2024 AMC 8 Problems/Problem 20
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?
![[asy] unitsize(4); pair P,Q,R,S,T,U,V,W; P=(0,30); Q=(30,30); R=(40,40); S=(10,40); T=(10,10); U=(40,10); V=(30,0); W=(0,0); draw(W--V); draw(V--Q); draw(Q--P); draw(P--W); draw(T--U); draw(U--R); draw(R--S); draw(S--T); draw(W--T); draw(P--S); draw(V--U); draw(Q--R); dot(P); dot(Q); dot(R); dot(S); dot(T); dot(U); dot(V); dot(W); label("$P$",P,NW); label("$Q$",Q,NW); label("$R$",R,NE); label("$S$",S,N); label("$T$",T,NE); label("$U$",U,NE); label("$V$",V,SE); label("$W$",W,SW); [/asy]](http://latex.artofproblemsolving.com/4/b/c/4bc4c5b02dfaf07a435a3a5098a847d0af49d6c1.png)
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. 
