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2

cos ± =
2 r r'

The cosine of a right angle is zero; therefore two circles are orthogonal if and only if:

r 2 + r' 2 = (O ’ O' )
2




which is the Pythagorean theorem.


Radical axis of two circles

The radical axis of two circles is the locus of the points which have the same
power with respect to both circles:

OP 2 ’ r 2 = O'P 2 ’ r' 2

Let X be the intersection of the radical axis with the line joining both centres of the circles.
Then:
RAMON GONZALEZ CALVET
90



OX 2 ’ r 2 = O'X 2 ’ r' 2 ’ OX 2 ’ O'X 2 = r 2 ’ r' 2

(OX + O'X ) · (OX ’ O'X ) = r 2 ’ r' 2

O + O' 
«
2¬X ’ · · OO' = r ’ r'
2 2

2


Since OX and O'X are proportional to OO', the inner and geometric products are
equivalent and we may isolate X:

O + O' + ( r 2 ’ r' 2 ) OO' ’1
X=
2

Now let us search which kind of geometric figure is the radical axis. Arranging as before
the terms, we have:

O + O' 
«
2 ¬P ’ · · OO' = r ’ r'
2 2

2


Introducing the point X we arrive at:

2 (P ’ X ) · OO' = 0

That is, the radical axis is a line passing through X and perpendicular to the line joining
both centres. If the circles intersect, then the radical axis is the straight line passing
through the intersection points.
The radical centre of three circles is the intersection of their radical axis. Let O, O'
and O'' be the centre of three circles with radii r, r' and r''. Let us denote with X (as above)
the intersections of the radical axis of the first and second circles with the line OO', and
with Y the intersection of the radical axis of the second and third circles with the line
joining its centres O'O''. The radical centre P must fulfil the former equation for both
radical axis:

± O' 2 ’ O 2 + r 2 ’ r' 2
 P · OO' = X · OO' =
± (P ’ X ) · OO' = 0  2

 ’
(P ’ Y ) · O'O'' = 0  P · O'O'' = Y · O'O'' = O'' ’ O' + r' ’ r''
2 2 2 2


 2

Now we have encountered the coefficients of linear combination of an imaginary
decomposition. This process is explained in page 74 for the calculation of the orthocentre.
However there is a difference: we wish not to resolve the vector into orthogonal
components but from the known coefficients we wish to reconstruct the vector P. Then:

P · O'O'' P · OO'
P= OO' ’ O'O''
OO' § O'O'' OO' § O'O''
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 91

The substitution of the known values of the coefficients results in the following formula
for the radical centre which is invariant under cyclic permutation of the circles:

((O'' ’ r'' 2 ) OO' + (O 2 ’ r 2 ) O'O'' + (O' 2 ’ r' 2 ) O''O )
P = (2 OO' § O'O'' )
’1 2




Exercises

9.1 Let A, B, C be the vertices of any triangle. The point M moves according to the
equation:

MA2 + MB2 + MC2 = k

Which geometric locus does the point M describe as a function of the values of the real
parameter k?

9.2 Prove that the geometric locus of the points P such that the ratio of distances from P to
two distinct points A and B is a constant k is a circle. Calculate the centre and the radius of
this circle.

9.3 Prove the Simson™s theorem: the feet of the perpendiculars from a point D upon the
sides of a triangle ABC are aligned if and only if D lies on the circumscribed circle.

9.4 The Bretschneider's theorem: let a, b, c and d be the successive sides of a
quadrilateral, m and n its diagonals and ± and γ two opposite angles. Show that the
following law of cosines for a quadrilateral is fulfilled:

m 2 n 2 = a 2 c 2 + b 2 d 2 ’ 2 a b c d cos(± + γ )

9.5 Draw three circles passing through each vertex of a triangle and the midpoints of the
concurrent sides. Then join the centre of each circle and the midpoint of the opposite side.
Show that the three segments obtained in this way intersect in a unique point.

9.6 Prove that the inversion is an opposite conformal transformation, that is, it preserves
the angles between curves, but changes their sign.
RAMON GONZALEZ CALVET
92

10. CROSS RATIOS AND RELATED TRANSFORMATIONS

Complex cross ratio

The complex cross ratio of any four points A, B, C and D on the plane is defined
as the quotient of the following two single ratios:

( A B C D ) = ( A C D ) ( B C D ) ’1 = AC AD ’1 BD BC ’1

If the four points are aligned, then the cross ratio is a real number, otherwise it is a
complex number. Denoting the angles CAD and CBD as ± and β respectively, the cross
ratio is written as:

AC AD BD BC
( ABC D)= Figure 10.1
2 2
AD BC

AC BD
exp[(± ’ β ) e12 ]
=
AD BC

If the four points are located in this order
on a circle (figure 10.1), the cross ratio is a
positive real number, because the inscribed angles
± and β intercept the same arc CD and hence are
equal, so:

AC BD
( ABC D)=
AD BC

But if one of the points C or D is located between A and B (figure 10.2), then ± and β
have distinct arcs CD with opposite orientation. Since ± and β are the half of these arcs, it
follows that ± ’β = π and the cross ratio is a
negative real number: Figure 10.2

AC BD
( ABC D)= ’
AD BC

Analogously, if the four points are aligned
and C or D is located between A and B, the cross
ratio is negative1.
Other definitions of the cross ratio differ
from this one given here only in the order in
which the vectors are taken. This question is
equivalent to study how the value of the cross ratio changes under permutations of the
points. For four points, there are 24 possible permutations, but only six different values for
the cross ratio. The calculations are somewhat laborious and I shall only show one case.
1
A straight line is a circle with infinite radius.
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 93

The reader may prove any other case as an exercise. I also indicate whether the
permutations are even or odd, that is, whether they are obtained by an even or an odd
number of exchanges of any two points from the initial definition ( A B C D ):

(ABCD)=(BADC)=(CDAB)=(DCBA)=r
(1) even
1
(BACD)=(ABDC)=(DCAB)=(CDBA)=
(2) odd
r
( A C B D ) = ( B D A C ) = ( C A D B ) = ( D B C A ) = 1 ’ r odd
(3)
1
(CABD)=(DBAC)=(ACDB)=(BDCA)=
(4) even
1’ r
r ’1
(BCAD)=(ADBC)=(DACB)=(CBDA)=
(5) even
r
r
(CBAD)=(DABC)=(ADCB)=(BCDA)=
(6) odd
r ’1

Let us prove, for example the equalities (3):

( A C B D ) = AB AD ’1 CD CB ’1 = ( AC + CB ) AD ’1 ( CB + BD ) CB ’1 =
(3)

