Question 14

We will visualize 3D shapes made of cubes by looking at their projections from the front, top, and side. When multiple views are given, we assume the overall dimensions of the object are the maximum required by any view, and the letter shapes are preserved.

Step 1 — Analyze the initial 'C' shape

Let us define the initial shape made of 8 cubes. The problem states it looks like a 'C' from the front. The 'C' diagram is 2 units wide and 3 units high. To make it with 8 cubes, we can assume it's 2 units deep in the leftmost column.

Let the coordinates of the cubes be $(x, y, z)$, where $x$ is width, $y$ is height, and $z$ is depth. The front view is the projection onto the $x-y$ plane. The initial 8 cubes are: Layer 1 (front, $z=0$): $(0,0,0), (1,0,0), (0,1,0), (0,2,0), (1,2,0)$ Layer 2 (back, $z=1$): $(0,0,1), (0,1,1), (0,2,1)$

The front view (projection onto $x-y$ plane) is: $\begin{array}{cc} \text{X} & \text{X} \\ \text{X} & \\ \text{X} & \text{X} \end{array}$ This matches the 'C' shape in the diagram.

Now, let us find the side view (projection onto $y-z$ plane, looking from the right). We consider all $(y,z)$ pairs present in the 8 cubes: $(0,0), (1,0), (2,0) \quad \text{from } (0,0,0), (0,1,0), (0,2,0), (1,0,0), (1,2,0)$ $(0,1), (1,1), (2,1) \quad \text{from } (0,0,1), (0,1,1), (0,2,1)$ The side view is a 3 units high by 2 units deep rectangle: $\begin{array}{cc} \text{X} & \text{X} \\ \text{X} & \text{X} \\ \text{X} & \text{X} \end{array}$ This is a rectangle.

Next, let us find the top view (projection onto $x-z$ plane, looking from the top). We consider all $(x,z)$ pairs present in the 8 cubes: $(0,0), (1,0) \quad \text{from } (0,0,0), (0,1,0), (0,2,0), (1,0,0), (1,2,0)$ $(0,1) \quad \text{from } (0,0,1), (0,1,1), (0,2,1)$ The top view is 2 units wide by 2 units deep, in an 'L' shape: $\begin{array}{cc} \text{X} & \text{X} \\ \text{X} & \end{array}$ This is an 'L' shape.

Step 2 — Construct a shape with 'C' front and 'A' top views

We need to make a shape that looks like 'C' from the front and 'A' from the top. The 'C' diagram is 2 units wide, but the 'A' diagram is 3 units wide. To resolve this conflict, we assume the overall dimensions of the object are the maximum required by any view. So, the object will be 3 units wide, 3 units high, and 3 units deep. We will interpret the 'C' and 'A' as 3x3 versions of their shapes.

Let $C_{xy}$ be the set of $(x,y)$ coordinates for the 3x3 'C' front view: $C_{xy} = \{ (0,0), (1,0), (2,0), (0,1), (0,2), (1,2), (2,2) \}$ Let $A_{xz}$ be the set of $(x,z)$ coordinates for the 3x3 'A' top view: $A_{xz} = \{ (0,0), (1,0), (2,0), (0,1), (2,1), (0,2), (2,2) \}$ To find the number of cubes in the shape, we place a cube at $(x,y,z)$ if and only if $(x,y) \in C_{xy}$ AND $(x,z) \in A_{xz}$. This gives the maximum number of cubes that satisfy both projections.

Let us list the cubes:

• For $x=0$:

  • $(0,0) \in C_{xy}$: $z$ can be $0,1,2$ (from $A_{xz}$ for $x=0$). So, $(0,0,0), (0,0,1), (0,0,2)$ are cubes. (3 cubes)
  • $(0,1) \in C_{xy}$: $z$ can be $0,1,2$ (from $A_{xz}$ for $x=0$). So, $(0,1,0), (0,1,1), (0,1,2)$ are cubes. (3 cubes)
  • $(0,2) \in C_{xy}$: $z$ can be $0,1,2$ (from $A_{xz}$ for $x=0$). So, $(0,2,0), (0,2,1), (0,2,2)$ are cubes. (3 cubes) Total for $x=0$: $3+3+3 = \mathbf{9}$ cubes.

