Where is the hole in this argument asserting the constructibility of all regular polygons? Some engineers have a so-called "general" method for constructing any (regular) polygon with the classical instruments only, given the length of its side (they may recognise that it appears to be inaccurate for some polygons, say for the $7$-gon, but this is beside the point of my question, as I hope you see later on).
They proceed by constructing a segment $AB$ equal to the given length, then bisecting it with a perpendicular line intersecting $AB$ at its midpoint $M$. From one end of the segment, say $A$, an angle $S\widehat AM=45°$ is constructed, where $S$ is the intersection point of the other arm of the angle and the perpendicular bisector. On point $B$, an angle $H\widehat BM=60°$ is similarly constructed. The points $S$ and $H$ are clearly the circumcentres of a square and a hexagon of side $AB$ respectively. All is fine up till now.
Then they proceed to bisect the segment $HS$ to get its midpoint $P$, which they assert to be the circumcentre of a pentagon of side $AB$. Since $HP=PS$, they mark off points above $H$ using the distance $HP$, and claim that these points give the circumcentres of any $n$-gon with $n\ge7$.
Of course, this is impossible according to the theorem of constructibility of Gauss. For example (and from now I shall focus on the $7$-gon wlog), the regular heptagon cannot be so constructed. It follows that even though all the steps of the construction (with one possible exception) appear to be justified, there must be something wrong with the reasoning somewhere. In particular, one suspects the highlighted step above as a possible source of an extraneous assumption, but I cannot quite pinpoint why this step is not justified. What exactly is the problem with this step (or any other in the argument, assuming it is not indeed this step as I think) in clear terms? In particular,

how can one make such an engineer see that there is something wrong with this construction, by pointing out some flaw in one or more of the steps therein?

Thank you.
 A: Let $E_n$ indicate the "engineer's circumcenter" of the $n$-gon with side $\overline{AB}$. Consider the case of $n=12$:

Although we might reasonably believe that $E_{12}$ looks too high to be the center of the polygon, we must admit that there could be some inaccuracies in the drawing. Fine. Let's define 
$$a :=|ME_4| = |MA| \qquad b := |ME_6| \qquad c := |E_5E_6| = \frac12(b-a)$$
By the engineer's construction, we find that
$$|ME_{12}| = |ME_{6}|+|E_6E_{12}| = b + 6 c = b + 3(b-a) = 4b - 3 a = a \left( 4 \sqrt{3} - 3 \right) \tag{1}$$ 
where I've incorporated $b/a = \sqrt3$, a well-known ratio from the $30^\circ$-$60^\circ$-$90^\circ$ triangle. Yet, the diagram makes clear (in a way that doesn't depend upon accurate drawing) that the distance from $M$ to the $12$-gon's center is actually 
$$2 a + b = a\left(2 + \sqrt3\right) \tag{2}$$ 
Consequently, if the engineer's construction were correct, then we would have
$$4\sqrt{3} - 3 = 2 + \sqrt{3} \qquad\to\qquad \sqrt{3} = \frac{5}{3} \tag{3}$$
 which is, of course, untrue. (Proof: $(5/3)^2 = 25/9 \neq 3$. Or, you know, recall that $\sqrt3$ is irrational. Whatevs.) $\square$
A: If the reason given is 'this works by extrapolation', then ask them to extrapolate the other direction: construct a point $P'$ on the segment $SM$ with $|SP'|=|SP|$ and ask them if they think $P'$ is the center of the equilateral triangle $ABH$ (which it would have to be, by construction); it should be fairly clear that it's not (and this can easily be seen with a ruler and some quick measurements using only the given diagram, since one just has to measure the distances from $P'$ to $H$ and to $A$, say).
Alternately, you may be able to go one step further and argue that if this is the case, then surely the point $P''$ on $SM$ with $|SP''|=2|SP|$ must be $M$ itself, since it should be the center of the 'digon' on base $AB$, and then show that that's not the case.
Given either of these, it should be possible to argue that if the formula doesn't work exactly 'going down' then there should be no reason to believe that it works exactly 'going up'.
A: The distance from the midpoint of the side of a regular polygon to the circum-center (the apothem), does not increase linearly with an increase in the number of sides. Forty years of engineering and I never heard of this crude construction before.
A: Converting a comment to an answer, as requested. I'll paraphrase and expand the thoughts.

OP asks: "[W]hy [...] is it that [the midpoint of the circumcentres of the square and hexagon] is not the circumcentre of the regular pentagon [...]?" 
I respond: It's just not ... and there's no reason to even suspect that it should be. (OP counters that there is a reason: "intuition", and its fondness for the mean. Be that as it may ...)
The issue can be settled by explicit calculation. The distance from the center to the side (of length $1$) of a regular $n$-gon (ie, the apothem) is given by
$$\frac{1}{2}\tan\frac{\pi(n-2)}{2n}$$
For $n=4$, this is $1/2 = 0.5$; for $n=6$, it's $\sqrt{3}/2 = 0.8660\ldots$; for $n=5$, it's
$$\frac{1}{2}\sqrt{1+\frac{2}{\sqrt{5}}} = 0.68819\ldots$$
This is not the average of $1/2$ and $\sqrt{3}/2$. It's close —the average is $0.6830\ldots$— but it's not equal. Extrapolating to arbitrary $n$ only compounds the error. $\square$

I'll note that my previous answer avoids the messy calculation of the $5$-gon's apothem. By considering $n=12$, the inaccuracy of the construction is exposed using only the well-known elements of the $30^\circ$-$60^\circ$-$90^\circ$ triangle. 
