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2. A base of vectors for the plane

2.1 Draw, for example, two vectors of the first quadrant and mark u1, u2, v1 and v2 on the
Cartesian axis. Then calculate the area of the parallelogram from the areas of the
rectangles:

A = ( u1 + v1 ) ( u2 + v2 ) ’ u1 u2 ’ v1 v2 ’ 2 v1 u2 = u1 v2 ’ u2 v1

2.2 The area of a triangle is the half of the area of the parallelogram formed by any two
sides. Therefore:

A = ( 3 e1 + 5 e2 ) § ( ’2 e1 ’ 3 e2 ) = ( ’9 + 10 ) e12 = e12 ’ ¦A¦ = 1

a b c = ( a1 e1 + a2 e2 ) ( b1 e1 + b2 e2 ) ( c1 e1 + c2 e2 )
2.3

= [ a1 b1 + a2 b2 + e12 ( a1 b2 ’ a2 b1 ) ] ( c1 e1 + c2 e2 )

= e1 (a1 b1 c1 + a2 b2 c1 + a1 b2 c2 ’ a2 b1 c2) + e2 (a1 b1 c2 + a2 b2 c2 ’ a1 b2 c1 + a2 b1 c1)
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 209

= ( c1 e1 + c2 e2 ) [ b1 a1 + b2 a2 + e12 ( b1 a2 ’ b2 a1 ) ] =

= ( c1 e1 + c2 e2 ) ( b1 e1 + b2 e2 ) ( a1 e1 + a2 e2 ) = c b a

u v = ( 2 e1 + 3 e2 ) ( ’3 e1 + 4 e2 ) = 6 + 17 e12
2.4

u v = 325 ±(u, v) =1.2315=70º33'36"

2.5 For this base we have:

3
e1 · e2 = | e1 | | e2 | cos π/3 = 1 | e1 § e2 | = | e1 | | e2 | | sin π/3 | =

Applying the original definitions of the modulus of a vector one finds:

u = 52
u2 = ( 2 e1 + 3 e2 )2 = 4 e12 + 9 e22 + 12 e1 · e2 = 4 + 36 + 12 = 52 ’

v = 49
v2 = (’3 e1 + 4 e2 )2 = 9 e12 + 16 e22 ’ 24 e1 · e2 = 9 + 64 ’24 = 49 ’

u v = ( 2 e1 + 3 e2 ) (’3 e1 + 4 e2 ) = ’6 e12 + 12 e22 + 8 e1 e2 ’9 e2 e1 =

u v = 2548
= 42 ’ e1 · e2 + 17 e1 § e2 = 41 + 17 e1 § e2 ’

u§v
u·v 41 17 3
cos ± = sin± = ±
= =
uv uv
2548 2548

±(u, v) = 0.6228=35º 41' 5" oriented with the sense from e1 to e2 .

When e1 and e2 are not perpendicular, e1 e2 ≠ ’ e2 e1 and e12 has not the meaning of a pure
area but the outer product e1 § e2 with an area of 3 . Also observe that u v = u v .

2.6 In order to find the components of v for the new base {u1, u2}, we must resolve the
vector v into a linear combination of u1 and u2 .

v = c1 u1 + c2 u2

3 (’ 3) ’ (’ 5) 5 u1 § v 2 (’ 5) ’ (’ 1) 3
v § u2
c1 = = = ’16 c2 = = =7
u1 § u 2 ’1 u1 § u 2 ’1

so in the new base v = (’16, 7).


3. The complex numbers

3.1 z t = ( 1 + 3 e12 ) ( ’2 + 2 e12 ) = ’2 ’6 + ( 2 ’6 ) e12 = ’8 ’4 e12
RAMON GONZALEZ CALVET
210

3.2 Every complex number z can be written as product of two vectors a and b, its modulus
being the product of the moduli of both vectors:
2 22
’ |z|=|a||b| ’ ¦z¦ = a b = a b b a = z z*
z=ab
4
3.3 The equation x ’1 = 0 is solved by extraction of the fourth roots:

x2 = 1 ’ x1 = 1, x2 = ’1
4
x =1
x2 = ’1 ’ x3 = e12 , x4 = ’e12

3.4 Passing to the polar form we have:

( ) ( )
n n
= ’ 2 ’π / 4
2π /4

= ’ 2n = 2n
2n 2n

’ n π/4 ’ n π/4 +π
nπ /4 n π/4



Therefore the arguments must be equal except for k times 2π:


nπ nπ
=’ +π + 2 k π ’ =π + 2 k π ’ n=2+4k
4 4 2

3 3 3
3.5 The three cubic roots of “3 + 3 e12 are 18 π / 4 , 18 11 π / 12 , 18 19 π / 12 .

3.6 Using the formula of the equation of second degree we find: z1 = 2’3 e12, z2 = 1+e12.

3.7 We suppose that the complex analytic extension f has a real part a of the form:

a = sin x K(y) with K(0) = 1

Applying the first Cauchy-Riemann condition, we find the imaginary part b of f :

‚a ‚b
b = cos x « K ( y ) dy
= cos x K ( y ) = ’
‚x ‚y

Applying the second Cauchy-Riemann condition:

‚a ‚b
= sin x « K ( y ) dy
= sin x K' ( y ) = ’
‚y ‚x

we arrive at a differential equation for K(y) whose solution is the hyperbolic cosine:

