Ratio of Lengths of Arms of Pentagram

Theorem

Consider a pentagram.

Let $AC$ be the length of one of the lines which span the pentagram and define it.

Let $B$ be one of the points where $AC$ intersects one of the other such lines such that $AB > AC$.

Then:

$\dfrac {AC} {AB} = \phi$

where $\phi$ denotes the golden mean.

Proof

Follows directly from Straight Lines Subtending Two Consecutive Angles in Regular Pentagon cut in Extreme and Mean Ratio.

$\blacksquare$

This theorem requires a proof. In particular: Flesh this out 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

The Ratio of Lengths of Arms of Pentagram was known to the Pythagoreans.

Sources

• 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $1 \cdotp 61803 \, 39887 \, 49894 \, 84820 \, 45868 \, 34365 \, 63811 \, 77203 \, 09179 \, 80576 \ldots$
• 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): golden section
• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): golden section