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• 1983 AIME Problems/Problem 14
  ...we can find the third side of the triangle using [[Stewart's Theorem]] or similar approaches. We get <math>AC = \sqrt{56}</math>. Now we have a kite <math>AQ ...=\frac{BR}{NR},</math> triangles <math>BNR</math> and <math>AMR</math> are similar. If we let <math>y=BN</math>, we have <math>AM=3BN=3y</math>.
  13 KB (2,151 words) - 17:48, 27 May 2024
• 1983 AIME Problems/Problem 15
  This solution, while similar to Solution 2, is arguably more motivated and less contrived. === Solution 4 (coordinate geometry) ===
  20 KB (3,497 words) - 15:37, 27 May 2024
• 1984 AIME Problems/Problem 3
  ...that go through <math>P</math>, all four triangles are [[similar triangles|similar]] to each other by the <math>AA</math> postulate. Also, note that the lengt Alternatively, since the triangles are similar by <math>AA</math>, then the ratios between the bases and the heights of ea
  4 KB (726 words) - 13:39, 13 August 2023
• 1985 AIME Problems/Problem 4
  A small [[square (geometry) | square]] is constructed inside a square of [[area]] 1 by dividing each s ...c{n-1}{\sqrt{1985}}</math>. Notice that <math>\triangle CEL</math> is also similar to <math>\triangle CDF</math> by <math>AA</math> similarity. Thus, <math>\f
  3 KB (484 words) - 21:40, 2 March 2020
• 1986 AIME Problems/Problem 9
  ...ll three smaller triangles and the larger triangle are [[similar triangles|similar]] (<math>\triangle ABC \sim \triangle DPD' \sim \triangle PEE' \sim \triang By similar triangles, <math>BE'=\frac{d}{510}\cdot450=\frac{15}{17}d</math> and <math>
  11 KB (1,850 words) - 18:07, 11 October 2023
• 1987 AIME Problems/Problem 15
  Because all the [[triangle]]s in the figure are [[similar]] to triangle <math>ABC</math>, it's a good idea to use [[area ratios]]. In [[Category:Intermediate Geometry Problems]]
  5 KB (838 words) - 18:05, 19 February 2022
• 1988 AIME Problems/Problem 12
  This problem is quite similar to the 1992 AIME problem 14, which also featured concurrent cevians and is [[Category:Intermediate Geometry Problems]]
  4 KB (727 words) - 23:37, 7 March 2024
• 1989 AIME Problems/Problem 15
  ...onally we now see that triangles <math>FPE</math> and <math>CPB</math> are similar, so <math>FE \parallel BC</math> and <math>\frac{FE}{BC} = \frac{1}{3}</mat [[Category:Intermediate Geometry Problems]]
  13 KB (2,091 words) - 00:20, 26 October 2023
• 1989 AIME Problems/Problem 10
  By identical logic, we can find similar expressions for the sums of the other two cotangents: [[Category:Intermediate Geometry Problems]]
  8 KB (1,401 words) - 21:41, 20 January 2024
• 2006 AMC 10B Problems/Problem 15
  Rhombus <math>ABCD</math> is similar to rhombus <math>BFDE</math>. The area of rhombus <math>ABCD</math> is <mat [[Category:Introductory Geometry Problems]]
  3 KB (445 words) - 22:01, 20 August 2022
• 1991 AIME Problems/Problem 12
  By similar logic, we have <math>APOS</math> is a cyclic quadrilateral. Let <math>AP = ...iangle ABD \sim \triangle OQP</math> by <math>AA</math> [[similar triangle|similar]]ity. From here, it's clear that
  8 KB (1,270 words) - 23:36, 27 August 2023
• 1991 AIME Problems/Problem 2
  ...r of the rectangle would have length <math>\frac{1}{168}\cdot{5}</math> by similar triangles. If you add the two lengths together, it is <math>\frac{167}{168} [[Category:Intermediate Geometry Problems]]
  4 KB (595 words) - 12:51, 17 June 2021
• 1992 AIME Problems/Problem 12
  ...square]]s. To take a bite, a player chooses one of the remaining [[square (geometry) | squares]], then removes ("eats") all squares in the quadrant defined by This game is similar to an AoPS book.
  2 KB (443 words) - 22:41, 22 December 2021
• 1992 AIME Problems/Problem 9
  ...the equations in <math>(1)</math> without directly resorting to trig. From similar triangles, [[Category:Intermediate Geometry Problems]]
  5 KB (874 words) - 10:27, 22 August 2021
• 2006 AMC 10B Problems/Problem 20
  ...has legs <math>2</math> and <math>20</math>. Aha! The two triangles are similar by SAS, with one triangle having side lengths <math>100</math> times the ot ...and 2 is one of the lengths of the adjacent sides. Those two triangles are similar because <math>AD</math> and <math>AB</math> are perpendicular. <math> \frac
  4 KB (594 words) - 15:45, 30 July 2023
• 2006 AMC 10B Problems/Problem 23
  We use mass points (similar to above). Let the triangle be <math>ABC</math> with cevians (lines to oppo [[Category:Introductory Geometry Problems]]
  5 KB (861 words) - 00:53, 25 November 2023
• 2006 AMC 10B Problems/Problem 24
  == Solution 1 (Similar Triangles) == ...point <math>E</math>. Triangles <math>EBO</math> and <math>ECP</math> are similar, and by symmetry, so are triangles <math>EAO</math> and <math>EDP</math>. T
  4 KB (558 words) - 14:38, 6 April 2024
• 1994 AIME Problems/Problem 10
  Using similar right triangles, we identify that <math>CD = \sqrt{AD \cdot BD}</math>. Let [[Category:Intermediate Geometry Problems]]
  3 KB (534 words) - 16:23, 26 August 2018
• 1994 AIME Problems/Problem 6
  ...agon to the area of a regular triangle. Since the ratio of the area of two similar figures is the square of the ratio of their side lengths, we see that the r Picturing the diagram in your head should give you an illustration similar to the one above. The distance from parallel sides of the center hexagon is
  4 KB (721 words) - 16:14, 8 March 2021
• 1995 AIME Problems/Problem 12
  Similar to Solution 1, <math>\angle APC</math> is the dihedral angle we want. WLOG, [[Category:Intermediate Geometry Problems]]
  8 KB (1,172 words) - 21:57, 22 September 2022

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