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

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

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

Latest revision as of 09:31, 9 November 2024

Contents

Problem

Solution 1

Solution 2

Solution 3

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

Solution Explained

Solution 3

Video Solution

Video Solution

Video Solution (CREATIVE ANALYSIS!!!)

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

See also

@@ Line 2: / Line 2: @@
 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?
-[[File:2019AMC8Prob12.png]]
+<!--- [[File:2019AMC8Prob12.png]] -->
+<asy>
+unitsize(2cm);
+pair x, y, z, trans;
+int i;
-<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>
+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>
+<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==
@@ Line 14: / Line 43: @@
 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 <math>A</math>'s opposite is <math>\boxed{\textbf{(A)}\ R}</math>.
+==Video Solution by Math-X (Simple Visualization!!!)==
+https://youtu.be/IgpayYB48C4?si=uPWa04P5Bi6wEZB-&t=3752
+~Math-X
 ==Solution Explained==
@@ Line 35: / Line 72: @@
 ~Education, the Study of Everything
-==Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)==
+==Video Solution by The Power of Logic(1 to 25 Full Solution)==
 https://youtu.be/Xm4ZGND9WoY