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• Logarithm
  ...would be a pain to have to calculate any time you wanted to use it (say in a comparison of large numbers). Its natural logarithm though (partly due to ...ly 7 digits before the decimal point. Comparing the logs of the numbers to a given precision can allow easier comparison than computing and comparing th
  4 KB (680 words) - 12:54, 16 October 2023
• Law of Cosines
  ...sines''' is a theorem which relates the side-[[length]]s and [[angle]]s of a [[triangle]]. It can be derived in several different ways, the most common ...h>, <math>b</math> and <math>c</math> opposite [[angle]]s of measure <math>A</math>, <math>B</math> and <math>C</math>, respectively, the Law of Cosines
  6 KB (1,003 words) - 00:02, 20 May 2024
• Law of Sines
  The '''Law of Sines''' is a useful identity in a [[triangle]], which, along with the [[law of cosines]] and the [[law of tan ...h>\triangle ABC</math>, where <math>a</math> is the side opposite to <math>A</math>, <math>b</math> opposite to <math>B</math>, <math>c</math> opposite
  4 KB (658 words) - 16:19, 28 April 2024
• Homothety
  ...ath> and factor <math>k</math> sends point <math>A</math> to a point <math>A' \ni HA'=k\cdot HA</math> This is denoted by <math>\mathcal{H}(H, k)</math> ...is <math>\frac{OD'}{OD}</math>. Additionally, <math>ABCD</math> and <math>A'B'C'D'</math> are homothetic with respect to <math>O</math>.
  3 KB (532 words) - 01:11, 11 January 2021
• 1985 IMO Problems/Problem 5
  ...gle <math>ABC</math> and intersects the segments <math>AB</math> and <math>BC</math> again at distinct points <math>K</math> and <math>N</math> respectiv .../math> is the Miquel Point of quadrilateral <math>ACNK</math>, so there is a spiral similarity centered at <math>M</math> that takes <math>KN</math> to
  3 KB (496 words) - 13:35, 18 January 2023
• Power of a Point Theorem
  ...e lengths of the [[line segment]]s formed when two [[line]]s [[intersect]] a [[circle]] and each other. ...[secant line|secant]] (middle figure). In this case, we have <math> AB^2 = BC\cdot BD </math>.
  5 KB (827 words) - 17:30, 21 February 2024
• Incenter
  ...r is the center of the [[incircle]]. Every [[nondegenerate]] triangle has a unique incenter. ...nd <math>AB</math>; hence it is equidistant from <math>BC</math> and <math>BC</math> and <math>CA</math> and therefore lies on an angle bisector of <math
  2 KB (381 words) - 19:38, 24 November 2011
• Triangle Inequality
  The '''Triangle Inequality''' says that in a [[nondegenerate]] [[triangle]] <math>ABC</math>: <math>AB + BC > AC</math>
  2 KB (268 words) - 03:02, 3 January 2021
• Rhombus
  ...eral]] all of whose sides are [[congruent (geometry) | congruent]]. It is a special type of [[parallelogram]], and its properties (aside from those pro * If all of a rhombus' [[angle]]s are [[right angle]]s, then the rhombus is a [[square (geometry) | square]].
  3 KB (490 words) - 15:30, 22 February 2024
• Kite
  ...ne. [[Quadrilateral]] <math>ABCD</math> is a kite if and only if <math>AB=BC</math> and <math>CD=DA</math>. Thus, there are two types of quadrilaterals
  739 bytes (112 words) - 01:17, 18 January 2020
• 2006 AIME I Problems
  ...h> \overline{AC} </math> is perpendicular to <math> \overline{CD}, AB=18, BC=21, </math> and <math> CD=14. </math> Find the perimeter of <math> ABCD. </ ...th> and let <math> S </math> be the sum of the elements of <math> \mathcal{A}. </math> Find the number of possible values of <math> S. </math>
  7 KB (1,173 words) - 03:31, 4 January 2023
• 2006 AIME I Problems/Problem 14
  A tripod has three legs each of length <math>5</math> feet. When the tripod i triple O=(0,0,0),T=(0,0,5),C=(0,3,0),A=(-3*3^.5/2,-3/2,0),B=(3*3^.5/2,-3/2,0);
  6 KB (980 words) - 21:45, 31 March 2020
• 2006 AIME I Problems/Problem 5
  ...n be written as <math> a\sqrt{2}+b\sqrt{3}+c\sqrt{5}, </math> where <math> a, b, </math> and <math> c </math> are [[positive]] [[integer]]s. Find <math> <cmath> a\sqrt{2}+b\sqrt{3}+c\sqrt{5} = \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2
  3 KB (439 words) - 18:24, 10 March 2015
• 2006 AIME I Problems/Problem 3
  ...th>a_n</math> must be divisible by <math>7.</math> But <math>a_n</math> is a nonzero digit, so the only possibility is <math>a_n = 7.</math> This gives ...uation to get <cmath>35=14b+7</cmath> <cmath>b=2.</cmath> Now, since <math>a=7</math>, <math>b=2</math>, and <math>c=5</math>, our number <math>N=100a+1
  4 KB (622 words) - 03:53, 10 December 2022
• 2006 AIME I Problems/Problem 1
  ...[[perpendicular]] to <math>\overline{CD}</math>, <math>AB=18</math>, <math>BC=21</math>, and <math>CD=14</math>. Find the [[perimeter]] of <math>ABCD</ma pair C=(0,0), D=(0,-14),A=(-(961-196)^.5,0),B=IP(circle(C,21),circle(A,18));
  2 KB (217 words) - 21:43, 2 February 2014
• 2021 JMC 10
  <math>\textbf{(A) } 3 \qquad\textbf{(B) } 7 \qquad\textbf{(C) } 8 \qquad\textbf{(D) } 9 \qqu <math>\textbf{(A) }\pi-e \qquad\textbf{(B) }2\pi-2e\qquad\textbf{(C) }2e\qquad\textbf{(D) }2
  12 KB (1,784 words) - 16:49, 1 April 2021
• 2006 AMC 12B Problems
  \text {(A) } - 2006 \qquad \text {(B) } - 1 \qquad \text {(C) } 0 \qquad \text {(D) } <math>\text {(A) } - 72 \qquad \text {(B) } - 27 \qquad \text {(C) } - 24 \qquad \text {(D)
  13 KB (2,058 words) - 12:36, 4 July 2023
• 2005 AMC 12A Problems
  {{AMC12 Problems|year=2005|ab=A}} (\mathrm {A}) \ 1 \qquad (\mathrm {B}) \ 2 \qquad (\mathrm {C})\ 5 \qquad (\mathrm {D})
  13 KB (1,971 words) - 13:03, 19 February 2020
• 2004 AMC 12A Problems
  {{AMC12 Problems|year=2004|ab=A}} <math>\text{(A) } 0.0029 \qquad \text{(B) } 0.029 \qquad \text{(C) } 0.29 \qquad \text{(D)
  13 KB (1,953 words) - 00:31, 26 January 2023
• 2002 AMC 12A Problems
  {{AMC12 Problems|year=2002|ab=A}} <math> \mathrm{(A) \ } \frac{7}{2}\qquad \mathrm{(B) \ } 4\qquad \mathrm{(C) \ } 5\qquad \mat
  12 KB (1,792 words) - 13:06, 19 February 2020

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