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> | ||
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| + | |||
| + | ==Official Video Solution (Effective, easy and simple!)== | ||
| + | https://www.youtube.com/watch?v=UwicmTsBvuU | ||
| + | ~TheMathGeek | ||
| + | |||
==Solution 1== | ==Solution 1== | ||
Latest revision as of 10:00, 23 December 2024
Contents
- 1 Problem
- 2 Official Video Solution (Effective, easy and simple!)
- 3 Solution 1
- 4 Solution 2
- 5 Solution 3 (arduous, stubborn and not recommended)
- 6 Video Solution 1 by Math-X (First understand the problem!!!)
- 7 Video Solution (A Clever Explanation You’ll Get Instantly)
- 8 Video Solution (Under 3 minutes)🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥
- 9 Video Solution by Power Solve
- 10 Video Solution by NiuniuMaths (Easy to understand!)
- 11 Video Solution 2 by OmegaLearn.org
- 12 Video Solution 3 by SpreadTheMathLove
- 13 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 14 Video Solution by Interstigation
- 15 Video Solution by Dr. David
- 16 Video Solution by WhyMath
- 17 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)
Official Video Solution (Effective, easy and simple!)
https://www.youtube.com/watch?v=UwicmTsBvuU ~TheMathGeek
Solution 1
The only equilateral triangles that can be formed are through the diagonals of the faces of the square. From P you have 
possible vertices that are possible to form a diagonal through one of the faces. Therefore, there are possible triangles. So the answer is 
~Math645
~andliu766
~e___
Solution 2
Each other compatible point must be an even number of edges away from P, so the compatible points are R, V, and T. Therefore, we must choose two of the three points, because P must be a point in the triangle. So, the answer is 
-ILoveMath31415926535
Solution 3 (arduous, stubborn and not recommended)
List them out- you get 
, 
, and 
. Therefore, the answer is
-Anonymussrvusdmathstudent234234
Video Solution 1 by Math-X (First understand the problem!!!)
https://youtu.be/BaE00H2SHQM?si=QSxNpXGLosdIpffx&t=5954
~MATH-X
Video Solution (A Clever Explanation You’ll Get Instantly)
https://youtu.be/5ZIFnqymdDQ?si=_-gBZbXx4rn3nLnx&t=2912 ~hsnacademy
Video Solution (Under 3 minutes)🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥
~please like and subscribe
Video Solution by Power Solve
https://www.youtube.com/watch?v=7_reHSQhXv8
Video Solution by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=V-xN8Njd_Lc
~NiuniuMaths
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
Video Solution by Interstigation
https://youtu.be/ktzijuZtDas&t=2353
Video Solution by Dr. David
Video Solution by WhyMath
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. 
