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2021 Fall AMC 10A Problems/Problem 24

Revision as of 15:18, 7 December 2021 by MRENTHUSIASM (talk | contribs) (→‎Solution 1)

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 $2.$ Without the loss of generality, we 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("$0$",midpoint(A--E),blue,Fill(1.5,2,white)); label("$1$",midpoint(E--H),blue,Fill(1.5,2,white)); label("$1$",midpoint(H--D),blue,Fill(1.5,2,white)); label("$0$",midpoint(G--C),blue,Fill(1.5,2,white)); label("$0$",midpoint(G--H),blue,Fill(1.5,2,white)); label("$1$",midpoint(B--F),blue,Fill(1.5,2,white)); label("$1$",midpoint(F--G),blue,Fill(1.5,2,white)); label("$0$",midpoint(E--F),blue,Fill(1.5,2,white)); label("$1$",midpoint(A1--E1),blue,Fill(1.5,2,white)); label("$0$",midpoint(E1--F1),blue,Fill(1.5,2,white)); label("$0$",midpoint(B1--F1),blue,Fill(1.5,2,white)); label("$1$",midpoint(F1--G1),blue,Fill(1.5,2,white)); label("$1$",midpoint(G1--C1),blue,Fill(1.5,2,white)); label("$0$",midpoint(G1--H1),blue,Fill(1.5,2,white)); label("$0$",midpoint(H1--D1),blue,Fill(1.5,2,white)); label("$1$",midpoint(E1--H1),blue,Fill(1.5,2,white)); label("The label of $\overline{AE}$ is $0.$",(D.x-0.25,D.y-0.5),blue,align=right); label("The label of $\overline{AE}$ is $1.$",(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 $\overline{AE}$ and $\overline{BF},$ 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, A2, B2, C2, D2, E2, F2, G2, H2, A3, B3, C3, D3, E3, F3, G3, H3, 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; A2 = A1+V; B2 = B1+V; C2 = C1+V; D2 = D1+V; E2 = E1+V; F2 = F1+V; G2 = G1+V; H2 = H1+V; A3 = A2+V; B3 = B2+V; C3 = C2+V; D3 = D2+V; E3 = E2+V; F3 = F2+V; G3 = G2+V; H3 = H2+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); draw(A2--B2--C2--D2--cycle^^A2--E2^^B2--F2^^C2--G2^^E2--F2--G2); draw(H2--D2^^H2--E2^^H2--G2,dashed); draw(A3--B3--C3--D3--cycle^^A3--E3^^B3--F3^^C3--G3^^E3--F3--G3); draw(H3--D3^^H3--E3^^H3--G3,dashed); dot("$A$",A,W,linewidth(4)); dot("$B$",B,(1,0),linewidth(4)); dot("$C$",C,SE,linewidth(4)); dot("$D$",D,SW,linewidth(4)); dot("$E$",E,N,linewidth(4)); dot("$F$",F,NE,linewidth(4)); dot("$G$",G,NE,linewidth(4)); dot("$H$",H,NW,linewidth(4)); dot("$A$",A1,W,linewidth(4)); dot("$B$",B1,(1,0),linewidth(4)); dot("$C$",C1,SE,linewidth(4)); dot("$D$",D1,SW,linewidth(4)); dot("$E$",E1,N,linewidth(4)); dot("$F$",F1,NE,linewidth(4)); dot("$G$",G1,NE,linewidth(4)); dot("$H$",H1,NW,linewidth(4)); dot("$A$",A2,W,linewidth(4)); dot("$B$",B2,(1,0),linewidth(4)); dot("$C$",C2,SE,linewidth(4)); dot("$D$",D2,SW,linewidth(4)); dot("$E$",E2,N,linewidth(4)); dot("$F$",F2,NE,linewidth(4)); dot("$G$",G2,NE,linewidth(4)); dot("$H$",H2,NW,linewidth(4)); dot("$A$",A3,W,linewidth(4)); dot("$B$",B3,(1,0),linewidth(4)); dot("$C$",C3,SE,linewidth(4)); dot("$D$",D3,SW,linewidth(4)); dot("$E$",E3,N,linewidth(4)); dot("$F$",F3,NE,linewidth(4)); dot("$G$",G3,NE,linewidth(4)); dot("$H$",H3,NW,linewidth(4)); label("$1$",midpoint(A--B),red,Fill(1.5,2,white)); label("