K' ( y ) = « K ( y ) dy 
 ’ K(y) = cosh y
K (0) = 1 

Hence the analytical extension of the sine is:
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 211


sin(x + e12 y) = sin x cosh y + e12 cos x sinh y

The analytical extensions of the trigonometric functions may be also obtained from the
exponential function. In the case of the cosine, we have:

exp( e12 z ) + exp( ’ e12 z ) exp(’ y )(cos x + e12 sin x ) + exp( y )(cos x ’ e12 sin x )
cos z = =
2 2

cos z = cos x cosh y ’ e12 sin x sinh y

3.8 Let us calculate some derivatives at z = 0:

f (z ) = log(1 + exp(’ z )) f(0) = log 2

’ exp(’ z ) ’1 1
f' (z ) = f' (0) = ’
=
1 + exp(’ z ) exp(z ) + 1 2

exp(z ) 1
f'' (z ) = f'' (0) =
(exp(z ) + 1) 2
4

exp(z ) 2 exp(2 z )
f''' (z ) = f''' (0) = 0

(exp(z ) + 1) (exp(z ) + 1)3
2




exp(z ) 2 exp(2 z ) 6 exp(3z )
(0) = ’ 1
4 exp( 2 z )
(z ) = ’ ’ + IV
IV
f
f
(exp(z ) + 1)2 (exp(z ) + 1)3 (exp(z ) + 1)3 (exp(z ) + 1)4 8

Hence the Taylor series is:

z z2 z4
f (z ) = log 2 ’ + ’ + ...
248

Observe that f (e12 π ) = log(0) is divergent. This is the singularity nearest the origin.
Therefore the radius of convergence of the series is π.

3.9 Separating fractions:

1 1 1
f (z ) = = ’
z 2 + 2 z ’ 8 6 (z ’ 2 ) 6 (z + 4 )

and developing the second fraction:
RAMON GONZALEZ CALVET
212

1
1« 
2 3
z ’ 2 « z ’ 2 « z ’ 2
1 1 6 ¬1 ’ · + ... ·
= = = +¬ · ’¬
z ’2 6¬ ·
z+4 6+ z’2 6 6 6
 
1+
6
n (z ’ 2 )
n

1
f (z ) = ’ ‘ (’ 1)
we find:
6( z ’ 2 ) n = 0 6 n+2

This Lauren series is convergent in the annulus 0 <¦z ’ 2¦< 6, which contains the
required annulus 1 <¦z ’ 2¦< 4 .

3.10 Taking into account that (1 ’ x) ’1 = 1 + x + x2 + ... , the function defined by the series
is:

1 1 1
f (z ) = ‘ n = ’1=
n =1 4 ( z + 1) 4z +3
1
n
1’
4 (z + 1)

Now we see that at z = ’3/4 the function has a pole. Therefore the radius of convergence
of this series (centred at z = ’1) is ¦’3/4 ’( ’1)¦=1/4.

3.11 From the successive derivatives of the sines calculated at z = 0, one obtains the
Taylor series for sin z:

z 2 n +1

z3 z5
’ ... = ‘ (’ 1)
n
sin z = z ’ +
(2 n + 1)!
3! 5 ! n =0



so the Lauren series for the given function is:

z 2 n ’1
sin z ∞ 1 z z3
= ‘ (’ 1)
n
=’+ ’ ...
(2 n + 1)! z 3! 5!
z2 n =0



Since the pole z = 0 is the unique singularity (see that the analytical extension of the real
sine in the exercise 3.7 has no singularities), the annulus of convergence is 0<¦z¦< ∞ .

3.12 If f(z) is analytic then the derivative at each point is unique and we can write for two
different directions dz1 and dz2:

df1 = dz1 f'(z) df2 = dz2 f'(z)

According to the relationship between the complex and vectorial planes, we can multiply
by e1 at the left in order to turn the complex differentials into vectors:
e1 e1
df = da + db e12 ’ df = da e1 + db e 2 dz = dx + dy e12 ’ dz = dx e1 + dy e 2

Now let us calculate the geometric product of the vector differentials:

df1 df2 = dz1 f'(z) dz2 f™(z) = dz1 dz2 [f™(z)]* f'(z) =¦f'(z)¦2 dz1 dz2
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 213


Since¦f'(z)¦2 is real, both geometric products have the same argument, ±(df1, df2) = ±(dz1,
dz2), and hence the transformation is conformal.


4. Transformations of vectors

4.1 If d is the direction vector of the reflection, the vector v' reflected of v with respect to
this direction is given by:

v' = d ’1 v d

If v'' is obtained from v' by a rotation through an angle ±, and z is a unitary complex
number with argument ± /2 then:

± ±
z = cos + e12 sin
v'' = z ’1 v' z
2 2

Joining both equations:

v'' = z ’1 d ’1 v d z = c ’1 v c

That is, one obtains a reflection with respect another direction with vector c = d z, the
product of the direction vector of the initial reflection and the complex number with half
argument.

4.2 Let w'' be the transformed vector of w by two consecutive reflections with respect
different directions u and v:

w'' = v ’1 w' v = v ’1 u ’1 w u v

Then we can write:

w'' = z ’1 w z z=uv z being a complex number

so that it is equivalent to a rotation with an angle equal to the double of that formed by
both direction vectors.

