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Difference between revisions of "2019 AMC 8 Problems/Problem 12"

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==Problem==
 
==Problem==
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The faces of a cube are painted in six different colors: red (R), white (W), green (G), brown (B), aqua (A), and purple (P). Three views of the cube are shown below. What is the color of the face opposite the aqua face?
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The faces of a cube are painted in six different colors: red <math>(R)</math>, white <math>(W)</math>, green <math>(G)</math>, brown <math>(B)</math>, aqua <math>(A)</math>, and purple <math>(P)</math>. Three views of the cube are shown below. What is the color of the face opposite the aqua face?
  
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[[File:2019AMC8Prob12.png]]
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<!--- [[File:2019AMC8Prob12.png]] -->
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<asy>
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unitsize(2cm);
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pair x, y, z, trans;
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int i;
  
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<math>\textbf{(A) }\text{red}\qquad\textbf{(B) }\text{white}\qquad\textbf{(C) }\text{green}\qquad\textbf{(D) }\text{brown}\qquad\textbf{(E) }\text{purple}</math>
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x = dir(-5);
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y = (0.6,0.5);
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z = (0,1);
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trans = (2,0);
 +
 
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for (i = 0; i <= 2; ++i) {
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draw(shift(i*trans)*((0,0)--x--(x + y)--(x + y + z)--(y + z)--z--cycle));
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draw(shift(i*trans)*((x + z)--x));
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draw(shift(i*trans)*((x + z)--(x + y + z)));
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draw(shift(i*trans)*((x + z)--z));
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}
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label(rotate(-3)*"$R$", (x + z)/2);
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label(rotate(-5)*slant(0.5)*"$B$", ((x + z) + (y + z))/2);
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label(rotate(35)*slant(0.5)*"$G$", ((x + z) + (x + y))/2);
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label(rotate(-3)*"$W$", (x + z)/2 + trans);
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label(rotate(50)*slant(-1)*"$B$", ((x + z) + (y + z))/2 + trans);
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label(rotate(35)*slant(0.5)*"$R$", ((x + z) + (x + y))/2 + trans);
 +
 
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label(rotate(-3)*"$P$", (x + z)/2 + 2*trans);
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label(rotate(-5)*slant(0.5)*"$R$", ((x + z) + (y + z))/2 + 2*trans);
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label(rotate(-85)*slant(-1)*"$G$", ((x + z) + (x + y))/2 + 2*trans);
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</asy>
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<math> \textbf{(A)} \text{ red}\qquad\textbf{(B)} \text{ white}\qquad\textbf{(C)} \text{ green}\qquad\textbf{(D)} \text{ brown}\qquad\textbf{(E)} \text{ purple} </math>
  
 
==Solution 1==
 
==Solution 1==
−
<math>B</math> is on the top, and <math>R</math> is on the side, and <math>G</math> is on the right side. That means that (image 2)<math>W</math> is on the left side. From the third image, you know that <math>P</math> must be on the bottom since <math>G</math> is sideways. That leaves us with the back, so the back must be <math>A</math>. The front is opposite of the back, so the answer is <math>\boxed{\textbf{(A)}\ R}</math>.~heeeeeeeheeeee
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<math>B</math> is on the top, and <math>R</math> is on the side, and <math>G</math> is on the right side. That means that (image <math>2</math>) <math>W</math> is on the left side. From the third image, you know that <math>P</math> must be on the bottom since <math>G</math> is sideways. That leaves us with the back, so the back must be <math>A</math>. The front is opposite of the back, so the answer is <math>\boxed{\textbf{(A)}\ R}</math>.
  
 
==Solution 2==
 
==Solution 2==
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Looking closely we can see that all faces except for <math>A</math> are connected with <math>R</math>. Thus the answer is <math>\boxed{\textbf{(A)}\ R}</math>.
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Looking closely, we can see that all faces are connected with <math>R</math> except for <math>A</math>. Thus, the answer is <math>\boxed{\textbf{(A)}\ R}</math>.
 +
 
 +
It is A, just draw it out!
 +
~phoenixfire
  
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==Solution 3==
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From pic 1 and 2, we know that the G's opposite is W. From pic 1 and 3, the B's opposite is P. So the <math>A</math>'s opposite is <math>\boxed{\textbf{(A)}\ R}</math>.
  
