I'll take a stab at it...
You can reduce the gain by dividing the input signal by two, which, technically, doesn't reduce the gain but, rather, reduces the input signal, which, in turn, reduces the output.
However, the input resistance is QR; therefore, your divider must look like QR in the end.
Per Thevenin, you open I and short V.
Shorting your input signal to ground puts the upper resistor of your divider in parallel with the lower, giving Rin = R(upper)||R(lower).
Hence, you'll need to set Rin = QR.
The way I see it, QR is the same as the source resistance, so unless you're driving the BPF with an op amp, a low Z source, you'll need to add it to R(upper), first.
Incidentally, when I think of a BPF, the circuit you gave is not what I think of; I'm wondering about your chosen circuit's significance.