• For $x=1$:

  • $(1,0) \in C_{xy}$: $z$ can be $0$ (from $A_{xz}$ for $x=1$). So, $(1,0,0)$ is a cube. (1 cube)
  • $(1,1) \notin C_{xy}$: No cubes for $(1,1)$.
  • $(1,2) \in C_{xy}$: $z$ can be $0$ (from $A_{xz}$ for $x=1$). So, $(1,2,0)$ is a cube. (1 cube) Total for $x=1$: $1+1 = \mathbf{2}$ cubes.

• For $x=2$:

  • $(2,0) \in C_{xy}$: $z$ can be $0,1,2$ (from $A_{xz}$ for $x=2$). So, $(2,0,0), (2,0,1), (2,0,2)$ are cubes. (3 cubes)
  • $(2,1) \in C_{xy}$: $z$ can be $0,1,2$ (from $A_{xz}$ for $x=2$). So, $(2,1,0), (2,1,1), (2,1,2)$ are cubes. (3 cubes)
  • $(2,2) \in C_{xy}$: $z$ can be $0,1,2$ (from $A_{xz}$ for $x=2$). So, $(2,2,0), (2,2,1), (2,2,2)$ are cubes. (3 cubes) Total for $x=2$: $3+3+3 = \mathbf{9}$ cubes.

The total number of cubes is: $9 + 2 + 9$ $= \mathbf{20 \text{ cubes}}$

Step 3 — Construct a shape with 'C' front, 'A' top, and 'F' side views

We need to make a shape that looks like 'C' from the front, 'A' from the top, and 'F' from the side. The overall dimensions of the object will be 3 units wide, 3 units high, and 3 units deep (to accommodate the 'A' and 'C' views). We interpret the 'F' diagram as a 3x3 version of its shape, stretched to fill the 3 units of depth.

Let $C_{xy}$ be the 3x3 'C' front view. Let $A_{xz}$ be the 3x3 'A' top view. Let $F_{yz}$ be the set of $(y,z)$ coordinates for the 3x3 'F' side view. The original 'F' is 3 high and 2 deep. If we scale it to 3 deep, it means the 'F' shape is present in all 3 layers of depth. $F_{yz} = \{ (0,0), (1,0), (2,0), (0,1), (2,1), (0,2), (1,2), (2,2) \}$ This is a 3x3 square with the middle-middle cell $(1,1)$ missing.

To find the number of cubes, we place a cube at $(x,y,z)$ if and only if $(x,y) \in C_{xy}$ AND $(x,z) \in A_{xz}$ AND $(y,z) \in F_{yz}$.

Let us list the cubes:

• For $x=0$:

  • $(0,0) \in C_{xy}$: $z$ can be $0,1,2$ (from $A_{xz}$ for $x=0$).
    • $(0,0,0)$: $(0,0) \in F_{yz}$. So, $(0,0,0)$ is a cube.
    • $(0,0,1)$: $(0,1) \in F_{yz}$. So, $(0,0,1)$ is a cube.
    • $(0,0,2)$: $(0,2) \in F_{yz}$. So, $(0,0,2)$ is a cube. (3 cubes)
  • $(0,1) \in C_{xy}$: $z$ can be $0,1,2$ (from $A_{xz}$ for $x=0$).
    • $(0,1,0)$: $(1,0) \in F_{yz}$. So, $(0,1,0)$ is a cube.
    • $(0,1,1)$: $(1,1) \notin F_{yz}$. No cube.
    • $(0,1,2)$: $(1,2) \in F_{yz}$. So, $(0,1,2)$ is a cube. (2 cubes)
  • $(0,2) \in C_{xy}$: $z$ can be $0,1,2$ (from $A_{xz}$ for $x=0$).
    • $(0,2,0)$: $(2,0) \in F_{yz}$. So, $(0,2,0)$ is a cube.
    • $(0,2,1)$: $(2,1) \in F_{yz}$. So, $(0,2,1)$ is a cube.
    • $(0,2,2)$: $(2,2) \in F_{yz}$. So, $(0,2,2)$ is a cube. (3 cubes) Total for $x=0$: $3+2+3 = \mathbf{8}$ cubes.

