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  • *[[ACM International Collegiate Programming Contest]] [http://icpc.baylor.edu/icpc * ACM Pacific Northwest American Region- Alaska, Hawaii, British Columbia, Washin
    7 KB (932 words) - 22:32, 13 June 2024
  • ...so we can split it in half (<math>\triangle ABM</math> and <math>\triangle ACM</math>).
    7 KB (1,149 words) - 12:36, 26 November 2024
  • <cmath>(\sin \angle ACM)(\sin \angle BAM)(\sin \angle CBM) = (\sin \angle CAM)(\sin \angle ABM)(\si
    7 KB (1,082 words) - 01:08, 30 September 2024
  • ...ruct the bisector of arc AB above AB. Call it X. <math>\angle ANM = \angle ACM = 45^{\circ} </math>. Now <math>\angle ANB = 90^{\circ}</math> which means
    4 KB (729 words) - 07:23, 23 May 2024
  • ..., and <math>AB = AC</math>, so by SSS, <math>\triangle ABM \cong \triangle ACM</math>. Thus <math>\angle AMC = \dfrac{\pi}{2}</math> and <math>\angle DMP
    3 KB (488 words) - 13:05, 15 December 2022
  • of the triangle <math>\triangle ACM</math>, and <math>I_2</math> be the incenter, and
    4 KB (740 words) - 17:03, 14 November 2024
  • <cmath>[ACM] = [ BCM] = \frac {1}{2}.</cmath>
    59 KB (10,203 words) - 03:47, 30 August 2023
  • Applying the Law of Sines to triangles <math>ABM</math> and <math>ACM</math> gives
    20 KB (3,565 words) - 10:54, 1 May 2024
  • ...e <math>r=\sqrt{14}</math>, and the inradius of <math>ABM</math> and <math>ACM</math> are <math>r_1=r_2= xr</math>, so,
    14 KB (2,213 words) - 03:58, 2 February 2025
  • ...rc</math>. Now by the Law of Sines on triangles <math>ABM</math> and <math>ACM</math>, we have <cmath>\frac{AB}{\sin 150^\circ} = \frac{AM}{\sin 20^\circ}
    5 KB (744 words) - 15:35, 28 June 2021
  • Using right <math>\triangle{FCE}</math>, since <math>\angle{ACM} = \angle{CFE}</math>, <math>\tan{\theta} = \frac{CE}{FE}</math>. Hence, pl
    31 KB (5,093 words) - 08:55, 17 December 2024
  • Since <math>\angle ACM = 40^\circ</math>, <math>\angle ACP = 140^\circ</math>, so, using the fact
    2 KB (384 words) - 12:37, 11 December 2024
  • ...}{\triangle{PQA}}=\frac{\frac{AP}{AB}\triangle{ABM}+\frac{AQ}{AC}\triangle{ACM}}{\frac{AP\cdot AQ}{AB \cdot AC}}=</math>
    3 KB (578 words) - 10:38, 30 January 2021
  • <cmath>\angle BDA = \angle ACB = \angle ACM \implies \triangle ABD \sim \triangle AMC \implies</cmath>
    30 KB (5,152 words) - 00:30, 2 January 2025