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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 24"

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For simplicity purposes, we name this cube <math>ABCDEFGH</math> by vertices, as shown below.
 
For simplicity purposes, we name this cube <math>ABCDEFGH</math> by vertices, as shown below.
 
<asy>
 
<asy>
 +
/* Made by MRENTHUSIASM */
 
size(150);
 
size(150);
  
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We apply casework to the label of <math>\overline{AE},</math> as shown below.
 
We apply casework to the label of <math>\overline{AE},</math> as shown below.
 
<asy>
 
<asy>
 +
/* Made by MRENTHUSIASM */
 
size(1200,150);
 
size(1200,150);
  

Revision as of 14:33, 7 December 2021

Problem

Each of the $12$ edges of a cube is labeled $0$ or $1$. Two labelings are considered different even if one can be obtained from the other by a sequence of one or more rotations and/or reflections. For how many such labelings is the sum of the labels on the edges of each of the $6$ faces of the cube equal to $2$?

$\textbf{(A) } 8 \qquad\textbf{(B) } 10 \qquad\textbf{(C) } 12 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 20$

Solution 1

For simplicity purposes, we name this cube $ABCDEFGH$ by vertices, as shown below. [asy] /* Made by MRENTHUSIASM */ size(150); pair A, B, C, D, E, F, G, H; A = (0,1); B = (1,1); C = (1,0); D = (0,0); E = (0.3,1.3); F = (1.3,1.3); G = (1.3,0.3); H = (0.3,0.3); draw(A--B--C--D--cycle^^A--E^^B--F^^C--G^^E--F--G); draw(H--D^^H--E^^H--G,dashed); dot("$A$",A,1.5*W,linewidth(4)); dot("$B$",B,1.5*(1,0),linewidth(4)); dot("$C$",C,1.5*SE,linewidth(4)); dot("$D$",D,1.5*SW,linewidth(4)); dot("$E$",E,1.5*N,linewidth(4)); dot("$F$",F,1.5*NE,linewidth(4)); dot("$G$",G,1.5*NE,linewidth(4)); dot("$H$",H,1.5*NW,linewidth(4)); [/asy] Note that for each face of this cube, two edges are labeled $0$ and two edges are labeled $1.$ For all twelve edges of this cube, we conclude that six edges are labeled $0,$ and six edges are labeled $1.$

We apply casework to face $ABCD.$ Recall that there are $\binom42=6$ ways to label its edges:

  1. Opposite edges have the same label.

    2. There are $2$ ways to label the edges of $ABCD.$ We will consider one of the ways, then multiply the count by $2:$ Assume that $\overline{AB},\overline{BC},\overline{CD},\overline{DA}$ are labeled $1,0,1,0,$ respectively.

