Create the page "Similar(geometry)" on this wiki! See also the search results found.

• Triangle
  ...ctions of the sides summing to 180 degrees. Triangles exist in Euclidean [[geometry]], and are the simplest possible polygon. In [[physics]], triangles are not ...des and is also [[equiangular]]. Note that all equilateral triangles are [[similar]]. All the angles of equilateral triangles are <math>60^{\circ}</math>
  4 KB (628 words) - 17:17, 17 May 2018
• American Scholastic Mathematics Association
  It features questions with probability, counting, arithmetic, algebra, and geometry. It is 35 minutes long and contains 7 questions. The difficulty of this math competition is similar to MOEMS (easier problems) and AMC 8 (harder problems)
  1 KB (153 words) - 13:11, 14 May 2019
• Ptolemy's Inequality
  ...math>P </math> such that the [[triangles]] <math>APB, \; DCB </math> are [[similar]] and have the same [[orientation]]. In particular, this means that ...rity]], we also know that the triangles <math>ABD, \; PBC </math> are also similar, which implies that
  3 KB (602 words) - 09:01, 7 June 2023
• 2011 AMC 10B Problems/Problem 22
  [[Category: Introductory Geometry Problems]] Notice that the smaller pyramid on top of the cube is similar to the larger pyramid. Thus, letting <math>x</math> be the edge length of t
  4 KB (691 words) - 18:38, 19 September 2021
• Homothety
  * The resulting image of a polygon from a homothety is [[similar]] to the original polygon. * https://brilliant.org/wiki/euclidean-geometry-homothety/ (contains sample problems and related proofs)
  3 KB (532 words) - 01:11, 11 January 2021
• Power of a Point Theorem
  Draw extra lines to create similar triangles! (Hint: Draw <math>AD</math> on all three figures. Draw another l * [[Geometry]]
  5 KB (828 words) - 23:10, 25 June 2024
• 2006 AIME I Problems/Problem 7
  ...by attaching this triangle with the next trapezoid. Parallel lines give us similar triangles, so we know the proportion of this triangle to the previous trian ...ratio we seed is <math>\frac{33(ay+yx)}{11xy}.</math> Finally note that by similar triangles <math>\frac{x}{x+a} =\frac{y}{y+b} \implies bx = ya.</math> There
  4 KB (709 words) - 01:50, 10 January 2022
• 2006 AMC 12A Problems/Problem 17
  (Similar to Solution 1) [[Category:Introductory Geometry Problems]]
  6 KB (958 words) - 23:29, 28 September 2023
• 2006 AMC 10A Problems/Problem 16
  ...\sim \triangle AEO_2 \sim \triangle AFC</math>. Using the first pair of [[similar triangles]], we write the proportion: [[Category:Introductory Geometry Problems]]
  5 KB (732 words) - 23:19, 19 September 2023
• 2006 AMC 10A Problems/Problem 17
  There are many different similar ways to come to the same conclusion using different [[right triangle|45-45- [[Category:Introductory Geometry Problems]]
  6 KB (1,066 words) - 00:21, 2 February 2023
• 2006 AMC 10A Problems/Problem 19
  How many non-[[similar]] triangles have angles whose degree measures are distinct positive integer [[Category:Introductory Geometry Problems]]
  2 KB (259 words) - 03:10, 22 June 2023
• 2005 AIME II Problems/Problem 8
  ...<math>O_3</math> to <math>\overline{AB}</math> be <math>T</math>. From the similar [[right triangle]]s <math>\triangle HO_1T_1 \sim \triangle HO_2T_2 \sim \tr ...=10-4=6</math>, and <math>O_1O_2=14</math>. But <math>O_1O_3O_3'</math> is similar to <math>O_1O_2O_2'</math> so <math>O_3O_3'=\frac{6}{14} \cdot 10=\frac{30}
  4 KB (693 words) - 13:03, 28 December 2021
• 2005 AIME II Problems/Problem 12
  === Solution 3 (similar triangles)=== ...>, and <math>KA = EB</math> (90 degree rotation), and now we can bash on 2 similar triangles <math>\triangle GAK \sim \triangle GHO</math>.
  13 KB (2,080 words) - 21:20, 11 December 2022
• 2005 AIME II Problems/Problem 15
  ...y=ax </math> contains the center of a circle that is externally [[tangent (geometry)|tangent]] to <math> w_2 </math> and internally tangent to <math> w_1. </ma ...stance from this tangent point to the origin is <math>\sqrt{69}.</math> By similar triangles, the slope of this line is then <math>\frac{\sqrt{69}}{5\sqrt{3}}
  12 KB (2,000 words) - 13:17, 28 December 2020
• 2005 AIME II Problems/Problem 14
  import olympiad; import cse5; import geometry; size(150); == Solution 2 (Similar Triangles)==
  13 KB (2,129 words) - 18:56, 1 January 2024
• 2004 AIME I Problems/Problem 14
  ...riangle]]s <math>\triangle CDA</math> and <math>\triangle CEB</math> are [[similar]] [[right triangle]]s. By the Pythagorean Theorem <math>CD=8\cdot\sqrt{6}< [[Category:Intermediate Geometry Problems]]
  4 KB (729 words) - 01:00, 27 November 2022
• 2004 AIME I Problems/Problem 11
  ...se the plane cut is parallel to the base of our solid, <math>C</math> is [[similar]] to the uncut solid and so the height and slant height of cone <math>C</ma <math>V</math> and <math>C</math> are similar cones, because the plane that cut out <math>C</math> was parallel to the ba
  5 KB (839 words) - 22:12, 16 December 2015
• 2004 AIME I Problems/Problem 10
  ...{A'B'C'} = r_{ABC} - 1 = 5</math>, and since the ratio of the areas of two similar figures is equal to the square of the ratio of their corresponding lengths, ...dn't know the formula for the distance from a point to a line, you can use similar triangles to get the ratio:
  5 KB (836 words) - 07:53, 15 October 2023
• 2004 AIME II Problems/Problem 12
  We use a similar argument with the line <math>DO</math>, and find the height from the top of [[Category:Intermediate Geometry Problems]]
  3 KB (431 words) - 23:21, 4 July 2013
• 2004 AIME II Problems/Problem 7
  ...theta)=\frac{BC'}{17}=\frac53</math>, so <math>BC'=\frac{85}{3}</math>. By similar triangles, <math>CC'=\frac{3}{17}BC'=\frac{15}{3}</math>, so <math>BC=\frac ...le, and so are B'DG and C'FG. Since C'F is 3, then using the properties of similar triangles GF is 51/8. DF is 22, so DG is 125/8. Finally, DB can to calculat
  9 KB (1,501 words) - 05:34, 30 October 2023

View (previous 20 | next 20) (20 | 50 | 100 | 250 | 500)