4.3 If the product of each transformed vector v' by the initial vector v is equal to a
complex number z2 (v' and v always form a constant angle) then:

v v' = z2

v' = v ’1 z2 = z* v ’1 z = z ’1 | z | 2 v ’1 z = | z | 2 ( z ’1 v z ) ’1

which represents an inversion with radius | z |2 followed by a rotation with an angle equal
to the double of the argument of z. As the algebra shows, both elemental transformations
commute.
RAMON GONZALEZ CALVET
214




5. Points and straight lines

5.1 If A, B, C and D are located following this order on the perimeter of the parallelogram,
then AB=DC:

’ B’A=C’D ’ D=C’B+A
AB = DC

D = (2, ’5) ’ (4, ’3) + (2, 4) = (0, 2)

¦AB § BC¦ = ¦( 2 e1 ’7 e2 ) § ( ’2 e1 ’2 e2 )¦ = ¦’18 e12¦ = 18
The area is:

5.2 The Euler™s theorem:

AD BC + BD CA + CD AB = ( D ’A ) ( C ’B ) + ( D ’B ) ( A ’C ) + ( D ’C ) ( B ’A )

=CA’AC+AB’BA+BC’CB=2(C§A+A§B+B§C)=

= 2 ( B ’ A ) § ( C ’ B ) = 2 AB § BC

This product only vanishes if A, B and C are collinear. On the other hand we see that the
product is the oriented area of the triangle ABC.

5.3 a) The area of the triangle ABC is the outer product of two sides:

AB = B ’ A = (4, 4) ’ (2, 2) = 2 e1 + 2 e2 AC = C ’ A = ( 4, 2) ’ (2, 2) = 2 e1

AB § AC = ( 2 e1 + 2 e2 ) § 2 e1 = 4 e2 § e1

AB § AC = 4 e 2 § e1 = 4 e 2 e1 sin 60 o = 4 3

b) The distance from A to B is the length of the vector AB, etc:

AB2 = ( 2 e1 + 2 e2 )2 = 4 e12 + 4 e22 + 8 e1 · e2 = 4 + 16 + 16 cos π/3 = 28

AC2 = ( 2 e1 )2 = 4 BC2 = (’2 e2 )2 = 4 e22 = 16

28
d( A, B ) =  AB = d( B, C ) = BC = 4 d( C, A) =  CA = 2

5.4 A side of the trapezoid is a vectorial sum of the other three sides:

AD = AB + BC + CD

AD2 = (AB +BC +CD )2 = AB2 +BC2 +CD2 +2 AB · BC + 2 AB · CD + 2 BC · CD

= AB2 + BC2 + CD2 + 2 AB · BC ’ 2¦AB¦¦CD¦+ 2 BC · CD

where the fact that AB and CD be vectors with the same direction and contrary sense has
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 215

been taken into account. Arranging terms:

AD2 ’ AB2 ’ BC2 ’ CD2 + 2 ¦AB¦¦CD¦ = 2 AB · BC + 2 BC · CD

AD2 ’ AB2 ’ BC2 ’ CD2 + 2 ¦AB¦¦CD¦ = 2 BC · ( AB + CD )

Without loss of generality we will suppose that ¦AB¦>¦CD¦:

¦AB + CD¦ = ¦AB¦’¦CD¦

Since the angle ABC is the supplement of the angle formed by the vectors AB and BC, we
have:

AD2 ’AB2 ’BC2 ’CD2 + 2 ¦AB¦¦CD¦ = ’2 ¦BC¦ ( ¦AB¦’¦CD¦ ) cos ABC

whence we obtain the angle ABC:

’ AD 2 + AB 2 + BC 2 + CD 2 ’ 2 AB CD
cos ABC =
( )
AB ’ CD
2 BC

The trapezoid can only exist for the range ’1< cos ABC <1, that is:

2 ¦BC¦ ( ¦AB¦’¦CD¦ ) > ’AD2 + AB2 + BC2 + CD2 ’2 ¦AB¦¦CD¦ >

> ’2 ¦BC¦ (¦AB¦’¦CD¦)

R=(1’p’q)O+pP+qQ ’ RP = ( 1 ’ p ’ q ) OP + q QP
5.5

’ RP § PQ = ( 1 ’ p ’ q ) OP § PQ

whence it follows that: Area RPQ = ( 1 ’ p ’ q ) Area OPQ

5.6 The direction vector of the straight line r is AB:

AB = B ’ A = (5, 4) ’ (2, 3) = 3 e1 + e2 AC = C ’ A = (1, 6) ’ (2, 3) = ’ e1 + 3 e2

The distance from the point C to the line r is:

(’ e1 + 3 e 2 ) § (3 e1 + e 2 )
AC § AB
d (C , r ) = = = 10
AB 10

The angle between the vectors AB and AC is deduced by means of the sine and cosine:

AB … AC AB § AC
cos ± = e12 sin± =
=0 = e12
AB AC AB AC

Therefore, ± = π/2. The angle between two lines is always comprised from ’π/2 to π/2
RAMON GONZALEZ CALVET
216

because a rotation of 2π around the intersection point does not alter the lines. When the
angle exceeds these boundaries, you may add or subtract π.

5.7 a) Three points D, E, F are aligned if they are linearly dependent, that is, if the
determinant of the coordinates vanishes.

D = (1 ’ xD ’ yD ) O + xD P + yD Q

E = (1 ’ xE ’ yE ) O + xE P + yE Q

F = (1 ’ xF ’ yF ) O + xF P + yF Q

1 ’ xD ’ yD xD yD
1 ’ xE ’ yE yE = 0
xE
1 ’ xF ’ yF xF yF

where the barycentric coordinate system is given by the origin O and points P, Q (for
example the Cartesian system is determined by O = (0, 0), P = (1, 0) and Q = (0, 1)). The
transformed points D', E' and F' have the same coordinates expressed for the base O', P'
and Q'. Then the determinant is exactly the same, so that it vanishes and the transformed
points are aligned. Therefore any straight line is transformed into another straight line.
b) Let O', P' and Q' be the transformed points of O, P and Q by the given affinity:

O' = (o1 , o 2 ) P' = ( p1 , p 2 ) Q' = (q1 , q 2 )

and consider any point R with coordinates ( x, y ):

