Create the page "Altitudes" on this wiki! See also the search results found.

Page title matches

• A game board is constructed by shading two of the regions formed by the altitudes of an equilateral triangle as shown. What is the probability that the tip of the spinner will come to rest in a shaded region? Express your answer as a common fraction.
  25 bytes (3 words) - 15:18, 30 October 2024

Page text matches

• Geometric inequality
  ...PB+PC\ge 2(PD+PE+PF)</math> where <math>D, E, F</math> are the foot of the altitudes from <math>P</math> to <math>BC, AC,</math> and <math>AB</math>, respective
  7 KB (1,300 words) - 01:11, 28 October 2024
• Orthic triangle
  ...<math>D</math>, <math>E</math>, and <math>F</math> denote the feet of the altitudes from <math>A</math>, <math>B</math>, and <math>C</math>, respectively. Then ...e ABC</math>. Similarly, <math>E</math> and <math>F</math> are feet of the altitudes from <math>B</math> and <math>C</math>, respectively. Then <math>\triangle
  8 KB (1,408 words) - 09:39, 10 July 2024
• Area
  ...<math>H=\frac{(h_a^{-1}+h_b^{-1}+h_c^{-1})}{2}</math> and the triangle has altitudes <math>h_a</math>, <math>h_b</math>, <math>h_c</math>.
  7 KB (1,185 words) - 20:48, 25 November 2024
• 2005 AIME II Problems/Problem 14
  ...the altitudes from E and D to AB and AC respectively. Call the feet of the altitudes R and S respectively. Now we start by drawing altitudes from D and E onto AC, labeling them as F and G, and labelling <math>\angle
  14 KB (2,340 words) - 16:38, 21 August 2024
• 2004 AIME II Problems/Problem 13
  3 KB (486 words) - 22:15, 7 April 2023
• 1983 AIME Problems/Problem 14
  ...t since triangles <math>AQP</math> and <math>BPR</math> are isosceles, the altitudes are also bisectors, so let <math>QM=MP=PN=NR=x</math>.
  14 KB (2,351 words) - 21:06, 8 December 2024
• 2006 AMC 10B Problems/Problem 19
  ...th>BE</math> are the bases and <math>OA</math> and <math>CO</math> are the altitudes, respectively, <math>[DBO] = [EBO] = \frac{\sqrt{3}-1}{2}</math>. The area
  5 KB (873 words) - 15:39, 29 May 2023
• 1999 AIME Problems/Problem 15
  ...e coordinates of <math>O</math>, we need to find the intersection point of altitudes <math>BE</math> and <math>AD</math>. The equation of <math>BE</math> is sim
  7 KB (1,175 words) - 13:26, 3 September 2024
• 2003 AIME II Problems/Problem 4
  ...<math>WXYZ</math> and <math>ABCD</math> are parallel, we can say that the altitudes of tetrahedron <math>ABCD</math> and <math>WXYZ</math> are in the ratio <ma
  3 KB (563 words) - 17:36, 30 July 2022
• Inradius
  Add in the incircle and drop the altitudes from the incenter to the sides of the triangle. Also draw the lines <math>\
  2 KB (321 words) - 22:54, 20 June 2024
• Furman University Wylie Mathematics Tournament/1996 Senior Exam/Solutions
  Notice that <math> CF=CE </math> and are the altitudes of the equilateral triangles. Thus, <math> CF=CE=\sqrt{3}. </math> Now, <ma
  9 KB (1,364 words) - 15:59, 21 July 2006
• 1970 IMO Problems/Problem 5
  ...o the plane <math>ABC</math> in the tetrahedron is the intersection of the altitudes of <math>\triangle ABC</math>. Prove that ...h>ADC</math> are also right. Let <math>H</math> be the intersection of the altitudes
  4 KB (794 words) - 03:40, 7 December 2024
• 2006 Romanian NMO Problems
  ...ath>. The points <math>A_1</math> and <math>B_1</math> are the feet of the altitudes from <math>A</math> and <math>B</math>, and <math>H</math> is the orthocent
  11 KB (1,779 words) - 14:57, 7 May 2012
• 2006 Romanian NMO Problems/Grade 7/Problem 3
  ...ath>. The points <math>A_1</math> and <math>B_1</math> are the feet of the altitudes from <math>A</math> and <math>B</math>, and <math>H</math> is the orthocent
  2 KB (294 words) - 22:24, 26 November 2023
• 2005 AMC 10A Problems/Problem 25
  .... This is applied twice, using different pairs of bases, and corresponding altitudes for height.).
  6 KB (897 words) - 17:55, 1 December 2024
• Acute triangle
  1 KB (194 words) - 13:09, 20 February 2024
• Obtuse triangle
  ...he triangle. (There is no triangle with exactly one altitude or all three altitudes outside the triangle.)
  1 KB (193 words) - 14:20, 12 June 2022
• Circumcenter
  ...ath> are also altitudes. Thus, the problem is reduced to proving that the altitudes of a triangle are concurrent. This can be done using Ceva's Theorem.
  2 KB (390 words) - 14:14, 22 November 2024
• Orthocenter
  ...center''' of a [[triangle]] is the point of intersection of its [[altitude|altitudes]]. It is [[mathematical convention | conventionally]] denoted <math>H</mat The lines highlighted are the [[altitude|altitudes]] of the triangle, they meet at the orthocenter.
  6 KB (875 words) - 02:22, 19 February 2025
• 1970 Canadian MO Problems
  Given a triangle <math>ABC</math> with angle <math>A</math> obtuse and with altitudes of length <math>h</math> and <math>k</math> as shown in the diagram, prove
  4 KB (604 words) - 04:32, 8 October 2014

View (previous 20 | next 20) (20 | 50 | 100 | 250 | 500)