Question

Five geometric blocks with their dimensions are given below. Calculate the length of the path as shown in the image below, between points A and B.

Hint:

Once you flatten the path onto one 2D plane, check whether the horizontal distance of either endpoint from a cylinder centre exactly equals the radius of that cylinder. If it does, that tangent segment is just a straight vertical drop, and you can skip the general tangent-length formula entirely.

Answer 1

Correct Answer: 6.07

Step 1: Understanding the Question:
The problem asks for the minimum length of a path (resembling a taut belt or string) wrapped around and between five geometric blocks from point A to point B.
By inspecting the given dimensions, we can determine the heights of key elements:
1. Point A is located on the vertical edge of the leftmost block at a height of 1 cm.
2. Point B is located at the top corner of the rightmost block, which has a height of 1 cm.
3. Both cylinders and the intermediate blocks have a height of 1 cm.
This vertical alignment implies that the entire red path lies on a horizontal plane at a height of $z = 1\text{ cm}$.
Consequently, we can simplify this three-dimensional problem into a two-dimensional geometry problem on a flat plane.

Step 2: Key Formula or Approach:
We set up a 2D Cartesian coordinate system on the horizontal plane $z = 1\text{ cm}$:
Let the long axis of the blocks be the $y$-axis (extending from left to right) and the transverse width be the $x$-axis.
The key formulas required are:
1. Distance between two points:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
2. Length of a tangent segment from an external point to a circle:
\[ L_{\text{tangent}} = \sqrt{d^2 - R^2} \]
3. Length of a circular arc:
\[ L_{\text{arc}} = R \times \theta \]

Step 3: Detailed Explanation:
1. Let us determine the coordinates of the key points on the plane:
- Point A is located at the front corner of the first block: $A = (1, 0)$.
- The first cylinder has its center at $C_1 = (0.5, 3.0)$ with a radius $R = 0.5\text{ cm}$.
- The second cylinder touches the first one and is centered at $C_2 = (0.5, 4.0)$ with a radius $R = 0.5\text{ cm}$.
- Point B is at the back corner of the final cube: $B = (0, 5.5)$.
2. Calculating the tangent segment from A to Cylinder 1:
- The distance from $A(1, 0)$ to the center $C_1(0.5, 3.0)$ is:
\[ d_1 = \sqrt{(1 - 0.5)^2 + (0 - 3)^2} = \sqrt{0.25 + 9.00} = \sqrt{9.25}\text{ cm} \]
- The length of the tangent segment $T_A$ from A to the contact point on Cylinder 1 is:
\[ L_{T_A} = \sqrt{d_1^2 - R^2} = \sqrt{9.25 - 0.25} = \sqrt{9.00} = 3\text{ cm} \]
- Since $T_A$ is exactly $(1, 3)$, this segment is a straight vertical line of length $3\text{ cm}$.
3. Calculating the tangent segment from B to Cylinder 2:
- The distance from $B(0, 5.5)$ to the center $C_2(0.5, 4.0)$ is:
\[ d_2 = \sqrt{(0 - 0.5)^2 + (5.5 - 4.0)^2} = \sqrt{0.25 + 2.25} = \sqrt{2.5}\text{ cm} \]
- The length of the tangent segment $T_B$ from B to the contact point on Cylinder 2 is:
\[ L_{T_B} = \sqrt{d_2^2 - R^2} = \sqrt{2.5 - 0.25} = \sqrt{2.25} = 1.5\text{ cm} \]
- This tangent point $T_B$ is located at $(0, 4.0)$.
4. Calculating the arc lengths:
- The path transitions smoothly at the contact point of the two cylinders.
- On Cylinder 1, the path wraps from $T_A(1, 3)$ (angle $0^{\circ}$ relative to $C_1$) to the contact point $(0.5, 3.5)$ (angle $90^{\circ}$). This is a quarter circle:
\[ L_{\text{arc1}} = 0.5 \times \frac{\pi}{2} = 0.25 \times 3.14 = 0.785\text{ cm} \]
- On Cylinder 2, the path wraps from the contact point $(0.5, 3.5)$ (angle $-90^{\circ}$ relative to $C_2$) to $T_B(0, 4.0)$ (angle $180^{\circ}$). This also forms a quarter-circle arc:
\[ L_{\text{arc2}} = 0.5 \times \frac{\pi}{2} = 0.25 \times 3.14 = 0.785\text{ cm} \]
5. Summing the components to find the total length:
\[ L_{\text{total}} = 3.0 + 0.785 + 0.785 + 1.5 = 6.07\text{ cm} \]

Step 4: Final Answer:
The total length of the path between points A and B is 6.07 cm.

Answer 2

Concept:

• When a straight tangent line is drawn from an external point to a circle, if the horizontal offset of that point from the centre of the circle exactly equals the radius, the point already lies on the vertical tangent line of the circle - so the tangent segment is simply a vertical distance, with no square root needed.
• Checking this special condition first, before applying the general tangent-length formula, can save a full calculation step at each cylinder.

Step 1: Flatten the path onto one 2D plane.
Since A, B, and both cylinders sit at the same height $z = 1\text{ cm}$, project everything onto a horizontal plane, with the long axis of the blocks as the $y$-axis and the transverse width as the $x$-axis. This gives $A = (1, 0)$, cylinder 1 centre $C_1 = (0.5, 3.0)$ with radius $0.5\text{ cm}$, cylinder 2 centre $C_2 = (0.5, 4.0)$ with radius $0.5\text{ cm}$, and $B = (0, 5.5)$.

Step 2: Check the horizontal offset of A from $C_1$.
Horizontal offset $= 1 - 0.5 = 0.5\text{ cm}$, exactly equal to the radius. So A already lies on the vertical tangent line of the cylinder, touching the circle at its rightmost point $(1, 3.0)$. The tangent segment is simply vertical:
$L_{T_A} = 3.0 - 0 = 3\text{ cm}$

Step 3: Check the horizontal offset of B from $C_2$.
Horizontal offset $= |0 - 0.5| = 0.5\text{ cm}$, again equal to the radius. B lies on the vertical tangent line of its own cylinder, touching at $(0, 4.0)$:
$L_{T_B} = 5.5 - 4.0 = 1.5\text{ cm}$

Step 4: Add the arc lengths where the path wraps the two touching cylinders.
The path turns a quarter circle around each cylinder at the point where they meet:
$L_{arc} = R \times \dfrac{\pi}{2} = 0.5 \times \dfrac{22}{7 \times 2} \approx 0.5 \times 1.57 = 0.785\text{ cm}$ (using $\pi \approx 3.14$)
Both arcs together contribute $0.785 + 0.785 = 1.57\text{ cm}$.

Step 5: Add all four pieces.
Total length $= 3 + 0.785 + 0.785 + 1.5 = 6.07\text{ cm}$.

Final Answer: 6.07 cm