Constructing a Nine-Sided Regular Polygon Using Unmarked Straightedge and

Compass and Its Proof with the Archimedes Method

Rodolfo A. Nieves Rivas

fesol7luzley@gmail.com

Abstract

This paper presents a method for constructing a regular polygon of nine

sides or regular enneagon. Then we prove the accuracy of such construction by using the

method of Archimedes. Hence, concluding with the trisection of an angle of 60°.

Keywords: Regular enneagon; method of Archimedes; trisection of an angle

Introduction

This article presents a method for constructing a nine sided regular polygon or regular

enneagon. We use for such construction the software Geogebra which keeps the conditions

of Euclidean and elementary geometry which demands the exclusive use of unmarked

straightedge and compass. Then to prove the accuracy of this construction we use the

method of Archimedes. Thus ending the trisection of an angle of 60°

Main Theorem: The interception point of the common side of two central supplementary

angles and the line segment that joints the midpoint of the chord of one of them to the

midpoint arc of the other, determines an inscribed trisector angle whose vertex is the end

point of the diameter of the circumscribed circumference, whose radii is equal to the sides

of the trisected central angle.

Construction of Figure 1

Step One: With the compass centered at point A and opening AC construct a circumference

of radius: AC and diameter EC.

Step Two: With the straightedge and compass construct a perpendicular line to point: A

determining in the circumference point B.

Step Three: With the compass centered at points: C and E construct two circumferences of

radius: AC and AE respectively. Determining points: M and D in the circumference of

diameter: CE.

Step Four: With the straightedge draw a parallel straight line: DF to the straight line: BA.

Thus, determining point: F in the line segment: AC.

Step Five: With the straightedge draw segments: ED and DC, constructing an inscribed

rectangular triangle. Then, draw the line segment: DA, determining two supplementary

angles: 120 ° and 60 ° respectively. Where: EAD = 120 ° and DAC = 60 ° with

the Common: DA

Step Six: With the straightedge and compass bisect angles: EAD=120 ° and the angle

DAC= 60 ° Determining in this way points: M and Q respectively on the circumference of

diameter: EC and besides you get point: P on the chord: ED.

Step Seven: With the straightedge draw line segments: BF and PQ. Thus determining the

point: G in common side: DA supplementary angles: EAD = 120 ° and DAC = 60 ° Step Eight: With the straightedge draw a segment of line from: the point: E passing through point G

and extends until point: H on the circumference of diameter: EC. Then join points: H to A.

Determining the angles: DAH= 20 ° and HAC= 40 ° where: DAH + HAC = DAC =

60 °.

Proof: Using the Method of Archimedes. (See figures 7, 8, 9 and 10)

Figure 1

Construction of a Regular Polygon of Nine Sides (See figures 2 and 3)

Figure 2

Figure 3

Method to Trisect Supplementary Angles: Based on the following theorem: If two supplementary angles are trisected. Then the two

trisectrices adjacent to the common side of these two supplementary angles determine an

angle of 60 °.

Example: 60 ° and 120 ° (See figure 4)

Figure 4

Remark: The CAD triangle is equilateral.

Example: 30 ° and 150 ° (See figure 5)

Figure 5

Remark: The CAD triangle is equilateral.

Example: 72 º and 108 º (See Figure 6)

Figure 6

Remark: The CAD triangle is equilateral.

Method Applied to the Construction of an Angle of 20 ° (See figure 7)

Construction of Figure 7

Step One: With the straightedge and compass construct two perpendicular straight lines

with point of interception: A. Then with the compass centered in point: A and opening of

radius: AC construct a circumference determining points: B, D, E and C with the

interception of the circumference and the perpendicular straight lines.

Step Two: With the compass centered in point: C and opening: CA as radius, construct

another circumference determining with both, points: F and G. Then, join points: F and G

with a segment of line perpendicular to the diameter: DC, determining point: H in the

radius: AC.

Step Three: With the straightedge join points: D to point: F and then point: C getting an

inscribed angle in the circumference of radius: AC. Then join points: F to point: A getting

two supplementary angles: DAF= 120º and FAC=60º with common side: FA

Step Four: With the straightedge and compass bisect two angles: DAF and FAC

determining point: K in the chord: DF and point: L in the arc: FC.

Step Five: With the straightedge join points: K and L determining point: I in the line

segment: FA common of the two supplementary angles: DAF and FAC and then join

points: B to H passing the line segment: BH by point: I

Step Six: With the straightedge draw a segment line: DI from point: D until point: H

extending to get point: J in the arc: FC, determining an inscribed angle: JDC = 20º

Figure 7

Proof using the Method of Archimedes (See figures 8, 9 and 10)

Figure 8

Figure 9

Conclusion and Discussion:

With the results in the construction and proof of an angle of twenty degrees is determined a

real solution or zero of a cubic equation of the form:

If: Cos3x = 4 (cosx)3-3Cosx

Where:X3-3x = 1

Then:X1 = 2Cos20 = 1.879385242 ...

And when:X3 - 3x = -1

Then:X2 = 2Cos40 = 1.53208886 ...

And for:2x3-6x = 0

Then:X3 = Square root of three

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