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(I asked this question in the general Mathematics forum, but I have been advised to post it here instead -- or as well.)

In David Wells's "Curious and Interesting Puzzles", Penguin, 1992, his Puzzle 38 is taken from a work (unspecified) by Abu'l-Wafa Al-Buzjani (but I suspect it could be "A Book on Those Geometric Constructions Which Are Necessary for a Craftsman"), reproduced, apparently, in J.L. Berggren: "Episodes in the Mathematics of Medieval Islam", Springer, 1986.

"Construct an equilateral triangle inside a square, so that one vertex is at a corner of the square and the other two vertices are on the opposite sides."

This is one of the three constructions provided in the solutions:

enter image description here

Let $M$ be the midpoint of $CD$.

Construct $MB$.

Construct an arc centre $B$ and radius $AB$ to cut $MB$ at $N$.

Produce $DN$ to $H$.

$DH$ is then one side of the equilateral triangle, where $DG = DH$ is one of the other sides.

Except it's not. $GH$ is longer than $DH$.

Analysing the angles, it turns out that $\angle CDH = \arctan \frac {3 - \sqrt 5} 4$, which is about $10.8$ degrees.

So clearly this is a mistake. (Wonderful though Wells's books are, they are often riddled with errors, from simple typos and misattributions through to bad mathematics.)

I have been unable to find online copy of either Abu'l-Wafa Al-Buzjani's work or J.L. Berggren's (and at this stage I am unable to hunt it down in a library, and unwilling to get a copy of my own), so I have not been able to find out whether the mistake is Wells's (mistranscribing the construction), or whether it has been sitting there all this time in Abu'l-Wafa Al-Buzjani and nobody has noticed it, or halfway between the two.

Is anyone able to throw any light on what is shown in those source works -- and if the error is in there as well, has anyone else ever noticed this?

Or even: am I the one to analyse this all wrong?

(Also, can anyone suggest the appropriate tags to place this under? I've hunted around for good ones, but I can only find the one. Please bear with me, I'm new here, and I'm afraid I'm an autodidact.)

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  • $\begingroup$ @Nick I embedded it in a cartesian plane with $A$ at $(0,0)$ and expressed $MB$ and arc $AN$ as equations in $x$ and $y$, then determined where they intersected. From there the tangent of $\angle CDH$ seemed to evaluate as I put it. When I plotted it on GeoGebra, my findings were corroborated by evaluating it to $10.81$ degrees, consistent with $\arctan \frac {2-\sqrt 5} 4$. What's your working to get $\ (2-\sqrt 3)$? I need to see where I made a mistake. $\endgroup$
    – Prime Mover
    Jul 25, 2020 at 22:26
  • $\begingroup$ @Nick Find my analysis here: proofwiki.org/wiki/… $\endgroup$
    – Prime Mover
    Jul 25, 2020 at 22:28
  • $\begingroup$ @AlexandreEremenko How did you reach that? $\endgroup$
    – Prime Mover
    Jul 25, 2020 at 22:57
  • $\begingroup$ @Nick Sorry I made a mistake in my first comment, it should say "$10.81$ degrees, consistent with $\arctan \frac {3-\sqrt 5} 4$" not "$\arctan \frac {2-\sqrt 5} 4$" $\endgroup$
    – Prime Mover
    Jul 25, 2020 at 23:03
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    $\begingroup$ I looked through Berggren's book. He gives (pp. 107-111) five geometric constructions from Abū al-Wafā's On Those Parts of Geometry Needed by Craftsmen, but this one is not among them. Nor does it appear in the exercises to that chapter or in the chapter on trigonometry. $\endgroup$
    – Conifold
    Jul 26, 2020 at 3:23

2 Answers 2

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Wells' book is available for download on PdfDrive. Puzzle #38 appears in the section of problems attributed to Abū al-Wafā and is preceeded by the comment that "he is best known for his study of geometrical dissections and of constructions with a rusty compass, meaning a compass which is so stiff that it can be used with only one opening". In the solution section he adds "Abul Wafa gave five different solutions. Here are three of them". There is no citation except for "[Berggren, 1986]" at the bottom of the page after the three solutions. As indicated in the OP, this apparently refers to Berggren's Episodes in the Mathematics of Medieval Islam. Similar citation appears under #45. Some others (#41,#42) have "[Wells, 1975]", which is his own article On Gems and Generalisations in Games and Puzzles magazine.

Berggren's book is available on ResearchGate (one can download from it after registering). In it section 8 of chapter 3 is entitled Geometry with a Rusty Compass and features five problems from Abū al-Wafā's On Those Parts of Geometry Needed by Craftsmen on pp. 107-111. They are:

  1. To construct at the endpoint A of a segment AB a perpendicular to that segment, without prolonging the segment beyond A.
  2. To divide a line segment into any number of equal parts.
  3. To bisect a given angle.
  4. To construct a square in a given circle.
  5. To construct in a given circle a regular pentagon with a compass opening equal to the radius of the circle.

At the end he adds "Abū al-Wafā”s treatise contains a wealth of beautiful constructions for regular ngons, including exact constructions for n = 3, 4, 5, 6, 8, 10. It also gives a verging construction for n = 9 which goes back to Archimedes and the approximation for n = 7 that gives the side of a regular heptagon in a circle as equal to half the side of an inscribed equilateral triangle." Some of these do appear as Wells's "puzzles" (#44,#45), but apparently not #38 or several others. Nor does it appear as an exercise to the chapter 3, or in chapter 5 on trigonometry, where Abū al-Wafā is also featured. Presumably, Wells got the missing ones from somewhere, perhaps even directly from Geometry Needed by Craftsmen, but he does not say.

