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2018 AMC 10A Problems/Problem 13
...with <math>AC</math> as <math>E</math>. Now, <math>\triangle ACB</math> is similar to <math>\triangle ADE</math> by <math>AA</math> similarity. Setting up the Like in Solution 3, we can make use of coordinate geometry to solve this problem. Because the length of the crease is constant no matt

8 KB (1,193 words) - 15:12, 3 July 2023
2018 AMC 10A Problems/Problem 23
==Solution 3 (Similar Triangles)== ...is <math>\frac{5}{3}(2)=\frac{10}{3}</math>. Now, let's extend this larger similar right triangle to the left until it hits the side of length <math>3</math>.

16 KB (2,454 words) - 13:30, 23 September 2023
2018 AMC 10A Problems/Problem 24
...ath>. Therefore substituting we have that <math>BG=\frac{5a}{6}</math>. By similar triangles, we have that <math>DF=\frac{5a}{12}</math>, and the height of th [[Category:Intermediate Geometry Problems]]

8 KB (1,170 words) - 17:13, 3 March 2024
2018 AMC 10A Problems/Problem 15
...math>, and <math>XB</math>. Now observe that <math>\triangle XYZ</math> is similar to <math>\triangle XAB</math> by SAS. [[Category:Introductory Geometry Problems]]

2 KB (342 words) - 17:10, 15 October 2023
2018 AMC 10A Problems/Problem 9
All of the triangles in the diagram below are similar to isosceles triangle <math>ABC</math>, in which <math>AB=AC</math>. Each o ...s comprised of the <math>7</math> small isosceles triangles and a triangle similar to <math>ADE</math> with side length ratio <math>3:4</math> (so an area rat

7 KB (1,065 words) - 14:44, 3 July 2023
2018 AMC 12A Problems
...ways can a student schedule <math>3</math> mathematics courses -- algebra, geometry, and number theory -- in a <math>6</math>-period day if no two mathematics All of the triangles in the diagram below are similar to isosceles triangle <math>ABC</math>, in which <math>AB=AC</math>. Each o

15 KB (2,380 words) - 18:52, 7 April 2022
2018 AMC 12B Problems/Problem 16
...rt6}{2}-\frac{\sqrt6}{2}i,-\frac{\sqrt6}{2}+\frac{\sqrt6}{2}i}</math> by a similar process.</li><p> [[Category:Intermediate Geometry Problems]]

11 KB (1,708 words) - 12:01, 18 March 2023
2018 AMC 12B Problems/Problem 25
...gles <math>\triangle{P_1SO_1}</math> and <math>\triangle{MSO_2}</math> are similar by AA similarity. And, since <math>O_1P_1</math> is <math>4</math> and <mat [[Category:Intermediate Geometry Problems]]

13 KB (2,080 words) - 19:09, 21 October 2023
2018 AMC 10B Problems/Problem 24
...>\overline {CD}</math> into a <math>1:1</math> ratio (This can be shown by similar triangles.). Thus, it must also divide transversal <math>\overline {AC}</ma ...is similar to the triangle with a base of <math>YC = \frac12.</math> Using similar triangles, we calculate the base to be <math>\frac{1}{4}</math> and the hei

17 KB (2,562 words) - 03:38, 6 September 2022
2018 AMC 12B Problems/Problem 13
==Solution 1 (Similar Triangles)== ==Solution 2 (Similar Triangles)==

6 KB (1,081 words) - 22:53, 22 July 2022
2017 Indonesia MO Problems/Problem 1
...ath> to <math>g</math> and from <math>D</math> to <math>g</math>. By using similar steps, if <math>g</math> passes through <math>B</math>, the distance from < Using similar triangle ratios, we have <math>DX_3 = \tfrac{bx}{a}</math> and <math>X_1 =

5 KB (984 words) - 00:01, 2 October 2018
2018 AIME I Problems/Problem 13
...^2+c^2-a^2}{2bc}}2} = \sqrt{\dfrac{(a-b+c)(a+b-c)}{4bc}}.</cmath> Applying similar logic to <math>\sin \tfrac B2</math> and <math>\sin\tfrac C2</math> and sim ==Solution 2 (Similar to Official MAA)==

13 KB (2,200 words) - 21:36, 6 January 2024
2018 AIME I Problems/Problem 8
First of all, draw a good diagram! This is always the key to solving any geometry problem. Once you draw it, realize that <math>EF=2, FA=16</math>. Why? Beca The equilateral triangle of side length <math>10</math> is similar to our large equilateral triangle of <math>24</math>. And the height of the

2 KB (298 words) - 10:44, 5 July 2022
2018 AIME II Problems/Problem 12
...rline{DB}</math> bisects diagonal <math>\overline{AC}</math>. This is very similar to what happens if <math>ABCD</math> were a parallelogram with <math>AB = C ==Solution 8 (Simple Geometry)==

18 KB (2,912 words) - 13:12, 24 January 2024
CyberMath Academy
...ents with the essential concepts and techniques in Pre-Algebra, Algebra I, Geometry, Number Theory and Combinatorics. We will have a specific emphasis on probl ...covers topics in trigonometry, advanced algebra, precalculus and advanced geometry, but excludes calculus. Our curriculum also includes some additional challe

6 KB (788 words) - 19:10, 22 December 2022
2008 iTest Problems/Problem 38
...is <math>4(\tfrac{6}{r} + 6 + 6r)</math> and that the equation above has a similar form. [[Category:Intermediate Geometry Problems]]

1 KB (176 words) - 17:08, 15 March 2021
1963 AHSME Problems/Problem 37
...hen <math>P</math> is between <math>P_1</math> and <math>P_7</math>. Using similar logic, <math>P</math> must be between <math>P_3</math> and <math>P_5</math> [[Category:Intermediate Geometry Problems]]

1 KB (225 words) - 18:53, 15 July 2018
1973 AHSME Problems/Problem 35
...Now, extend <math>OR</math> to meet the circle again at <math>T</math>. By similar reasoning, <math>MT = s</math>. Furthermore, since <math>\angle MOT, \angle [[Category:Intermediate Geometry Problems]]

7 KB (1,181 words) - 20:41, 17 April 2022
2008 iTest Problems/Problem 96
...C</math>. If triangles <math> ABC</math> and <math>XYZ</math> are directly similar, then the area of <math>\triangle XYZ</math> may be written in the form <ma [[Category:Olympiad Geometry Problems]]

3 KB (558 words) - 20:13, 4 January 2019
2019 AIME II Problems/Problem 7
...AHG\sim \triangle BID\sim \triangle EFC\sim \triangle ABC</math>, so using similar triangle ratios, we find that <math>BI=HA=30</math>, <math>BD=HG=55</math>, ...riangle EXD</math> and <math>\triangle BID\sim \triangle HIZ</math>. Using similar triangles, we get that

7 KB (1,053 words) - 14:58, 14 January 2024

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