Difference between revisions of "2024 AMC 8 Problems/Problem 20"
(→Solution 1: by Math 645) |
(→Problem) |
||
| Line 7: | Line 7: | ||
==Solution 1== by Math 645 | ==Solution 1== by Math 645 | ||
| − | The only equilateral triangles that can be formed are through the diagonals of the faces of the square with length sqrt(2). From P you have 3 possible vertices that are possible to form a diagonal through one of the faces. So there are 3 possible triangles. So the answer is | + | The only equilateral triangles that can be formed are through the diagonals of the faces of the square with length sqrt(2). From P you have 3 possible vertices that are possible to form a diagonal through one of the faces. So there are 3 possible triangles. So the answer is \textbf{(D) }3 \qquad |
==Video Solution 1 by Math-X (First understand the problem!!!)== | ==Video Solution 1 by Math-X (First understand the problem!!!)== | ||
Revision as of 16:07, 26 January 2024
Contents
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== by Math 645
The only equilateral triangles that can be formed are through the diagonals of the faces of the square with length sqrt(2). From P you have 3 possible vertices that are possible to form a diagonal through one of the faces. So there are 3 possible triangles. So the answer is \textbf{(D) }3 \qquad
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