R = ( x, y ) = ( 1 ’ x ’ y ) O + x P + y Q

Then R', the transformed point of R, is:

R' = ( 1 ’ x ’ y ) O' + x P' + y Q' = ( 1 ’ x ’ y ) ( o1 , o2 ) + x ( p1 , p2 ) + y ( q1 , q2 ) =

= ( x ( p1 ’ o1 ) + y (q1 ’ o1 ) + o1 , x ( p2 ’ o2 ) + y ( q2 - o2 ) + o2 ) = ( x', y' )

where we see that the coordinates x' and y' of R' are linear functions of the coordinates of
R:
x' = x ( p1 ’ o1 ) + y ( q1 ’ o1 ) + o1

y' = x ( p2 ’ o2 ) + y ( q2 - o2 ) + o2

In matrix form:

« x'  « p1 ’ o1 q1 ’ o1  « x  « o1 
¬ ·=¬ ·¬ · + ¬ ·
¬ y' · ¬ p ’ o q2 ’ o2 · ¬ y · ¬ o2 ·
 2    
2



Because every linear (and non degenerate) mapping of coordinates can be written in this
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 217

regular matrix form, now we see that it is always an affinity.

c) Let us consider any three non aligned points A, B, C and their coordinates:

A = ( 1 ’ x A ’ y A ) O + x A P + y AQ

B = ( 1 ’ x B ’ y B ) O + x B P + y BQ

C = ( 1 ’ xC ’ y C ) O + xC P + yC Q

In matrix form:

« A  «1 ’ x A ’ y A y A  «O 
xA
¬·¬ ·¬ ·
¬ B · = ¬1 ’ x B ’ y B xB yB · ¬ P ·
¬ C · ¬1 ’ x ’ y yC · ¬ Q ·
xC
  
C C



A certain point D is expressed with coordinates whether for { O, P, Q } or { A, B, C }:

D = (1 ’ b ’ c ) A + b B + c C = (1 ’ x D ’ y D ) O + x D P + y D Q

In matrix form:

«O  « A
¬· ¬·
D = (1 ’ xD ’ yD y D )¬ P · = ( 1 ’ b ’ c b c)¬ B ·
xD
¬Q · ¬C ·
 

«1 ’ x A ’ y A yA 
xA
¬ ·
(1 ’ xD ’ yD y D ) = ( 1 ’ b ’ c b c ) ¬1 ’ x B ’ y B
xD xB yB ·
¬1 ’ x ’ y yC ·
xC
 
C C



which leads to the following system of equations:

± x D = ( 1 ’ b ’ c) x A + b x B + c xC

 y D = ( 1 ’ b ’ c) y A + b y B + c yC

An affinity does not change the coordinates x, y of A, B, C and D, but only the point base -
{O', P', Q'} instead of {O, P, Q}-. Therefore the solution of the system of equations for b
and c is the same. Then we can write:

D' = ( 1 ’ b ’ c ) A' + b B' + c C'

d) If the points D, E, F and G are the consecutive vertices in a parallelogram then:

DE = GF ’ G = D ’ E + F
RAMON GONZALEZ CALVET
218


The affinity preserves the coordinates expressed in any base {D, E, F}. Then the
transformed points form also a parallelogram:

G' = D' ’ E' + F' ’ D'E' = G'F'

e) For any three aligned points D, E, F the single ratio r is:

DE DF ’1 = r ’ ’ E=(1’r)D+rF
DE = r DF

The ratio r is a coordinate within the straight line DF and it is not changed by the affinity:
’1
E' = ( 1 ’ r ) D' + r F' ’ D'E' D'F' =r

5.8 This exercise is the dual of the problem 2. Then I have copied and pasted it changing
the words for a correct understanding.
a) Three lines D, E, F are concurrent if they are linearly dependent, that is, if the
determinant of the dual coordinates vanishes:

D = (1 ’ xD ’ yD ) O + xD P + yD Q

E = (1 ’ xE ’ yE ) O + xE P + yE Q

F = (1 ’ xF ’ yF ) O + xF P + yF Q

1 ’ xD ’ yD xD yD
1 ’ xE ’ yE yE = 0
xE
1 ’ xF ’ yF xF yF

where the dual coordinate system is given by the lines O, P and Q. For example, the
Cartesian system is determined by O = [0, 0] (line ’x ’y +1=0), P = [1, 0] (line x = 0) and
Q = [0, 1] (line y = 0). The transformed lines D', E' and F' have the same coordinates
expressed for the base O', P' and Q'. Then the determinant also vanishes and the
transformed lines are concurrent. Therefore any pencil of lines is transformed into another
pencil of lines.
b) Let O', P' and Q' be the transformed lines of O, P and Q by the given transformation:

O' = [o1 , o 2 ] P' = [ p1 , p 2 ] Q' = [q1 , q 2 ]

and consider any line R with dual coordinates [x, y]:

R = [x, y] = ( 1 ’ x ’ y ) O + x P + y Q

Then R', the transformed line of R, is:

R' = ( 1 ’ x ’ y ) O' + x P' + y Q' = ( 1 ’ x ’ y ) [ o1 , o2 ] + x [ p1 , p2 ] + y [ q1 , q2 ] =

= [ x ( p1 ’ o1 ) + y (q1 ’ o1 ) + o1 , x ( p2 ’ o2 ) + y ( q2 ’ o2 ) + o2 ] = [ x', y' ]
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 219


where we see that the coordinates x' and y' of R' are linear functions of the coordinates of
R:

x' = x ( p1 ’ o1 ) + y ( q1 ’ o1 ) + o1

y' = x ( p2 ’ o2 ) + y ( q2 - o2 ) + o2

In matrix form:

