Difference between revisions of "2024 AMC 8 Problems/Problem 20"
Niuniumaths (talk | contribs) (→Video Solution by Power Solve) |
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~andliu766 | ~andliu766 | ||
+ | ==Video Solution by NiuniuMaths (Easy to understand!)== | ||
+ | https://www.youtube.com/watch?v=V-xN8Njd_Lc | ||
+ | |||
+ | ~NiuniuMaths | ||
==Video Solution by Power Solve== | ==Video Solution by Power Solve== | ||
https://www.youtube.com/watch?v=7_reHSQhXv8 | https://www.youtube.com/watch?v=7_reHSQhXv8 |
Revision as of 01:12, 27 January 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Video Solution by NiuniuMaths (Easy to understand!)
- 4 Video Solution by Power Solve
- 5 Video Solution 1 by Math-X (First understand the problem!!!)
- 6 Video Solution 2 by OmegaLearn.org
- 7 Video Solution 3 by SpreadTheMathLove
- 8 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 9 See Also
Problem
Any three vertices of the cube , shown in the figure below, can be connected to form a triangle. (For example, vertices , , and can be connected to form isosceles .) How many of these triangles are equilateral and contain as a vertex?
Solution 1
The only equilateral triangles that can be formed are through the diagonals of the faces of the square with length . From P you have possible vertices that are possible to form a diagonal through one of the faces. So there are possible triangles. So the answer ~Math 645
~andliu766
Video Solution by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=V-xN8Njd_Lc
~NiuniuMaths
Video Solution by Power Solve
https://www.youtube.com/watch?v=7_reHSQhXv8
Video Solution 1 by Math-X (First understand the problem!!!)
~Math-X
Video Solution 2 by OmegaLearn.org
Video Solution 3 by SpreadTheMathLove
https://www.youtube.com/watch?v=Svibu3nKB7E
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=Xg-1CWhraIM
See Also
2024 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.