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

(→‎Solution)
(→‎Solution)
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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 <math>0</math>s and <math>1</math>s.) I found that to be very helpful when solving this problem.
 
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 <math>0</math>s and <math>1</math>s.) I found that to be very helpful when solving this problem.
 +
==Solution==
 +
Since we want the sum of the edges of each face to be <math>2</math>, we need there to be <math>2</math> <math>1</math>s and <math>2</math> <math>0</math>s on each face. Through experimentation, we find that either <math>2, 4,</math> or all of them have <math>1</math>s adjacent to <math>1</math>s and <math>0</math>s adjacent to <math>0</math> on each face. WLOG, let the first face (counterclockwise) be <math>0,0,1,1</math>. In this case we are trying to have all of them be adjacent to each other. First face: <math>0,0,1,1</math>. Second face: <math>2</math> choices: <math>1,0,0,1</math> or <math>0,0,1,1</math>. After that, it is basically forced and everything will fall in to place. Since we assumed WLOG, we need to multiply <math>2</math> by <math>4</math> to get a total of <math>8</math> different arrangements.
 +
 +
Secondly: <math>4</math> of the faces have all of them adjacent and <math>2</math> of the faces do not: WLOG counting counterclockwise, we have <math>0,0,1,1</math>. Then, we choose the other face next to it. There are two cases, which are <math>0,1,0,1</math> and <math>1,0,1,0</math>. Therefore, this subcase has <math>4</math> different arrangements. Then, we can choose the face at front to be <math>1,0,1,0</math>. This has <math>4</math> cases. The sides can either be <math>0,1,1,0</math> or <math>1,1,0,0</math>. Therefore, we have another <math>8</math> cases.
 +
 +
Summing these up, we have <math>8+4+8 = 20</math>. Therefore, our answer is <math>\boxed {\textbf{(E) }20}</math>
 +
 +
~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 <math>0</math>s and <math>1</math>s.) I found that to be very helpful when solving this problem.
 +
 +
==Solution 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);
 +
</asy>
 +
 +
We see that each face has to have 2 1's and 2 0's. We can try all the cases, starting 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>
 +
 +
In this case, we can completely fill in the rest of the cube.
 +
 +
<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 <math>2</math> diametrically opposite vertices to put <math>3</math> <math>1</math>'s on the connecting edges. As a result, this case has <math>\frac{8}{2}=4</math> 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>
 +
 +
In this case, we have different ways to fill out <math>BC</math> and <math>CD</math>
 +
====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("$$", C--G, SE);
 +
label("$0$", D--H, SW);
 +
</asy>
 +
 +
This goes to:
 +
 +
<asy>
 +
pair A, B, C, D, E, F, G, H;
 +
A = (0, 0);
 +
B = (12.071,0);
 +
C = (12.071,12.071);
 +
D = (0,12.071);
 +
E = (3.536,3.536);
 +
F = (8.536,3.536);
 +
G = (8.536,8.536);
 +
H = (3.536,8.536);
 +
 +
draw(A--B--C--D--A--E--F--G--H--E--F--B--C--G--H--D);
 +
label("$A$", A, SW);
 +
label("$B$", B, SE);
 +
label("$C$", C, NE);
 +
label("$D$", D, NW);
 +
label("$E$", E, NW);
 +
label("$F$", F, NE);
 +
label("$G$", G, SE);
 +
label("$H$", H, SW);
 +
 +
label("$0$", A--B, S);
 +
label("$0$", B--C, W); // Breaks for some reason when I put it to the east
 +
label("$1$", C--D, N);
 +
label("$1$", D--A, W);
 +
label("$0$", E--F, S);
 +
label("$1$", F--G, W); // Same here
 +
label("$1$", G--H, N);
 +
label("$0$", H--E, W);
 +
label("$1$", A--E, NW);
 +
label("$1$", B--F, NE);
 +
label("$0$", C--G, SE);
 +
label("$0$", D--H, SW);
 +
</asy>
 +
 +
====Case 2.2====
 +
<asy>
 +
pair A, B, C, D, E, F, G, H;
 +
A = (0, 0);
 +
B = (12.071,0);
 +
C = (12.071,12.071);
 +
D = (0,12.071);
 +
E = (3.536,3.536);