= AC AD ’1 CB CB ’1 + AC AD ’1 BD CB ’1 + CB AD ’1 CB CB ’1 + CB AD ’1 BD CB ’1

In the last term, under the reflection in the direction CB, the complex number AD ’1 BD
becomes conjugate, that is, these vectors are exchanged:

= AC AD ’1 + AC AD ’1 BD CB ’1 + CB AD ’1 + BD AD ’1

= ( AC + CB + BD ) AD ’1 ’ AC AD ’1 BD BC ’1 = AD AD ’1 ’ r = 1 ’ r

Let us see that this value is the same for all the cross ratios of the case (3):

( B D A C ) = BA BC ’1 DC DA ’1 = AB CB ’1 CD AD ’1 = AB AD ’1 CD CB ’1

=(ACBD)

where in the third step the permutative property has been applied. Also by using this
property one obtains:

( C A D B ) = CD CB ’1 AB AD ’1 = CD AD ’1 AB CB ’1 = AB AD ’1 CD CB ’1

=(ACBD)

( D B C A ) = DC DA ’1 BA BC ’1 = BA DA ’1 DC BC ’1 = AB AD ’1 CD CB ’1

=(ACBD)
RAMON GONZALEZ CALVET
94

Harmonic characteristic and ranges

From the cross ratio, one can find a quantity independent of the symbols of the
points and only dependent on their locations. I call this quantity the harmonic
characteristic -because of the obvious reasons that follow now- denoting it as [A B C D].
The simplest way to calculate it (but not the unique) is through the alternated addition of
all the permutations of the cross ratio, condensed in the alternated sum of the six different
values:

(ABCD)’(BACD)’(ACBD)+(CABD)+(BCAD)’(CBAD)=

r ’1
1 1 r
’ (1 ’ r ) +
=r’ + ’
1’ r r ’1
r r

(2 r ’ 1 ) ( r + 1 ) ( r ’ 2 )
=
r (1 ’ r)

When two points are permuted, this sum still changes the sign. The square of this
sum is invariant under every permutation of the points, and is the suitable definition of the
harmonic characteristic:

[ A B C D ] = [( A B C D ) ’ ( B A C D ) ’ ( A C B D ) + (C A B D ) + ( B C A D ) ’ ( A B C D )]
2




(2 r ’ 1 )2 ( r + 1 )2 ( r ’ 2 )2
[ A B C D] =
r 2 (1 ’ r)
2




The first notable property of the harmonic characteristic is the fact that it vanishes for a
harmonic range of points, that is, when the cross ratio (A B C D) takes the values 1/2, ’1
or “2 dependent on the denominations of the points. The second notable property is the
fact that the harmonic characteristic of a degenerate range of points (two or more
coincident points) is infinite.
Perhaps, the reader believes that other harmonic characteristics may be obtained in
many other ways, that is, by using other combinations of the permutations of the cross
ratio. However, other attempts lead whether to the same function of r (or a linear
dependence) or to constant values. For example, we can take the sum of all the squares,
which is also invariant under any permutation of the points to find:

(A B C D) 2 + (B A C D)2 + (A C B D)2 + (C A B D)2 + (B C A D)2 + (C B A D)2

(r ’ 1)2 + r 2 = [ A B C D ] + 21
1 1
= r + 2 + (1 ’ r ) +
2
+
2

(1 ’ r )2 (r ’ 1)2
r2 2
r

On the other hand, M. Berger (vol. I, p. 127) has used a function that is also
linearly dependent on the harmonic characteristic:
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 95

4(r2 ’ r +1 )
3
[ A B C D]
=1+
27 r 2 ( 1 ’ r )
2
27

Other useless possibilities are the sum of all the values of the permutations of the cross
ratio, which is identical to zero, or their product giving always a unity result.
Four points A, B, C and D so ordered on a line are said to be a harmonic range if
the segments AB, AC and AD are in harmonic progression, that is, the inverse vectors
AB’1, AC ’1 and AD ’1 are in arithmetic progression:

AB ’1 ’ AC ’1 = AC ’1 ’ AD ’1 AC AD ’1 BD BC ’1 = 2


The cross ratio of a harmonic range is 2, but if this order is altered then it can be “1 or 1/2.
Note that this definition also includes harmonic ranges of points on a circle. On the other
hand, all the points of a harmonic range always lie on a line or a circle.
In order to construct the fourth point D forming a harmonic range with any three
given points A, B and C (ordered in this way on a line) one must follow these steps (figure
10.3):
1) Draw the line passing through A, B and C. Trace two any distinct lines R and S also
different from ABC and passing through A.
2) Draw any line passing through B. This
line will cut the line R in the point P and the Figure 10.3
line S in the point X.
3) Draw the line passing through C and X.
The intersection of this line with R will be
denoted by Q. Draw also the line passing
through C and P. This line intersects the
line S in the point Y.
4) Now trace the line passing through Y and
Q, which will cut the line ABC in the
searched point D, which completes the
harmonic range.
For a clearer displaying I have
drawn some segments instead of lines, although the construction is completely general
only for lines.
If A, B and C are not aligned, the point D which completes the harmonic range lies
on a circle passing through the three given points (figure 10.4). In order to determine D
one must make an inversion (which preserves the cross ratio if this is real, as shown
below) with centre on the circle ABC, which transforms this circle into a line. In this line
we determine the point D™ forming a harmonic range with A™, B™, C™ and then, by means
of the same inversion, the point D is obtained. In the drawing method the radius and the
exact position of the centre of inversion do not play any role whenever it lie on the circle
ABC. Then we shall draw any line outside the circle, project A, B and C into this line,
make the former construction and obtain D by projecting D™ upon the circle.
The cross ratio becomes conjugated under circular inversion. In order to prove this
statement, let us see firstly how the single ratio is transformed. If A', B', C' and D' are the
transformed points of A, B, C and D under an inversion with centre O and any radius, we
RAMON GONZALEZ CALVET
96

have:

A'C' A'D' ’1 = (OC' ’ OA' ) (OD' ’ OA' ) ’1 = (OC ’1 ’ OA ’1) (OD ’1 ’ OA ’1) ’1

= OC ’1 ( OA ’ OC ) OA ’1 [ OD ’1 ( OA ’ OD ) OA ’1 ] ’1

= OC ’1 CA OA ’1 [ OD ’1 DA OA ’1 ] ’1 = OC ’1 CA OA ’1 OA DA ’1 OD

= OC ’1 AC AD ’1 OD

Figure 10.4
Analogously we have:

B'D' B'C' ’1 = OD ’1 BD BC ’1 OC

The product of both expressions is the cross
ratio:

( A' B' C' D' ) = A'C' A'D' ’1 B'D' B'C' ’1

= OC ’1 AC AD ’1 BD BC ’1 OC

= OC ’1 ( A B C D ) OC = ( A B C D )*

So, the cross ratio of four points on a line or a circle, which is real, is preserved, otherwise
it becomes conjugated.
Finally let us see that the single ratio is the limit of the cross ratio for one point at
infinity:
AC AD ’1 BD BC ’1 = BD BC ’1 = ( B D C )
(∞ B C D) = lim
A’ ∞

Analogously ( A ∞ C D ) = ( A C D ), etc.


