Construction of Regular Pentagon using Rusty Compass

Theorem

Using a straightedge and rusty compass, it is possible to inscribe a regular pentagon inside a circle.

Construction

Let the rusty compass be set to the radius $AD$ of the circle $\CC$ whose center is at $D$.

Let $ADO$ be a diameter of $\CC$.

Using Construction of Perpendicular using Rusty Compass, construct a straight line at right angles to $AD$ from the endpoint $A$.

Mark off $AE$ on this perpendicular so that $AE = AD$.

Bisect $AD$ at $Z$ and draw $ZE$.

On $ZE$, mark off $ZH = DA$ and bisect $ZH$ at $T$.

Construct a straight line at right angles to $EZ$ at $T$.

Let this perpendicular meet $DA$ which has been produced to $I$.

Construct a circle whose center is at $I$ with radius $AD$.

Let this circle meet circle $\CC$ at $L$ and $M$.

Bisect $MO$ and $LO$ and construct perpendiculars to $MO$ and $NO$ respectively at these points of bisection.

Let these perpendiculars meet $\CC$ at $N$ and $P$.

The vertices of the required regular pentagon are $L$, $M$, $N$, $O$ and $P$.

Proof

This theorem requires a proof. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

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 ($\text {940}$ – $\text {998}$): $45$