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\begin{figure}\begin{center}\BoxedEPSF{Pentaflake.epsf scaled 700}\end{center}\end{figure}

\begin{figure}\begin{center}\BoxedEPSF{PentaflakeDistances.epsf scaled 730}\end{center}\end{figure}

A Fractal with 5-fold symmetry. As illustrated above, five Pentagons can be arranged around an identical Pentagon to form the first iteration of the pentaflake. This cluster of six pentagons has the shape of a pentagon with five triangular wedges removed. This construction was first noticed by Albrecht Dürer (Dixon 1991).

For a pentagon of side length 1, the first ring of pentagons has centers at Radius

d_1=2r={\textstyle{1\over 2}}(1+\sqrt{5}\,)R =\phi R,
\end{displaymath} (1)

where $\phi$ is the Golden Ratio. The Inradius $r$ and Circumradius $R$ are related by
r=R\cos({\textstyle{1\over 5}}\pi)={\textstyle{1\over 4}}(\sqrt{5}+1)R,
\end{displaymath} (2)

and these are related to the side length $s$ by
s=2\sqrt{R^2-r^2}={\textstyle{1\over 2}}R\sqrt{10-2\sqrt{5}}.
\end{displaymath} (3)

The height $h$ is
h=s\sin({\textstyle{2\over 5}}\pi)={\textstyle{1\over 4}}s\sqrt{10+2\sqrt{5}}={\textstyle{1\over 2}}\sqrt{5}\,R,
\end{displaymath} (4)

giving a Radius of the second ring as
d_2=2(R+h)=(2+\sqrt{5}\,)R=\phi^3 R.
\end{displaymath} (5)

Continuing, the $n$th pentagon ring is located at
\end{displaymath} (6)

Now, the length of the side of the first pentagon compound is given by

\end{displaymath} (7)

so the ratio of side lengths of the original pentagon to that of the compound is
{s_2\over s}={R\sqrt{5+2\sqrt{5}}\over {\textstyle{1\over 2}}R\sqrt{10-2\sqrt{5}}} = 1+\phi.
\end{displaymath} (8)

We can now calculate the dimension of the pentaflake fractal. Let $N_n$ be the number of black pentagons and $L_n$ the length of side of a pentagon after the $n$ iteration,

$\displaystyle N_n$ $\textstyle =$ $\displaystyle 6^n$ (9)
$\displaystyle L_n$ $\textstyle =$ $\displaystyle (1+\phi)^{-n}.$ (10)

The Capacity Dimension is therefore
$\displaystyle d_{\rm cap}$ $\textstyle =$ $\displaystyle -\lim_{n\to\infty} {\ln N_n\over\ln L_n} = {\ln 6\over\ln(1+\phi)}={\ln 2+\ln 3\over\ln(1+\phi)}$  
  $\textstyle =$ $\displaystyle 1.861715\ldots.$ (11)

See also Pentagon


Dixon, R. Mathographics. New York: Dover, pp. 186-188, 1991.

mathematica.gif Weisstein, E. W. ``Fractals.'' Mathematica notebook Fractal.m.

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© 1996-9 Eric W. Weisstein