# Inscribing Equilateral Triangle inside Square with a Coincident Vertex/Mistake

## Source Work

1992: David Wells: *Curious and Interesting Puzzles*:

- The Puzzles:
- Abul Wafa ($\text {940}$ – $\text {998}$): $38$
- Solution

- Abul Wafa ($\text {940}$ – $\text {998}$): $38$

## Mistake

*Abul Wafa gave five different solutions. Here are three of them. ...*

*... Join $B$ to the midpoint, $M$, of $DC$. Draw an arc with centre $B$ and radius $BA$ to cut $BM$ at $N$. Let $DN$ cut $CB$ at $H$. Then $H$ is one of the vertices sought.*

## Correction

The construction is incorrect.

$DH$ is shorter than $GH$.

Let $\Box ABCD$ be embedded in a Cartesian plane such that:

\(\ds A\) | \(=\) | \(\ds \tuple {0, 0}\) | ||||||||||||

\(\ds B\) | \(=\) | \(\ds \tuple {a, 0}\) | ||||||||||||

\(\ds C\) | \(=\) | \(\ds \tuple {a, a}\) | ||||||||||||

\(\ds D\) | \(=\) | \(\ds \tuple {0, a}\) |

By Equation of Straight Line in Plane and some algebra, the equation for the straight line $MB$ is:

- $(1): \quad y = 2 \paren {a - x}$

By Equation of Circle in Cartesian Plane and some algebra, the equation for the circle with center at $B$ and radius $a$ is:

- $(2): \quad y^2 = 2 a x - x^2$

Hence their point of intersection $M$ is found by solving the simultaneous equations $(1)$ and $(2)$:

\(\ds 4 \paren {a - x}^2\) | \(=\) | \(\ds 2 a x - x^2\) | substituting for $y$ in $(2)$ | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds 4 a^2 - 8 a x + 4 x^2\) | \(=\) | \(\ds 2 a x - x^2\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds 5 x^2 - 10 a x + 4 a^2\) | \(=\) | \(\ds 0\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds \dfrac {10 a \pm \sqrt {100 a^2 - 80 a^2} } {2 \times 5}\) | Solution to Quadratic Equation | ||||||||||

\(\ds \) | \(=\) | \(\ds a \paren {1 \pm \dfrac {\sqrt 5} 5}\) | simplification |

We are interested in the negative square root in this expression, as the positive square root corresponds to the point on $BM$ for negative $y$.

Thus we have:

\(\ds y\) | \(=\) | \(\ds 2 \paren {a - x}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds a \paren {2 - 2 \paren {1 - \dfrac {\sqrt 5} 5} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \dfrac {2 a \sqrt 5} 5\) |

Thus we can calculate the tangent of $\angle CDH$:

\(\ds \tan \angle CDH\) | \(=\) | \(\ds \dfrac {a - \frac {2 a \sqrt 5} 5} {a \paren {1 - \frac {\sqrt 5} 5} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \dfrac {5 - 2 \sqrt 5} {\paren {5 - \sqrt 5} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \dfrac {\paren {5 - 2 \sqrt 5} \paren {5 + \sqrt 5} } {\paren {5 - \sqrt 5} \paren {5 + \sqrt 5} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \dfrac {15 - 5 \sqrt 5} {25 - 5}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \dfrac {3 - \sqrt 5} 4\) |

But from Tangent of 15 Degrees:

- $\tan 15 \degrees = 2 - \sqrt 3$

So:

- $\angle CDH \ne \tan 15 \degrees$

(in fact it is about $10.8 \degrees$).

Hence from Inscribing Equilateral Triangle inside Square with a Coincident Vertex: Lemma:

- $\triangle DGH$ is not an equilateral triangle.

$\blacksquare$

It appears that this mistake originates with David Wells, in his *Curious and Interesting Puzzles*.

This construction cannot be found in J.L. Berggren's work *Episodes in the Mathematics of Medieval Islam*, from which Wells claims to have sourced it.

It also appears that it may not have been the case that Abu'l-Wafa Al-Buzjani actually gave five different solutions.

Wells seems to have got his material from J.L. Berggren's *Episodes in the Mathematics of Medieval Islam*, as stated above, but that work does not contain any such constructions for this result.

It is not clear where Wells actually got the information about those $5$ constructions, but it certainly was not from Berggren.

Berggren's work, in section $8$ of chapter $3$, entitled *Geometry with a Rusty Compass*, features five problems from Abū al-Wafā's *On Those Parts of Geometry Needed by Craftsmen* on pp. $107$ - $111$. They are:

- To construct at the endpoint $A$ of a segment $AB$ a perpendicular to that segment, without prolonging the segment beyond $A$.

- To divide a line segment into any number of equal parts.

- To bisect a given angle.

- To construct a square in a given circle.

- 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 $n$gons, 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 in Wells's *Curious and Interesting Puzzles* (numbers $43$, $44$ and $45$), but apparently he does not include $38$ 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 *On Those Parts of Geometry Needed by Craftsmen*, but he does not say.

## Sources

- 1986: J.L. Berggren:
*Episodes in the Mathematics of Medieval Islam* - 1992: David Wells:
*Curious and Interesting Puzzles*... (previous) ... (next): Abul Wafa ($\text {940}$ – $\text {998}$): $38$