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,NW);
label("
",Q,NW);
label("
",R,NE);
label("
",S,N);
label("
",T,NE);
label("
",U,NE);
label("
",V,SE);
label("
",W,SW);
[/asy]
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.