000RM.dvi

126 Basic geometric constructions

4.3 Equal subdivisions of a segment ...........

Here is an algorithm to divide a given segment inton< 2 k+1equal parts,
making use of the binary representation ofn, which contains not more
thank+1digits.
Construct a squareABCDon the given segment. By repeated bisec-
tions ofCD, introduce the pointsC 1 ,C 2 , ...,Cksuch that

DCi=

1

2 i

DC, i=1, 2 ,...k.

We also putC 0 =C.
Letn< 2 k+1. Its binary representation has no more thank+1digits.
Suppose it hasm+1nonzero binary digits.
Along the directionCD, relabel those points corresponding to the
nonzerodigits, asQ 0 ,Q 1 , ...,Qm.^1
LetP 0 be the (orthogonal) projection ofQ 0 onAB.
For eachj=1,...,m, construct the segmentsPj− 1 DandAQj, and
mark the projection of their intersection onABas the pointPj.
ThenPmis the point which dividesABintonequal parts.

Here is the case forn=13with binary representation 11012.

A B=P 0

D Q^2 Q^1 C=Q 0

P 2 P 1
It would be interesting if this procedure can be modified to give a
simple construction of subdivision points with ratiom : ninstead of
1:n.

(^1) For example, ifn=13, which has binary representation 11012 , we relabelC 0 asQ 0 ,C 2 asQ 1 , and
C 2 asQ 2.