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• 2006 AIME II Problems/Problem 12
  Because the ratio of the area of two similar figures is the square of the ratio of the corresponding sides, <math>[GBC] ==Solution 2: Analytic Geometry/Coord Bash==
  6 KB (1,033 words) - 02:36, 19 March 2022
• Primary Mathematics World Contest
  ...similar to [[AMC 10]]/[[AMC 12]] level, with an emphasis on number theory, geometry and logic.
  1 KB (183 words) - 13:57, 15 October 2018
• 2004 AMC 12A Problems/Problem 18
  ...>\overline{AB}</math> is constructed inside the square, and the [[tangent (geometry)|tangent]] to the semicircle from <math>C</math> intersects side <math>\ove ...h>CDE</math> are in arithmetic progression. Thus it is [[similar triangles|similar]] to the triangle <math>3 - 4 - 5</math> and since <math>DC = 2</math>, <ma
  5 KB (738 words) - 13:11, 27 March 2023
• Lexington Math Tournament
  ...nd is replaced with an Accuracy Round, but the rest of the event follows a similar format. ...ach competition, but some examples of past mini-events are the estimathon, geometry bee, science bowl (or ball), and Tetris.
  3 KB (484 words) - 20:04, 12 March 2024
• Concurrence/Problems
  Hallie is teaching geometry to Warren. She tells him that the three medians, the three angle bisectors, To find the similar formula for <math>CE</math>, we just switch the signs of <math>BC^2</math>
  3 KB (458 words) - 15:44, 1 December 2015
• 2000 AMC 12 Problems/Problem 21
  ...e angle chasing shows that the two right triangles are [[similar triangles|similar]]. Thus the ratio of the sides of the triangles are the same. Since <math>A Since triangle <math>A</math> is similar to the large triangle, it has <math>h_A = a(\frac{c}{b}) = \frac{ac}{b}</ma
  5 KB (804 words) - 01:22, 13 May 2024
• 2007 AMC 10A Problems/Problem 21
  ...th> be the surface area of the inner square. The ratio of the areas of two similar figures is equal to the square of the ratio of their sides. As the diagonal [[Category:Introductory Geometry Problems]]
  4 KB (609 words) - 14:41, 3 December 2023
• 1951 AHSME Problems
  In connection with proof in geometry, indicate which one of the following statements is ''incorrect'': <math> \textbf{(A)}\ \text{the triangles are similar in opposite pairs}\qquad\textbf{(B)}\ \text{the triangles are congruent in
  23 KB (3,641 words) - 22:23, 3 November 2023
• 2002 AMC 12B Problems/Problem 20
  ...ht triangles <math>\triangle MON</math> and <math>\triangle XOY</math> are similar by Leg-Leg with a ratio of <math>\frac{1}{2}</math>, so <math>XY=2(MN)=\box [[Category:Introductory Geometry Problems]]
  3 KB (447 words) - 15:02, 17 August 2023
• 2002 AMC 12B Problems/Problem 25
  Similar to Solution 1, we proceed to get the area of the circle satisfying <math>f( [[Category:Introductory Geometry Problems]]
  2 KB (365 words) - 14:48, 7 March 2022
• 2007 AMC 10A Problems/Problem 18
  ...at triangles <math>\triangle AMH</math> and <math>\triangle GMC</math> are similar because <math>\overline{AH} \parallel \overline{CG}</math>. Also, since <ma The triangles <math>CBN</math> and <math>CGH</math> are clearly similar with ratio <math>1:2</math>, hence <math>BN=2</math> and thus <math>AN=6</m
  6 KB (867 words) - 00:17, 20 May 2023
• 2004 AMC 12B Problems/Problem 19
  Above, <math>E,F,</math> and <math>G</math> are points of [[tangent (geometry)|tangency]]. By the Two Tangent Theorem, <math>BF = BE = 18</math> and <mat ...erline{FH} = 3x</math>.Triangles <math>OFH</math> and <math>BEH</math> are similar so <math>\frac{\overline{OF}}{\overline{BE}} = \frac{\overline{FH}}{\overl
  3 KB (520 words) - 19:12, 20 November 2023
• 2007 AMC 10B Problems/Problem 11
  <math>\triangle ADC</math> is similar to <math>\triangle ACR</math>, so <cmath>\frac {2\sqrt{2}}{3} = \frac {3}{2 [[Category:Introductory Geometry Problems]]
  5 KB (851 words) - 22:02, 26 July 2021
• 2002 AMC 10A Problems/Problem 25
  ...F</math> is similar to triangle <math>BEC</math>, we can use properties of similar triangles and find that <math>DE = 12 \cdot \frac{5}{13} = \frac{60}{13}</m [[Category:Introductory Geometry Problems]]
  8 KB (1,308 words) - 07:05, 19 December 2022
• Isogonal conjugate
  One can prove a similar theorem in the case <math>P</math> outside <math>\triangle ABC.</math> ...cated on the ray <math>OA_1</math> so that <math>\triangle OAA_1</math> is similar to <math>\triangle OA_2A</math>. Points <math>B_2</math> and <math>C_2</mat
  54 KB (9,416 words) - 08:40, 18 April 2024
• 2008 AMC 12B Problems/Problem 11
  ...c 1 8</math> that of the entire cone (mountain). These cones are obviously similar so the radius and height of the small cone must be <math>\sqrt[3]{\frac1 8} [[Category:Introductory Geometry Problems]]
  2 KB (251 words) - 22:12, 8 May 2021
• 2008 AIME I Problems/Problem 14
  We let <math>AC=x</math>. From similar triangles, we have that <math>PC=\frac{x\sqrt{x^2+18x}}{x+9}</math> (Use Py ...nt on <math>x</math> is that it must be greater than <math>0</math>. Using similar triangles, we can deduce that <math>PA=\frac{9x}{x+9}</math>. Now, apply la
  8 KB (1,333 words) - 00:18, 1 February 2024
• 2008 AIME II Problems/Problem 5
  Similar to solution 1; Notice that it forms a right triangle. Remembering that the [[Category:Intermediate Geometry Problems]]
  8 KB (1,338 words) - 23:15, 28 November 2023
• 2008 AIME II Problems/Problem 11
  ...eta) = \frac {3}{4}</math>. Therefore, <math>XC = \frac {64}{3}</math>. By similar reasoning in triangle <math>QBY</math>, we see that <math>BY = \frac {4r}{3 ...- (\overline{PZ})^2} = \sqrt{(16+r)^2-(16-r)^2} = 8\sqrt{r}</cmath> Using similar similarity as was done to find <math>\overline{BY}</math> we have <math>\fr
  6 KB (1,065 words) - 20:12, 9 August 2022
• 1998 APMO Problems/Problem 4
  ...ath> are similar triangles, so <math>AFEN</math> and <math>ACBM</math> are similar figures. It follows that [[Category:Olympiad Geometry Problems]]
  2 KB (345 words) - 18:42, 13 April 2008

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