isogonal conjugate]]
  2 KB (364 words) - 00:42, 19 April 2024
• 1993 USAMO Problems/Problem 2
  E lies on the isoptic cubic of ABCD, so it has an isogonal conjugate in ABCD.
  4 KB (679 words) - 04:42, 1 October 2024
• 2013 IMO Problems/Problem 3
  By isogonal conjugacy, <math>I_aA_1,I_bB_1I_cC_1</math> concur at the Bevan point <math
  3 KB (553 words) - 08:41, 17 January 2016
• 2011 USAMO Problems/Problem 5
  ...pect to triangle <math>BCR</math>, <math>P</math> and <math>Q_1</math> are isogonal conjugates. Therefore, <math>Q_1</math> and <math>Q_2</math> lie on the ref *[[Isogonal conjugate]]
  4 KB (660 words) - 00:04, 15 February 2024
• Barycentric coordinates
  == Product of isogonal segments == [[File:Isogonal formulas.png|350px|right]]
  33 KB (6,737 words) - 16:24, 22 November 2024
• 2012 AIME II Problems/Problem 15
  .... Thus points <math>F</math> and <math>M</math> map to each other, and are isogonal. It follows that <math>AF</math> is a symmedian of <math>\triangle{ABC}</ma
  15 KB (2,516 words) - 16:28, 17 September 2024
• 2012 USAMO Problems/Problem 5
  Thus <math>B'</math> is the intersection of the isogonal of <math>B</math> with respect to <math>\angle P</math> Analogously, <math>A'</math> is the intersection of the isogonal of <math>A</math> with respect to <math>\angle P</math>
  5 KB (767 words) - 21:32, 2 May 2023
• 2015 USAJMO Problems/Problem 5
  Note that lines <math>AC, AX</math> are isogonal in <math>\triangle ABD</math>, so an inversion centered at <math>A</math> w ...th> are isogonal in <math>\triangle ABD</math> and <math>CD, CE</math> are isogonal in <math>\triangle CDB</math>. From the law of sines it follows that
  2 KB (354 words) - 09:20, 23 April 2023
• Mock AIME 3 2010 Problems
  1 be the isogonal
  14 KB (2,904 words) - 17:24, 16 May 2017
• 2017 AIME II Problems/Problem 12
  -Isogonal -Isogonal
  6 KB (908 words) - 01:35, 23 January 2024
• 2019 USAJMO Problems/Problem 3
  Since <math>P</math> is the isogonal conjugate of <math>N</math>, <math>\measuredangle PEA = \measuredangle MEC
  4 KB (654 words) - 19:13, 12 April 2021
• 2019 USAMO Problems/Problem 2
  Since <math>P</math> is the isogonal conjugate of <math>N</math>, <math>\measuredangle PEA = \measuredangle MEC
  4 KB (620 words) - 21:51, 18 October 2022
• Kimberling’s point X(24)
  ...= \angle AUF \implies</math> points <math>H</math> and <math>P</math> are isogonal conjugate with respect <math>\triangle UVW.</math>
  10 KB (1,751 words) - 14:34, 25 November 2022

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