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• 2003 AIME II Problems
  ...> so that <math>AD = BD = 15.</math> Given that the area of triangle <math>CDM</math> may be expressed as <math>\frac {m\sqrt {n}}{p},</math> where <math>
  7 KB (1,127 words) - 09:02, 11 July 2023
• 1989 AIME Problems/Problem 12
  ...ath>N</math>. <math>d</math> is the [[median]] of triangle <math>\triangle CDM</math>. The formula for the length of a median is <math>m=\sqrt{\frac{2a^2+ Now that we know the sides of <math>\triangle CDM</math>, we proceed to find the length of <math>d</math>.
  2 KB (376 words) - 13:49, 1 August 2022
• 1995 AIME Problems/Problem 9
  Let <math>x=\angle CAM</math>, so <math>3x=\angle CDM</math>. Then, <math>\frac{\tan 3x}{\tan x}=\frac{CM/1}{CM/11}=11</math>. Ex ...y the [[Midpoint]] Theorem. So, <math>\angle HBA=x</math> and <math>\angle CDM=\angle CHB=\angle HDA= 3x</math>.
  7 KB (1,181 words) - 13:47, 3 February 2023
• 2003 AIME II Problems/Problem 11
  ...> so that <math>AD = BD = 15.</math> Given that the area of triangle <math>CDM</math> may be expressed as <math>\frac {m\sqrt {n}}{p},</math> where <math> ...orthogonal projection from <math>C</math> to <math>AB.</math> Thus, <math>[CDM]=\frac{(DM)(MN)} {2}</math>, <math>MN=BN-BM</math>, and <math>[ABC]=\frac{2
  5 KB (772 words) - 19:47, 1 August 2023
• 2005 AMC 10B Problems/Problem 14
  ...point of <math> \overline{BD}</math>. What is the area of <math> \triangle CDM</math>? ...}{2}</math> Since <math>\overline{MF}</math> is the height <math>\triangle CDM</math>, the area is <math>\boxed{\textbf{(C) }\frac{\sqrt{3}}{2}}</math>.
  5 KB (882 words) - 22:12, 30 April 2024
• 2005 AMC 10B Problems
  ...point of <math> \overline{BD}</math>. What is the area of <math> \triangle CDM</math>?
  12 KB (1,874 words) - 21:20, 23 December 2020
• Euler line
  <cmath>z+x = \frac {[CDM]}{2[CAM]} = \frac {1}{2(1 + n)}.</cmath>
  59 KB (10,203 words) - 04:47, 30 August 2023
• 2003 USAMO Problems/Problem 4
  ...BM = \angle DCM</math> if and only if triangles <math>BCM</math> and <math>CDM</math> are similar, that is ...<math>\angle CMB = \angle DMC</math>, triangles <math>BCM</math> and <math>CDM</math> are similar. Consequently, <math>\frac{CM}{DM} = \frac{BM}{CM}</math
  5 KB (820 words) - 02:39, 10 January 2023
• 2011 AMC 10B Problems/Problem 18
  ...<math>\angle AMD \cong \angle CDM \longrightarrow \angle CMD \cong \angle CDM</math>. Use the [[Base Angle Theorem]] to show <math>\overline{DC} \cong \o
  5 KB (782 words) - 14:29, 1 April 2024
• 2011 AMC 12B Problems/Problem 10
  ...>\angle AMD = \angle CDM</math>, so <math>\angle AMD = \angle CMD = \angle CDM</math>, so <math>\bigtriangleup CMD</math> is isosceles, and hence <math>CM
  781 bytes (119 words) - 10:11, 5 August 2019
• 2014 IMO Problems/Problem 4
  ...BAC}</math> should be equal to <math>\angle {BDN}</math> and <math>\angle {CDM}</math>. We can quickly prove that the triangles <math>ABC</math>, <math>AP
  13 KB (1,811 words) - 14:36, 1 June 2024
• 1977 IMO Problems
  ...e <math>ABCD</math> we construct the equilateral triangles <math>ABK, BCL, CDM, DAN.</math> Prove that the midpoints of the four segments <math>KL, LM, MN
  3 KB (466 words) - 16:54, 29 January 2021
• 1977 IMO Problems/Problem 1
  ...e <math>ABCD</math> we construct the equilateral triangles <math>ABK, BCL, CDM, DAN.</math> Prove that the midpoints of the four segments <math>KL, LM, MN
  1 KB (199 words) - 16:40, 29 January 2021
• 2022 AIME II Problems/Problem 11
  ...<cmath>S_{\bigtriangleup{ABM}}+S_{\bigtriangleup{AB'M}}+S_{\bigtriangleup{CDM}}+S_{\bigtriangleup{CD'M}}+S_{\bigtriangleup{B'C'M}}</cmath> ...ector of <math>\angle CDA</math> and <math>DC = DQ</math>, <math>\triangle CDM \cong \triangle QDM</math>.
  14 KB (2,254 words) - 18:26, 8 February 2024