The homography (Möbius transformation)

The homography is defined as the geometric transformation that preserves the
complex cross ratio of four points on the plane. In order to specify a homography one
must gives only the images A', B' and C' of three points A, B, and C because the fourth
image D' is determined by the conservation of their cross ratio:

( A' B' C' D' ) = ( A B C D )

A'C' A'D' ’1 B'D' B'C' ’1 = AC AD ’1 BD BC ’1

Arranging vectors we write:

A'D' ’1 B'D' = A'C' ’1 AC AD ’1 BD BC ’1 B'C' = A'C' ’1 AC BC ’1 B'C' AD ’1 BD

After conjugation we find:
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 97



B'D' A'D' ’1 = BD AD ’1 B'C' BC ’1 AC A'C' ’1

B'D' A'D' ’1 = BD AD ’1 z z = BC ’1 B'C' A'C' ’1 AC
where

D'B' D'A' ’1 = DB DA ’1 z ’ ( D' B' A' ) = ( D B A ) z

Hence the homography multiplies the single ratio by a complex number z. The modulus of
the single ratio increases proportionally to the modulus of z and the angle BDA is
increased in the argument of z:

z = ¦z¦ exp[± e12 ]

D'B' DB
=z
D'A' DA

angle B'D'A' = angle BDA + ±

Then a homography can also be determined by specifying the images A' and B' of two
points A and B, and a complex number z. The image D' of D is calculated by isolation in
the former equation:

( D'A' + A'B' ) D'A' ’1 = DB DA ’1 z

A'B' D'A' ’1 = DB DA ’1 z ’ 1

D'A' = ( z DB DA ’1 ’ 1 ) ’1 A'B'

D' = A' + ( 1 ’ z DB DA ’1 ) ’1 A'B'

The homography preserves the complex cross ratio of any four points.
Analogously to the former equality, we find for any three points P, Q, R :

P' = A' + ( 1 ’ z PB PA ’1 ) ’1 A'B'

Q' = A' + ( 1 ’ z QB QA ’1 ) ’1 A'B'

R' = A' + ( 1 ’ z RB RA ’1 ) ’1 A'B'

D'Q' = [ ( 1 ’ z QB QA ’1 ) ’1 ’ ( 1 ’ z DB DA ’1 ) ’1 ] A'B' =

= z ( ’DB DA ’1 + QB QA ’1 ) ( 1 ’ z DB DA ’1 ) ’1 ( 1 ’ z QB QA ’1 ) ’1 A'B'

D'R' = z ( ’DB DA ’1 + RB RA ’1 ) ( 1 ’ z DB DA ’1 ) ’1 ( 1 ’ z RB RA ’1 ) ’1 A'B'

P'Q' = z ( ’PB PA ’1 + QB QA ’1 ) ( 1 ’ z PB PA ’1 ) ’1 ( 1 ’ z QB QA ’1 ) ’1 A'B'
RAMON GONZALEZ CALVET
98



P'R' = z ( ’PB PA ’1 + RB RA ’1 ) ( 1 ’ z PB PA ’1 ) ’1 ( 1 ’ z RB RA ’1 ) ’1 A'B'

In the cross ratio, many factors are simplified:

D'Q' D'R' ’1 P'R' P'Q' ’1 = ( ’DB DA ’1 + QB QA ’1 ) ( ’DB DA ’1 + RB RA ’1 ) ’1

( ’PB PA ’1 + RB RA ’1 ) ( ’PB PA ’1 + QB QA ’1 ) ’1

Writing the first factor as product of factors we obtain:
’1 ’1 ’1 ’1
’DB DA + QB QA = ’ ( DA + AB ) DA + ( QA + AB ) QA =

= ’AB DA ’1 + AB QA ’1 = AB DA ’2 ( ’DA QA2 + DA2 QA ) QA ’2 =

= AB DA ’2 DA ( ’QA + DA ) QA QA ’2

= AB DA ’1 QD QA ’1

In the same way each of the other three factors can be written as product of factors:
’1 ’1 ’1 ’1
’DB DA + RB RA = AB DA RD RA
’1 ’1 ’1 ’1
’PB PA + RB RA = AB PA RP RA
’1 ’1 ’1 ’1
’PB PA + QB QA = AB PA QP QA

Then the cross ratio will be:

D'Q' D'R' ’1 P'R' P'Q' ’1 = ( AB DA ’1 QD QA ’1 ) ( RA RA ’1 DA AB ’1 )

( AB PA ’1 RP RA ’1 ) ( QA QP ’1 PA AB ’1 )

where the brackets are not needed but only show the factors they come from. Applying the
permutative property and simplifying, we arrive at:

D'Q' D'R' ’1 P'R' P'Q' ’1 = DQ DR ’1 PR PQ ’1

This proves that the cross ratio of every set of four point is preserved under a
homography. Four points lying on a line or a circle have a real cross ratio. Then the
homography is a circular transformation since it transforms circles into circles in a general
sense (taking the lines as circles with infinite radius).
The product of two circular inversions is always a homography because the cross
ratio becomes conjugate under inversion and thus it is preserved under an even number of
inversions. Let us calculate the homography resulting from a composition of two
inversions. Let A be any point, A' be the transformed point under a first inversion with
centre O and radius r, and A" be the transformed point of A' under a second inversion with
centre P and radius s, which is consequently the transformed point of A under the
homography:
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 99



OA' = r2 OA ’1 PA'' = s2 PA' ’1

From where it follows that:

s2 PA'' ’1 = PA' = OA' ’ OP = r2 OA ’1 ’ OP

PA'' ’1 = s ’2 ( r2 OA ’1 ’ OP )

PA'' = s2 (r2 OA ’1 ’ OP ) ’1 = s2 [( r2 ’OP OA ) OA ’1] ’1 = s2 OA (r2 ’OP OA ) ’1

This equation allows to determine the image A" of any point A having the centre O and P
and the radius r and s as data. For instance, let us calculate the homography corresponding
to:

O = ( 0, 0 ) r=2 P = ( 3, 0 ) s=3

OP = 3 e1

In order to specify the homography we determine the images of three points:

A = ( 0, 2 ) B = ( 1, 1 ) C = ( 1, 0 )

