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Five Disks Problem

\begin{figure}\begin{center}\BoxedEPSF{FiveDisks.epsf}\end{center}\end{figure}

Given five equal Disks placed symmetrically about a given center, what is the smallest Radius $r$ for which the Radius of the circular Area covered by the five disks is 1? The answer is $r=\phi-1=1/\phi=0.6180339\ldots$, where $\phi$ is the Golden Ratio, and the centers $c_i$ of the disks $i=1$, ..., 5 are located at

\begin{displaymath}
c_i=\left[{\matrix{{1\over\phi}\cos\left({2\pi i\over 5}\rig...
...\pi i\over 5}\right)\cr}}\right].\hrule width 0pt height 4.2pt
\end{displaymath}

The Golden Ratio enters here through its connection with the regular Pentagon. If the requirement that the disks be symmetrically placed is dropped (the general Disk Covering Problem), then the Radius for $n=5$ disks can be reduced slightly to 0.609383... (Neville 1915).

See also Arc, Disk Covering Problem, Flower of Life, Seed of Life


References

Ball, W. W. R. and Coxeter, H. S. M. ``The Five-Disc Problem.'' In Mathematical Recreations and Essays, 13th ed. New York: Dover, pp. 97-99, 1987.

Neville, E. H. ``On the Solution of Numerical Functional Equations, Illustrated by an Account of a Popular Puzzle and of its Solution.'' Proc. London Math. Soc. 14, 308-326, 1915.




© 1996-9 Eric W. Weisstein
1999-05-26