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• 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
• 1995 AIME Problems/Problem 9
  In a similar fashion, we encode the angles as complex numbers, so if <math>BM=x</math>, [[Category:Intermediate Geometry Problems]]
  7 KB (1,181 words) - 13:47, 3 February 2023
• 1995 AIME Problems/Problem 4
  ...th>O_9A_9 = \frac{2 \cdot O_6A_6 + 1 \cdot O_3A_3}{3} = 5</math> (consider similar triangles). Applying the [[Pythagorean Theorem]] to <math>\triangle O_9A_9P == Solution 2 (Analytic Geometry) ==
  3 KB (605 words) - 11:30, 5 May 2024
• 1996 AIME Problems/Problem 14
  ...gh the interiors of how many of the <math>1\times 1\times 1</math> [[cube (geometry) | cube]]s? ...(plane of <math>x=0</math> is not considered since <math>m \ne 0</math>). Similar arguments for slices along <math>y</math>-planes and <math>z</math>-planes
  5 KB (923 words) - 21:21, 22 September 2023
• 1996 AIME Problems/Problem 4
  ...= 49</math>, and so the sides of the shadow are <math>7</math>. Using the similar triangles in blue, [[Category:Intermediate Geometry Problems]]
  2 KB (257 words) - 17:50, 4 January 2016
• 1998 AIME Problems/Problem 12
  In a similar vein, using LoC on <math>\Delta PEQ</math> and <math>\Delta CEQ,</math> res [[Category:Intermediate Geometry Problems]]
  5 KB (876 words) - 20:27, 9 June 2022
• 1998 AIME Problems/Problem 6
  There are several [[similar triangles]]. <math>\triangle PAQ\sim \triangle PDC</math>, so we can write [[Category:Intermediate Geometry Problems]]
  2 KB (254 words) - 19:38, 4 July 2013
• 1999 AIME Problems/Problem 8
  ...responding angles we see that all of the triangles are [[similar triangles|similar]], so they are all equilateral triangles. We can solve for their side lengt [[Category:Intermediate Geometry Problems]]
  3 KB (445 words) - 19:40, 4 July 2013
• 2000 AIME I Problems/Problem 8
  By [[similar triangle]]s, we find that the dimensions of the liquid in the first cone to ...of the container is <math>100\pi</math>. The cone formed by the liquid is similar to the original, but scaled down by <math>\frac{3}{4}</math> in all directi
  4 KB (677 words) - 16:33, 30 December 2023
• 2001 AIME I Problems/Problem 7
  ...> is <math>\frac{43}{63}</math>, which is the scale factor between the two similar triangles, and thus <math>DE = \frac{43}{63} \times 20 = \frac{860}{63}</ma ...ratio of <math>FP:PA</math> is <math>20:43</math>. Therefore, the smaller similar triangle <math>ADE</math> is <math>43/63</math> the height of the original
  9 KB (1,540 words) - 08:31, 1 December 2022
• 2002 AIME I Problems/Problem 11
  We can use a similar trick as with reflections in 2D: Imagine that the entire space is divided i [[Category:Intermediate Geometry Problems]]
  3 KB (591 words) - 15:11, 21 August 2019

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