OA = 2 e2 OB = e1 + e2 OC = e1

OP OA = 6 e12 OP OB = 3 + 3 e12 OP OC = 3

’ 27 e1 + 18 e 2 ’ 9 e1 + 18 e 2
PA'' = PB'' = PC'' = 9 e1
13 5

« 12 18  « 6 18 
A'' = ¬ , B'' = ¬ , ·
· C'' = (12, 0)
 13 13  5 5 


Projective cross ratio

Figure 10.5
The projective cross ratio of four
points A, B, C and D with respect to a
point X is defined as the cross ratio of
the pencil of lines XA, XB, XC and XD,
which is the cross ratio of the
intersection points of these lines with
any crossing line (figure 10.5). As it will
be shown, the projective cross ratio is
independent of the crossing line The
intersection points A™, B™, C™ and D™ are
RAMON GONZALEZ CALVET
100

the projections with centre X of the points A, B, C and D upon the crossing line. Due to the
alignment of these projections, the projective cross ratio is always a real number:

{ X, A B C D } = ( A' B' C' D' ) = A'C' A'D' ’1 B'D' B'C' ’1

The projected points A', B', C' and D' are dependent on the crossing line and this
expression is not suitable for calculations. Note that the segment A'C' is the base of the
triangle XA'C', B'D' is the base of the triangle XB'D', etc. Since the altitudes of all these
triangles are the distance from X to the crossing line, their bases are proportional to their
areas, so we can write:

XA' § A'C' XB' § B'D '
{ X , A B C D} =
XA' § A'D' XB' § B'C'

The area of a triangle is the half of the outer product of any two sides. For example
for the triangle XA'C' we have:

XA' § A'C' = XA' § ( A'X + XC' ) = XA' § XC'

Then, the cross ratio becomes:

XA' § XC' XB' § XD'
{ X , A B C D} =
XA' § XD' XB' § XC'

The simplification of the modulus gives:

sin A'XC' sin B'XD' sin AXC sin BXD
{ X , A B C D} = =
sin A'XD' sin B'XC' sin AXD sin BXC

because the angles A'XC' and AXC are identical, B'XD' and BXD are also identical, etc.
Now it is obvious that the projective cross ratio does not depend on the crossing line, but
only on the angles between the lines of the pencil. If we introduce the modulus of XA, XB,
etc., we find the outer products of these vectors:

XA § XC XB § XD
{ X , A B C D} =
XA § XD XB § XC

Now, the arbitrary crossing line of the pencil has disappeared in the formula, which has
become suitable for calculations with Cartesian coordinates2. When the points A, B, C and
D are aligned, the complex and projective cross ratios have the same value independently
of the point X, because the segments AC, BD, AD and BC have the same direction, which
can be taken as the crossing line:
2
The reader will find an application of this formula for the projective cross ratio to the error
calculus in the electronic article "Estimation of the error committed in determining the place from
where a photograph was taken", Ramon González, Josep Homs, Jordi Solsona (1999)
http://www.terra.es/personal/rgonzal1 (translation of the paper published in the Butlletí de la
Societat Catalana de Matemàtiques 12 [1997] 51-71).
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 101



{X, A B C D} = ( A B C D )

If A, B, C and D lie on a circle, the complex cross ratio is equal to the projective
cross ratio taken from any point X on that circle. To prove this statement, make an
inversion with centre X. Since the circle ABCDX passes through the centre of inversion, it
is transformed into a line, where the points A', B', C' and D' (transformed of A, B, C and D
by the inversion) lie with the same cross ratio. Now observe that X, A and A™ are aligned,
also X, B and B™, etc. So the projective cross ratio is equal to the complex cross ratio.
From the arguments given above, it follows that the cross ratio of a pencil of lines
is the quotient of the outer products of their direction (or normal) vectors independently of
their modulus:
v § v XC v XB § v XD n XA § n XC n XB § n XD
{ X , A B C D} = XA =
v XA § v XD v XB § v XC n XA § n XD n XB § n XC

Instead of XA, XB, XC and XD, I shall denote by A, B, C and D the pencil of lines
passing through X. Then the projective cross ratio {X, ABCD} is written as (ABCD). Let
us suppose that these lines have the following dual coordinates expressed in the lines base
{O, P, Q}:

A = [xA , yA] B = [xB , yB] C = [xC , yC] D =[xD , yD]

Then the direction vector of the line A is:

v A = (1 ’ x A ’ y A )v O + x A v P + y A v Q

But the direction vectors of the base fulfil:

vO + v P + vQ = 0

Hence:

v A = (’ 1 + 2 x A + y A ) v P + (’ 1 + x A + 2 y A ) v Q

and analogously:

v B = (’ 1 + 2 x B + y B ) v P + (’ 1 + x B + 2 y B ) v Q

v C = (’ 1 + 2 x C + y C ) v P + (’ 1 + x C + 2 y C ) v Q

v D = (’ 1 + 2 x D + y D ) v P + (’ 1 + x D + 2 y D ) v Q

The outer product of vA and vC is:
RAMON GONZALEZ CALVET
102

v A § v C = ( x C ’ x A ’ ( y C ’ y A ) + 3 ( x A y C ’ x C y A )) v P § v Q

which leads to the fact that the cross ratio only depends on the dual coordinates:

v A § vC v B § v D
( A B C D) = =
v A § v D v B § vC

( x C ’ x A ’ ( y C ’ y A ) + 3 ( x A y C ’ x C y A )) ( x D ’ x B ’ ( y D ’ y B ) + 3 ( x B y D ’ x D y B ))
=
( x D ’ x A ’ ( y D ’ y A ) + 3 ( x A y D ’ x D y A )) ( x C ’ x B ’ ( y C ’ y B ) + 3 (x B y C ’ x C y B ))


(x A ’ 1 / 3, y A ’ 1 / 3) § (x C ’ 1 / 3, y C ’ 1 / 3) ( x B ’ 1 / 3, y B ’ 1 / 3) § ( x D ’ 1 / 3, y D ’ 1 / 3)
=
(x B ’ 1 / 3, y B ’ 1 / 3) § (x C ’ 1 / 3, y C ’ 1 / 3) ( x A ’ 1 / 3, y A ’ 1 / 3) § ( x D ’ 1 / 3, y D ’ 1 / 3)


(x A ’ x C , y A ’ y C ) § (x C ’ 1 / 3, y C ’ 1 / 3) ( x B ’ x D , y B ’ y D ) § ( x D ’ 1 / 3, y D ’ 1 / 3)
=
(x B ’ x C , y B ’ y C ) § (x C ’ 1 / 3, y C ’ 1 / 3) ( x A ’ x D , y A ’ y D ) § ( x D ’ 1 / 3, y D ’ 1 / 3)

And taking into account the fact that the dual points corresponding to the lines of the
pencil are aligned in the dual plane, it follows that:

( A B C D ) = (x A ’ x C , y A ’ y C ) (x B ’ x C , y B ’ y C ) (x B ’ x D , y B ’ y D ) (x A ’ x D , y A ’ y D )’1
’1




= AC BC ’1 BD AD ’1 = AC AD ’1 BD BC ’1

where the four dual vectors are proportional. That is: the projective cross ratio of a pencil
of lines is the cross ratio of the corresponding dual points. Obviously, by means of the
duality principle the dual proposition must also hold: the cross ratio of four aligned points
is equal to the projective cross ratio of the dual pencil of the corresponding lines on the
dual plane.


