A thin wire is used to construct all the edges of a cube of 1 m side by bending,
cutting and soldering the wire. If the wire is 12 m long, what is the minimum number
of cuts required to construct the wire frame to form the cube?
Options:
(A) 3
(B) 4
(C) 6
(D) 12
Official Answer Key:
A
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2025 Q.4)**
**1. Geometric & Graph Theoretic Formulation**:
- **Target Structure**: Cube wireframe has $V = 8$ vertices and $E = 12$ edges, each of length $1\text{ m}$.
- **Vertex Degree**: Each vertex of a cube connects $3$ incident edges (degree $d = 3$, odd degree).
- **Continuous Wire Properties**: Total wire length = $12\text{ m}$. Each uncut continuous wire path can contribute at most $2$ odd-degree endpoints.
**2. Step-by-Step Derivation**:
- By the Euler Path Theorem (Handshaking Lemma):
$\text{Minimum number of continuous paths } k = \frac{\text{Number of odd-degree vertices}}{2} = \frac{8}{2} = 4$
- To cover all $12$ edges of the cube without overlaps, the wire must be divided into at least $4$ separate continuous paths.
- To cut a single continuous $12\text{ m}$ wire into $4$ pieces:
$\text{Minimum cuts required} = k - 1 = 4 - 1 = 3 \text{ cuts}$
**3. Official Verification**:
The verified official IIT answer is **Option (A)** (`3`).
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