Where is the hole in this argument asserting the constructibility of all regular polygons?

Question

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.

Answer 1

Converting a comment to an answer, as requested. I'll paraphrase and expand the thoughts.

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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. $\square$

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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.

Answer 2

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'.

Answer 3

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.

Answer 4

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.) $\square$

Answer 5

Let the length of $AB$ be $s.$ For integers $n \geq 4,$ let $C_n$ be the point given by the engineer's construction as the alleged circumcenter of an $n$-gon (polygon with $n$ sides) including side $AB.$ In particular, according to the construction, $S = C_4,$ $P = C_5,$ and $H = C_6.$

Let $h_n$ be the distance from $C_n$ to $M$ (where $M$ is the midpoint of $AB$). Let $\theta_n = \angle AC_nB,$ that is, $\theta_n$ is the angle at the apex of the isoceles triangle formed by the alleged circumcenter of the $n$-gon and the side of the desired $n$-gon.

Following the steps of the construction, we have $h_4 = \frac s2,$ $h_6 = \frac s2\sqrt3,$ and in general, for $n\geq 4,$ $$h_{n+1} = h_n + \frac12(h_6 - h_4) = h_n + \frac s4(\sqrt3 - 1).$$ We also have $$\theta_n = 2 \arctan\left(\frac {s}{2h_n}\right).$$

Since the location of the alleged circumcenter is as far as the construction was given, let's suppose the next step to construct the $n$-gon is to replicate the isoceles triangle $\triangle AC_nB$ adjacent to the original triangle, for example, construct $\triangle A'C_nB$ so that $A'$ is on the opposite side of $BC_n$ from $A$ and $\triangle A'C_nB \cong \triangle AC_nB.$ Then $A'B$ is another side of the alleged $n$-gon that is supposed to be constructed by this method. We then repeat this step until we have built all the sides of the $n$-gon.

If the construction is accurate, we should expect the sequence of sides to close upon itself when there are $n$ sides. In fact, the sequence of sides will close exactly (forming an exact regular $n$-gon) if the apex angles of $n$ isoceles triangles fill in the entire circle around the center $C_n$, that is, if
$n\theta_n = 360 \text{ degrees}.$ But let's work out the values of $\theta_n$ (measured in degrees) in a few cases and see what actually happens:

\begin{array}{ccccc} n & \dfrac{2h_n}{s} & \theta_n & n\theta_n & \dfrac{360}{\theta_n}\\ \hline 4 & 1 & 90 & 360 & 4 \\ 5 & 1.366025404 & 72.41204623 & 362.0602311 & 4.971548503 \\ 6 & 1.732050808 & 60 & 360 & 6 \\ 7 & 2.098076211 & 50.96746918 & 356.7722842 & 7.063328939 \\ 8 & 2.464101615 & 44.17732616 & 353.4186093 & 8.148976665 \\ 9 & 2.830127019 & 38.9208091 & 350.2872819 & 9.249550775 \\ 10 & 3.196152423 & 34.74731825 & 347.4731825 & 10.36051178 \\ 11 & 3.562177826 & 31.36174493 & 344.9791943 & 11.47895312 \\ 12 & 3.92820323 & 28.56475913 & 342.7771096 & 12.60294191 \\ 13 & 4.294228634 & 26.21776217 & 340.8309082 & 13.73114905 \\ 14 & 4.660254038 & 24.22181756 & 339.1054458 & 14.86263362 \\ 15 & 5.026279442 & 22.50462422 & 337.5693633 & 15.99671234 \\ 16 & 5.392304845 & 21.01223175 & 336.1957081 & 17.13287785 \\ 17 & 5.758330249 & 19.70362898 & 334.9616927 & 18.27074598 \\ 18 & 6.124355653 & 18.54712081 & 333.8481746 & 19.41002076 \\ 19 & 6.490381057 & 17.51784696 & 332.8390923 & 20.55047066 \\ 20 & 6.856406461 & 16.59604739 & 331.9209479 & 21.69191202 \\ \end{array}

The results are perfect for $n = 4$ and $n = 6.$

For $n = 5,$ we end up with two triangles overlapping by about two degrees at $C_5.$ That may be good enough for a line drawing, especially if we make sure to erase any parts of the overlapping sides that appear to be outside the pentagon.

For $n = 7,$ we end up with a gap of more than three degrees after constructing the seventh isoceles triangle. Again, for a line drawing that's merely for looking at, we can fudge the result and just extend the two sides adjacent to the gap until they meet. For $n = 8,$ the gap is more than six degrees; for $n = 9,$ the gap is getting close to ten degrees. I wonder when the people who look at the resulting line drawings will start to notice the discrepancy?

Going onward in the table, watch the values of $360/\theta_n,$ which indicate how many of the constructed isoceles triangles will "fit" around the point $C_n.$ The integer part of $360/\theta_n$ represents how many triangles will fit without overlapping at all, and the fractional part represents how much of the apex angle of one more such triangle will "fit" between the others without overlapping. For example, the value of $360/\theta_5$ indicates that there is a slight overlap when we construct the fifth side of the alleged pentagon.

What this column shows is that if we perform the construction accurately, when we have constructed the $11$th side of the alleged $11$-gon there is a gap almost large enough to fit half of another isoceles triangle. The alleged $15$-gon has very nearly enough space to fit a $16$th side, and the alleged $16$-gon has space for a $17$th side with room to spare.

The discrepancies just get worse as we try more sides. The alleged $23$-gon winds up with at least $25$ sides (two more than it is supposed to have). The construction that is supposed to give $30$ sides gives $33.$

You cannot fix these errors by drawing more carefully. The errors are inherent in the trigonometry of the construction. For come purposes, your engineer may not care about the two-degree overlap of the pentagon construction, but surely it is not acceptable to get $17$ sides when you wanted $16$ or $33$ when you wanted $30.$

If we're going to go to so much trouble to get such bad results, we might as well just use a calculator to find the inradius of the polygon, lay off that distance along the perpendicular bisector of $AB,$ and use a protractor to construct the central angles around that point.

What this construction might be good for is as an approximate shortcut for the construction of the pentagon (for which an exact method also exists), the heptagon, and (if accuracy really doesn't matter that much) perhaps the octagon. Even so, does anyone make engineering drawings with classical instruments nowadays, or is this just a historical curiosity?

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The preceding part of this answer shows that the method gets results that (I hope) would be regarded as bad even for rough-drawing purposes, but it does not point out an error in any particular step (other than the fact that at the end we have the wrong result). So let's consider what kind of result a correct construction would have to get.

If the side of a regular $n$-gon is $s,$ the angle subtended by that side at the circumcenter of the polygon is $\alpha_n,$ and the inradius of the polygon is $r_n,$ then $$ r_n = \frac s2 \cot\left( \frac{\alpha_n}{2} \right). $$ But $\alpha_n = \frac{2\pi}{n}$ (that is, $\frac{360}{n}$ when we measure the angle in degrees). The engineer's construction implicitly says that $r_n$ is a linear function of $n,$ at least for $n \geq 4,$ which is to say that $\cot(\pi/n)$ is a linear function of $n.$ That's simply wrong; if it were correct, trigonometric functions would be much easier to calculate than is actually the case. The claimed linear relationship is only approximately true, and even the approximation is only good over a small interval of values.

So the incorrect step specifically is when we lay off a distance of some multiple of $HP$ as if $\cot(\pi/n)$ were a linear function of $n.$ It's a decent approximation (for some purposes) for the first few multiples of $HP,$ but really only for a few multiples, and then it quickly gets worse until you end up constructing the wrong polygon altogether.