The points at the infinity and homogeneous coordinates

The points at the infinity play the same role as the points with finite coordinates in
the projective geometry. Under a projectivity they can be mapped mutually without
special distinction. In order to avoid the algebraic inconsistencies of the infinity, the
homogeneous coordinates are defined as proportional to the barycentric coordinates with
a no fixed homogeneous constant. The following expressions are equivalent and represent
the same point:

(x, y) = (1 ’ x ’ y, x, y) = k (1 ’ x ’ y, x, y)

Then for instance:

(2, 3) = (’4, 2, 3 ) = (’20, 10, 15)
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 103



A point has finite coordinates if and only if the sum of the homogeneous
coordinates are different from zero. In this case we can normalise the coordinates
dividing by this sum in order to obtain the barycentric coordinates, and hence the
Cartesian coordinates by omitting the first coordinate. On the other hand, the sum of the
homogeneous coordinates sometimes vanishes and then they cannot be normalised (the
normalisation implies a division by zero yielding infinite values). In this case, we are
regarding a point at the infinity, whose algebraic operations should be performed with
the finite values of the homogeneous coordinates3. For example:

(2, ’1, ’1) = (2 ∞, ’∞, ’∞) = (∞, ∞)

This is the point at the infinity in the direction of the bisector of the first quadrant. On the
other hand, points at the infinity are equivalent to directions and hence to vectors. For
example:

v = 2 O ’ P ’ Q = ’OP’OQ = ’2 e1 ’2 e2

Every vector proportional to v also represents this direction and the point at the infinity
given above.


Perspectivity and projectivity

A perspectivity is a central projection of a range of points on a line upon another
line. Since the cross ratio of a pencil of lines is independent of the crossing line, it follows
from the construction (figure 10.6) that a
perspectivity preserves the projective cross
ratio:

{O, A B C D} = {O, A' B' C' D'}

The composition of two perspectivities with
different centres also maps ranges4 of points
to other ranges preserving their cross ratio.
This is the most general transformation
Figure 10.6
which preserves alignment and the projective
cross ratio. A projectivity (also projective
transformation) is defined as the geometric transformation which maps aligned points to
aligned points preserving their cross ratio. A theorem of projective geometry states that
any projectivity is expressible as the sequence of not more than three perspectivities. If the
projectivity relates ranges on two distinct lines, two perspectivities suffice.

3
In fact, the calculations are equally feasible with ∞ if we understand it strictly as a number.
Then the equality 2∞ = ∞ should be wrong. We could simplify by ∞, e.g.: (2 ∞, ’∞, 1) = (2, ’1,
0) is a finite point and (2 ∞, ’∞, ’∞) = (2, ’1, ’1) a point at the infinity.
4
A range of points is defined as a set of aligned points.
RAMON GONZALEZ CALVET
104

A projectivity is determined by giving three aligned points and their corresponding
images, which must be also aligned. On the other hand, a projectivity is determined by
giving four points no three of which are collinear (a quadrilateral) and their non-aligned
images (another quadrilateral). Since the projectivity preserves the alignment and the
incidence, we must take a linear transformation of the coordinates. However, the affinity
is already a linear transformation of coordinates but only of the independent coordinates.
The more general linear transformation must account for all the three barycentric
coordinates. Thus the projectivity is given by a 3—3 matrix:

«1 ’ x' ’ y'  h02  «1 ’ x ’
« h00 y
h01
¬ · ¬ ·¬ ·
= k ¬ h10
x' h11 h12 · ¬ x
¬ · · with det hij ≠ 0
¬ · ¬h h22 · ¬ ·
y' h21 y
   20  

The coefficients hij being given, the product of the matrices on the right hand side
does not warrantee a set of coordinates with sum equal to 1. This means that the matrix hij
of a perspectivity is defined except by the proportionality constant k (which depends on
the coordinates), and the transformed point is obtained with homogeneous coordinates
instead of barycentric ones. Nevertheless, this matrix maps collinear points to collinear
points. This follows immediately from the matrices equalities:

« k (1 ’ x A' ’ y A' ) l (1 ’ x B' ’ y B' ) m (1 ’ x C' ’ y C' ) « h00 h02 
h01
¬ ·¬ ·
· = ¬ h10 h12 · …
k x A' l x B' m x C' h11
¬
¬ · ¬h h22 ·
k y A' l y B' m y C' h21
   20 

«1 ’ x A ’ y A 1 ’ x B ’ y B 1 ’ xC ’ yC 
¬ ·
xA xB xC
¬ ·
¬ ·
yA yB yC
 

A set of points are collinear if the determinant of the barycentric coordinates vanishes.
Since det hij ≠ 0, the implication of the alignment is bi-directional:

1 ’ x A' ’ y A' 1 ’ x B' ’ y B' 1 ’ x C' ’ y C' 1 ’ x A ’ y A 1 ’ x B ’ y B 1 ’ xC ’ yC
=
klm x A' x B' x C' xA xB xC det hij
y A' y B' y C' yA yB yC

As an example, let us transform the origin of coordinates under the following
projectivity:

«1 ’ 1 2 «1 «1
¬ ·¬ · ¬ ·
(0, 0) ’ (1, 2, 1) = « 1 , 2 , 1  = « 1 , 1 
2 3 1· ¬0· = ¬ 2· ¬ ·¬ ·
¬ ’
4 4 4 2 4
¬ 1 2 0· ¬0· ¬1·
    

Analogously the transformed point of (1, 0) is:
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 105

« 1 ’ 1 2  « 0  « ’ 1
¬ ·¬ · ¬ ·
(1, 0) ’ (’ 1, 3, 2) = « 3 , 1 
2 3 1· ¬1· = ¬ 3 · ¬ ·
¬ ’
4 2
¬ 1 2 0· ¬0· ¬ 2 ·
    

Also we may transform the point (0, ∞) at the infinity, which has the homogeneous
coordinates (’1, 0, 1):

« 1 ’ 1 2  « ’ 1 « 1 
¬ ·¬ · ¬ ·
(0, ∞ ) ’ (1, ’ 1, ’ 1) = (1, 1)
2 3 1 · ¬ 0 · = ¬ ’ 1·
¬ ’
¬ 1 2 0 · ¬ 1 · ¬ ’ 1·
    