    We apply casework to the label of $\overline{AE},$ as shown below. [asy] /* Made by MRENTHUSIASM */ size(1200,150); pair A, B, C, D, E, F, G, H, A1, B1, C1, D1, E1, F1, G1, H1, V; A = (0,1); B = (1,1); C = (1,0); D = (0,0); E = (0.3,1.3); F = (1.3,1.3); G = (1.3,0.3); H = (0.3,0.3); V = (3,0); A1 = A+V; B1 = B+V; C1 = C+V; D1 = D+V; E1 = E+V; F1 = F+V; G1 = G+V; H1 = H+V; draw(A--B--C--D--cycle^^A--E^^B--F^^C--G^^E--F--G); draw(H--D^^H--E^^H--G,dashed); draw(A1--B1--C1--D1--cycle^^A1--E1^^B1--F1^^C1--G1^^E1--F1--G1); draw(H1--D1^^H1--E1^^H1--G1,dashed); dot("$A$",A,1.5*W,linewidth(4)); dot("$B$",B,1.5*(1,0),linewidth(4)); dot("$C$",C,1.5*SE,linewidth(4)); dot("$D$",D,1.5*SW,linewidth(4)); dot("$E$",E,1.5*N,linewidth(4)); dot("$F$",F,1.5*NE,linewidth(4)); dot("$G$",G,1.5*NE,linewidth(4)); dot("$H$",H,1.5*NW,linewidth(4)); dot("$A$",A1,1.5*W,linewidth(4)); dot("$B$",B1,1.5*(1,0),linewidth(4)); dot("$C$",C1,1.5*SE,linewidth(4)); dot("$D$",D1,1.5*SW,linewidth(4)); dot("$E$",E1,1.5*N,linewidth(4)); dot("$F$",F1,1.5*NE,linewidth(4)); dot("$G$",G1,1.5*NE,linewidth(4)); dot("$H$",H1,1.5*NW,linewidth(4)); label("$1$",midpoint(A--B),red,Fill(1.5,2,white)); label("$0$",midpoint(B--C),red,Fill(1.5,2,white)); label("$1$",midpoint(C--D),red,Fill(1.5,2,white)); label("$0$",midpoint(D--A),red,Fill(1.5,2,white)); label("$1$",midpoint(A1--B1),red,Fill(1.5,2,white)); label("$0$",midpoint(B1--C1),red,Fill(1.5,2,white)); label("$1$",midpoint(C1--D1),red,Fill(1.5,2,white)); label("$0$",midpoint(D1--A1),red,Fill(1.5,2,white)); label("$1$",midpoint(A--E),blue,Fill(1.5,2,white)); label("$0$",midpoint(E--F),blue,Fill(1.5,2,white)); label("$0$",midpoint(B--F),blue,Fill(1.5,2,white)); label("$1$",midpoint(F--G),blue,Fill(1.5,2,white)); label("$1$",midpoint(G--C),blue,Fill(1.5,2,white)); label("$0$",midpoint(G--H),blue,Fill(1.5,2,white)); label("$0$",midpoint(H--D),blue,Fill(1.5,2,white)); label("$1$",midpoint(E--H),blue,Fill(1.5,2,white)); label("$0$",midpoint(A1--E1),blue,Fill(1.5,2,white)); label("$1$",midpoint(E1--H1),blue,Fill(1.5,2,white)); label("$1$",midpoint(H1--D1),blue,Fill(1.5,2,white)); label("$0$",midpoint(G1--C1),blue,Fill(1.5,2,white)); label("$0$",midpoint(G1--H1),blue,Fill(1.5,2,white)); label("$1$",midpoint(B1--F1),blue,Fill(1.5,2,white)); label("$1$",midpoint(F1--G1),blue,Fill(1.5,2,white)); label("$0$",midpoint(E1--F1),blue,Fill(1.5,2,white)); label("The label of $\overline{AE}$ is $1.$",(D.x-0.25,D.y-0.5),blue,align=right); label("The label of $\overline{AE}$ is $0.$",(D1.x-0.25,D1.y-0.5),blue,align=right); [/asy] We have $2\cdot2=4$ such labelings for this case.

  2. Opposite edges have different labels.

    3. There are $4$ ways to label the edges of $ABCD.$ We will consider one of the ways, then multiply the count by $4:$ Assume that $\overline{AB},\overline{BC},\overline{CD},\overline{DA}$ are labeled $1,1,0,0,$ respectively.

    We apply casework to the labels of $\overline{BF}$ and $\overline{AE},$ as shown below.

    DIAGRAM WILL BE DONE TOMORROW.

    We have $4\cdot4=16$ such labelings for this case.

Therefore, we have $4+16=\boxed{\textbf{(E) } 20}$ such labelings in total.

~MRENTHUSIASM

Solution 2

Since we want the sum of the edges of each face to be $2$, we need there to be two $1$s and two $0$s on each face. Through experimentation, we find that either $2, 4,$ or all of them have $1$s adjacent to $1$s and $0$s adjacent to $0$ on each face. WLOG, let the first face (counterclockwise) be $0,0,1,1$. In this case we are trying to have all of them be adjacent to each other. First face: $0,0,1,1$. Second face: $2$ choices: $1,0,0,1$ or $0,0,1,1$. After that, it is basically forced and everything will fall in to place. Since we assumed WLOG, we need to multiply $2$ by $4$ to get a total of $8$ different arrangements.