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  • $\begingroup$ Yes I've got Wells's book, that's where I got that incorrect construction from. What I was particularly interested in was whether the 3rd construction for puzzle 38 could in fact be found in either Berggren or Abul Wafa. As for getting access to the Berggren book, ordinary peasants like me can't join ResearchGate because it's elitist and exclusionary, so I'm up a gum tree there -- but I refuse to fork out £40 for an overpriced texbook. $\endgroup$
    – Prime Mover
    Jul 26, 2020 at 11:07
  • $\begingroup$ Thanks indeed for your research, though, I'm just terminally irritated by publications which publish such utter balderdash as Wells increasingly appears to be doing. Does he never ever check his work? No wonder he failed his degree. $\endgroup$
    – Prime Mover
    Jul 26, 2020 at 11:15
  • $\begingroup$ Incidentally, he would not have included the construction for Wells's $38$ in that section $3$, because it was not made using a rusty compass. Would $38$ have be included in a different section? As I say, I (enragingly) cannot access any of the works behind ResearchGate's firewall because they limit access to students and academics, or people with professional authority. To damned hell with them. $\endgroup$
    – Prime Mover
    Jul 26, 2020 at 11:36
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    $\begingroup$ @PrimeMover Try this site. I looked at all references to Abū al-Wafā in Berggren's index. $\endgroup$
    – Conifold
    Jul 26, 2020 at 22:41
  • $\begingroup$ Thank you, but: Bad token: e62f45bd5c94efa172f6c4c3f731bed6 Time: 2020-07-27 08:18:12 GB 87.75.58.14 $\endgroup$
    – Prime Mover
    Jul 27, 2020 at 5:18
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Denote $\beta=\angle MBC,\;\alpha=\angle DMN,\;\gamma=\angle CDH$. We have: $$\tan\beta=1/2,\quad\cos\alpha=-\sin\beta=-1/\sqrt{5},\quad\sin\alpha=\cos\beta=2/\sqrt{5}.$$ Take $DM=1$, then $MN=\sqrt{5}-2$, and let $x=DN$. By the rule of cosines, $$x^2=1^2+(\sqrt{5}-2)^2+2(\sqrt{5}-2)/\sqrt{5}=12-24/\sqrt{5}.$$ Then by the rule of sines, $$\sin\gamma=\frac{(\sqrt{5}-2)(2/\sqrt{5})}{\sqrt{12-24/\sqrt{5}}}\approx0.1875,$$ and the calculator gives $\gamma\approx 10.8$ degrees.

Remark. An internet search finds a Russian translation of this book of Abu'l Wafa,

Абу-л-Вафа ал-Бузджани. Книга о том, что необходимо ремесленнику из геометрических построений. Физико-математические науки в странах Востока, 1966, 1, 56-140.

This is a collection of translations. But this book I could not find on Internet.

Remark: since the true solution involves a construction of $2-\sqrt{3}$ with a compass and ruler, which is easy, it is plausible that Abu'l Wafa had a correct construction which was distorted by some scribe, or translator, or by Wells himself. But there is no answer to this unless we get some translation of Abu'l Wafa.

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  • $\begingroup$ But it is still apparent that the triangle in question is not equilateral? For $\triangle DHG$ to be equilateral, it must be that $\angle CDH = 15^\circ$. $\endgroup$
    – Prime Mover
    Jul 25, 2020 at 23:29
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    $\begingroup$ Yes, of course. I suppose Abu'l Wafa made this mistake (which is kind of strange) and David Wells did not notice. But I do not see how to check this. Probably Abu'l Wafa is not available in English. I do read Russian, but do not see how to obtain the Russian translation. $\endgroup$
    – Alexandre Eremenko
    Jul 25, 2020 at 23:37
  • $\begingroup$ I've looked again, and I see there's a factor of $2$ missing in your use of the cosine rule, I think it should be $x^2 = 1^2 + (\sqrt 5 - 2)^2 + 2 (\sqrt 5 - 2)/ \sqrt 5$ which seems to evaluate to $x = \sqrt {12 - 16/\sqrt 5}$ but my arithmetic went wrong after this -- but the bottom line is that the correct angle does seem to be $10.8^\circ$. $\endgroup$
    – Prime Mover
    Jul 25, 2020 at 23:43
  • $\begingroup$ All I know is that the construction as given in Wells is wrong (as we have established). What I do not know is whether it was wrong in Abul Wafa also, and copied by Wells via Berggren and checked by neither of them (and nobody else in the last 1000 years either), or Abul Wafa (and possible Berggren) was correct, but that Wells copied it wrong and created a faulty construction which is not the same as the one Abul Wafa built. In other words: am I the first in history to notice this, or is it just Wells's mistake? $\endgroup$
    – Prime Mover
    Jul 25, 2020 at 23:51
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    $\begingroup$ Thanks, I corrected my mistake, and your first calculation was really correct. Sorry. $\endgroup$
    – Alexandre Eremenko
    Jul 26, 2020 at 0:17

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