Can a 7th grader solve this?

Imagine putting a voltage source or voltage probe between $A$ and $D$ . By symmetry, the voltage across the $BC$ resistor is always zero. That resistor is effectively an open circuit, and can be taken out. Then the equivalent resistance $R$ is:

$\frac{1}{R} = \frac{1}{2 r} + \frac{1}{2 r} + \frac{1}{r} \\ R = \frac{r}{2}$

Easy problem, just have to twist your minds, First, calculate equivalent resistance between A AND D considering A AND B to series and then parallel connection and finally between A and C as series and again parallel and so our answer is: $\frac{r}{2}$

The reconfiguration of the network makes this relationship more clear:

The effective resistance is given by the three parallel branches $DA, DBA,DCA$ :

$\displaystyle \frac{1}{R_{eq}} = \frac{1}{r} + \frac{1}{2r}+ \frac{1}{2r}$

Thus:

$\displaystyle R_{eq} = \frac{r}{2}$