Difference between revisions of "2024 AMC 8 Problems/Problem 20"

(Video Solution 3 by SpreadTheMathLove)
(Solution 1: by Math 645)
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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==
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==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 == D (3). ==
  
 
==Video Solution 1 by Math-X (First understand the problem!!!)==
 
==Video Solution 1 by Math-X (First understand the problem!!!)==

Revision as of 15:53, 26 January 2024

Problem

Any three vertices of the cube $PQRSTUVW$, shown in the figure below, can be connected to form a triangle. (For example, vertices $P$, $Q$, and $R$ can be connected to form isosceles $\triangle PQR$.) How many of these triangles are equilateral and contain $P$ as a vertex?

$\textbf{(A)}0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }6$

==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 == D (3). ==

Video Solution 1 by Math-X (First understand the problem!!!)

https://youtu.be/N_9qlD9pgL0

~Math-X

Video Solution 2 by OmegaLearn.org

https://youtu.be/m1iXVOLNdlY

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