Circular resistance?

Let $ADBCA$ be the wire of resistance $100$ ohms where $A,B$ and $C,D$ are diametrically opposite points. So the wires $ADB$ and $ACB$ will have resistances $50$ ohms each and these two wires are joined in parallel between $A \ and \ B$ . Hence the equivalent resistance is given by :

$R_{eqv}=\dfrac{50 \times 50}{50+50} = \dfrac{2500}{100}=\boxed{25 \ ohms}$

Using $R=\frac{\rho L}{A}$ it is clear that $R\propto L$ so halving the distance halves the resistance and both paths have $R=\frac{100\Omega}{2}=50\Omega$ .

Since these paths are in parallel we use the reciprocal addition giving:

$\frac{1}{R_{AB}}=\frac{1}{50\Omega}+\frac{1}{50\Omega}=\frac{2}{50\Omega}=\frac{1}{25\Omega}\Rightarrow R_{AB}=\boxed{25\Omega}$

Parallel therefore 50*50/100

By Kirchoffs law Resutant 1/R = 1/R1 + 1/R2