® x '  ® p1 ’ o1 q1 ’ o1  ® x  ® o1 
 y ' =  p ’ o  y  + o 
q2 ’ o2 
°» °2 ° » ° 2»
»
2



Because any linear mapping (non degenerate) of dual coordinates can be written in this
regular matrix form, now we see that it is always an affinity.

c) Let us consider any three non concurrent lines A, B, C and their coordinates:

A = ( 1 ’ x A ’ y A ) O + x A P + y AQ

B = ( 1 ’ x B ’ y B ) O + x B P + y BQ

C = ( 1 ’ xC ’ y C ) O + xC P + yC Q

In matrix form:

® A ®1 ’ x A ’ y A y A  ®O 
xA
 B  = 1 ’ x ’ y y B  P
xB
   
B B

C  1 ’ x C ’ y C y C  Q 
xC
°» ° »° »

A certain line D is expressed with dual coordinates whether for {O, P, Q } or {A, B, C }:

D = ( 1 ’ b ’ c) A + b B + c C = ( 1 ’ x D ’ y D ) O + x D P + y D Q

In matrix form:

®O  ® A
y D ]  P  = [1 ’ b ’ c b c ]  B 
D = [1 ’ x D ’ y D xD
 
Q  C 
°» °»

®1 ’ x A ’ y A yA
xA
y D ] = [1 ’ b ’ c b c ] 1 ’ x B ’ y B yB 
[1 ’ x D ’ y D xD xB
 
1 ’ x C ’ y C yC 
xC
° »
RAMON GONZALEZ CALVET
220

which leads to the following system of equations:

± x D = (1 ’ b ’ c ) x A + b x B + c x C

 y D = (1 ’ b ’ c ) y A + b y B + c y C

An affinity does not change the
Figure 16.1
coordinates x, y of A, B, C and D,
but only the lines base -{O', P', Q'}
instead of {O, P, Q}-. Therefore the
solution of the system of equations
for b and c is the same. Then we can
write:

D' = (1 ’ b ’ c ) A' + b B' + c C'

d) The affinity maps parallel points
into parallel points (points aligned
with the point (1/3, 1/3), point at the
infinity in the dual plane). If the
lines D, E, F and G are the
consecutive vertices in a dual
parallelogram (figure 16.1) then:

DE = GF ’ G = D ’ E + F

Where DE is the dual vector of the intersection point of the lines D and E, and GF the
dual vector of the intersection point of G and F. Obviously, the points EF and DG are also
parallel because from the former equality it follows:

EF = DG

The affinity preserves the coordinates expressed in any base {D, E, F}. Then the
transformed lines form also a dual parallelogram:

G' = D' ’ E' + F' ’ D'E' = G'F'

e) For any three concurrent lines D, E, F the single dual ratio r is:

DE DF ’1 = r ’ ’ E=(1’r)D+rF
DE = r DF

The ratio r is a coordinate within the pencil of lines DF and it is not changed by the
affinity:
’1
E' = ( 1 ’ r ) D' + r F' ’ D'E' D'F' =r

f) Let us consider four concurrent lines A, B, C and D with the following dual coordinates
expressed in the lines base {O, P, Q}:

A = [xA , yA] B = [xB , yB] C = [xC , yC] D = [xD , yD]
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 221


Then the direction vector of the line A is:

v A = ( 1 ’ x A ’ y A ) vO + x A v P + y A vQ

But the direction vectors of the base fulfils:

vO + v P + vQ = 0

Then:
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 is:

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

Then the cross ratio only depends on the dual coordinates but not on the direction of the
base vectors (see chapter 10):

v A § vC v B § v D
( ABCD ) = =
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 ))

Hence it remains invariant under an affinity. Each outer product can be written as an outer
product of dual vectors obtained by subtraction of the coordinates of each line and the
infinite line:

® 1 ® 1
1 1
v A § v C = 3  x A ’ , y A ’  §  x C ’ , y C ’  = 3 LA § LC
3 3» ° 3 3»
°

where LA = A ’ L is the dual vector going from the line L at the infinity to the line A, etc.
Then we can write this useful formula for the cross ratio of four lines:
RAMON GONZALEZ CALVET
222


( ABCD ) = LA § LC LB § LD
LA § LD LB § LC

This exercise links with the section Projective cross ratio in the chapter 10.

5.9 To obtain the dual coordinates of the first line, solve the identity:

x “ y “ 1 ≡ a' ( “ x “ y + 1) + b' x + c' y ’ a' = ’1 c' = ’2
b' = 0

1 2
a= b=0 c=
’ ’ [0, 2/3]
3 3

For the second line:

x “ y + 3 ≡ a' ( “ x “ y + 1) + b' x + c' y ’ a' = 3 b' = 4 c' = 2

3 4 2
a= b= c=
’ ’ [4/9, 2/9]
9
9 9

Both lines are aligned in the dual plane with the line at the infinity (whose dual
coordinates are [1/3, 1/3]) since the determinant of the coordinates vanishes:

1/ 3 0 2/3
2/9 4/9 2/9 = 0
1/ 3 1/ 3 1/ 3

Therefore they are parallel (this is also trivial from the general equations).

