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

 
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==Problem==
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== Problem ==
  
 
Any three vertices of the cube <math>PQRSTUVW</math>, shown in the figure below, can be connected to form a triangle. (For example, vertices <math>P</math>, <math>Q</math>, and <math>R</math> can be connected to form isosceles <math>\triangle PQR</math>.) How many of these triangles are equilateral and contain <math>P</math> as a vertex?
 
Any three vertices of the cube <math>PQRSTUVW</math>, shown in the figure below, can be connected to form a triangle. (For example, vertices <math>P</math>, <math>Q</math>, and <math>R</math> can be connected to form isosceles <math>\triangle PQR</math>.) How many of these triangles are equilateral and contain <math>P</math> as a vertex?
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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>
  
==Solution 1==
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== 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
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~Math645, andliu766, e___
~andliu766
 
~e___
 
  
==Solution 2==
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== 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 <math>{3 \choose 2} = \boxed{\textbf{(D) }3}</math>
 
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>
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-ILoveMath31415926535
 
-ILoveMath31415926535
  
 +
== Solution 3 (arduous, stubborn and not recommended) ==
  
==Solution 3 (arduous, stubborn and not recommended)==
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List them out- you get <math>PRV</math>, <math>PRT</math>,<math>PTQ</math>, and <math>PVT</math>. Therefore, the answer is <math>\boxed{\textbf{(F) 4}}</math>
  
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>
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-Anonymussrvusdmathstudent234234
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 +
== Solution 4 (Easy) ==
 +
 
 +
After looking at the cube, we realize that an equilateral triangle can only be formed by three lines that form a diagonal along a face of the cube (such as <math>PV</math>). Because the problem has a condition that one of the triangle's vertex must be on <math>P</math>, the three diagonals that can be formed are <math>PT, PR</math>, and <math>PV</math>.
  
 +
Now, we can choose any of 2 out of the 3 lines we have listed, and connect any of them with another line (for example, if we choose <math>PT</math> and <math>PR</math>, the third diagonal is <math>RT</math>). Thus, there are 3 ways to choose the 3 diagonals, so the answer is <math>3</math>, or <math>\boxed{\textbf{(D) }3}</math>
 +
 +
~TechnoDragon
  
-Anonymussrvusdmathstudent234234
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== Video Solution 1 by Central Valley Math Circle (Goes through full thought process) ==
 +
 
 +
https://youtu.be/JVPNQJL1Cqc
 +
 
 +
~mr_mathman
 +
 
 +
== Video Solution 2 by Math-X (First understand the problem) ==
 +
 
 +
https://youtu.be/BaE00H2SHQM?si=QSxNpXGLosdIpffx&t=5954
 +
 
 +
== Video Solution 3 (A Clever Explanation You’ll Get Instantly) ==
  
==Video Solution (A Clever Explanation You’ll Get Instantly)==
 
 
https://youtu.be/5ZIFnqymdDQ?si=_-gBZbXx4rn3nLnx&t=2912
 
https://youtu.be/5ZIFnqymdDQ?si=_-gBZbXx4rn3nLnx&t=2912
 
 
~hsnacademy
 
~hsnacademy
  
==Video Solution 1 by Math-X (First understand the problem!!!)==
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== Video Solution 4 (Under 3 minutes) ==
https://youtu.be/BaE00H2SHQM?si=QSxNpXGLosdIpffx&t=5954
 
  
==Video Solution (Under 3 minutes)🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥==
 
 
https://youtu.be/8X3Fwhp5-d8
 
https://youtu.be/8X3Fwhp5-d8
  
~please like and subscribe
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== Video Solution 5 by Power Solve ==
  
==Video Solution by Power Solve==
 
 
https://www.youtube.com/watch?v=7_reHSQhXv8
 
https://www.youtube.com/watch?v=7_reHSQhXv8
  
==Video Solution by NiuniuMaths (Easy to understand!)==
+
== Video Solution 6 by NiuniuMaths (Easy to understand) ==
 +
 
 
https://www.youtube.com/watch?v=V-xN8Njd_Lc
 
https://www.youtube.com/watch?v=V-xN8Njd_Lc
  
~NiuniuMaths
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== Video Solution 7 by OmegaLearn.org ==
  
