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2006 AMC 10A Problems/Problem 19
How many non-[[similar]] triangles have angles whose degree measures are distinct positive integer [[Category:Introductory Geometry Problems]]

2 KB (259 words) - 03:10, 22 June 2023
2005 AIME II Problems/Problem 8
...<math>O_3</math> to <math>\overline{AB}</math> be <math>T</math>. From the similar [[right triangle]]s <math>\triangle HO_1T_1 \sim \triangle HO_2T_2 \sim \tr ...=10-4=6</math>, and <math>O_1O_2=14</math>. But <math>O_1O_3O_3'</math> is similar to <math>O_1O_2O_2'</math> so <math>O_3O_3'=\frac{6}{14} \cdot 10=\frac{30}

4 KB (693 words) - 13:03, 28 December 2021
2005 AIME II Problems/Problem 12
=== Solution 3 (similar triangles)=== ...>, and <math>KA = EB</math> (90 degree rotation), and now we can bash on 2 similar triangles <math>\triangle GAK \sim \triangle GHO</math>.

13 KB (2,080 words) - 21:20, 11 December 2022
2005 AIME II Problems/Problem 15
...y=ax </math> contains the center of a circle that is externally [[tangent (geometry)|tangent]] to <math> w_2 </math> and internally tangent to <math> w_1. </ma ...stance from this tangent point to the origin is <math>\sqrt{69}.</math> By similar triangles, the slope of this line is then <math>\frac{\sqrt{69}}{5\sqrt{3}}

12 KB (2,000 words) - 13:17, 28 December 2020
2005 AIME II Problems/Problem 14
import olympiad; import cse5; import geometry; size(150); == Solution 2 (Similar Triangles)==

13 KB (2,129 words) - 18:56, 1 January 2024
2004 AIME I Problems/Problem 14
...riangle]]s <math>\triangle CDA</math> and <math>\triangle CEB</math> are [[similar]] [[right triangle]]s. By the Pythagorean Theorem <math>CD=8\cdot\sqrt{6}< [[Category:Intermediate Geometry Problems]]

4 KB (729 words) - 01:00, 27 November 2022
2004 AIME I Problems/Problem 11
...se the plane cut is parallel to the base of our solid, <math>C</math> is [[similar]] to the uncut solid and so the height and slant height of cone <math>C</ma <math>V</math> and <math>C</math> are similar cones, because the plane that cut out <math>C</math> was parallel to the ba

5 KB (839 words) - 22:12, 16 December 2015
2004 AIME I Problems/Problem 10
...{A'B'C'} = r_{ABC} - 1 = 5</math>, and since the ratio of the areas of two similar figures is equal to the square of the ratio of their corresponding lengths, ...dn't know the formula for the distance from a point to a line, you can use similar triangles to get the ratio:

5 KB (836 words) - 07:53, 15 October 2023
2004 AIME II Problems/Problem 12
We use a similar argument with the line <math>DO</math>, and find the height from the top of [[Category:Intermediate Geometry Problems]]

3 KB (431 words) - 23:21, 4 July 2013
2004 AIME II Problems/Problem 7
...theta)=\frac{BC'}{17}=\frac53</math>, so <math>BC'=\frac{85}{3}</math>. By similar triangles, <math>CC'=\frac{3}{17}BC'=\frac{15}{3}</math>, so <math>BC=\frac ...le, and so are B'DG and C'FG. Since C'F is 3, then using the properties of similar triangles GF is 51/8. DF is 22, so DG is 125/8. Finally, DB can to calculat

9 KB (1,501 words) - 05:34, 30 October 2023
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 (2,149 words) - 18:44, 5 February 2024
1983 AIME Problems/Problem 15
This solution, while similar to Solution 2, is arguably more motivated and less contrived. === Solution 4 (coordinate geometry) ===

19 KB (3,221 words) - 01:05, 7 February 2023
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

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