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Step-by-Step Solution
Step 1: Identify the resistance of each side
The total resistance of the wire is given as $18\,\Omega$. When this wire is bent into an equilateral triangle, each side will have equal length (and hence equal resistance). Since there are 3 sides in an equilateral triangle, the resistance of each side is:
$\displaystyle R_\text{side} = \frac{18\,\Omega}{3} = 6\,\Omega.$
Step 2: Determine the paths between two vertices
To find the resistance between any two vertices, say A and B, there are two paths:
The direct side AB with a resistance of $6\,\Omega.$
The indirect path A → C → B, which consists of two sides in series. Hence that route has a resistance of $6\,\Omega + 6\,\Omega = 12\,\Omega.$
Step 3: Calculate the combined (effective) resistance
The direct side (6 Ω) and the two-series path (12 Ω) are in parallel. The equivalent resistance $R_\text{eq}$ is given by the parallel combination formula:
$$
\frac{1}{R_\text{eq}} = \frac{1}{6} + \frac{1}{12}
= \frac{2}{12} + \frac{1}{12}
= \frac{3}{12}
= \frac{1}{4}.
$$
Thus, $R_\text{eq} = 4\,\Omega.$
Step 4: State the final answer
Therefore, the resistance between any two vertices of the equilateral triangle formed by the wire is $4\,\Omega.$