5.10 The point (2, 1) is the intersection of the lines x “ 2 = 0 and y “ 1 = 0, whose dual
coordinates are:


x“2=0 [1/5, 2/5]


y“1=0 [1/2, 0]

The difference of dual coordinates gives a dual direction vector for the point:

v = [1/2, 0] “ [1/5, 2/5] = [3/10, “4/10]

and hence we obtain the dual continuous and general equations for the point:

b ’ 1/ 2 c
= 4 b + 3c = 2

’4
3

The point (“3, “1) is the intersection of the lines x + 3 = 0 and y + 1 = 0, whose dual
coordinates are:
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 223


x+3=0 [4/10, 3/10]


y+1=0 [1/4, 2/4]

The difference of dual coordinates gives a dual direction vector for the point:

w = [1/4, 2/4] “ [4/10, 3/10] = [“3/20, 4/20]

and hence we obtain the dual continuous and general equations for the point:

b ’ 1/ 4 c ’ 2 / 4
8b + 6c = 5
= ”
’3 4

Note that both dual direction vectors v and w are proportional and the points are parallel in
a dual sense. Hence the points are aligned with the centroid (1/3, 1/3) of the coordinate
system.


6. Angles and elemental trigonometry

6.1 Let a, b and c be the sides of a triangle taken anticlockwise. Then:

c2 = ( a + b )2 = a2 + b2 + 2 a · b
a+b+c=0 ’ c=’a’b ’

c2 = a2 + b2 + 2 ¦a¦ ¦b¦ cos(π ’ γ ) = a2 + b2 ’ 2 ¦a¦¦b¦ cos γ (law of
cosines)

The area s of a triangle is the half of the outer product of any pair of sides:

2s=a§b=b§c=c§a

¦a¦¦b¦ sin(π ’ γ ) e12 = ¦b¦¦c¦ sin(π ’ ±) e12 = ¦c¦¦a¦ sin(π ’ β ) e12

sinγ sin± sinβ
= = (law of sines)
c a b

From here one quickly obtains:

a+b a’b
sin± + sinβ sin± ’ sinβ
= =
sin± sinβ
a b

By dividing both equations and introducing the identities for the addition and subtraction
of sines we arrive at:
RAMON GONZALEZ CALVET
224

±+β ± ’β
2 sin
cos
a + b sin± + sinβ 2 2
= =
±+β ± ’β
a ’ b sin± ’ sinβ
2 cos sin
2 2

±+β
tg
a+b 2
= (law of tangents)
±’β
a’b
tg
2

6.2 Let us substitute the first cosine by the half angle identity and convert the addition of
the two last cosines into a product:
β ’ 2γ + ±
±’β β ’±
cos(± ’ β ) + cos(β ’ γ ) + cos(γ ’ ± ) = 2 cos 2 ’ 1 + cos cos
2 2 2

Let us extract common factor and convert the addition of cosines into a product:

±’β ±’β ± ’ 2γ + β
« 
= 2 cos + cos · ’1
¬ cos
2 2 2
 

±’β ± ’γ γ ’β
= 2 cos ’1
cos cos
2 2 2

±’β β ’γ γ ’±
= 2 cos ’1
cos cos
2 2 2

The identity for the sines is proved in a similar way.

6.3 Using the De Moivre™s identity:

cos 4± + e12 sin 4± ≡ (cos ± + e12 sin ± )
4




After developing the right hand side we find:

sin 4± ≡ 4 cos3± sin± ’ 4 cos± sin3± cos 4± ≡ cos4± ’ 6 cos2± sin2± + sin4±

4 tg ± ’ 4 tg 3 ±
And dividing both identities: tg 4± =
1 ’ 6 tg 2 ± + tg 4 ±

6.4 Let ±, β and γ be the angles of the triangle PAB with vertices A, B and P respectively.
The angle γ embracing the arc AB is constant for any point P on the arc AB. By the law of
sines we have:

PA PB PC
= =
sin± sinβ sinγ
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 225

The sum of both chords is:

sin± + sinβ
PA + PB = PC
sinγ

Converting the sum of sines into a product we find:

±+β ±’β π ’γ ±’β
2 sin cos 2 sin cos
2 2 = PC 2 2
PA + PB = PC
sinγ sinγ

Since γ is constant, the maximum is attained for cos(± /2 ’ β /2)=1, that is, when the
triangle PAB is isosceles and P is the midpoint of the arc AB.

6.5 From the sine theorem we have:

±+β ±’β ±’β
2 sin cos cos
a+b sin± + sinβ 2 2= 2
= =
γ γ γ
sinγ
c
2 sin cos sin
2 2 2

Following the same way, we also have:

±+β ±’β ±+β
2 cos sin sin
a’b sin± ’ sinβ 2 2= 2
= =
γ γ γ
sinγ
c
2 sin cos cos
2 2 2

6.6 Take a as the base of the triangle and draw the altitude. It divides a in two segments
which are the projections of the sides b and c on a:

a = b cos γ + c sinβ and so forth.

6.7 From the double angle identities we have:

± ± ± ±
cos ± ≡ cos 2 ’ sin 2 ≡ 1 ’ 2 sin 2 ≡ 2 cos 2 ’ 1
2 2 2 2

From the last two expressions the half-angle identities follow:

± 1 ’ cos ± ± 1 + cos ±
≡± ≡±
sin cos
2 2 2 2

Making the quotient:
RAMON GONZALEZ CALVET
226

1 ’ cos ± 1 ’ cos ± sin±
tg ± ≡ ± ≡ ≡
1 + cos ± sin± 1 + cos±


7. Similarities and single ratio

7.1 First at all draw the triangle and see that the homologous vertices are:

P = ( 0, 0 ) P' = ( 4, 2 )
Q = ( 2, 0 ) Q' = ( 2, 0 )
R = ( 0, 1 ) R' = ( 5, 1 )

QR = ’2 e1 + e2 RP = ’e2
PQ = 2 e1

P'Q' = ’2 e1 + 2 e2 R'P' = ’e1 + e2
Q'R' = 3 e1 + e2

r = P'Q' PQ ’1 = Q'R' QR ’1 = R'P' RP ’1 = ’1 ’ e12 = 2 5π / 4

2 and the angle between the directions of homologous sides is
The size ratio is ¦r¦ =
5π/4.

