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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.

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    $\begingroup$ For those looking for a diagram: imgur.com/a/7xNZkXk. I'm not sure if this is the type of answer you're looking for - but geometry software or a little trigonometry shows that the angles aren't correct, but for $n=5$ or $n=7$ they're not far off: for $n=8$ the angle constructed is around $44.18^\circ$ instead of the correct $45^\circ$. $\endgroup$ – B. Mehta Jun 1 '18 at 23:22
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    $\begingroup$ An engineer should be able to determine (perhaps by careful measurement of a large enough figure) that $$\angle APB \approx 72.412^\circ \neq 72^\circ = \frac15\cdot 360^\circ$$ so that the construction already fails for $n=5$. $\endgroup$ – Blue Jun 1 '18 at 23:42
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    $\begingroup$ "how can one make an engineer see .... which they can easily grasp?" Well, an engineer could just as easily ask "how can one make a mathematician see that nobody practical should care if it is not exact as in the real world accuracy only exists to a margin of error; how can we put it in practical terms they can grasp that they are wasting time and their lives on things that can't matter". Let's not condescend, shall we? Engineers know damned well it is an approximation. They just don't see why it should matter. (And to be fair, you haven't explained why it should.) $\endgroup$ – fleablood Jun 2 '18 at 0:35
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    $\begingroup$ @Allawonder: It's not so much that the construction provides valid $4$-gon and $6$-gon centers, but that it builds upon them. The centers of a square and hexagon are "easy" cases; the construction's value is in providing (approximations of) the "hard" cases. $\endgroup$ – Blue Jun 2 '18 at 18:56
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    $\begingroup$ @Allawonder: As for disqualifying my answer for containing "an irrelevancy" ... I may be answering your question, but I'm addressing the Math.SE community. My epilogue puts my two answers in context for the general reader. $\endgroup$ – Blue Jun 2 '18 at 18:58
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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.

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  • $\begingroup$ @Blue This is all well and good -- except for the epilogue. First, the computation is not that messy. Second, as I said before, we all already know that the construction is problematic, so that it is not a question of accuracy. I wanted to know which step of the construction assumed an unjustified step, directly. Your first answer, by producing a contradiction, only shows what we all knew before -- that there is something wrong with the argument. However, it does not tell us where the fault is, neither does it tell us exactly what this fault is. $\endgroup$ – Allawonder Jun 2 '18 at 17:16
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    $\begingroup$ @Allawonder: Either accept the answer or don't. I no longer care to try and determine what will satisfy you. $\endgroup$ – Blue Jun 2 '18 at 17:17
  • $\begingroup$ @Blue It is not a question of what will satisfy me. Anybody can see that the first answer entirely misses the point of the question (and I entirely take the blame for that), so appending the epilogue to this more relevant answer seems to me to be, er..., what-shall-I-call-it? I hope you got the point, anyway. PS. The first sentence is tautological, so it is vacuously true. $\endgroup$ – Allawonder Jun 2 '18 at 17:38
  • $\begingroup$ @Jean-ClaudeArbaut Right off the bat, the first statement is false. It does lead to at least one regular polygon, for example the square whose side is $AB$ -- or you dispute this? $\endgroup$ – Allawonder Jun 2 '18 at 18:08
  • $\begingroup$ @Jean-ClaudeArbaut I have responded to a similar comment by the same person in the thread under the question. $\endgroup$ – Allawonder Jun 2 '18 at 19:01
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Let $E_n$ indicate the "engineer's circumcenter" of the $n$-gon with side $\overline{AB}$. Consider the case of $n=12$:

enter image description here

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.) $\square$

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    $\begingroup$ @Allawonder: Once you have received answers (especially multiple ones), you should not change the fundamental nature of the question. Rather, post a new one. You can edit the question to provide a link to the revised version (and you should also have a link in the new one that points back here). $\endgroup$ – Blue Jun 2 '18 at 14:34
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    $\begingroup$ BTW: As for which step is wrong: My comment to the question itself indicates that the construction of point $P$ (the ostensible center of the pentagon) is already inaccurate. The construction fails at its very first substantive step. (I consider construction of $45^\circ$ & $60^\circ$ angles merely preliminaries.) The center of the pentagon is not the midpoint of the centers of the square and hexagon; it's passably close at small scales —which is evidently why it's part of engineering lore— but it's wrong. Extrapolating to $n$-gons is unjustified, and merely compounds the error. $\endgroup$ – Blue Jun 2 '18 at 15:16
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    $\begingroup$ "But why [...] is it that $P$ is not the circumcentre of the regular pentagon of side $AB$." That's like asking "Why is it that $2+2$ is not $22$?" It's just not ... and there's no reason to even suspect that it should be. The distance from the center to the side (of length $s$) of a regular $n$-gon (ie, the apothem) is given by $$\frac{s}{2}\tan\frac{\pi(n-2)}{2n}$$ For $n=4$, this is $s/2$; for $n=6$, it's $s\sqrt{3}/2$. For $n=5$, it's $$\frac{s}{2}\sqrt{1+\frac{2}{\sqrt{5}}}$$ which is not the average of $s/2$ and $s\sqrt{3}/2$. $\endgroup$ – Blue Jun 2 '18 at 15:35
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    $\begingroup$ @Allawonder: I doubt the construction's creator intended to make a claim of exactitude, but it's safe to say that construction asserts that the apothem of the pentagon is very nearly approximated by the arithmetic mean of the apothems of the square and the hexagon. $\endgroup$ – Blue Jun 2 '18 at 17:11
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    $\begingroup$ @Allawonder: The only reason to believe that $P$ is the center of the pentagon seems to be an assumption that the center distances of $n$-gons for $n=4,5,6,\ldots$ "ought to be" an arithmetic progression. Blue's argument very nicely shatters that assumption, leaving no reason that I can see for thinking $P$ ought to be the center of a pentagon either. $\endgroup$ – Henning Makholm Jun 2 '18 at 20:55
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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'.

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  • $\begingroup$ I have to thank you for the answer, but there are a couple of problems. In your first paragraph, you conclude that $P'$ is not the centre of the $\triangle ABH$, but it indeed seems to be, for the angle $AP'B=120°$, as expected. Also, one might object to your second paragraph by saying there is no polygon of less than three sides, so it is meaningless to extrapolate downwards beyond $P'$. Finally, it is not clear how the argument in your last paragraph goes, since this seems to work well for the constructible polygons. As I said in the question, where I suspect something to have crept in is... $\endgroup$ – Allawonder Jun 2 '18 at 14:39
  • $\begingroup$ ...in the bisection of $SH$, which seems reasonable but evidently doesn't produce correct results for the $7$-gon, without loss of generality. I shall edit the question to make clearer exactly where I think the problem may be with the construction. Thanks once again. $\endgroup$ – Allawonder Jun 2 '18 at 14:41
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    $\begingroup$ @Allawonder: Why do you say that $AP'B=120^\circ$? That's certainly not obvious. Indeed, by my calculations (following the method in David K's answer in the case $n=3$) that angle would be about $115.25^\circ$. The construction described doesn't work for all the constructible polygons; it only works for the square and the hexagon. $\endgroup$ – Eric Wofsey Jun 2 '18 at 21:27
  • $\begingroup$ @EricWofsey That's right. I realised this only later. I had been a bit too hasty. $\endgroup$ – Allawonder Jun 11 '18 at 13:55
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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?


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.

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  • $\begingroup$ Thank you, even though a large segment of your answer seems unnecessary as it is irrelevant (the engineer may already know it is inaccurate, so that that wasn't my question), so that the last segment might have sufficed well as it immediately shows that the function connecting the inradius to the number of sides is nonlinear as the construction assumes. Thanks once again. The question is now closed. $\endgroup$ – Allawonder Jun 11 '18 at 14:06
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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.

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  • $\begingroup$ Concerning your last sentence, see, for example, Parker and Pickup, Engineering Drawing, 3rd edition, pp. 11, 12. $\endgroup$ – Allawonder Jun 1 '18 at 23:51
  • $\begingroup$ I remember that book and even had classes given by Parker, I believe. Well, that was around 1970, in Bristol England. This was back in the day when pi = 22/7. These rough estimates were fine for aircraft lofting, but not so good for high precision jet turbines. $\endgroup$ – Phil H Jun 2 '18 at 0:10
  • $\begingroup$ Many engineering and technology schools still train their students by that book. My question was about how to convince such people (contra the authors, e.g.) that there is something the matter with the construction. $\endgroup$ – Allawonder Jun 2 '18 at 0:21
  • $\begingroup$ Like I said in my answer, the apothem does not increase linearly with an in crease in the number of sides. For a circle, the radius increases linearly with a linear increase in circumference. But chord length/arc length ratio varies so the linearity is lost. I can see how they could sell it as a close approximation as that's all it is. Pickup and Parker, the authors, must be long gone by now. $\endgroup$ – Phil H Jun 2 '18 at 0:55
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    $\begingroup$ Of course I don't, but I can tell you from experience, I've seen a lot of parts made that didn't go together for want of a better understanding of geometry. $\endgroup$ – Phil H Jun 2 '18 at 1:16

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