Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 1)
(→‎Solution 1)
 
(24 intermediate revisions by 17 users not shown)
Line 27: Line 27:
  
 
<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>
 +
 +
 +
==Official Video Solution (Effective, easy and simple!)==
 +
https://www.youtube.com/watch?v=UwicmTsBvuU
 +
~TheMathGeek
 +
  
 
==Solution 1==
 
==Solution 1==
  
 
The only equilateral triangles that can be formed are through the diagonals of the faces of the square. From P you have <math>3</math> possible vertices that are possible to form a diagonal through one of the faces. Therefore, there are <math>3</math> possible triangles. So the answer is <math>\boxed{\textbf{(D) }3}</math>
 
The only equilateral triangles that can be formed are through the diagonals of the faces of the square. From P you have <math>3</math> possible vertices that are possible to form a diagonal through one of the faces. Therefore, there are <math>3</math> possible triangles. So the answer is <math>\boxed{\textbf{(D) }3}</math>
−
~Math645 ~ e___ from Evergreen Middle School
+
~Math645
 
~andliu766
 
~andliu766
 +
~e___
  
 
==Solution 2==
 
==Solution 2==
Line 39: Line 46:
  
 
-ILoveMath31415926535
 
-ILoveMath31415926535
 +
 +
 +
==Solution 3 (arduous, stubborn and not recommended)==
 +
 +
List them out- you get <math>PRV</math>, <math>PRT</math>, and <math>PVT</math>. Therefore, the answer is <math>\boxed{\textbf{(D) }3}</math>
 +
 +
 +
-Anonymussrvusdmathstudent234234
 +
  
 
==Video Solution 1 by Math-X (First understand the problem!!!)==
 
==Video Solution 1 by Math-X (First understand the problem!!!)==
 
https://youtu.be/BaE00H2SHQM?si=QSxNpXGLosdIpffx&t=5954
 
https://youtu.be/BaE00H2SHQM?si=QSxNpXGLosdIpffx&t=5954
  
−
~Math-X
+
~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)🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥==
 +
https://youtu.be/8X3Fwhp5-d8
 +
 
 +
~please like and subscribe
 +
 
 +
==Video Solution by Power Solve==
 +
https://www.youtube.com/watch?v=7_reHSQhXv8
  
 
==Video Solution by NiuniuMaths (Easy to understand!)==
 
==Video Solution by NiuniuMaths (Easy to understand!)==
Line 49: Line 77:
  
 
~NiuniuMaths
 
~NiuniuMaths
−
==Video Solution by Power Solve==
 
−
https://www.youtube.com/watch?v=7_reHSQhXv8
 
  
 
==Video Solution 2 by OmegaLearn.org==
 
==Video Solution 2 by OmegaLearn.org==
Line 61: Line 87:
  
 
https://www.youtube.com/watch?v=Xg-1CWhraIM
 
https://www.youtube.com/watch?v=Xg-1CWhraIM
 +
==Video Solution by Interstigation==
 +
https://youtu.be/ktzijuZtDas&t=2353
 +
 +
==Video Solution by Dr. David==
 +
https://youtu.be/yDM_2aGYRZU
 +
 +
==Video Solution by WhyMath==
 +
https://youtu.be/XsC-x3b4mxE
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2024|num-b=19|num-a=21}}
 
{{AMC8 box|year=2024|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 10:00, 23 December 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?

[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]

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


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 $3$ possible vertices that are possible to form a diagonal through one of the faces. Therefore, there are $3$ possible triangles. So the answer is $\boxed{\textbf{(D) }3}$ ~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 ${3 \choose 2} = \boxed{\textbf{(D) }3}$

-ILoveMath31415926535


Solution 3 (arduous, stubborn and not recommended)

List them out- you get $PRV$, $PRT$, and $PVT$. Therefore, the answer is $\boxed{\textbf{(D) }3}$


-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)🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥

https://youtu.be/8X3Fwhp5-d8

~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

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

Video Solution by Interstigation

https://youtu.be/ktzijuZtDas&t=2353

Video Solution by Dr. David

https://youtu.be/yDM_2aGYRZU

Video Solution by WhyMath

https://youtu.be/XsC-x3b4mxE

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_8_Problems/Problem_20&oldid=237626"