Create the page "Similar(geometry)" on this wiki! See also the search results found.

• 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
• 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

View (previous 20 | next 20) (20 | 50 | 100 | 250 | 500)