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

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==Solution 2==
 
==Solution 2==
  
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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 <math>{1000 \choose 2} = \boxed{\textbf{(D) }3}</math>
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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 <math>{3 \choose 2} = \boxed{\textbf{(D) }3}</math>
  
 
-ILoveMath31415926535
 
-ILoveMath31415926535
 +
  
 
==Solution 3 (arduous, stubborn and not recommended)==
 
==Solution 3 (arduous, stubborn and not recommended)==

Revision as of 12:59, 11 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]

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

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

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