This case exemplifies how points at the infinity can be mapped under a projectivity to
points with finite coordinates and vice versa. As a comparison, the affinity always maps
points at the infinity to points at the infinity (although it does not leave them invariant).
That is, the affinity always maps parallel lines to parallel lines, while under a projectivity
parallel lines can be mapped to concurrent lines and vice versa5. Parallel lines are
concurrent in a point at the infinity, which does not play any special role in the
projectivity.
Let us see that four non collinear points (a quadrilateral) and their non collinear
images (another quadrilateral) determine a unique projectivity. Let us suppose that the
images of the points (0, 0), (0, 1), (1, 0) and (1, 1) are (a1, a2), (b1, b2), (c1, c2) and (d1, d2)
where no three of which are collinear. The barycentric coordinates of these points are:

(0, 0) = (1, 0, 0) (1, 0) = (0, 1, 0) (0, 1) = (0, 0, 1)

a0 = 1’ a1 ’ a2
(a1, a2) = (a0, a1, a2) with etc.

The substitution into the matrix equality produces a system of three equations with three
unknowns k, l and m:

m c 0  « ’ 1
« d 0  « k a0 l b0
¬·¬ ·¬ ·
¬ d 1 · = ¬ k a1 l b1 m c1 · ¬ 1 ·
¬d · ¬k a m c2 · ¬ 1 ·
l b2
 2  2  

Since the points (0, 0), (1, 0) and (0, 1) are not collinear, the matrix is regular and the
system has a unique solution, which proves that a projectivity is determined by the image
of the base quadrilateral. If the vertices of the initial quadrilateral are not the base points,
the matrix is calculated as the product of the inverse matrix of the projectivity which maps
the base quadrilateral to the initial quadrilateral multiplied by the matrix of the projectivity
which maps the point base to the final quadrilateral.
Let us calculate the projectivity which maps the quadrilateral with vertices (1, 1)-
(2, 3)-(3, 1)-(1, 0) to the quadrilateral (4, 1)-(6, 0)-(4, 0)-(6, 1). The projectivity which

5
Of course, the affinity is a special kind of projectivity.
RAMON GONZALEZ CALVET
106

maps the base quadrilateral (0, 0)-(1, 0)-(0, 1)-(1/3, 1/3) to the first quadrilateral is given
by:

«0 « ’ k ’4l ’ 3 m  «1
¬· ¬ ·¬ ·
¬1· = ¬ k 2l 3 m · ¬1·
¬0· ¬ k m · ¬1·
3l
   

5 1 1
k= l=’ m=
whose solution is:
4 2 4

Taking the values k = 5, l = ’2 m =1, we find the matrix:

« ’ 5 8 ’ 3 «7 5 6 
¬ · 1¬ ·
¬ 5 ’ 4 3 · whose inverse matrix is ¬5 5 0 ·
20 ¬
¬ 5 ’6 1 · ·
 ’ 5 5 ’ 10 
 

The projectivity which maps the base quadrilateral to the final quadrilateral is:

« ’ 6  « ’ 4k ’5l ’ 3 m  « 1
¬ ·¬ ·¬ ·
¬ 6 · = ¬ 4k 6l 4 m · ¬ 1·
¬1· ¬ k 0 · ¬ 1·
0
   

whose solution is: k = 1, l =1, m = ’1 and the square matrix:

«’ 4 ’ 5 3 
¬ ·
6 ’ 4·
¬4
¬1 0·
0
 

Then, the projectivity which maps the initial quadrilateral to the final one is the product:

« ’ 4 ’ 5 3  « 7 5 6  « ’ 68 ’ 30 ’ 54 
¬ ·¬ ·¬ ·
6 ’ 4 · ¬ 5 5 0 · = ¬ 78
¬4 30 64 ·
¬1 0 · ¬ ’ 5 5 ’ 10 · ¬ 7 6·
0 5
   

For example, the point (2, 5/3) is transformed into:

« ’ 68 ’ 30 ’ 54  « ’ 8  « 94 
¬ ·¬ · ¬ · « 62 2
64 · ¬ 6 · = ¬ ’ 124 · = ¬ , ’ ·
¬ 78 30
 13 13 
¬7 6 ·¬ 5 · ¬ 4 ·
5
    

A projectivity may be also defined in an alternative way: let ABCD be a
quadrilateral and A'B'C'D' another quadrilateral. Then the mapping:
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 107


D=aA+bB+cC D' = a' A' + b' B' + c' C'

is a projectivity if the coefficients a', b', c' are homogeneous linear functions of a, b, c. For
example in the foregoing example the quadrilateral (1, 1)-(2, 3)-(3, 1)-(1, 0) is mapped to
the quadrilateral (4, 1)-(6, 0)-(4, 0)-(6, 1). Then we have:

1 1 5
(1, 0) = (1 ’ b ’ c ) (1, 1) + b ( 2, 3) + c (3, 1) b=’ c= a =1’ b ’ c =

2 4 4

(6, 1) = (1 ’ b' ’ c' ) ( 4, 1) + b' (6, 0) + c' ( 4, 0) ’ b' = 1 c' = ’1 a' = 1 ’ b' ’ c' = 1

Taking into account that for a = 1, b = 0, c = 0 ’ a' =1, b' = 0, c' = 0 and analogously for
each of the first three points, the linear mapping of coefficients is a diagonal matrix which
can be written simply:

4
a' = b' = ’2 b c' = ’ 4 c
a
5

If we wish to calculate the transformed point of (2, 5/3) we calculate firstly a, b and c:

« 5 1 1 1
P = ¬ 2, · = (1 ’ b ’ c ) (1, 1) + b (2, 3) + c (3, 1) b= c= a=

 3 3 3 3
4 2 4
a' = b' = ’ c' = ’

15 3 3

Normalising the coefficients, we find the transformed point P':

(4, 1) + 10 (6, 0) + 20 (4, 0) = « 62 , ’ 2 
4
P' = ’ ¬ ·
26 26 26  13 13 

This example shows the normalisation. In general, we say that the following
transformation is a projectivity:

D = (1’ x ’ y) O + x P + y Q ’ D' = (1 ’ x' ’ y' ) O' + x' P' + y' Q'

lx my
x' = y' =
k (1 ’ x ’ y ) + l x + m y k (1 ’ x ’ y ) + l x + m y

Let us prove that a projectivity defined in this algebraic way leaves invariant the
projective cross ratio (as evident by construction from perspectivities). Under a
projectivity, four collinear points A, B, C and D are transformed into another four collinear
points A', B', C' and D' . Their cross ratio is the quotient of differences of any coordinate
(it is not needed that x and y are Cartesian coordinates) because these differences for x are
proportional to the differences for y due to the alignment:
RAMON GONZALEZ CALVET
108

l [k ( x C ’ x A ) + (k ’ m ) ( x A y C ’ x C y A )]
x' C ’ x' A =
[k (1 ’ x C ’ y C ) + l x C + m y C ] [k (1 ’ x A ’ y A ) + l x A + m y A ]

