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 "2019 AMC 8 Problems/Problem 12"
(→‎Solution 3)
(2 intermediate revisions by 2 users not shown)
Line 4: Line 4:
 
[[File:2019AMC8Prob12.png]]
 
[[File:2019AMC8Prob12.png]]
  
−
<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>
+
<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==
Line 14: Line 14:
 
It is A, just draw it out!
 
It is A, just draw it out!
 
~phoenixfire
 
~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 <math>A</math>'s opposite is <math>\boxed{\textbf{(A)}\ R}</math>.
  
 
==Video Solution by Math-X (Simple Visualization!!!)==
 
==Video Solution by Math-X (Simple Visualization!!!)==
Line 19: Line 22:
  
 
~Math-X
 
~Math-X
−
 
  
 
==Solution Explained==
 
==Solution Explained==

Revision as of 02:08, 5 March 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?

2019AMC8Prob12.png

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

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(Problem 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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_12&oldid=216471"