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• 2014 IMO Problems/Problem 4
  ...ath>NED</math>, <math>MDF</math> and <math>FCP</math> must be two pairs of similar triangles. Therefore we must prove angles <math>CBM</math> and <math>ANC, A ...ac {BF}{FM}</math>, so triangles <math>BFM</math> and <math>NEC</math> are similar. So the angles <math>CBM</math> and <math>ANC, BCN</math> and <math>AMB</ma
  13 KB (1,811 words) - 14:36, 1 June 2024
• 2012 UNCO Math Contest II Problems/Problem 7
  ...he centers of the circle and the point where the tangents meet, we get two similar triangles in ratio <math>1:3</math>. Let the hypotenuse of the smaller tria [[Category:Intermediate Geometry Problems]]
  2 KB (336 words) - 12:57, 2 January 2020
• 2009 UNCO Math Contest II Problems/Problem 8
  ...he two parallel lines, we know that the two triangles with known areas are similar by angle-angle-angle similarity. Noticing that the areas of the triangles a [[Category:Intermediate Geometry Problems]]
  1 KB (174 words) - 02:20, 29 January 2019
• 2007 UNCO Math Contest II Problems/Problem 9
  ...ve <math>BL = OM = \frac{BH}{3}</math>. This means the scale factor of the similar triangles <math>ABC</math> and <math>A_1B_1B</math> is <math>\frac {1}{3}</ [[Category:Intermediate Geometry Problems]]
  3 KB (412 words) - 18:49, 29 January 2018
• 2006 UNCO Math Contest II Problems/Problem 5
  In general, if <math>G</math> and <math>H</math> are similar triangles with sides <math>s_1</math> and <math>s_2</math> , then <math>\fr [[Category:Introductory Geometry Problems]]
  1 KB (184 words) - 03:20, 13 January 2019
• 1988 AHSME Problems/Problem 2
  Triangles <math>ABC</math> and <math>XYZ</math> are similar, with <math>A</math> corresponding to <math>X</math> and <math>B</math> to Since the triangles are similar, we know that the corresponding sides of the triangle are in ratio to each
  950 bytes (142 words) - 06:21, 31 August 2015
• 1987 AHSME Problems/Problem 2
  <math>\triangle DBE</math> is similar to <math>\triangle ABC</math> by AA, so <math>\overline{DE}</math> = 1 by s [[Category: Introductory Geometry Problems]]
  1 KB (134 words) - 18:37, 23 April 2017
• 1986 AHSME Problems/Problem 16
  is similar to <math>\triangle PCA</math>. The length of <math>PC</math> is [[Category: Intermediate Geometry Problems]]
  1 KB (210 words) - 23:51, 10 February 2018
• 1986 AHSME Problems/Problem 19
  ...the first part of her route. For her <math>y</math>-coordinate, we can use similar logic to find that the coordinate is <math>\sqrt{3} + 0 - \frac{\sqrt{3}}{2 [[Category: Intermediate Geometry Problems]]
  2 KB (240 words) - 20:45, 9 October 2017
• 1986 AHSME Problems/Problem 27
  <math>ABE</math> and <math>DCE</math> are similar isosceles triangles. It remains to find the square of the ratio of their si [[Category: Intermediate Geometry Problems]]
  1 KB (219 words) - 15:54, 2 August 2016
• 2015 AMC 10A Problems/Problem 21
  Similar to solution 1, <math>\triangle CTD</math> is an isosceles right triangle. < [[Category: Introductory Geometry Problems]]
  4 KB (689 words) - 20:16, 18 June 2024
• AA similarity
  ...if two pairs of corresponding angles are congruent, then the triangles are similar. * [[Similarity (geometry)]]
  915 bytes (147 words) - 01:18, 24 December 2022
• 2015 AMC 10B Problems/Problem 13
  ==Solution 2 (very similar to Solution 1)== [[Category: Introductory Geometry Problems]]
  2 KB (315 words) - 17:14, 2 August 2022
• 2015 AIME I Problems/Problem 15
  ...o positions is equal to 10<math>w</math>:6<math>w</math> = 10:6. Through a similar argument, the areas between each set of vertical lines also maintains a rat ..., with <math>T'</math> the center of this circular cross section. Then, by similar triangles, <math>T'M' = \frac{3}{4} y</math> and thus <cmath>A'B' = 2A' M'
  9 KB (1,407 words) - 19:37, 17 February 2024
• 2015 AIME I Problems/Problem 7
  ...th>, and triangles <math>\Delta BHJ</math> and <math>\Delta JFC</math> are similar to <math>\Delta EDC</math> since they are <math>1-2-\sqrt{5}</math> triangl ...ngth in terms of <math>x,y</math> since <math>AB=AN+NH+HB</math>. By using similar triangles as in the first part, we have
  6 KB (1,105 words) - 21:02, 9 November 2023
• 2015 AIME I Problems/Problem 4
  ...erages.) Thus, by the Mean Geometry Theorem, <math>\triangle BMN</math> is similar to both <math>\triangle BAD</math> and <math>\triangle BEC</math>, which me ...>. Additionally, <math>MX = \frac{1}{2} (2\sqrt{3}) = \sqrt{3}</math> from similar triangles meaning we can now just do pythagorean theorem on right triangle
  5 KB (852 words) - 00:12, 8 May 2024
• 2015 AIME I Problems/Problem 13
  Similar to Solution <math>2</math>, so we use <math>\sin{2\theta}=2\sin\theta\cos\t [[Category:Intermediate Geometry Problems]]
  9 KB (1,351 words) - 17:26, 16 January 2024
• 2015 AIME I Problems/Problem 11
  ...yields <math>AI=</math><math>\sqrt{x^2-2xy+64}</math>. <math>ADI</math> is similar to <math>AEB</math> by <math>AA</math>, so we [[Category: Intermediate Geometry Problems]]
  5 KB (906 words) - 17:43, 27 September 2023
• 2015 AIME II Problems/Problem 7
  Similar triangles can also solve the problem. ...Q</math> is parallel to <math>BC</math> meaning <math>\Delta APQ</math> is similar to <math>\Delta ABC</math>.
  7 KB (1,180 words) - 14:08, 14 February 2023
• 2015 AIME II Problems/Problem 15
  ...a lot of time left in your mocking of this AIME, go ahead and use analytic geometry. ...s k=\frac{5}{3}</math>. We can do some more length chasing using triangles similar to <math>O_1BN</math> to get that <math>AK = AL = \frac{24}{15}</math>, <ma
  31 KB (5,086 words) - 19:15, 20 December 2023

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