Equivalent resistance between the adjacent corners of a regular n-sided polygon of uniform wire of resistance R would be :

When, a uniform wire of resistance $\mathrm{R}$ is shaped into a regular $\mathrm{n}$-sided polygon, the resistance of each side will be, <br/><br/>$\frac{\mathrm{R}}{\mathrm{n}}=\mathrm{R}_1$ <br/><br/>Let $R_1 $ and $ R_2$ be the resistance between adjacent corners of a regular polygon <br/><br/>$\therefore$ The resistance of $(\mathrm{n}-1)$ sides, $\mathrm{R}_2=\frac{(\mathrm{n}-1) \mathrm{R}}{\mathrm{n}}$ <br/><br/>Since two parts are parallel, therefore, <br/><br/>$$ \begin{aligned} & \mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{R}_1 \mathrm{R}_2}{\mathrm{R}_1+\mathrm{R}_2}=\frac{\left(\frac{\mathrm{R}}{\mathrm{n}}\right)\left(\frac{\mathrm{n}-1}{\mathrm{n}}\right) \mathrm{R}}{\left(\frac{\mathrm{R}}{\mathrm{n}}\right)+\left(\frac{\mathrm{n}-1}{\mathrm{n}}\right) \mathrm{R}} \\\\ & \Rightarrow \mathrm{R}_{\mathrm{eq}}=\frac{(\mathrm{n}-1) \mathrm{R}^2}{\mathrm{n}^2} \times \frac{\mathrm{n}}{\mathrm{R}+\mathrm{nR}-\mathrm{R}} \\\\ &\Rightarrow \mathrm{R}_{\mathrm{eq}}=\frac{(\mathrm{n}-1) \mathrm{R}}{\mathrm{n}^2} \end{aligned} $$