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• Homothety
  ...ath>k</math> sends point <math>A</math> to a point <math>A' \ni HA'=k\cdot HA</math> This is denoted by <math>\mathcal{H}(H, k)</math>.
  3 KB (533 words) - 12:51, 2 September 2024
• 1984 AIME Problems/Problem 6
  ...,F</math>), we see that <math>\triangle AEM\cong \triangle CFM</math> by [[HA congruency]]; hence <math>M</math> lies on the line. The coordinates of <ma
  6 KB (1,022 words) - 18:29, 22 January 2024
• 1991 AIME Problems/Problem 9
  ...tice that <math>\alpha \cdot \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 inf
  10 KB (1,590 words) - 13:04, 20 January 2023
• Euler line
  <cmath>q = \frac {\vec {AH}}{\vec {HA'}} = \frac {\cos \alpha}{\cos \beta \cdot \cos \gamma} = \tan \beta \cdot \
  59 KB (10,203 words) - 03:47, 30 August 2023
• 2003 AMC 10A Problems/Problem 22
  ...th>. By Pythagorean theorem, <math>(HA)^2\; =\; (HB)^2\; +\; (BA)^2\;...\;HA=10</math>. <math>HA\: =\: 10 =\: \frac{2}{5}*(GA)</math>.
  9 KB (1,446 words) - 11:43, 18 September 2024
• De Longchamps point
  ...<math>\omega</math> swap <math>A</math> and <math>D \implies R_\omega^2 = HA \cdot HD.</math> Well known that <math>HA = – 2R \cos A.</math>
  10 KB (1,782 words) - 19:35, 28 September 2024
• 2009 AIME II Problems/Problem 13
  ...on from <math>O</math> to line <math>AB</math>. Then, by definition, <math>HA=HB=r\sin\left(\dfrac{\angle AOB}{2}\right)</math>. Thus, <math>AB = 2r\sin\
  8 KB (1,279 words) - 21:32, 3 October 2024
• 2011 AIME I Problems/Problem 2
  <math>HA=EF=3\sqrt{21}-12</math>
  5 KB (873 words) - 03:33, 17 December 2023
• 1980 USAMO Problems/Problem 1
  ...math>k</math>. Thus if the first object has true weight x, then <math> x = hA + k, a = hx + k</math>.
  915 bytes (172 words) - 13:46, 26 March 2023
• 1964 AHSME Problems/Problem 35
  ...is the altitude to the side of length <math>14</math>, the ratio <math>HD:HA</math> is: ...and side (<math>HD</math>) is less than the ratios' right hand side (<math>HA</math>).
  6 KB (967 words) - 11:42, 24 November 2024
• 1970 Canadian MO Problems/Problem 2
  ...h>(a,h)</math> and <math>(b,k)</math> are base-altitude pairs so <math>k = ha/b</math>. Therefore <math>a - b > k - h</math>, so <math>a + h > b + k</mat
  1 KB (236 words) - 19:32, 5 June 2018
• 1964 AHSME Problems
  ...he altitude to the side length <math>14</math>, what is the ratio <math>HD:HA</math>?
  19 KB (2,907 words) - 13:16, 20 February 2020
• University of California, Santa Barbara
  ...sity of California system. The main campus is located on a 1,022-acre (414 ha) site near Goleta, California, United States, 8 miles (13 km) from Santa Ba
  570 bytes (86 words) - 19:33, 9 September 2015
• 1964 AHSME Problems/Problem 15
  ...se <math>a</math>, so the height <math>h</math> satisfies <math>\frac{1}{2}ha = T</math>. This means <math>h = \frac{2T}{a}</math>. Becuase the triangl
  2 KB (260 words) - 20:01, 23 July 2019
• 2015 IMO Problems/Problem 3
  <math>M \longleftrightarrow Q</math>. Let <math>HA</math> and <math>HQ</math> intersect with nine-point circle <math>T</math> ...math> because <math>T</math> and <math>N</math> are the midpoints of <math>HA</math> and <math>HQ</math>, respectively.
  2 KB (422 words) - 14:02, 1 June 2024
• 2018 AMC 10B Problems/Problem 17
  ...= \frac{8 - x}{2}</math>. By the Pythagorean theorem, <math>PH^2 + PA^2 = HA^2</math>, so <math>\left(\frac{6 - x}{2} \right)^2 + \left(\frac{8 - x}{2}
  3 KB (581 words) - 19:37, 2 September 2024
• 2018 AIME II Problems
  ...side lengths <math>AB = CD = EF = GH = 10</math> and <math>BC = DE = FG = HA = 11</math> is formed by removing 6-8-10 triangles from the corners of a <m
  9 KB (1,385 words) - 23:26, 20 January 2024
• 2018 AIME I Problems/Problem 13
  <math>r(a+b+c) = ha = r_a (b+c-a)\implies </math>
  13 KB (2,200 words) - 20:36, 6 January 2024
• 2018 AIME II Problems/Problem 9
  ...side lengths <math>AB = CD = EF = GH = 10</math> and <math>BC = DE = FG = HA = 11</math> is formed by removing 6-8-10 triangles from the corners of a <m
  5 KB (782 words) - 03:49, 27 August 2024
• 2001 JBMO Problems/Problem 2
  ...have <math>HA' = HB',</math> so by SAS Congruency, we have <math>\triangle HA'Y \cong \triangle HB'Y,</math> making <math>A'Y = YB'</math> and <math>A'YC
  3 KB (394 words) - 23:27, 13 August 2018

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