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2011 AMC 12A Problems/Problem 22
Using a method similar to before, if a point is <math>60</math>-ray partitional, then we must be a [[Category:Intermediate Geometry Problems]]

10 KB (1,819 words) - 20:39, 1 July 2024
2011 AMC 12B Problems/Problem 20
We can calculate the distances with coordinate geometry. (Note that <math>XA = XB = XC</math> because <math>X</math> is the circumc ...<math>\frac{XB}{XC} = \frac{BE}{CE} = 1</math>, <math>XB = XC</math>. In a similar fashion <math>XA = XB = XC = R</math>, where <math>R</math> is the circumci

8 KB (1,200 words) - 19:31, 7 August 2023
2001 AMC 10 Problems/Problem 21
...</math>. Examining the cross section of the cone and cylinder, we find two similar triangles. Hence, <math> \frac{12-2r}{12}=\frac{2r}{10} </math> which we so ==Solution 3 (Very similar to solution 2 but explained more)==

5 KB (697 words) - 23:14, 10 February 2024
2011 AIME I Problems/Problem 4
Note: This is similar to Solution 2 after the first four lines ...ivalent to the value of <math>\cos\left(\frac{\beta}{2}\right)</math> by a similar argument as above. Then rearranging yields <math>IC = b\cdot\frac{\sin\left

9 KB (1,523 words) - 12:08, 15 July 2024
2011 AIME I Problems/Problem 14
...t(1+\sqrt{2}\right)=0.\end{align*}</cmath> By the [https://www.cuemath.com/geometry/distance-between-two-lines/ distance between parallel lines formula], a cor ...til they intersect. Denote their intersection as <math>I_1</math>. Through similar triangles & the <math>45-45-90</math> triangles formed, we find that <math>

8 KB (1,344 words) - 18:39, 9 February 2023
2011 AIME I Problems/Problem 8
...et <math>\{u, v, w, x, y, z\} = \{AU, AV, CW, CX, BY, BZ\}</math>. Then by similar triangles ...h_c = \frac{40\sqrt{221}}{30}, h_a = \frac{40\sqrt{221}}{23}</math>. From similar triangles we can see that <math>\frac{27h}{h_a}+\frac{27h}{h_c} \le 27 \rig

6 KB (1,077 words) - 21:47, 12 April 2022
2011 AIME II Problems/Problem 13
==Solution 8 (similar to solution 4)== [[Category:Intermediate Geometry Problems]]

13 KB (2,055 words) - 05:25, 9 September 2022
2011 USAMO Problems/Problem 3
If <math>r-t=s-t=0</math>, then <math>P</math> must be similar to <math>P'</math> and the conclusion is obvious. [[Category:Olympiad Geometry Problems]]

11 KB (1,925 words) - 12:07, 31 August 2023
2011 AMC 10B Problems/Problem 9
[[Category: Introductory Geometry Problems]] ...of sides (in similar figures) because area is a second-degree property of similar figures. So like solution 3, the ratio of sides is <math>\sqrt{\frac{1}{3}}

4 KB (587 words) - 22:08, 31 August 2023
2011 USAMO Problems/Problem 5
...through <math>R</math>, contradicting <math>Q_1 Q_2 \parallel AB</math>. A similar contradiction occurs if <math>Q_1 Q_2 \parallel CD</math> but <math>Q_1 Q_2 [[Category:Olympiad Geometry Problems]]

4 KB (660 words) - 01:04, 15 February 2024
1996 AHSME Problems/Problem 30
import olympiad; import geometry; size(150); defaultpen(linewidth(0.8)); ...ngle ADE</math>. Hence triangles <math>ABQ</math> and <math>EDQ</math> are similar. Therefore, <cmath>\frac{AQ}{EQ}=\frac{BQ}{DQ}=\frac{AB}{ED}=\frac{3}{5}.</

10 KB (1,515 words) - 13:09, 20 December 2023
Alcumus
AoPS generates high-quality solutions for the students to learn and to do similar problems easily. *[[Geometry]]

3 KB (533 words) - 10:55, 7 February 2023
1950 AHSME Problems/Problem 47
...gle formed by taking away the rectangle and the two small portions left is similar to the big triangle.Hence, the proportions of the heights is equal to the p [[Category:Introductory Geometry Problems]]

2 KB (293 words) - 02:24, 11 June 2024
2012 AMC 10A Problems/Problem 15
[[Category: Introductory Geometry Problems]] Triangle <math>EAB</math> is similar to triangle <math>EHI</math>; line <math>HI = 1/2</math>

6 KB (935 words) - 23:41, 13 September 2023
2012 AMC 12A Problems/Problem 16
==Solution 9 (Similar Triangles)== [[Category:Introductory Geometry Problems]]

9 KB (1,496 words) - 02:40, 2 October 2022
2012 AMC 12B Problems/Problem 17
...In other words, the longer leg is 3 times the shorter leg in any triangle similar to <math>\triangle AGB</math>. ...Triangles <math>\triangle EKM</math> and <math>\triangle MKF</math>, being similar to <math>\triangle AGB</math>, also have legs in a 1:3 ratio, therefore, <m

12 KB (2,183 words) - 21:05, 23 December 2023
University of Northern Colorado Math Contest
...ey for the Final Round of the contest in late January. The Final Round is similar in spirit to the First Round, but the problems are more challenging. Studen ...e students to exercise their creativity and ingenuity to solve problems in geometry, algebra, combinatorics, probability, and number theory. All the old contes

2 KB (336 words) - 15:09, 2 January 2020
2012 AMC 10B Problems/Problem 23
We can approach this problem similar to the solution above, by attacking it on the Cartesian plane. As mentioned [[Category: Introductory Geometry Problems]]

5 KB (851 words) - 11:25, 21 April 2024
2012 AIME I Problems/Problem 8
...is a right triangular pyramid. Note also that pyramid <math>MPBQ</math> is similar to <math>DPCN</math> with scale factor <math>.5</math> and thus the volume ...nd the volume <math>[DNCMQB]</math> is a sum of areas of cross sections of similar triangles running parallel to face <math>ABFE.</math> Let <math>x</math> be

10 KB (1,592 words) - 13:35, 4 April 2024
1968 IMO Problems/Problem 1
...ACB}{2} = \angle ABC</math>, so <math>ABC</math> and <math>ADB</math> are similar by AA Similarity. Hence, <math>c^2 = b(a+b)</math>. Then proceed as in Solu [[Category:Olympiad Geometry Problems]]

4 KB (729 words) - 08:47, 9 March 2019

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