Secondly: $4$ of the faces have all of them adjacent and $2$ of the faces do not: WLOG counting counterclockwise, we have $0,0,1,1$. Then, we choose the other face next to it. There are two cases, which are $0,1,0,1$ and $1,0,1,0$. Therefore, this subcase has $4$ different arrangements. Then, we can choose the face at front to be $1,0,1,0$. This has $4$ cases. The sides can either be $0,1,1,0$ or $1,1,0,0$. Therefore, we have another $8$ cases.

Summing these up, we have $8+4+8 = 20$. Therefore, our answer is $\boxed {\textbf{(E) }20}$

~Arcticturn

Remark: It is very easy to get disorganized when counting, so when doing this problem, make sure to draw a diagram of the cube. Labeling is a bit harder, since we often confuse one side with another. Try doing the problem by labeling sides on the lines (literally letting the lines pass through your $0$s and $1$s.) I found that to be very helpful when solving this problem.

Solution 3

[asy] pair A, B, C, D, E, F, G, H; A = (0, 0); B = (12.071,0); C = (12.071,12.071); D = (0,12.071); E = (3.536,3.536); F = (8.536,3.536); G = (8.536,8.536); H = (3.536,8.536); draw(A--B--C--D--A--E--F--G--H--E--F--B--C--G--H--D); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$E$", E, NW); label("$F$", F, NE); label("$G$", G, SE); label("$H$", H, SW); [/asy]

We see that each face has to have 2 1's and 2 0's. We can start with edges connecting to A.

Case 1

[asy] pair A, B, C, D, E, F, G, H; A = (0, 0); B = (12.071,0); C = (12.071,12.071); D = (0,12.071); E = (3.536,3.536); F = (8.536,3.536); G = (8.536,8.536); H = (3.536,8.536); draw(A--B--C--D--A--E--F--G--H--E--F--B--C--G--H--D); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$E$", E, NW); label("$F$", F, NE); label("$G$", G, SE); label("$H$", H, SW); label("$1$", A--B, S); label("$1$", A--D, W); label("$1$", A--E, NW); [/asy]

This goes to:

[asy] pair A, B, C, D, E, F, G, H; A = (0, 0); B = (12.071,0); C = (12.071,12.071); D = (0,12.071); E = (3.536,3.536); F = (8.536,3.536); G = (8.536,8.536); H = (3.536,8.536); draw(A--B--C--D--A--E--F--G--H--E--F--B--C--G--H--D); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$E$", E, NW); label("$F$", F, NE); label("$G$", G, SE); label("$H$", H, SW); label("$1$", A--B, S); label("$0$", B--C, W); // Breaks for some reason when I put it to the east label("$0$", C--D, N); label("$1$", D--A, W); label("$0$", E--F, S); label("$1$", F--G, W); // Same here label("$1$", G--H, N); label("$0$", H--E, W); label("$1$", A--E, NW); label("$0$", B--F, NE); label("$1$", C--G, SE); label("$0$", D--H, SW); [/asy]

We can see that we choose $2$ diametrically opposite vertices to put $3$ $1$'s on the connecting edges. As a result, this case has $\frac{8}{2}=4$ orientations.

Case 2

[asy] pair A, B, C, D, E, F, G, H; A = (0, 0); B = (12.071,0); C = (12.071,12.071); D = (0,12.071); E = (3.536,3.536); F = (8.536,3.536); G = (8.536,8.536); H = (3.536,8.536); draw(A--B--C--D--A--E--F--G--H--E--F--B--C--G--H--D); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$E$", E, NW); label("$F$", F, NE); label("$G$", G, SE); label("$H$", H, SW); label("$0$", A--B, S); label("$1$", A--D, W); label("$1$", A--E, NW); [/asy]

Filling out a bit more, we have:

[asy] pair A, B, C, D, E, F, G, H; A = (0, 0); B = (12.071,0); C = (12.071,12.071); D = (0,12.071); E = (3.536,3.536); F = (8.536,3.536); G = (8.536,8.536); H = (3.536,8.536); draw(A--B--C--D--A--E--F--G--H--E--F--B--C--G--H--D); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$E$", E, NW); label("$F$", F, NE); label("$G$", G, SE); label("$H$", H, SW); label("$0$", A--B, S); label("$1$", A--D, W); label("$1$", A--E, NW); label("$0$", H--E, W); label("$0$", D--H, SW); [/asy]

Let's try filling out $BC$ and $CD$ first.

