info prev up next book cdrom email home


\begin{figure}\begin{center}\BoxedEPSF{Pentagon.epsf scaled 1000}\end{center}\end{figure}

The regular convex 5-gon is called the pentagon. By Similar Triangles in the figure on the left,

{d\over 1}={1\over{1\over\phi}}=\phi,
\end{displaymath} (1)

where $d$ is the diagonal distance. But the dashed vertical line connecting two nonadjacent Vertices is the same length as the diagonal one, so
\end{displaymath} (2)

\end{displaymath} (3)

Solving the Quadratic Equation gives
{1\pm\sqrt{1+4}\over 2}
\end{displaymath} (4)

and taking the plus sign gives the Golden Ratio
\phi={\textstyle{1\over 2}}(1+\sqrt{5}\,).
\end{displaymath} (5)

\begin{figure}\begin{center}\BoxedEPSF{PentagonVertices.epsf scaled 600}\end{center}\end{figure}

The coordinates of the Vertices relative to the center of the pentagon with unit sides are given as shown in the above figure, with

$\displaystyle c_1$ $\textstyle =$ $\displaystyle \cos\left({2\pi\over 5}\right)= {\textstyle{1\over 4}}(\sqrt{5}-1)$ (6)
$\displaystyle c_2$ $\textstyle =$ $\displaystyle \cos\left({4\pi\over 5}\right)= {\textstyle{1\over 4}}(\sqrt{5}+1)$ (7)
$\displaystyle s_1$ $\textstyle =$ $\displaystyle \sin\left({2\pi\over 5}\right)= {\textstyle{1\over 4}}\sqrt{10+2\sqrt{5}}$ (8)
$\displaystyle s_2$ $\textstyle =$ $\displaystyle \sin\left({4\pi\over 5}\right)= {\textstyle{1\over 4}}\sqrt{10-2\sqrt{5}}.$ (9)

For a regular Polygon, the Circumradius, Inradius, Sagitta, and Area are given by
$\displaystyle R_n$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}a\csc\left({\pi\over n}\right)$ (10)
$\displaystyle r_n$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}a\cot\left({\pi\over n}\right)$ (11)
$\displaystyle x_n$ $\textstyle =$ $\displaystyle R_n-r_n={\textstyle{1\over 2}}a\tan\left({\pi\over 2n}\right)$ (12)
$\displaystyle A_n$ $\textstyle =$ $\displaystyle {\textstyle{1\over 4}}na^2\cot\left({\pi\over n}\right).$ (13)

Plugging in $n=5$ gives
$\displaystyle R$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}a\csc({\textstyle{1\over 5}}\pi)={\textstyle{1\over 10}} a\sqrt{50+10\sqrt{5}}$ (14)
$\displaystyle r$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}a\cot({\textstyle{1\over 5}}\pi)={\textstyle{1\over 10}} a\sqrt{25+10\sqrt{5}}$ (15)
$\displaystyle x$ $\textstyle =$ $\displaystyle {\textstyle{1\over 10}} a\sqrt{25-10\sqrt{5}}\,$ (16)
$\displaystyle A$ $\textstyle =$ $\displaystyle {\textstyle{5\over 4}} a^2\sqrt{25+10\sqrt{5}}.$ (17)

Five pentagons can be arranged around an identical pentagon to form the first iteration of the ``Pentaflake,'' which itself has the shape of a pentagon with five triangular wedges removed. For a pentagon of side length 1, the first ring of pentagons has centers at radius $\phi$, the second ring at $\phi^3$, and the $n$th at $\phi^{2n-1}$.

\begin{figure}\begin{center}\BoxedEPSF{Pentaflake1.epsf scaled 500}\end{center}\end{figure}

In proposition IV.11, Euclid showed how to inscribe a regular pentagon in a Circle. Ptolemy also gave a Ruler and Compass construction for the pentagon in his epoch-making work The Almagest. While Ptolemy's construction has a Simplicity of 16, a Geometric Construction using Carlyle Circles can be made with Geometrography symbol $2S_1+S_2+8C_1+0C_2+4C_3$, which has Simplicity 15 (De Temple 1991).

\begin{figure}\begin{center}\BoxedEPSF{PentagonConstruction.epsf scaled 1000}\end{center}\end{figure}

The following elegant construction for the pentagon is due to Richmond (1893). Given a point, a Circle may be constructed of any desired Radius, and a Diameter drawn through the center. Call the center $O$, and the right end of the Diameter $P_0$. The Diameter Perpendicular to the original Diameter may be constructed by finding the Perpendicular Bisector. Call the upper endpoint of this Perpendicular Diameter $B$. For the pentagon, find the Midpoint of $OB$ and call it $D$. Draw $DP_0$, and Bisect $\angle
ODP_0$, calling the intersection point with $OP_0$ $N_1$. Draw $N_1P_1$ Parallel to $OB$, and the first two points of the pentagon are $P_0$ and $P_1$ (Coxeter 1969).

Madachy (1979) illustrates how to construct a pentagon by folding and knotting a strip of paper.

See also Cyclic Pentagon, Decagon, Dissection, Five Disks Problem, Home Plate, Pentaflake, Pentagram, Polygon, Trigonometry Values Pi/5


Ball, W. W. R. and Coxeter, H. S. M. Mathematical Recreations and Essays, 13th ed. New York: Dover, pp. 95-96, 1987.

Coxeter, H. S. M. Introduction to Geometry, 2nd ed. New York: Wiley, pp. 26-28, 1969.

De Temple, D. W. ``Carlyle Circles and the Lemoine Simplicity of Polygonal Constructions.'' Amer. Math. Monthly 98, 97-108, 1991.

Dixon, R. Mathographics. New York: Dover, p. 17, 1991.

Dudeney, H. E. Amusements in Mathematics. New York: Dover, p. 38, 1970.

Madachy, J. S. Madachy's Mathematical Recreations. New York: Dover, p. 59, 1979.

Pappas, T. ``The Pentagon, the Pentagram & the Golden Triangle.'' The Joy of Mathematics. San Carlos, CA: Wide World Publ./Tetra, pp. 188-189, 1989.

Richmond, H. W. ``A Construction for a Regular Polygon of Seventeen Sides.'' Quart. J. Pure Appl. Math. 26, 206-207, 1893.

Wantzel, M. L. ``Recherches sur les moyens de reconnaître si un Problème de Géométrie peut se résoudre avec la règle et le compas.'' J. Math. pures appliq. 1, 366-372, 1836.

info prev up next book cdrom email home

© 1996-9 Eric W. Weisstein