Difference between revisions of "2024 AMC 8 Problems/Problem 20"
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<math>\textbf{(A)}0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }6</math> | <math>\textbf{(A)}0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }6</math> | ||
| − | ==Solution 1== | + | ==Solution 1== |
| − | The only equilateral triangles that can be formed are through the diagonals of the faces of the square with length sqrt | + | The only equilateral triangles that can be formed are through the diagonals of the faces of the square with length <math>\sqrt{2}</math>. From P you have <math>3</math> possible vertices that are possible to form a diagonal through one of the faces. So there are <math>3</math> possible triangles. So the answer <math>\boxed{\textbf{(D) }3}</math> |
| + | ~Math 645 | ||
| + | |||
| + | ~andliu766 | ||
==Video Solution 1 by Math-X (First understand the problem!!!)== | ==Video Solution 1 by Math-X (First understand the problem!!!)== | ||
Revision as of 17:21, 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
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 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