 +
F = (8.536,3.536);
 +
G = (8.536,8.536);
 +
H = (3.536,8.536);
 +
 +
draw(A--B--C--D--A--E--F--G--H--E--F--B--C--G--H--D);
 +
label("$A$", A, SW);
 +
label("$B$", B, SE);
 +
label("$C$", C, NE);
 +
label("$D$", D, NW);
 +
label("$E$", E, NW);
 +
label("$F$", F, NE);
 +
label("$G$", G, SE);
 +
label("$H$", H, SW);
 +
 +
label("$0$", A--B, S);
 +
label("$$", 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 2 ways to go from here. 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("$$", C--G, SE);
 +
label("$0$", D--H, SW);
 +
</asy>
 +
 +
This goes to:
 +
 +
<asy>
 +
pair A, B, C, D, E, F, G, H;
 +
A = (0, 0);
 +
B = (12.071,0);
 +
C = (12.071,12.071);
 +
D = (0,12.071);
 +
E = (3.536,3.536);
 +
F = (8.536,3.536);
 +
G = (8.536,8.536);
 +
H = (3.536,8.536);
 +
 +
draw(A--B--C--D--A--E--F--G--H--E--F--B--C--G--H--D);
 +
label("$A$", A, SW);
 +
label("$B$", B, SE);
 +
label("$C$", C, NE);
 +
label("$D$", D, NW);
 +
label("$E$", E, NW);
 +
label("$F$", F, NE);
 +
label("$G$", G, SE);
 +
label("$H$", H, SW);
 +
 +
label("$0$", A--B, S);
 +
label("$0$", B--C, W); // Breaks for some reason when I put it to the east
 +
label("$1$", C--D, N);
 +
label("$1$", D--A, W);
 +
label("$1$", E--F, S);
 +
label("$1$", F--G, W); // Same here
 +
label("$0$", G--H, N);
 +
label("$0$", H--E, W);
 +
label("$1$", A--E, NW);
 +
label("$0$", B--F, NE);
 +
label("$1$", C--G, SE);
 +
label("$0$", D--H, SW);
 +
</asy>
 +
 +
We can see that this is just case 1, but with the <math>1</math>'s flipped to <math>0</math>'s, and vice versa. Therefore, this should also have <math>4</math> Orientations.
 +
 +
=====Case 2.2.2=====
 +
 +
<asy>
 +
pair A, B, C, D, E, F, G, H;
 +
A = (0, 0);
 +
B = (12.071,0);
 +
C = (12.071,12.071);
 +
D = (0,12.071);
 +
E = (3.536,3.536);
 +
F = (8.536,3.536);
 +
G = (8.536,8.536);
 +
H = (3.536,8.536);
 +
 +
draw(A--B--C--D--A--E--F--G--H--E--F--B--C--G--H--D);
 +
label("$A$", A, SW);
 +
label("$B$", B, SE);
 +
label("$C$", C, NE);
 +
label("$D$", D, NW);
 +
label("$E$", E, NW);
 +
label("$F$", F, NE);
 +
label("$G$", G, SE);
 +
label("$H$", H, SW);
 +
 +
label("$0$", A--B, S);
 +
label("$$", 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("$0$", 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 just case 1, but with the <math>1</math>'s flipped to <math>0</math>'s, and vice versa. Therefore, this should also have <math>4</math> Orientations.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=23|num-a=25}}
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 01:25, 2 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

Since we want the sum of the edges of each face to be $2$, we need there to be $2$ $1$s and $2$ $0$s on each face. Through experimentation, we find that either four of them 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

Since we want the sum of the edges of each face to be $2$, we need there to be $2$ $1$s and $2$ $0$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 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); [/asy]

We see that each face has to have 2 1's and 2 0's. We can try all the cases, starting 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]

In this case, we can completely fill in the rest of the cube.

[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]

In this case, we have different ways to fill out $BC$ and $CD$

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]

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 2 ways to go from here. 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 just case 1, but with the $1$'s flipped to $0$'s, and vice versa. 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("$0$", 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 just case 1, but with the $1$'s flipped to $0$'s, and vice versa. Therefore, this should also have $4$ Orientations.

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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