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• Circle
  ===Circumference and Area=== ...circumference]] (distance around a circle) is <math>2 \pi r</math> and the area is <math>\pi r^2</math>. Both formulas involve the mathematical constant [
  9 KB (1,510 words) - 19:56, 16 January 2025
• Pi
  ...ics and appears in many surprising places. The number pi often shows up in problems in [[number theory]], particularly [[algebraic number theory]]. For example ...ea of the circle is <math>\pi</math>, and the area of the square is 4, the ratio of the amount of points inside the circle to the total number of points app
  8 KB (1,424 words) - 16:49, 17 February 2025
• Logarithm
  == Problems == <math>\ln a</math> can also be defined as the area under the curve <math>y=\frac{1}{x}</math> between 1 and a, or <math>\int^a
  5 KB (721 words) - 20:44, 9 March 2025
• Power of a Point Theorem
  == Problems == ...h>R</math>. If <math>AR:BR=1:4</math> and <math>CR:DR=4:9</math>, find the ratio <math>AB:CD</math> .
  5 KB (948 words) - 17:04, 21 February 2025
• Right triangle
  ...30^{\circ}-60^{\circ}-90^{\circ}</math> triangle]], which has sides in the ratio of <math>x:x\sqrt3:2x</math>. This triangle is analogous to an equilateral * The [[area]] of the triangle is equal to half of the product of the lengths of the leg
  4 KB (543 words) - 19:37, 31 January 2025
• 2006 AIME I Problems
  {{AIME Problems|year=2006|n=I}} [[2006 AIME I Problems/Problem 1|Solution]]
  7 KB (1,173 words) - 03:31, 4 January 2023
• 2006 AIME I Problems/Problem 14
  We will use <math>[...]</math> to denote volume (four letters), area (three letters) or length (two letters). We have the volume ratio <math>\frac {[TSBC]}{[TABC]} = \frac {[TS]}{[TA]} = \frac {4}{5}</math>.
  6 KB (980 words) - 21:45, 31 March 2020
• 2006 AIME I Problems/Problem 7
  ...B </math> is 11/5. Find the ratio of shaded region <math> D </math> to the area of shaded region <math> A. </math> Since the area of the triangle is equal to <math>\frac{1}{2}bh</math>,
  4 KB (709 words) - 01:31, 5 January 2025
• 2006 AMC 12B Problems
  {{AMC12 Problems|year=2006|ab=B}} [[2006 AMC 12B Problems/Problem 1|Solution]]
  13 KB (2,058 words) - 12:36, 4 July 2023
• 2006 AMC 12A Problems
  {{AMC12 Problems|year=2006|ab=A}} [[2006 AMC 12A Problems/Problem 1|Solution]]
  15 KB (2,223 words) - 13:43, 28 December 2020
• 2005 AMC 12A Problems
  {{AMC12 Problems|year=2005|ab=A}} [[2005 AMC 12A Problems/Problem 1|Solution]]
  13 KB (1,965 words) - 22:18, 7 September 2024
• 2003 AMC 12A Problems
  {{AMC12 Problems|year=2003|ab=A}} [[2003 AMC 12A Problems/Problem 1|Solution]]
  13 KB (1,955 words) - 21:06, 19 August 2023
• 2002 AMC 12A Problems
  {{AMC12 Problems|year=2002|ab=A}} [[2002 AMC 12A Problems/Problem 1|Solution]]
  12 KB (1,792 words) - 13:06, 19 February 2020
• 2000 AMC 12 Problems
  {{AMC12 Problems|year=2000|ab=}} [[2000 AMC 12 Problems/Problem 1|Solution]]
  13 KB (1,948 words) - 10:35, 16 June 2024
• 2003 AMC 12B Problems
  {{AMC12 Problems|year=2003|ab=B}} [[2003 AMC 12B Problems/Problem 1|Solution]]
  13 KB (1,987 words) - 18:53, 10 December 2022
• 2006 AMC 12A Problems/Problem 21
  What is the ratio of the area of <math>S_2</math> to the area of <math>S_1</math>? <math>S_1</math> is a circle with a radius of <math>7</math>. So, the area of <math>S_1</math> is <math>49\pi </math>.
  2 KB (280 words) - 17:35, 17 September 2023
• 2006 AMC 10A Problems
  {{AMC10 Problems|year=2006|ab=A}} [[2006 AMC 10A Problems/Problem 1|Solution]]
  13 KB (2,028 words) - 16:32, 22 March 2022
• 2006 AMC 10A Problems/Problem 17
  ...th>. In addition, <math>AH=AC=2</math>, and <math>AD=3</math>. What is the area of quadrilateral <math>WXYZ</math> shown in the figure? ...em]] we can get that each side is <math>\sqrt{\frac{1^2}{2}}</math> so the area of the middle square would be <math>(\sqrt{\frac{1^2}{2}})^2=(\sqrt{\frac{1
  6 KB (1,106 words) - 10:20, 4 November 2024
• 2005 AIME II Problems/Problem 14
  From here, we can use Heron's Formula to find the altitude. The area of the triangle is <math>\sqrt{21*6*7*8} = 84</math>. We can then use simil ==Solution 4 (Ratio Lemma and Angle Bisector Theorem)==
  14 KB (2,340 words) - 16:38, 21 August 2024
• 2004 AIME I Problems/Problem 11
  ...s of the painted surfaces of <math> C </math> and <math> F </math> and the ratio between the [[volume]]s of <math> C </math> and <math> F </math> are both e ...rac13 \pi r^2 h = \frac13 \pi 3^2\cdot 4 = 12 \pi</math> and has [[surface area]] <math>A = \pi r^2 + \pi r \ell</math>, where <math>\ell</math> is the [[s
  5 KB (839 words) - 22:12, 16 December 2015

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