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• 2002 AIME I Problems/Problem 2
  ...e triangle is therefore <math>2\sqrt{3}+2</math> and we use simplification similar to as showed above, and we reach the result <math>\frac{1}{2} \cdot (\sqrt{ [[Category:Intermediate Geometry Problems]]
  2 KB (287 words) - 19:54, 4 July 2013
• 2003 AIME II Problems/Problem 14
  ...<math>c-b=8\sqrt{3}/3,</math> and thus <math>c=18\sqrt{3}/3</math>. Using similar methods (or symmetry), we determine that <math>D=(10\sqrt{3}/3,10)</math>, This is similar to solution 2 but faster and easier.
  9 KB (1,461 words) - 15:09, 18 August 2023
• 2003 AIME II Problems/Problem 11
  By AA similarity, <math>\Delta EMB</math> is similar to <math>\Delta ACB</math>, <math>\frac{EM}{AC}=\frac{MB}{CB}</math>. Solvi [[Category: Intermediate Geometry Problems]]
  5 KB (772 words) - 19:47, 1 August 2023
• 2003 AIME II Problems/Problem 6
  ...hus <math>[A'DE]=\frac{1}{9}[A'B'C']=\frac{1}{9}[ABC].</math> We can apply similar reasoning to the other small triangles in <math>A'B'C'</math> not contained [[Category: Intermediate Geometry Problems]]
  5 KB (787 words) - 17:38, 30 July 2022
• 2003 AIME II Problems/Problem 4
  ...th>Z</math> on <math>ABD</math>. Clearly, the two regular tetrahedrons are similar, so if we can find the ratio of the sides, we can find the ratio of the vol [[Category: Intermediate Geometry Problems]]
  3 KB (563 words) - 17:36, 30 July 2022
• 2002 AIME II Problems/Problem 14
  ...h>O</math> on <math>\overline{AP}</math> is drawn so that it is [[Tangent (geometry)|tangent]] to <math>\overline{AM}</math> and <math>\overline{PM}</math>. Gi ...ower of a point]]. Using the fact that the ratio of corresponding sides in similar triangles is equal to the ratio of their perimeters, we have
  4 KB (658 words) - 19:15, 19 December 2021
• 2002 AIME II Problems/Problem 13
  Notice now that <math>\triangle{PBQ}</math> is similar to <math>\triangle{EBA}</math>. Therefore, Also, <math>\triangle{PRA}</math> is similar to <math>\triangle{DBA}</math>. Therefore,
  6 KB (935 words) - 13:23, 3 September 2021
• 2001 AIME II Problems/Problem 15
  ...is at <math>(8,8,8)</math>. Using a little visualization (involving some [[similar triangles]], because we have parallel lines) shows that the tunnel meets th [[Category:Intermediate Geometry Problems]]
  4 KB (518 words) - 15:01, 31 December 2021
• 2001 AIME II Problems/Problem 13
  ...ine{CD}</math> at <math>M</math>. Note that <math>\triangle{DAB}</math> is similar to <math>\triangle{BMC}</math> by AA similarity, since <math>\angle{ABD}=\a ...th>x=12</math>.So <math>AM=DM=20</math> similarly, we use the same pair of similar triangle we get <math>\frac{CM}{BM}=\frac{BM}{DM}</math>, we get that <math
  4 KB (743 words) - 03:32, 23 January 2023
• 2001 AIME II Problems/Problem 12
  ...math>\frac 12</math> of the original one. Since the ratio of the volume of similar polygons is the cube of the ratio of their corresponding lengths, it follow [[Category:Intermediate Geometry Problems]]
  2 KB (380 words) - 00:28, 5 June 2020
• 2001 AIME II Problems/Problem 7
  ...STR \sim \triangle UQV</math>. Since the ratio of corresponding lengths of similar figures are the same, we have ...we find that <math>A_{1}Q = 60</math>, <math>A_{1}R = 90</math>. Using the similar triangles, <math>RA_{2} = 45</math>, <math>QA_{3} = 20</math>, so <math>A_{
  7 KB (1,112 words) - 02:15, 26 December 2022
• 2001 AIME II Problems/Problem 6
  ...hs are positive, we discard the other root). The ratio of the areas of two similar figures is the square of the ratio of their corresponding side lengths, so [[Category:Intermediate Geometry Problems]]
  4 KB (772 words) - 19:31, 6 December 2023
• 2000 AIME II Problems/Problem 6
  ...e segment into three partitions with lengths <math>x_1, 75, x_2</math>. By similar triangles, we easily find that <math>\frac{x - 75}{100} = \frac{x_1+x_2}{10 [[Category:Intermediate Geometry Problems]]
  3 KB (433 words) - 19:42, 20 December 2021
• 2005 AMC 10B Problems/Problem 14
  ...ve that <math>\triangle AYC</math> and <math>\bigtriangleup MXC</math> are similar. By the [[Pythagorean theorem]], <math>AY</math> is <math>\sqrt3</math>. [[Category:Introductory Geometry Problems]]
  5 KB (882 words) - 22:12, 30 April 2024
• 2005 IMO Shortlist Problems/G3
  ...ath>. It follows that triangles <math> \displaystyle ABK, LDA </math> are similar. Then since <math> \displaystyle ABCD </math> is a parallelogram, ...th>, this implies that triangles <math> \displaystyle KBC, CDL </math> are similar. This means that
  3 KB (453 words) - 10:53, 24 June 2007
• 1959 IMO Problems/Problem 5
  ...{MB} = \frac{AN}{NB} </math> since the triangles <math>ABN, BCM</math> are similar. Then <math>NM </math> bisects <math>ANB </math>. [[Category:Olympiad Geometry Problems]]
  4 KB (729 words) - 08:23, 23 May 2024
• Geometry/Introduction
  == Introductory Topics in Geometry == The following topics make a good introduction to [[geometry]].
  1 KB (122 words) - 16:25, 18 May 2021
• 1995 AHSME Problems/Problem 8
  <math>\triangle BDE</math> is similar to <math>\triangle BAC</math> by angle-angle similarity since <math>E=C = 9 [[Category:Introductory Geometry Problems]]
  1 KB (199 words) - 13:58, 5 July 2013
• Geometry/Olympiad
  An olympiad-level study of [[geometry]] involves familiarity with intermediate topics to a high level, a multitud === Synthetic geometry ===
  2 KB (242 words) - 10:16, 18 June 2023
• Equilateral triangle
  ...me angle measurements, so, by AA similarity, all equilateral triangles are similar). [[Category:Geometry]]
  1 KB (186 words) - 19:57, 15 September 2022

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