7.2 Let ABC be a right triangle being C the right angle. The altitude CD cutting the base
AB in D splits ABC in two right angle triangles: ADC and BDC. In order to simplify I
introduce the following notation:

DB = c ’ x
AB = c BC = a CA = b AD = x

The triangles CBA and DCA are oppositely similar because the angle CAD is common and
the other one is a right angle. Hence:

b x ’1 = ( c b ’1 )* = b ’1 c b2 = c x


The triangles DBC and CBA are also oppositely similar, because the angle DBC is
common and the other one is a right angle. Hence:

a ( c ’ x ) ’1 = ( c a ’1 )* = a ’1 c a2 = c ( c ’ x )


Summing both results the Pythagorean theorem is obtained:

a2 + b2 = c ( c ’ x ) + c x = c2

7.3 First at all we must see that the triangles ABM and BCM are oppositely similar. Firstly,
they share the angle BMC. Secondly, the angles MBC and BAM are equal because they
embrace the same arc BC. In fact, the limiting case of the angle BAC when A moves to B
is the angle MBA. Finally the angle BCM is equal to the angle ABM because the sum of
the angles of a triangle is π. The opposite similarity implies:

MA AB ’1 = BC ’1 MB MA = BC ’1 MB AB

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 227


MB AB ’1 = BC ’1 MC MC ’1 = AB MB ’1 BC ’1


Multiplying both expressions we obtain:

MA MC ’1 = BC ’1 MB AB2 MB ’1 BC ’1 = AB2 BC-2

7.4 Let Q be the intersection point of the segments PA and BC. The triangle PQC is
directly similar to the triangle PBA and oppositely similar to the triangle BQA:

BA PA ’1 = QC PC ’1 ’ QC = BA PA ’1 PC ’ ¦QC¦ = ¦BA¦ ¦PA¦’1 ¦PC¦

BQ BA ’1 = PA ’1 PB ’ BQ = PA ’1 PB BA ’ ’1
¦BQ¦ = ¦PA¦ ¦PB¦ ¦BA¦

Summing both expressions: ¦BC¦ = ¦BQ¦ + ¦QC¦ = ¦BA¦ ( ¦PB¦ + ¦PC¦ ) ¦PA¦ ’1

The three sides of the triangle are equal; therefore: ¦PA¦ = ¦PB¦ + ¦PC¦

7.5 Let us firstly calculate the homothety ratio k, which is the quotient of homologous
sides and the similarity ratio of both triangles:

AB ’1 A'B' = k

Secondly we calculate the centre of the homothety by isolation from:

OA' = OA k

OA' ’ OA = OA ( 1 ’ k )

AA' ( 1 ’ k ) ’1 = OA

Since the segment AA' is known, this equation allows us to calculate the centre O :

O = A ’ AA' ( 1 ’ k ) ’1 = A ’ AA' ( 1 ’ AB ’1 A'B' ) ’1

7.6 Draw the line passing through P and the centre
Figure 16.2
of the circle. This extended diameter cuts the circle
in the points R and R'. See that the angles R'RQ and
QQ'R' are supplementary because they intercept
opposite arcs of the circle. Then the angles PRQ
and R'Q'P are equal. Therefore the triangle QRP
and R'Q'P are oppositely similar and we have:

PR ’1 PQ = PR' PQ' ’1 ’ PQ PQ' = PR PR'

Since PR and PR' are determined by P and the
circle, the product PQ PQ' (the power of P) is constant independently of the line PQQ'.

7.7 The bisector d of the angle ab divides the triangle abc in two triangles, which are
RAMON GONZALEZ CALVET
228

obviously not similar! However we may apply the law of sines to both triangles to find:

n b
m a
= =
sin ad sin dm sin db sin nd

The angles nd and dm are supplementary and sin nd = sin dm. On the other hand, the
angles ad and db are equal because of the angle bisector. Therefore it follows that:

m n
=
a b

8. Properties of the triangles

8.1 Let A, B and C be the vertices of the given triangle with anticlockwise position. Let S,
T and U be the vertices of the three equilateral triangles drawn over the sides AB, BC and
CA respectively. Let P, Q and R be the centres of the triangles ABS, BCT and CAU
respectively. The side CU is obtained from AC through a rotation of 2π/3:

2π 2π
t = 12π / 3 = cos + e12 sin
CU = AC t with
3 3

CR is 2/3 of the altitude of the equilateral triangles ACU ; therefore is 1/3 of the
diagonal of the parallelogram formed by CA and CU:

CA + CU CA ( 1 ’ t )
CR = =
3 3

The same argument applies to the other equilateral triangles:

BC ( 1 ’ t ) AB ( 1 ’ t )
BQ = AP =
3 3

From where P, Q and R as functions of A, B and C are obtained:

AB ( 1 ’ t ) BC ( 1 ’ t ) CA ( 1 ’ t )
P = A+ Q=B+ R=C +
3 3 3

Let us calculate the vector PQ:

( C ’ B ’ B + A)( 1 ’ t )
PQ = Q ’ P = B ’ A +
3
A+ B +C
G=
Introducing the centroid:
3

PQ = AB + BG ( 1 ’ t ) = AG “ BG t

Analogously:
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 229


QR = BG “ CG t RP = CG “ AG z

Now we apply a rotation of 2π/3 to PQ in order to obtain QR:

