A notch filter is a system with transfer function
$$H(s) = \frac{s^2 + \omega_0^2}{s^2+ 2\omega_0\cos(\theta)s + \omega_0^2}$$
In "math form", this transfer function can be written as
$$ H(s) = \frac{s^2 + b_0}{s^2+ a_1s + a_0} \tag1$$
To implement such a transfer function, a Twin-T circuit can be used.
simulate this circuit– Schematic created using CircuitLab
If one tries to write up node equations for \$V_a, V_b, V_c\$ and solving to find the transfer function you get coefficients with far too many terms, making it impossible to analyze. To simplify this, we usually set all component values equal \$R_1 = R_2 = R_3 = R\$ and \$C_1 = C_2 = C_3 = C \$. Doing this and solving the node equations yields this transfer function
$$H(s) = \frac{s^2 + \frac{1}{RC}s + \left(\frac{1}{RC} \right)^2}{s^2 + \frac{4-3\alpha}{RC}s + \left(\frac{1}{RC} \right)^2}$$
where \$\alpha\$ is the potentiometer wiper position. In "math form", the transfer function is written as$$ H(s) = \frac{s^2 + b_1s + b_0}{s^2+ a_1s + a_0} \tag2$$
As one can see, equation (2) does not have the same form as equation (1), because the term \$\frac{1}{RC}s\$ is present in the numerator. However, if I instead set the component values to \$R_1 = R_2 = 2R_3 = R\$ and \$C_1 = C_2 = C_3/2 = C \$ then the transfer function becomes
$$H(s) = \frac{s^2 +\left(\frac{1}{RC} \right)^2}{s^2 + \frac{4-4\alpha}{RC}s + \left(\frac{1}{RC} \right)^2} \tag3 $$
which in "math form" is written as \$H(s) = \frac{s^2 + b_0}{s^2+ a_1s + a_0}\$ which does match the form in (1).
My question is: How does one obtain the insight to set \$R_3 = \frac{1}{2}R \$ and \$C_3 = 2C \$? Like what argument does one have (other than blind trial and error) to set the component values like this?