Case 2.1

[asy] pair A, B, C, D, E, F, G, H; A = (0, 0); B = (12.071,0); C = (12.071,12.071); D = (0,12.071); E = (3.536,3.536); F = (8.536,3.536); G = (8.536,8.536); H = (3.536,8.536); draw(A--B--C--D--A--E--F--G--H--E--F--B--C--G--H--D); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$E$", E, NW); label("$F$", F, NE); label("$G$", G, SE); label("$H$", H, SW); label("$0$", A--B, S); label("$$", B--C, W); // Breaks for some reason when I put it to the east label("$$", C--D, N); label("$1$", D--A, W); label("$0$", E--F, S); label("$$", F--G, W); // Same here label("$$", G--H, N); label("$0$", H--E, W); label("$1$", A--E, NW); label("$1$", B--F, NE); label("$$", C--G, SE); label("$0$", D--H, SW); [/asy]

This goes to:

[asy] pair A, B, C, D, E, F, G, H; A = (0, 0); B = (12.071,0); C = (12.071,12.071); D = (0,12.071); E = (3.536,3.536); F = (8.536,3.536); G = (8.536,8.536); H = (3.536,8.536); draw(A--B--C--D--A--E--F--G--H--E--F--B--C--G--H--D); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$E$", E, NW); label("$F$", F, NE); label("$G$", G, SE); label("$H$", H, SW); label("$0$", A--B, S); label("$0$", B--C, W); // Breaks for some reason when I put it to the east label("$1$", C--D, N); label("$1$", D--A, W); label("$0$", E--F, S); label("$1$", F--G, W); // Same here label("$1$", G--H, N); label("$0$", H--E, W); label("$1$", A--E, NW); label("$1$", B--F, NE); label("$0$", C--G, SE); label("$0$", D--H, SW); [/asy]

We can see that it consists of chains of three $1$'s, with the middle of each chain being opposite edges. As a result, this case has $\frac{12}{2}=6$ orientations.

Case 2.2

[asy] pair A, B, C, D, E, F, G, H; A = (0, 0); B = (12.071,0); C = (12.071,12.071); D = (0,12.071); E = (3.536,3.536); F = (8.536,3.536); G = (8.536,8.536); H = (3.536,8.536); draw(A--B--C--D--A--E--F--G--H--E--F--B--C--G--H--D); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$E$", E, NW); label("$F$", F, NE); label("$G$", G, SE); label("$H$", H, SW); label("$0$", A--B, S); label("$$", B--C, W); // Breaks for some reason when I put it to the east label("$$", C--D, N); label("$1$", D--A, W); label("$1$", E--F, S); label("$$", F--G, W); // Same here label("$$", G--H, N); label("$0$", H--E, W); label("$1$", A--E, NW); label("$0$", B--F, NE); label("$$", C--G, SE); label("$0$", D--H, SW); [/asy]

Oh no... We have different ways of filling out $FG$ and $GH$. More casework!