==Video Solution 2 by OmegaLearn.org==
 
 
https://youtu.be/m1iXVOLNdlY
 
https://youtu.be/m1iXVOLNdlY
  
==Video Solution 3 by SpreadTheMathLove==
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== Video Solution 8 by SpreadTheMathLove ==
 +
 
 
https://www.youtube.com/watch?v=Svibu3nKB7E
 
https://www.youtube.com/watch?v=Svibu3nKB7E
  
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
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== Video Solution 9 by CosineMethod [Fast and Easy] ==
  
 
https://www.youtube.com/watch?v=Xg-1CWhraIM
 
https://www.youtube.com/watch?v=Xg-1CWhraIM
==Video Solution by Interstigation==
+
 
 +
== Video Solution 10 by Interstigation ==
 +
 
 
https://youtu.be/ktzijuZtDas&t=2353
 
https://youtu.be/ktzijuZtDas&t=2353
  
==Video Solution by Dr. David==
+
== Video Solution 11 by Dr. David ==
 +
 
 
https://youtu.be/yDM_2aGYRZU
 
https://youtu.be/yDM_2aGYRZU
  
==See Also==
+
== Video Solution by 12 WhyMath ==
 +
 
 +
https://youtu.be/XsC-x3b4mxE
 +
 
 +
== Video Solution (Effective, easy and simple) ==
 +
 
 +
https://www.youtube.com/watch?v=UwicmTsBvuU
 +
~TheMathGeek
 +
 
 +
==Video Solution by Daily Dose of Math (Simple, Certified, and Logical)==
 +
 
 +
https://youtu.be/OwJvuq6F7sQ
 +
 
 +
~Thesmartgreekmathdude
 +
 
 +
== 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 17:56, 30 January 2025

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 $PRV$, $PRT$,$PTQ$, and $PVT$. Therefore, the answer is $\boxed{\textbf{(F) 4}}$

-Anonymussrvusdmathstudent234234

Solution 4 (Easy)

After looking at the cube, we realize that an equilateral triangle can only be formed by three lines that form a diagonal along a face of the cube (such as $PV$). Because the problem has a condition that one of the triangle's vertex must be on $P$, the three diagonals that can be formed are $PT, PR$, and $PV$.

Now, we can choose any of 2 out of the 3 lines we have listed, and connect any of them with another line (for example, if we choose $PT$ and $PR$, the third diagonal is $RT$). Thus, there are 3 ways to choose the 3 diagonals, so the answer is $3$, or $\boxed{\textbf{(D) }3}$

~TechnoDragon

Video Solution 1 by Central Valley Math Circle (Goes through full thought process)

https://youtu.be/JVPNQJL1Cqc

~mr_mathman

Video Solution 2 by Math-X (First understand the problem)

https://youtu.be/BaE00H2SHQM?si=QSxNpXGLosdIpffx&t=5954

Video Solution 3 (A Clever Explanation You’ll Get Instantly)

https://youtu.be/5ZIFnqymdDQ?si=_-gBZbXx4rn3nLnx&t=2912 ~hsnacademy

Video Solution 4 (Under 3 minutes)

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

Video Solution 5 by Power Solve

https://www.youtube.com/watch?v=7_reHSQhXv8

Video Solution 6 by NiuniuMaths (Easy to understand)

https://www.youtube.com/watch?v=V-xN8Njd_Lc

Video Solution 7 by OmegaLearn.org

https://youtu.be/m1iXVOLNdlY

Video Solution 8 by SpreadTheMathLove

https://www.youtube.com/watch?v=Svibu3nKB7E

Video Solution 9 by CosineMethod [Fast and Easy]

https://www.youtube.com/watch?v=Xg-1CWhraIM

Video Solution 10 by Interstigation

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

Video Solution 11 by Dr. David

https://youtu.be/yDM_2aGYRZU

Video Solution by 12 WhyMath

https://youtu.be/XsC-x3b4mxE

Video Solution (Effective, easy and simple)

https://www.youtube.com/watch?v=UwicmTsBvuU ~TheMathGeek

Video Solution by Daily Dose of Math (Simple, Certified, and Logical)

https://youtu.be/OwJvuq6F7sQ

~Thesmartgreekmathdude

See Also

2024 AMC 8 (ProblemsAnswer KeyResources)
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

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