$1$",midpoint(B--C),red,Fill(1.5,2,white)); label("$0$",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("$1$",midpoint(B1--C1),red,Fill(1.5,2,white)); label("$0$",midpoint(C1--D1),red,Fill(1.5,2,white)); label("$0$",midpoint(D1--A1),red,Fill(1.5,2,white)); label("$1$",midpoint(A2--B2),red,Fill(1.5,2,white)); label("$1$",midpoint(B2--C2),red,Fill(1.5,2,white)); label("$0$",midpoint(C2--D2),red,Fill(1.5,2,white)); label("$0$",midpoint(D2--A2),red,Fill(1.5,2,white)); label("$1$",midpoint(A3--B3),red,Fill(1.5,2,white)); label("$1$",midpoint(B3--C3),red,Fill(1.5,2,white)); label("$0$",midpoint(C3--D3),red,Fill(1.5,2,white)); label("$0$",midpoint(D3--A3),red,Fill(1.5,2,white)); label("$0$",midpoint(A--E),blue,Fill(0,0,white)); label("$0$",midpoint(B--F),blue,Fill(0,0,white)); label("$1$",midpoint(E--F),blue,Fill(1.5,2,white)); label("$1$",midpoint(E--H),blue,Fill(1.5,2,white)); label("$1$",midpoint(D--H),blue,Fill(0,0,white)); label("$0$",midpoint(F--G),blue,Fill(1.5,2,white)); label("$0$",midpoint(G--H),blue,Fill(1.5,2,white)); label("$1$",midpoint(G--C),blue,Fill(0,0,white)); label("$0$",midpoint(A1--E1),blue,Fill(0,0,white)); label("$1$",midpoint(B1--F1),blue,Fill(0,0,white)); label("$0$",midpoint(E1--F1),blue,Fill(1.5,2,white)); label("$1$",midpoint(E1--H1),blue,Fill(1.5,2,white)); label("$1$",midpoint(D1--H1),blue,Fill(0,0,white)); label("$0$",midpoint(F1--G1),blue,Fill(1.5,2,white)); label("$1$",midpoint(G1--H1),blue,Fill(1.5,2,white)); label("$0$",midpoint(G1--C1),blue,Fill(0,0,white)); label("$1$",midpoint(A2--E2),blue,Fill(0,0,white)); label("$0$",midpoint(B2--F2),blue,Fill(0,0,white)); label("$0$",midpoint(E2--F2),blue,Fill(1.5,2,white)); label("$1$",midpoint(E2--H2),blue,Fill(1.5,2,white)); label("$0$",midpoint(D2--H2),blue,Fill(0,0,white)); label("$0$",midpoint(F2--G2),blue,Fill(1.5,2,white)); label("$1$",midpoint(G2--H2),blue,Fill(1.5,2,white)); label("$1$",midpoint(G2--C2),blue,Fill(0,0,white)); label("$1$",midpoint(A3--E3),blue,Fill(0,0,white)); label("$0$",midpoint(B3--F3),blue,Fill(0,0,white)); label("$0$",midpoint(E3--F3),blue,Fill(1.5,2,white)); label("$0$",midpoint(E3--H3),blue,Fill(1.5,2,white)); label("$1$",midpoint(D3--H3),blue,Fill(0,0,white)); label("$1$",midpoint(F3--G3),blue,Fill(1.5,2,white)); label("$1$",midpoint(G3--H3),blue,Fill(1.5,2,white)); label("$0$",midpoint(G3--C3),blue,Fill(0,0,white)); label("The label of $\overline{AE}$ is $0.$",(D.x-0.25,D.y-0.5),blue,align=right); label("The label of $\overline{BF}$ is $0.$",(D.x-0.25,D.y-0.75),blue,align=right); label("The label of $\overline{AE}$ is $0.$",(D1.x-0.25,D1.y-0.5),blue,align=right); label("The label of $\overline{BF}$ is $1.$",(D1.x-0.25,D1.y-0.75),blue,align=right); draw(brace((G3.x+0.25,-0.5),(D2.x-0.25,-0.5),.3),blue); label("The label of $\overline{AE}$ is $1.$",(G2.x-0.25,D2.y-1),blue,align=right); label("The label of $\overline{BF}$ is $0.$",(G2.x-0.25,D2.y-1.25),blue,align=right); [/asy] 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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