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~phoenixfire
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==Video Solution by Math-X (Simple Visualization!!!)==
 +
https://youtu.be/IgpayYB48C4?si=uPWa04P5Bi6wEZB-&t=3752
 +
 
 +
~Math-X
 +
 
 +
==Solution Explained==
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https://youtu.be/gOZOCFNXMhE ~ The Learning Royal
 +
 
 +
==Solution 3==
 +
Associated video - https://www.youtube.com/watch?v=K5vaX_EzjEM
 +
 
 +
== Video Solution ==
 +
 
 +
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=VXBqE-jh2WA&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=13
 +
 
 +
==Video Solution==
 +
https://youtu.be/dru7MQO6jqs
 +
 
 +
~savannahsolver
 +
 
 +
==Video Solution (CREATIVE ANALYSIS!!!)==
 +
https://youtu.be/kD4V_InGI_g
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution by The Power of Logic(1 to 25 Full Solution)==
 +
https://youtu.be/Xm4ZGND9WoY
 +
 
 +
~Hayabusa1
  
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==Note==
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==See also==
 
{{AMC8 box|year=2019|num-b=11|num-a=13}}
 
{{AMC8 box|year=2019|num-b=11|num-a=13}}
−
Only two of the cubes are required to solve the problem.
 

Latest revision as of 09:31, 9 November 2024

Problem

The faces of a cube are painted in six different colors: red $(R)$, white $(W)$, green $(G)$, brown $(B)$, aqua $(A)$, and purple $(P)$. Three views of the cube are shown below. What is the color of the face opposite the aqua face?

[asy] unitsize(2cm); pair x, y, z, trans; int i; x = dir(-5); y = (0.6,0.5); z = (0,1); trans = (2,0); for (i = 0; i <= 2; ++i) { draw(shift(i*trans)*((0,0)--x--(x + y)--(x + y + z)--(y + z)--z--cycle)); draw(shift(i*trans)*((x + z)--x)); draw(shift(i*trans)*((x + z)--(x + y + z))); draw(shift(i*trans)*((x + z)--z)); } label(rotate(-3)*"$R$", (x + z)/2); label(rotate(-5)*slant(0.5)*"$B$", ((x + z) + (y + z))/2); label(rotate(35)*slant(0.5)*"$G$", ((x + z) + (x + y))/2); label(rotate(-3)*"$W$", (x + z)/2 + trans); label(rotate(50)*slant(-1)*"$B$", ((x + z) + (y + z))/2 + trans); label(rotate(35)*slant(0.5)*"$R$", ((x + z) + (x + y))/2 + trans); label(rotate(-3)*"$P$", (x + z)/2 + 2*trans); label(rotate(-5)*slant(0.5)*"$R$", ((x + z) + (y + z))/2 + 2*trans); label(rotate(-85)*slant(-1)*"$G$", ((x + z) + (x + y))/2 + 2*trans); [/asy]

$\textbf{(A)} \text{ red}\qquad\textbf{(B)} \text{ white}\qquad\textbf{(C)} \text{ green}\qquad\textbf{(D)} \text{ brown}\qquad\textbf{(E)} \text{ purple}$

Solution 1

$B$ is on the top, and $R$ is on the side, and $G$ is on the right side. That means that (image $2$) $W$ is on the left side. From the third image, you know that $P$ must be on the bottom since $G$ is sideways. That leaves us with the back, so the back must be $A$. The front is opposite of the back, so the answer is $\boxed{\textbf{(A)}\ R}$.

Solution 2

Looking closely, we can see that all faces are connected with $R$ except for $A$. Thus, the answer is $\boxed{\textbf{(A)}\ R}$.

It is A, just draw it out! ~phoenixfire

Solution 3

From pic 1 and 2, we know that the G's opposite is W. From pic 1 and 3, the B's opposite is P. So the $A$'s opposite is $\boxed{\textbf{(A)}\ R}$.

Video Solution by Math-X (Simple Visualization!!!)

https://youtu.be/IgpayYB48C4?si=uPWa04P5Bi6wEZB-&t=3752

~Math-X

Solution Explained

https://youtu.be/gOZOCFNXMhE ~ The Learning Royal

Solution 3

Associated video - https://www.youtube.com/watch?v=K5vaX_EzjEM

Video Solution

Solution detailing how to solve the problem: https://www.youtube.com/watch?v=VXBqE-jh2WA&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=13

Video Solution

https://youtu.be/dru7MQO6jqs

~savannahsolver

Video Solution (CREATIVE ANALYSIS!!!)

https://youtu.be/kD4V_InGI_g

~Education, the Study of Everything

Video Solution by The Power of Logic(1 to 25 Full Solution)

https://youtu.be/Xm4ZGND9WoY

~Hayabusa1

See also

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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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