Case 2.2.1

[asy] pair A, B, C, D, E, F, G, H; A = (0, 0); B = (12.071,0); C = (12.071,12.071); D = (0,12.071); E = (3.536,3.536); F = (8.536,3.536); G = (8.536,8.536); H = (3.536,8.536); draw(A--B--C--D--A--E--F--G--H--E--F--B--C--G--H--D); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$E$", E, NW); label("$F$", F, NE); label("$G$", G, SE); label("$H$", H, SW); label("$0$", A--B, S); label("$$", B--C, W); // Breaks for some reason when I put it to the east label("$$", C--D, N); label("$1$", D--A, W); label("$1$", E--F, S); label("$1$", F--G, W); // Same here label("$0$", G--H, N); label("$0$", H--E, W); label("$1$", A--E, NW); label("$0$", B--F, NE); label("$$", C--G, SE); label("$0$", D--H, SW); [/asy]

This goes to:

[asy] pair A, B, C, D, E, F, G, H; A = (0, 0); B = (12.071,0); C = (12.071,12.071); D = (0,12.071); E = (3.536,3.536); F = (8.536,3.536); G = (8.536,8.536); H = (3.536,8.536); draw(A--B--C--D--A--E--F--G--H--E--F--B--C--G--H--D); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$E$", E, NW); label("$F$", F, NE); label("$G$", G, SE); label("$H$", H, SW); label("$0$", A--B, S); label("$0$", B--C, W); // Breaks for some reason when I put it to the east label("$1$", C--D, N); label("$1$", D--A, W); label("$1$", E--F, S); label("$1$", F--G, W); // Same here label("$0$", G--H, N); label("$0$", H--E, W); label("$1$", A--E, NW); label("$0$", B--F, NE); label("$1$", C--G, SE); label("$0$", D--H, SW); [/asy]

We can see that this is the inverse of case 1 (Define inverse to mean swapping $1$'s for $0$'s and $0$'s for $1$'s). Therefore, this should also have $4$ orientations.

Case 2.2.2

[asy] pair A, B, C, D, E, F, G, H; A = (0, 0); B = (12.071,0); C = (12.071,12.071); D = (0,12.071); E = (3.536,3.536); F = (8.536,3.536); G = (8.536,8.536); H = (3.536,8.536); draw(A--B--C--D--A--E--F--G--H--E--F--B--C--G--H--D); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$E$", E, NW); label("$F$", F, NE); label("$G$", G, SE); label("$H$", H, SW); label("$0$", A--B, S); label("$$", B--C, W); // Breaks for some reason when I put it to the east label("$$", C--D, N); label("$1$", D--A, W); label("$1$", E--F, S); label("$0$", F--G, W); // Same here label("$1$", G--H, N); label("$0$", H--E, W); label("$1$", A--E, NW); label("$0$", B--F, NE); label("$$", C--G, SE); label("$0$", D--H, SW); [/asy]

This goes to:

[asy] pair A, B, C, D, E, F, G, H; A = (0, 0); B = (12.071,0); C = (12.071,12.071); D = (0,12.071); E = (3.536,3.536); F = (8.536,3.536); G = (8.536,8.536); H = (3.536,8.536); draw(A--B--C--D--A--E--F--G--H--E--F--B--C--G--H--D); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$E$", E, NW); label("$F$", F, NE); label("$G$", G, SE); label("$H$", H, SW); label("$0$", A--B, S); label("$1$", B--C, W); // Breaks for some reason when I put it to the east label("$0$", C--D, N); label("$1$", D--A, W); label("$1$", E--F, S); label("$0$", F--G, W); // Same here label("$1$", G--H, N); label("$0$", H--E, W); label("$1$", A--E, NW); label("$0$", B--F, NE); label("$1$", C--G, SE); label("$0$", D--H, SW); [/asy]

This is the inverse of case 2.1, so this will also have $6$ orientations.

Putting Them All Together

We see that if the $3$ edges connecting to $A$ has two $0$'s, and one $1$, it would have the same solutions as if it had two $1$'s, and one $0$. The solutions would just be inverted. As case 2.1 and case 2.2.2 are inverses, and case 2.2.1 has case 1 as an inverse, there would not be any additional solutions.

Similarly, if the $3$ edges connecting to $A$ has three $0$'s, it would be the same as the inverse of case 1, or case 2.2.1, resulting in no new solutions.

Putting all the cases together, we have $4+6+4+6=\boxed {\textbf{(E) }20}$ solutions.

~ConcaveTriangle

See Also

2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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 AMC 10 Problems and Solutions

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