Well, you're point 1) is spot on. Steady state would be back to room temperature. What I meant was the state that the liquid and the solid are the same temperature, while the ice is melting. A graph of the liquid temp would be a slope down, then a flat line, then a gradual return to room temp. By steady state I meant the flat line of constant temperature (0C if at STP).

Points 2) ... all moot. Ice shape A vs ice shape B - all else being equal. Remember the ice is submerged in the drink. And if the sphere is not, you pointed out yourself that it would loose heat (melt) even faster. I was only trying to prove that it doesn't melt less.

So, two identical glasses, both containing a cylindrical-shaped water-alcohol mix of the same ratio, both cooled to the same temperature, both in the same ambient conditions. However complex the system is, I think it's fair to say that the two cylinders of fluid at the boundary conditions will absorb heat at the same rate.

q = m·ΔHf

Is that formula wrong because it doesn't contain a shape factor? Does the energy to melt one gram of ice differ based on the shape of the ice? (That's rhetorical, I know it does not.) Are the boundary conditions of 0C water/alcohol/ice somehow different? Maybe only in how much solid to solid contact we have between ice and glass..