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• 2009 AMC 12A Problems/Problem 21
  <center><cmath>p(2009 + 9002\pi i) = p(2009) = p(9002) = 0</cmath></center> From the three zeroes, we have <math>p(x) = (x - (2009 + 9002\pi i))(x - 2009)(x - 9002)</math>.
  2 KB (322 words) - 10:25, 29 July 2020
• 2009 AMC 12A Problems/Problem 20
  pair D=MP("D",(0,0)),C=MP("C",(12,0)),A=MP("A",C+14*expi(145*pi/180),N),B=MP("B",A+(9,0),N),E=IP(A--C,B--D);MP("9",(A+B)/2,N);MP("12",(C+D) ...\triangle AEB \sim \triangle EDC</math> (which should be trivial given the two equal triangles) we have that
  3 KB (543 words) - 21:09, 23 October 2023
• 2009 AMC 12A Problems/Problem 19
  ...Bethany did the same with a regular heptagon (7 sides). The areas of the two regions were <math>A</math> and <math>B</math>, respectively. Each polygon ...isosceles triangle formed by the center of the polygon <math>S</math> and two consecutive vertices <math>X</math> and <math>Y</math>. We are given that <
  4 KB (630 words) - 21:27, 30 December 2023
• 2009 AMC 10A Problems/Problem 21
  ...e area of the large circle is <math>L = \pi R^2 = \pi r^2 (1+\sqrt 2)^2 = \pi r^2 (3+2\sqrt 2)</math>. The area of four small circles is <math>S = 4\pi r^2</math>. Hence their ratio is:
  3 KB (474 words) - 12:50, 29 September 2023
• 2009 AMC 10A Problems/Problem 19
  ...h> as it rolls once around the circumference of circle <math>A</math>. The two circles have the same points of tangency at the beginning and end of circle ...cumference of circle <math>B</math> with radius <math>r</math> is <math>2r\pi</math>. Since circle <math>B</math> makes a complete revolution and ''ends
  2 KB (276 words) - 09:57, 8 June 2021
• 2001 AMC 12 Problems/Problem 8
  ...^\circ</math> sector of a circle of radius <math>10</math> by aligning the two straight sides? The length of the red line is <math>\dfrac{252}{360}\cdot 2\pi \cdot 10 = 14\pi</math>. This is the circumference of a circle with radius <math>7</math>.
  2 KB (279 words) - 00:32, 30 December 2023
• 2009 AMC 12B Problems
  A rectangular yard contains two flower beds in the shape of congruent isosceles right triangles. The remain Kiana has two older twin brothers. The product of their ages is <math>128</math>. What is
  13 KB (2,030 words) - 03:04, 5 September 2021
• 2009 AMC 12B Problems/Problem 23
  ...34 + \frac34i</math> into polar form as <math>\frac{3\sqrt{2}}{4}e^{\frac{\pi}{4}i}</math>. Restated using geometric probabilities, we are trying to find ...ath>2</math> by <math>2</math> square centered at the origin. Graphing our two equations gives us the four lines <cmath>x-y=\frac{4}{3},</cmath> <cmath>x-
  2 KB (422 words) - 13:25, 20 January 2020
• 2009 AMC 12B Problems/Problem 24
  For how many values of <math>x</math> in <math>[0,\pi]</math> is <math>\sin^{ - 1}(\sin 6x) = \cos^{ - 1}(\cos x)</math>? ...in^{-1}(x) \leq \pi/2</math> and <math>\forall x: 0\leq \cos^{-1}(x) \leq \pi</math>.
  7 KB (1,287 words) - 07:09, 22 December 2022
• 2009 AIME I Problems/Problem 12
  ...>O</math> be center of the circle and <math>P</math>,<math>Q</math> be the two points of tangent such that <math>P</math> is on <math>BI</math> and <math> Since the ratios between corresponding lengths of two similar diagrams are equal, we can let <math>AD = 144, CD = 420</math> and
  12 KB (1,970 words) - 22:53, 22 January 2024
• 1966 AHSME Problems/Problem 4
  ...the radius of the smaller circle is <math>x</math>, so it's area is <math>\pi x^2</math>. ...be <math>2x^2 \pi</math>. Putting one over the other and dividing, you get two as the answer: or <math>\boxed{(B)}</math>.
  1 KB (191 words) - 22:09, 14 January 2018
• Liouville's Theorem (complex analysis)
  <cmath> \lvert f'(z_0) \rvert = \biggl\lvert \frac{1}{2\pi i} point <math>z \in \mathbb{C}</math>. Now for any two complex numbers <math>A</math>
  2 KB (412 words) - 20:30, 16 January 2024
• 2000 AMC 8 Problems
  ...e numbers lie in the interval between <math>\frac{5}{3}</math> and <math>2\pi?</math> Consider these two geoboard quadrilaterals. Which of the following statements is true?
  15 KB (2,165 words) - 18:29, 5 June 2024
• 2009 AIME II Problems/Problem 13
  ...ath> instead of <math>B</math>. <math>\angle AOC_1</math> = <math>\frac {\pi}{7}</math>. Using the [[Law of Cosines]], <math>\overline {AC_1}^2</math> = <math>8 - 8 \cos \frac {\pi}{7}</math>,
  8 KB (1,279 words) - 20:27, 17 May 2024
• 1965 IMO Problems
  Determine all values <math>x</math> in the interval <math>0\leq x\leq 2\pi </math> which satisfy the inequality .../math> is equal to <math>k</math>. Compute the ratio of the volumes of the two solids obtained.
  3 KB (497 words) - 12:39, 29 January 2021
• Fundamental group
  ...example, the fundamental group of a figure eight is the [[free group]] on two [[generator]]s, which is not abelian. However, the fundamental group of a c We say that two binary operations <math>\circ, \cdot</math> on a
  8 KB (1,518 words) - 20:11, 23 January 2017
• 2003 AMC 12B Problems/Problem 25
  ...ave to place the third point is a <math>60</math> degrees arc(if the first two are <math>60</math> degrees apart), with a <math>\frac{1}{6}</math> probabi ...ave to place the third point is a <math>120</math> degree arc(if the first two are the same point), with a <math>\frac{1}{3}</math> probability.
  4 KB (635 words) - 11:46, 1 September 2022
• 2010 AMC 12A Problems
  ...}\ \frac{0.4}{\pi} \qquad \textbf{(C)}\ 0.4 \qquad \textbf{(D)}\ \frac{4}{\pi} \qquad \textbf{(E)}\ 4</math> In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and
  12 KB (1,817 words) - 15:00, 12 August 2020
• 2010 AMC 12A Problems/Problem 13
  The image below shows the two curves for <math>k=4</math>. The blue curve is <math>x^2+y^2=k^2</math>, wh Clearly the only such integer is <math>k=1</math>, hence the two curves are only disjoint for <math>k=1</math> and <math>k=-1</math>.
  9 KB (1,622 words) - 20:53, 11 September 2023
• 2010 AMC 12A Problems/Problem 17
  ...math>\triangle ABC</math>, we get that <math>AC^2 = r^2+1^2-2r\cos{\frac{2\pi}{3}} = r^2+r+1</math>. Therefore, the area of <math>\triangle ACE</math> is Note: To verify that the quadratic <math>r^2-6r+1</math> has two positive roots, we can either solve for the roots directly or note that dis
  4 KB (690 words) - 10:13, 14 October 2022

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