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• Joining an ARML team
  ...and B teams -- those selected usually consist of USAMO qualifiers, the top 12 at the state tournament, and team veterans. (Of course, these groups are r ...Math Meet and other math competitions. The top 15 or so finishers at the NC State Math Contest (Comprehensive) and all USAMO qualifiers are among those
  22 KB (3,533 words) - 20:58, 2 June 2024
• 2000 AMC 12 Problems/Problem 21
  {{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #21]] and [[2000 AMC 10 Problems|2000 AMC 10 #19]]}} ...a} = \frac{bc}{a}</math> and <cmath>[B] = \frac{bh}{2} = \frac{bc^2}{2a} = nc^2 = n[C]</cmath>
  5 KB (804 words) - 01:22, 13 May 2024
• Mass points
  [[2016 AMC 12A Problems/Problem 12]] [[ 1988 AIME Problems/Problem 12]]
  5 KB (849 words) - 14:27, 8 August 2024
• 2009 AIME II Problems/Problem 15
  <cmath>= \sin\left(\frac{\widehat{AN} + \widehat{NC}}{2}\right)</cmath> ...h> and <math>E</math>, respectively, and let <math>MC = x</math> and <math>NC = y</math>. Since <math>A</math> is the midpoint of arc <math>MN</math>, <m
  11 KB (1,862 words) - 21:23, 23 May 2024
• 2011 AMC 12A Problems/Problem 13
  Triangle <math>ABC</math> has side-lengths <math>AB = 12, BC = 24,</math> and <math>AC = 18.</math> The line through the incenter of ...es of isosceles triangles that <math>MO = MB</math>. Similarly, <math>NO = NC</math>. The perimeter of <math>\triangle{AMN}</math> then becomes
  4 KB (683 words) - 03:12, 23 January 2023
• 2011 AIME I Problems/Problem 4
  Hence <math>MN=\tfrac 12 PQ</math>. Since <cmath>PQ=BP+AQ-AB=120+117-125=112,</cmath> so <math>MN=\b ...N = \sin \angle BIA = \sin (90^\circ + \tfrac 12 \angle BCA) = \cos \tfrac 12 \angle BCA = \cos \angle BCI</math>, we have that <math>MN = CI \cdot \sin
  9 KB (1,523 words) - 12:08, 15 July 2024
• 2000 AMC 8 Problems/Problem 25
  .../math>, and the bottom midpoint as <math>N</math>, we know that <math>DN = NC = 6</math>, and <math>BM = MC = 3</math>. <math>[\triangle ABM] = \frac{1}{2}\cdot 12\cdot 3 = 18</math>
  2 KB (325 words) - 01:07, 6 August 2024
• 2013 AIME I Problems/Problem 9
  A paper equilateral triangle <math>ABC</math> has side length <math>12</math>. The paper triangle is folded so that vertex <math>A</math> touches pair C = MP("C", (12,0), dir(-20));
  11 KB (1,778 words) - 20:30, 11 August 2024
• 2018 AIME II Problems/Problem 14
  <asy> size(200); import olympiad; defaultpen(linewidth(1)+fontsize(12)); ...>MB=4-x=\frac{14}{5}=BX</math>, <math>AM=x+3=\frac{21}{5}=AN</math>, <math>NC=8-AN=\frac{19}{5}=XC</math>, <math>AQ=\frac{21}{5}-y</math>, <math>PQ=\frac
  9 KB (1,518 words) - 14:32, 28 January 2024
• 2020 AMC 10B Problems/Problem 21
  ...- 2 \sqrt{2} + 2) \cdot (2 \sqrt{2} - 2) = 2x \sqrt{2} - 2x + 8 \sqrt{2} -12</math> ...2}{2} + 2x - 2x \sqrt{2} - 4 \sqrt{2} + 6 + 2x \sqrt{2} - 2x + 8 \sqrt{2} -12= 1</cmath>
  11 KB (1,850 words) - 23:33, 27 March 2024
• 2021 AIME I Problems/Problem 7
  ...e same constant <math>c = \frac{1}{k}</math>, we have that <math>mc \equiv nc \equiv 1 \pmod 4</math>. Then either <math>m \equiv n \equiv 1 \pmod 4</mat ..., we get the sets <math>\{2, 10, 18, 26\}, \{6, 14, 22, 30\}, \{4, 20\}, \{12, 28\}</math>. Some sets have been omitted; this is because they were counte
  9 KB (1,523 words) - 09:12, 3 December 2023
• 1982 AHSME Problems/Problem 10
  If <math>AB=12, BC=24</math>, and <math>AC=18</math>, then the perimeter of <math>\triangl pair B=origin, C=(24,0), A=intersectionpoints(Circle(B,12), Circle(C,18))[0], O=incenter(A,B,C), M=intersectionpoint(A--B, O--O+40*di
  2 KB (274 words) - 13:31, 16 July 2024