Let us suppose that the collinear points A, B, C and D lie on the line having the equation y
= s x + t, where the constants s and t play the role of a “slope” and “ordinate intercept”
although the x-axis and the y-axis be not orthogonal:

l ( x C ’ x A ) [k ’ (k ’ m ) t ]
x' C ’ x' A =
[k (1 ’ x C ’ y C ) + l x C + m y C ] [k (1 ’ x A ’ y A ) + l x A + m y A ]

Now we see that a projectivity changes every single ratio due to the denominators, but the
cross ratio is preserved because these denominators are cancelled in this case:

(x C' ’ x A' ) (x D' ’ x B' ) (x C ’ x A ) (x D ’ x B )
( A' B' C' D' ) = = (A B C D)
=
(x D' ’ x A' ) (x C' ’ x B' ) (x D ’ x A ) (x C ’ x B )

The projectivity as a tool for theorems demonstration

One of the most interesting applications of the projectivity is the demonstration of
theorems of projective nature owing to the fact that the projectivity preserves alignment
and incidence. For example, in the figure 10.3 I have displayed without proof the method
of construction of the fourth point which
completes a harmonic range from the other
three. Now let us apply to this figure a
projectivity which maps the point A to the
infinity. In this case the lines ABCD, R and S
become parallel (figure 10.7) and the cross ratio
becomes a single ratio:
B'D'
( A B C D ) = ( ∞ B' C' D' ) = =2
B'C'
Figure 10.7
By similarity of triangles one sees that B'D' =2
B'C', so that the cross ratio is 2 and
Figure 10.8
proves that the points A, B, C and
D in the figure 10.3 are a harmonic
range.
Another example is the
proof of the Pappus™ theorem: the
lines joining points with distinct
letters belonging to two distinct
ranges ABC and A'B'C' intersect in
three collinear points L, M and N
(figure 10.8). Applying a
perspectivity to make both ranges
parallel, now we see that the
triangles ABL and A'B'L are
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 109

similar so that:

A'B' AB A'B' AB
L= A+ L= B+
B' A'
and also
AB + A'B' AB + A'B' AB + A'B' AB + A'B'

We may gather both equations obtaining an expression more adequate to our purposes
which is function of the midpoints:

A'B' AB
A+ B A' + B'
L= +
AB + A'B' AB + A'B'
2 2

Analogously for M and N we obtain:

A'C' AC
A+C A' + C'
M= +
AC + A'C' AC + A'C'
2 2

B'C' BC
B+C B' + C'
N= +
BC + B'C' BC + B'C'
2 2

These expressions fulfil the following equality what implies that the three points are
aligned:
AB + A'B' BC + B'C'
M= L+ N
AC + A'C' AC + A'C'

To prove this, begin from

( AC + A'C' ) M = ( AB + A'B' ) L + ( BC + B'C' ) N ’

A'C' ( A + C ) + AC ( A' + C' ) = A'B' ( A + B ) + AB ( A' + B' ) + B'C' (B + C ) + BC (B' + C' )

Arranging A, B and C on the left hand side and A', B' and C' on the right hand side
and taking into account that ¦AC¦=¦AB¦+¦BC¦ and ¦A'C'¦=¦A'B'¦+¦B'C'¦ we obtain
the following vector equality:

’ B'C' AB + A'B' BC = ’ AB B'C' + BC A'B'

which is an identity taking into account that the lines ABC and A'B'C' are parallel and all
vectors have the same direction and sense.
The linear relation between L, M and N is very useful to calculate the cross ratio
(LMNR). In fact the coefficients of the midpoints are the relative altitudes of the point with
respect to both parallel lines. Since this cross ratio is also the quotient of the relative
altitudes, we have:
RAMON GONZALEZ CALVET
110



«  ’ AC
B'C' A'B'
¬ ·

¬ BC + B'C' AB + A'B' · AC + A'C' AC
(LMNR ) =   = = (OABC )
«  AB
’ AB B'C' A'C'
¬ ·
¬ BC + B'C' ’ AC + A'C' ·
AB + A'B'  

The single ratio is the cross ratio corresponding to O = ∞. This identity of cross ratios is
preserved under a projectivity and it is also valid for the upper scheme on the figure 10.8
when O is a point with finite coordinates. Analogously:

« 
B'C' A'B' A'C'
¬ ·

¬ BC + B'C' AB + A'B' · AC + A'C' A'C'
 
(LMNR' ) = = = (OA'B'C' )
«  A'B'
A'B' B'C' A'C'
¬ ·

AB + A'B' ¬ BC + B'C' AC + A'C' ·
 


The homology

The points A', B' are homologous of the points A and B with respect to the centre
O and the axis of homology given by the point F and the vector v (figure 10.9) if:
1) The centre O does not lie on the axis.
2) Every pair of homologous points and the homology centre are aligned, that is,
O, A and A' are aligned and also O, B and B'.
3) Every pair of homologous lines AB and A'B' intersects in a point on the
homology axis.
A homology6 is determined by an axis, a
centre O and a pair of homologous points A and Figure 10.9
A'. To obtain the homologous of any point B, we
draw the line AB, whose intersection with the
homology axis will be denoted as Z. Then we
draw the line A'Z and the intersection with the
line OB is the homologous point B'. By
construction, the homology axis and the lines
AB, A'B' and OZ are concurrent in Z. The
projective cross ratio of this pencil of lines is
independent of the crossing line where it is
measured. Let G and H be the intersections of
the homology axis with the lines OA and OB
respectively. Then we have:

{ Z, O A G A' } = { Z, O B H B' }


6
The word homology is used by H. Eves (A Survey of Geometry, p. 105) with a different
meaning. He names as homology a composite homothety, that is, a single homothety followed
by a rotation.
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 111

Since the four points of each set are collinear, their projective cross ratio is identical to
their complex cross ratio:

( O A G A' ) = ( O B H B' )

That is, the cross ratio of the four points O, B, H and B' is a real constant r for any point B
and is called homology ratio. Hence we can write:

