We are mixing pure EtOH with water and the initial mass of EtOH is: $$ M_{EtOH} = \rho_{100} * V_{100} \tag{1} $$ The mass of EtOH portion in Solution would be $$ M_{EtOH} = \mu * \rho_x * V_x \tag{2} $$ Equating (1) nad (2) $$ \rho_{100} V_{100} = \mu \rho_x V_x \tag{3} $$ or $$ \frac{V_{100}}{V_x}= \mu \frac{\rho_x} {\rho_{100}} \tag{4} $$ From ABV definition: $$ \sigma = \frac{V_{100}}{V_x} \tag{5} $$ From (4) nad (5) $$ \sigma = \mu \frac{\rho_x}{\rho_{100}} \tag{5} $$ Equivalently, $$ \mu = \sigma \frac{\rho_{100}}{\rho_{x}} \tag{6} $$ Now, Let’s mix two solutions $ (V_1, \sigma_1) + (V_2, \sigma_2) -> (V_3, \sigma_3)$ By definition of ABW, $$ \mu_3 = \frac{M_1^{EtOH} + M_2^{EtOH}} {M_{tot}} $$ or, $$ \mu_3 = \frac{V_1\rho_1\mu_1 + V_2\rho_2\mu_2}{V_1\rho_1 + V_2\rho_2} \tag{7} $$ Substitute (6) into (7), $$ \frac{\sigma_3 \rho_{100}}{\rho_3} = \frac{ (V_1 \rho_1 \sigma_1 \rho_{100}) / \rho_1 + ( V_2 \rho_2 \sigma_2 \rho_{100}) / \rho_2 } {(V_1 \rho_1 + V_2 \rho_2)} \tag{8} $$ Reducing both parts on common variables gives: $$ \frac{\sigma_3}{\rho_3} = \frac{V_1 \sigma_1 + V_2 \sigma_2}{V_1 \rho_1 + V_2 \rho_2} \tag{9} $$ In practical case, we know initial Volume and ABVs for two reagents and product: $ V_1, \sigma_1, \sigma_2, \sigma_3,$. Usually, We need to find $V_2$
Let's resolve (9) in repect to $V_2$: $$ V_2 = \frac{\sigma_1 \rho_3 - \sigma_3 \rho_1} {\sigma_3 \rho_2 - \sigma_2 \rho_3} V_1 \tag{10} $$ Generals equation (10) can be reduced for case of mixing EtOH with pure water. In this case $\sigma_2=0$ and $ \rho_2 \approx 1 $: $$ V_2 = ( \frac{\sigma_1}{\sigma_3} \rho_3 - \rho_1) V_1 \tag{11} $$

The formulas (10) and (11) are working formula for our Alcohol calculators
where $\rho$ (density) is an empirical values interpolated from tables $\rho = \rho(\mu(\sigma)) $

Derivation EtOH and H20 Formula: