OP diluted to SG 1.110 when he should have diluted to SG 1.112 (he wasn't clear on the distinction between SG and density). Assuming he measured SG perfectly he would have gotten 19.67% instead of 20%. I'd say, though technically he did it incorrectly he got close enough for government work as we say around DC.
The gravimetric method suggested by ESBbrewer in No. 2 is spot on and, if you have a good balance, probably the best way to go.
The implication of No. 3 is that one realizes that a liter of 20% is going to weigh 1110.1 grams and that there are 222.02 grams of phosphoric acid in it. This can be obtained from 222.02/.85 = 261.2 grams of 85% acid which, as it has a density of 1.6856 means that 261.2/1.6856 = 154.96 mL of 85% acid will deliver that much phosphoric. Put that in a 1L volumetric flask and make up to the mark and Bob's your uncle. That post said 155.1 but I think that is sufficiently close to 154.96 that we could call it correct.
In No.6 you confuse SG and density but as the process of dilution essentially involves ratios the missing 0.998203 factor cancels out and again you get very close to the 'right' answer.
So all the methods suggested here are, for all practical purposes, correct. Perhaps the most naive would say that 20:85::1:4.25 so that I could add 3.25 parts water to 1 part 85% acid to get a 20% solution. In fact, a 29% solution would be obtained by doing that but even this result isn't that far off.
I'll also point out that it isn't the percentage strength of the solutions that is so important. It is the normality - the number of protons in a mL of them. When we do pH calculations we wind up, for a given desired mash pH, with a proton deficit number that must be satisfied by adding acid. If we have a proton deficit of, for example, 20 mEq it is trivial to compute the number of mL acid of given strength required to furnish those. It is simply mL = mEq/N where N is the normality. Phosphoric acid at 85% concentration is about 14.9 N, 20% is 2.3 N and 10% 1.1 N to mash pH of 5.5 and these normalities change very little for other mash pH's in the usual range. Thus it would take 20/1.1 = 18.2 mL of 10% acid to cover the 20 mEq deficit.
So even the very naive (or rather, let's say, lazy because the very naive would not know what we are talking about here) could make his dilution by adding 3.25 parts water to 1 part acid, obtain his 29% solution, standardize it against chalk and determine that it is actually 3.5 N. If he writes '3.5 N' on the bottle he is then set to make acid additions to his mash once he calculates the estimated proton deficit.
