Dividing a Segment into N parts:
Besteman Construction II
A Mathematical Droodle: What Is This About?

 

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

Explanation

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

Dividing a Segment into N parts:
Besteman Construction II

The applet illustrates an algorithm for dividing a segment in N equal parts found by a college student, Nathan Besteman, and reported in the recent issue of Mathematics Teacher.

 

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

The construction employs a well known property of angle bisectors in a triangle. In ΔACD, let DP be the angle bisector of angle D. Then

(1)AP/PC = AD/CD

It follows from (1) that if CD = N·AD, then AP = PC/N, or PC = N·AP, so that by adding AP we get AP = AC/(N+1). This suggests an iterative algorithm: for N = 2, there is a known construction of the midpoint B of AC by joining the points of intersection of two circles of equal radii centered at A and C. To trisect AC, reduce the radius of the circle at A to AB, which gives a new triangle ACD. The angle bisector of D will now divide AC in the ratio 1:2, such that AP = AC/3. Reduce the radius of the circle at A to AP and consider a new triangle ACD. The angles bisector will now divided AC in the ratio 1:3, and so on.

References

  1. N. Besteman and J. Ferdinands, Another Way to Divide a Line Segment into n Equal Parts, Mathematics Teacher, Vol. 98, No. 6 (Feb. 2005), pp. 428-433

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

71471367