CONSTRUCTION OF A REGULAR PENTAGON

Bruce David Wilner
December 2002

The German mathematician Carl Friedrich Gauss (1777-1855) derived constructions for regular polygons having k = ( 2^2^n )+ 1 sides, e.g., k = 5, 17, 257, 65537. It is instructive to examine the construction for a regular pentagon (k = 5). The following presentation explicates the construction for such a pentagon and provides details on supportive constructions, i.e., for a perpendicular to a line through a given point and for the bisectors of a line segment and an angle.

The following diagram illustrates the procedure, which is explained step-by-step in the subsequent text.

1. Construct circle O and diameter AB.

2. Construct perpendicular OC to AB, where C is a point on the circumference of O.

3. Bisect OC at D.

4. Bisect angle ODB; bisector intersects OB at point E.

5. Construct perpendicular EF to OB, where F is a point on the circumference of O.

6. Measure arc BF and mark off three equal arcs FG, GH, and HJ on circumference of O. (Any difference in length between these arcs and arc JB is an artifact of inaccuracies in the construction, widths of lines, etc.)

7. Note that side GH of pentagon FGHJB is perpendicular to diameter AB.

The proof of the validity of this construction is lengthy and is therefore omitted.

The supporting construction for a perpendicular to a given line through a given point is as follows:

1. Locate point C on line segment AB.

2. Fix compass at C and mark off equal distances CD and CE, where D lies on CA and E lies on CB.

3. Fix compass at D and sweep arc above C.

4. Fix compass at E and sweep equal arc above C; arcs intersect at point F.

5. CF is perpendicular to AB.

A proof of the validity of this construction is as follows:

The supporting construction for the bisector of a given line segment is as follows:

1. Fix compass at A and sweep off arc (of length greater than one-half length of AB), intersecting AB at C.

2. Fix compass at B and sweep off equal arc, intersecting AB at D; arcs intersect at points E and F.

3. EF bisects AB at G.

A proof of the validity of this construction is as follows:

Finally, the supporting construction for the bisector of a given angle is as follows:

1. Fix compass at B and sweep arc from BA to BC; arc intersects BA at D and BC at E.

2. Fix compass at D and sweep arc within space between A and C.

3. Fix compass at E and sweep equal arc within space between A and C; arcs intersect at point F.

4. BF bisects angle ABC.

A proof of the validity of this construction is as follows:

Bibliography

Ball, W. W. R., Mathematical Recreations and Essays (New York: Macmillan, 1962), pp. 95-96.