• For $x=1$:

  • $(1,0) \in C_{xy}$: $z$ can be $0$ (from $A_{xz}$ for $x=1$).
    • $(1,0,0)$: $(0,0) \in F_{yz}$. So, $(1,0,0)$ is a cube. (1 cube)
  • $(1,1) \notin C_{xy}$: No cubes for $(1,1)$.
  • $(1,2) \in C_{xy}$: $z$ can be $0$ (from $A_{xz}$ for $x=1$).
    • $(1,2,0)$: $(2,0) \in F_{yz}$. So, $(1,2,0)$ is a cube. (1 cube) Total for $x=1$: $1+1 = \mathbf{2}$ cubes.

• For $x=2$:

  • $(2,0) \in C_{xy}$: $z$ can be $0,1,2$ (from $A_{xz}$ for $x=2$).
    • $(2,0,0)$: $(0,0) \in F_{yz}$. So, $(2,0,0)$ is a cube.
    • $(2,0,1)$: $(0,1) \in F_{yz}$. So, $(2,0,1)$ is a cube.
    • $(2,0,2)$: $(0,2) \in F_{yz}$. So, $(2,0,2)$ is a cube. (3 cubes)
  • $(2,1) \in C_{xy}$: $z$ can be $0,1,2$ (from $A_{xz}$ for $x=2$).
    • $(2,1,0)$: $(1,0) \in F_{yz}$. So, $(2,1,0)$ is a cube.
    • $(2,1,1)$: $(1,1) \notin F_{yz}$. No cube.
    • $(2,1,2)$: $(1,2) \in F_{yz}$. So, $(2,1,2)$ is a cube. (2 cubes)
  • $(2,2) \in C_{xy}$: $z$ can be $0,1,2$ (from $A_{xz}$ for $x=2$).
    • $(2,2,0)$: $(2,0) \in F_{yz}$. So, $(2,2,0)$ is a cube.
    • $(2,2,1)$: $(2,1) \in F_{yz}$. So, $(2,2,1)$ is a cube.
    • $(2,2,2)$: $(2,2) \in F_{yz}$. So, $(2,2,2)$ is a cube. (3 cubes) Total for $x=2$: $3+2+3 = \mathbf{8}$ cubes.

The total number of cubes is: $8 + 2 + 8$ $= \mathbf{18 \text{ cubes}}$

Step 4 — Other letter combinations

We can think of many other letter combinations. For example, we could make a shape that looks like an 'L' from the front, an 'L' from the top, and an 'L' from the side.

Let's consider a simple example:

• Front view: 'L' (2 units wide, 2 units high)
• Top view: 'L' (2 units wide, 2 units deep)
• Side view: 'L' (2 units deep, 2 units high)

The overall dimensions would be 2x2x2. Let $L_{xy} = \{ (0,0), (1,0), (0,1) \}$. Let $L_{xz} = \{ (0,0), (1,0), (0,1) \}$. Let $L_{yz} = \{ (0,0), (1,0), (0,1) \}$.

A cube $(x,y,z)$ exists if $(x,y) \in L_{xy}$ AND $(x,z) \in L_{xz}$ AND $(y,z) \in L_{yz}$.

• $(0,0,0)$: $(0,0) \in L_{xy}$, $(0,0) \in L_{xz}$, $(0,0) \in L_{yz}$. So, $(0,0,0)$ is a cube.
• $(1,0,0)$: $(1,0) \in L_{xy}$, $(1,0) \in L_{xz}$, $(0,0) \in L_{yz}$. So, $(1,0,0)$ is a cube.
• $(0,1,0)$: $(0,1) \in L_{xy}$, $(0,0) \in L_{xz}$, $(1,0) \in L_{yz}$. So, $(0,1,0)$ is a cube.
• $(0,0,1)$: $(0,0) \in L_{xy}$, $(0,1) \in L_{xz}$, $(0,1) \in L_{yz}$. So, $(0,0,1)$ is a cube. All other combinations of $(x,y,z)$ will fail at least one condition. This shape would have 4 cubes.

Answer

(i) From the side, it looks like a rectangle. From the top, it looks like an 'L' shape. (ii) The shape would have 20 cubes. (iii) The shape would have 18 cubes. (iv) One example is a shape that looks like an 'L' from the front, an 'L' from the top, and an 'L' from the side.