( O B H B' ) = OH OB' ’1 BB' BH ’1 = r ≠ 1, 0

This equation leads to a simpler definition of the homology: B' is the homologous point of
B with respect a given axis and centre O (outside the axis) with ratio r if O, B and B' are
aligned and the cross ratio ( O B H B' ) is equal to r, being H the intersection point of the
line OB with the homology axis. Hence, a homology is determined by giving a centre O,
an axis and the homology ratio. If the homology ratio is the unity we have two degenerate
cases: whether B lies on O and B' is any point, or B' lies on H (on the homology axis) and
B is any point, that is, there is not one to one mapping, a not useful transformation. Then
we must impose a ratio distinct of the
Figure 10.10
unity. On the other hand, a null
homology ratio implies that BB' = 0 and
the homology becomes the identity.
In the special case r = 2 (when
the points A and A′ are harmonic
conjugates with respect O and H) the
transformation is called a harmonic
homology.
The first limit line is defined as
the set of points L on the plane having as
homologous points those at infinity
(figure 10.10):

( O L H ∞ ) = OH O∞ ’1 L∞ LH ’1 = r

The limit of the quotient of both distances tending to the infinity is the unity:

O∞ ’1 L∞ = lim OX ’1 LX = 1
X ’∞

OH LH ’1 = r LH OH ’1 = r ’1

Then:

Because all the points H lie on the homology axis, all the points L form a line, the limit
line.
The second limit line is defined as the points that are homologous of those at
infinity:

( O ∞ H L' ) = OH OL' ’1 ∞L' ∞H ’1 = r
RAMON GONZALEZ CALVET
112



OH OL' ’1 = r OL' OH ’1 = r ’1


Therefore the distance from the second limit line to the homology axis is equal to the
distance from the first limit line to the homology centre (figure 10.10), and both lines are
located whether between or outside both elements according to the value of r:

OL' = LH = r ’1 OH

Let us calculate the homologous point B' for every point B on the plane. Since H is
a changing point on the homology axis, we write it as a function of a fixed point F also on
the axis and its direction vector v (figure 10.9). At one time, the point H is the intersection
of the line OB and the homology axis:

H = O + b OB = F + k v k, b real

b = OF § v ( OB § v ) ’1
b OB ’ k v = OF ’

H = O + OF § v ( OB § v ) ’1 OB

Now we may calculate the segments OH and BH:

OH = OF § v ( OB § v ) ’1 OB

BH = BO + OF § v ( OB § v ) ’1 OB = [ OF § v ( OB § v ) ’1 ’ 1 ] OB =

= ( OF § v ’ OB § v ) ( OB § v ) ’1 OB = BF § v ( OB § v ) ’1 OB

By substitution in the cross ratio we obtain:

OH OB' ’1 BB' BH ’1 = OH BH ’1 BB' OB' ’1 = OF § v ( BF § v ) ’1 BB' OB' ’1 = r

BB' OB' ’1 = r v § BF ( v § OF ) ’1

and isolating OB' as a function of OB and other known data:

( ’OB + OB' ) OB' ’1 = r v § BF ( v § OF ) ’1 ’OB OB' ’1 = r v § BF ( v § OF ) ’1 ’ 1


OB' ’1 = OB ’1 [ 1 ’ r v § BF ( v § OF ) ’1 ] = OB ’1 ( v § OF ’ r v § BF ) ( v § OF ) ’1

Inverting the vectors we find:

OB' = OB v § OF ( v § OF ’ r v § BF ) ’1 = OB ( 1 ’ r v § BF ( v § OF ) ’1 ) ’1

This equation allows to calculate the homologous point B' of any point B using the data of
the homology: the centre O, the ratio r, and a direction vector v and a point F of the axis.
The homology preserves the projective cross ratio of any set of points. Let A', B',
C', D' and E' be the transformed points of A, B, C, D and E under the homology. Then:
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 113



{ E, A B C D } = { E', A' B' C' D' }

Let us prove this statement. Being A' and E' homologous points of A and E we have:

OA' = OA ( 1 ’ r v § AF ( v § OF ) ’1 ) ’1

OE' = OE ( 1 ’ r v § EF ( v § OF ) ’1 ) ’1

Thus the segment E'A' is:

E'A' = OA' ’ OE'

E'A' = OA ( 1 ’ r v § AF ( v § OF ) ’1 ) ’1 ’ OE ( 1 ’ r v § EF ( v § OF ) ’1 ) ’1 =

= [ OA ( 1 ’ r v § EF ( v § OF ) ’1 ) ’ OE ( 1 ’ r v § AF ( v § OF ) ’1 ) ]

[ 1 ’ r v § AF ( v § OF ) ’1 ] ’1 [ 1 ’ r v § EF ( v § OF ) ’1 ] ’1 =

= [ EA + r ( ’OA v § EF + OE v § AF ) ( v § OF ) ’1 ]

[ 1 ’ r v § AF ( v § OF ) ’1 ] ’1 [ 1 ’ r v § EF ( v § OF ) ’1 ] ’1 =

= [ EA ( 1 ’ r ) + r ( OA v § OE ’ OE v § OA ) ( v § OF ) ’1 ]

[ 1 ’ r v § AF ( v § OF ) ’1 ] ’1 [ 1 ’ r v § EF ( v § OF ) ’1 ] ’1 =

= [ EA ( 1 ’ r ) + r v OA § OE ( v § OF ) ’1 ]

[ 1 ’ r v § AF ( v § OF ) ’1 ] ’1 [ 1 ’ r v § EF ( v § OF ) ’1 ] ’1

For the other vectors analogous expressions are obtained:

E'B' = [ EB ( 1’ r ) + r v OB § OE ( v § OF ) ’1 ]

[ 1 ’ r v § BF ( v § OF ) ’1 ] ’1 [ 1 ’ r v § EF ( v § OF ) ’1 ] ’1

E'C' = [ EC ( 1’ r ) + r v OC § OE ( v § OF ) ’1 ]

[ 1 ’ r v § CF ( v § OF ) ’1 ] ’1 [ 1 ’ r v § EF ( v § OF ) ’1 ] ’1

E'D' = [ ED ( 1’ r ) + r v OD § OE ( v § OF ) ’1 ]

[ 1 ’ r v § DF ( v § OF ) ’1 ] ’1 [ 1 ’ r v § EF ( v § OF ) ’1 ] ’1

Let us calculate the outer product E'A' § E'C':
RAMON GONZALEZ CALVET
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E'A' § E'C' = [ EA § EC ( 1 ’ r )2 + ( EA § v OC § OE + v § EC OA § OE ) r ( 1 ’ r )

( v § OF ) ’1 ] [ 1 ’ r v § AF ( v § OF ) ’1 ] ’1 [ 1 ’ r v § CF ( v § OF ) ’1 ] ’1

[ 1 ’ r v § EF ( v § OF ) ’1 ] ’2

By substitution of the following identity, which may be proved by means of the geometric
algebra (see exercise 1.4):

EA § v OC § OE + v § EC OA § OE = EA § v EC § OE + v § EC EA § OE =

EA § EC v § OE

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