Inscribing equilateral triangle in square -- mistake in historical work by Abu'l-Wafa Al-Buzjani?

111 Views Asked by At

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?

1

There are 1 best solutions below

2
On BEST ANSWER

Your analysis is correct. Here is a corrected figure below. Rather than joining $B$ and $M$, the correct point of intersection is simply that of the circular arc $AC$ with the perpendicular bisector of $DC$.
enter image description here