Geometry with Trigonometry

Sec. 6.7 Coordinate treatment of harmonic ranges 97

In (c)−^12 =t−^12 ort−^12 =^12 , so eithert=0ort=1. Similarly eithers=0or
s=1. We rule out the case oft=1asthenλwould be undefined, and we rule out
the case ofs=1asthen−λwould be undefined. What remains ist=s=0andwe
excluded this by takingλ=0; it would imply thatZ 3 =Z 4 =Z 1.
Thus just one ofZ 3 ,Z 4 is in the segment[Z 1 ,Z 2 ]and the other is on the lineZ 1 Z 2
but outside this segment. HenceZ 3 andZ 4 divide{Z 1 ,Z 2 }internally and externally
in the same ratio. We recall that we then call(Z 1 ,Z 2 ,Z 3 ,Z 4 )a harmonic range.
We note above that there can be no solution forsift=^12 ; thus there is no cor-
respondingZ 4 whenZ 3 is the mid-pointZ 0 ofZ 1 andZ 2. Similarly there can be no
solution fortifs=^12 ; thus there is no correspondingZ 3 whenZ 4 isZ 0.

6.7.2 Interchangeofpairsofpoints

If the points Z 3 and Z 4 divide{Z 1 ,Z 2 }internally and externally in the same ratio, then
it turns out that the points Z 1 and Z 2 also divide{Z 3 ,Z 4 }internally and externally in
the same ratio.
Proof. For we had

x 3 =

1

1 +λ

x 1 +

λ

1 +λ

x 2 ,y 3 =

1

1 +λ

y 1 +

λ

1 +λ

y 2 ,

x 4 =

1

1 −λ

x 1 −

λ

1 −λ

x 2 ,y 4 =

1

1 −λ

y 1 −

λ

1 −λ

y 2.

Then

( 1 +λ)x 3 =x 1 +λx 2 ,

( 1 −λ)x 4 =x 1 −λx 2.

By addition and subtraction, we find that

x 1 =

1 +λ

2

x 3 +

1 −λ

2

x 4 ,

x 2 =

1 +λ

2 λ

x 3 +

λ− 1

2 λ

x 4 ,

and by a similar argument,

y 1 =

1 +λ

2

y 3 +

1 −λ

2

y 4 ,

y 2 =

1 +λ

2 λ

y 3 +

λ− 1

2 λ

y 4.

If we defineμby
1
1 +μ

=

1 +λ

2

,

so that

μ=^1 −λ

1 +λ

, μ

1 +μ

=^1 −λ

2

,