Division of Straight Line into Equal Parts using Rusty Compass

Theorem

Let $AB$ be a line segment.

Using a straightedge and rusty compass, it is possible to divide $AB$ into as many equal parts as required.

Construction

Let $n + 1$ divisions be required.

Using Construction of Perpendicular using Rusty Compass, construct straight line at right angles to $AB$ from the endpoints $A$ and $B$, but in opposite directions.

Mark off $n$ points along each of those perpendiculars, equally spaced by the length determined by the opening of your rusty compass.

Let the points from $A$ be $A_1, A_2, \dotsc, A_n$.

Let the points from $B$ be $B_1, B_2, \dotsc, B_n$.

Join the points:

$A_1$ to $B_n$

$A_2$ to $B_{n - 1}$

$\cdots$

$A_{n - 1}$ to $B_2$

$A_n$ to $B_1$

The divisions are seen where these lines intersect $AB$.

Historical Note

This construction was discussed by Abu'l-Wafa Al-Buzjani in a work of his from the $10$th century.

Sources

• 1986: J.L. Berggren: Episodes in the Mathematics of Medieval Islam
• 1992: David Wells: Curious and Interesting Puzzles ... (previous) ... (next): Abul Wafa ($\text {940}$ – $\text {998}$): $44$