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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 <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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<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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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>
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==Solution 1==
 
==Solution 1==
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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>.
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==Solution 2==
 
==Solution 2==
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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>.
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It is A, just draw it out!
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~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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 +
==Video Solution by Math-X (Simple Visualization!!!)==
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https://youtu.be/IgpayYB48C4?si=uPWa04P5Bi6wEZB-&t=3752
 +
 +
~Math-X
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 +
==Solution Explained==
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https://youtu.be/gOZOCFNXMhE ~ The Learning Royal
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 +
==Solution 3==
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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
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 +
==Video Solution by The Power of Logic(1 to 25 Full Solution)==
 +
https://youtu.be/Xm4ZGND9WoY
 +
 +
~Hayabusa1
 +
 +
==See also==
 
{{AMC8 box|year=2019|num-b=11|num-a=13}}
 
{{AMC8 box|year=2019|num-b=11|num-a=13}}

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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