Determine the equivalent resistance between the points a and b for...

For the first circuit, let the vertices of the triangle be $a$, $b$, and $c$ (top), and the central node be $o$. The circuit is symmetric about the vertical axis passing through $c$ and $o$. When a potential difference is applied across $a$ and $b$, the nodes $c$ and $o$ will be at the same potential. Since $V_c = V_o$, no current flows through the resistor connecting $c$ and $o$, so it can be removed from the circuit. The simplified circuit consists of three parallel branches between $a$ and $b$. The first branch is the direct resistor between $a$ and $b$ with resistance $r$. The second branch is the series combination of resistors $ac$ and $cb$, with resistance $r + r = 2r$. The third branch is the series combination of resistors $ao$ and $ob$, with resistance $r + r = 2r$. The equivalent resistance $R_a$ is given by $\dfrac{1}{R_a} = \dfrac{1}{r} + \dfrac{1}{2r} + \dfrac{1}{2r} = \dfrac{2}{r}$, which gives $R_a = \dfrac{r}{2}$.

For the second circuit, the structure is a hexagon with a central node $o$. Let the nodes on the perimeter be $a, c, d, b, f, e$ in clockwise order. The circuit is symmetric about the vertical axis passing through $o$ and the midpoints of resistors $cd$ and $ef$. If a voltage $2V$ is applied between $a$ and $b$ such that $V_a = V$ and $V_b = -V$, the potential along the vertical axis of symmetry is zero. The circuit can be split into two identical halves (left and right) in series. The left half consists of nodes $a, c, e$ and the grounded points (node $o$ and the midpoints of $cd$ and $ef$). Node $c$ is connected to ground via resistor $co$ (resistance $r$) and the left half of resistor $cd$ (resistance $r/2$). These two are in parallel, so the equivalent resistance from $c$ to ground is $R_{c,gnd} = \dfrac{r \times (r/2)}{r + r/2} = \dfrac{r}{3}$. Similarly, the equivalent resistance from node $e$ to ground is $R_{e,gnd} = \dfrac{r}{3}$. From node $a$, there are three parallel paths to ground. The first path is through $c$ to ground with resistance $R_{ac} + R_{c,gnd} = r + \dfrac{r}{3} = \dfrac{4r}{3}$. The second path is through $e$ to ground with resistance $R_{ae} + R_{e,gnd} = r + \dfrac{r}{3} = \dfrac{4r}{3}$. The third path is directly through $o$ to ground with resistance $R_{ao} = r$. The equivalent resistance of the left half, $R_{left}$, is given by $\dfrac{1}{R_{left}} = \dfrac{1}{4r/3} + \dfrac{1}{4r/3} + \dfrac{1}{r} = \dfrac{3}{4r} + \dfrac{3}{4r} + \dfrac{1}{r} = \dfrac{6}{4r} + \dfrac{1}{r} = \dfrac{3}{2r} + \dfrac{1}{r} = \dfrac{5}{2r}$. This gives $R_{left} = \dfrac{2r}{5}$. The total equivalent resistance between $a$ and $b$ is $R_b = 2 \times R_{left} = 2 \times \dfrac{2r}{5} = \dfrac{4r}{5}$.

Answer: (a) $\dfrac{r}{2}$, (b) $\dfrac{4r}{5}$