PQ t = ( AG “ BG t ) t = AG t “ BG t2

Since t is a third root of the unity (1 + t + t2 = 0) then:

PQ t = AG t + BG + BG t = AG t + BG t + CG t + BG ’ CG t

= 3 GG + BG “ CG t = BG “ CG t = QR

Through the same way we find:

QR t = RP RP t = PQ

Therefore P, Q and R form an equilateral triangle with centre in G, the centroid of the
triangle ABC:

P + Q + R A + B + C AB + BC + CA
(1 ’ t ) = G
= +
3 3 3

8.2 The substitution of the centroid G of the triangle ABC gives:

(A + B + C )
2 2
A+ B + C
®
= 3 P 2 ’ 2 P · (A + B + C ) +
3 PG = 3  P ’
2

3 3
° »
A2 + B 2 + C 2 + 2 A · B + 2 B · C + 2 C · A
= 3 P ’ 2 P · (A + B + C ) +
2

3
2 (’ A ’ B 2 ’ C 2 + A · B + B · C + C · A)
[ ]
2
= ( A ’ P ) + (B ’ P ) + (C ’ P ) +
2 2 2

3
(B ’ A) + (C ’ B ) + ( A ’ C )
2 2 2
= PA 2 + PB 2 + PC 2 ’
3
AB + BC + CA 2
2 2
= PA + PB + PC ’
2 2 2

3

8.3 a) Let us calculate the area of the triangle GBC:

GB § BC = [ ’ a A + ( 1 ’ b ) B ’ c C ] § BC = [ ’ a A + ( a + c ) B ’ c C ] § BC =

GB § BC
a=
= ( a AB + c CB ) § BC = a AB § BC ’
AB § BC

The proof is analogous for the triangles GCA and GAB.

b) Let us develop PG2 following the same way as in the exercise 8.2:
RAMON GONZALEZ CALVET
230


PG2 = [P ’ ( a A + b B + c C ) ]2 = P2 ’ 2 P · ( a A + b B + c C ) + ( a A + b B + c C )2

= ( a + b + c ) P2 ’ 2 a P · A ’ 2 b P · B ’ 2 c P · C + a2 A2 + b2 B2 + c2 C 2 +

+2abA·B+2bcB·C+2caC·A

= a (A ’ P )2 + b ( B ’ P )2 + c ( C ’ P )2 + a ( a ’ 1 ) A2 + b ( b ’ 1) B2

+ c ( c ’ 1 ) C2 + 2 a b A · B + 2 b c B · C + 2 c a C · A

= a PA2 + b PB2 + c PC2 ’ a ( b + c ) A2 ’ b ( c + a ) B2 ’ c ( a + b ) C 2

+2abA·B+2bcB·C+2caC·A

= a PA2 + b PB2 + c PC2 ’ a b ( B ’ A )2 ’ b c ( C ’ B )2 ’ c a ( A ’ C )2

= a PA2 + b PB2 + c PC2 ’ a b AB2 ’ b c BC2 ’ c a CA2

The Leibniz™s theorem is a particular case of the Apollonius™ lost theorem for a = b = c =
1/3.

8.4 Let us consider the vertices A, B, C and D ordered clockwise on the perimeter. Since
AP is AB turned π/3, BQ is BC turned π/3, etc, we have:

z = cos π/3 + e12 sin π/3

AP = AB z BQ = BC z CR = CD z DS = DA z

PR · QS = ( PA + AC + CR ) · ( QB + BD + DS )

= [ ( CD ’ AB ) z + AC ] · [ ( DA ’ BC ) z + BD ]

= [ ( ’AC + BD ) z + AC ] · [ ( ’AC ’ BD ) z + BD ] =

= [ ( ’AC + BD ) z ] · [ ( ’AC ’ BD ) z ] + [ (’AC + BD ) z ] · BD

+ AC · [ ( ’AC ’ BD ) z + BD ]

The inner product of two vectors turned the same angle is equal to that of these
vectors before the rotation. We use this fact for the first product. Also we must develop
the other products in geometric products and permute vectors and the complex number z:
z + z*
PR · QS = ( ’AC + BD ) · ( ’AC ’ BD ) + ( ’AC BD ’ BD AC ’ AC2 + BD2 )
2
AC 2 ’ BD 2
+ AC · BD =
2

Therefore ¦AC¦ = ¦BD¦ ” PR · QS = 0. The statement b) is proved through an
analogous way.
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 231


8.5 The median is the segment going from a vertex to the midpoint of the opposite side:
2 2
«B+C AB 2 AC 2 AB · AC
 « AB AC 
m =¬ ’ A· = ¬ + ·= + +
2
A
2 2 2 4 4 2


AB 2 AC 2 2 AB · AC ’ AB 2 ’ AC 2 AB 2 AC 2 BC 2
= + + = + ’
2 2 4 2 2 4

8.6 If E is the intersection point of the bisector of B with the line parallel to the bisector of
A, then the following equality holds:

« BA BC  « AB AC 
E =B+b¬ ·=C +a¬ ·
+ +
¬ BA · ¬ AB AC ·
BC 
  

where a, b are real. Arranging terms we obtain a vectorial equality:

« AB AC  « BA BC 
’ a¬ ·+b¬ · = BC
+ +
¬ AB · ¬ BA BC ·
AC 
  

The linear decomposition yields after simplification:

BC CA
a=’
AB + BC + CA

In the same way, if D is the intersection of the bisector of A with the line parallel to the
bisector of B we have:

« AB AC  « BA BC 
D = A+c¬ ·=C +d ¬ ·
+ +
¬ AB AC · ¬ BA BC ·
   

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