ñòð. 10 |

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 ) âˆ’ u1 u2 âˆ’ v1 v2 âˆ’ 2 v1 u2 = u1 v2 âˆ’ 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 ) âˆ§ ( âˆ’2 e1 âˆ’ 3 e2 ) = ( âˆ’9 + 10 ) e12 = e12 â‡’ ï£¦Aï£¦ = 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 âˆ’ a2 b1 ) ] ( c1 e1 + c2 e2 )

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

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

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

2.4

u v = 325 Î±(u, v) =1.2315=70Âº33'36"

2.5 For this base we have:

3

e1 Â· e2 = | e1 | | e2 | cos Ï€/3 = 1 | e1 âˆ§ e2 | = | e1 | | e2 | | sin Ï€/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 Â· e2 = 4 + 36 + 12 = 52 â‡’

v = 49

v2 = (âˆ’3 e1 + 4 e2 )2 = 9 e12 + 16 e22 âˆ’ 24 e1 Â· e2 = 9 + 64 âˆ’24 = 49 â‡’

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

u v = 2548

= 42 âˆ’ e1 Â· e2 + 17 e1 âˆ§ e2 = 41 + 17 e1 âˆ§ e2 â‡’

uâˆ§v

uÂ·v 41 17 3

cos Î± = sinÎ± = Â±

= =

uv uv

2548 2548

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

When e1 and e2 are not perpendicular, e1 e2 â‰ âˆ’ e2 e1 and e12 has not the meaning of a pure

area but the outer product e1 âˆ§ 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 (âˆ’ 3) âˆ’ (âˆ’ 5) 5 u1 âˆ§ v 2 (âˆ’ 5) âˆ’ (âˆ’ 1) 3

v âˆ§ u2

c1 = = = âˆ’16 c2 = = =7

u1 âˆ§ u 2 âˆ’1 u1 âˆ§ u 2 âˆ’1

so in the new base v = (âˆ’16, 7).

3. The complex numbers

3.1 z t = ( 1 + 3 e12 ) ( âˆ’2 + 2 e12 ) = âˆ’2 âˆ’6 + ( 2 âˆ’6 ) e12 = âˆ’8 âˆ’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

â‡’ |z|=|a||b| â‡’ ï£¦zï£¦ = a b = a b b a = z z*

z=ab

4

3.3 The equation x âˆ’1 = 0 is solved by extraction of the fourth roots:

x2 = 1 â‡’ x1 = 1, x2 = âˆ’1

4

x =1

x2 = âˆ’1 â‡’ x3 = e12 , x4 = âˆ’e12

3.4 Passing to the polar form we have:

( ) ( )

n n

= âˆ’ 2 âˆ’Ï€ / 4

2Ï€ /4

= âˆ’ 2n = 2n

2n 2n

â‡’

âˆ’ n Ï€/4 âˆ’ n Ï€/4 +Ï€

nÏ€ /4 n Ï€/4

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

nÏ€

nÏ€ nÏ€

=âˆ’ +Ï€ + 2 k Ï€ â‡’ =Ï€ + 2 k Ï€ â‡’ n=2+4k

4 4 2

3 3 3

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

3.6 Using the formula of the equation of second degree we find: z1 = 2âˆ’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 :

âˆ‚a âˆ‚b

b = cos x âˆ« K ( y ) dy

= cos x K ( y ) = â‡’

âˆ‚x âˆ‚y

Applying the second Cauchy-Riemann condition:

âˆ‚a âˆ‚b

= sin x âˆ« K ( y ) dy

= sin x K' ( y ) = âˆ’

âˆ‚y âˆ‚x

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

K' ( y ) = âˆ« K ( y ) dy ï£¼

ï£½ â‡’ K(y) = cosh y

K (0) = 1 ï£¾

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( âˆ’ e12 z ) exp(âˆ’ y )(cos x + e12 sin x ) + exp( y )(cos x âˆ’ e12 sin x )

cos z = =

2 2

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

3.8 Let us calculate some derivatives at z = 0:

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

âˆ’ exp(âˆ’ z ) âˆ’1 1

f' (z ) = f' (0) = âˆ’

=

1 + exp(âˆ’ 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

âˆ’

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

2

exp(z ) 2 exp(2 z ) 6 exp(3z )

(0) = âˆ’ 1

4 exp( 2 z )

(z ) = âˆ’ âˆ’ + 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 âˆ’ + âˆ’ + ...

248

Observe that f (e12 Ï€ ) = log(0) is divergent. This is the singularity nearest the origin.

Therefore the radius of convergence of the series is Ï€.

3.9 Separating fractions:

1 1 1

f (z ) = = âˆ’

z 2 + 2 z âˆ’ 8 6 (z âˆ’ 2 ) 6 (z + 4 )

and developing the second fraction:

RAMON GONZALEZ CALVET

212

1

1ï£« ï£¶

2 3

z âˆ’ 2 ï£« z âˆ’ 2ï£¶ ï£« z âˆ’ 2ï£¶

1 1 6 ï£¬1 âˆ’ ï£· + ... ï£·

= = = +ï£¬ ï£· âˆ’ï£¬

z âˆ’2 6ï£¬ ï£·

z+4 6+ zâˆ’2 6 ï£6ï£¸ ï£6ï£¸

ï£ ï£¸

1+

6

n (z âˆ’ 2 )

n

âˆž

1

f (z ) = âˆ’ âˆ‘ (âˆ’ 1)

we find:

6( z âˆ’ 2 ) n = 0 6 n+2

This Lauren series is convergent in the annulus 0 <ï£¦z âˆ’ 2ï£¦< 6, which contains the

required annulus 1 <ï£¦z âˆ’ 2ï£¦< 4 .

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

is:

âˆž

1 1 1

f (z ) = âˆ‘ n = âˆ’1=

n =1 4 ( z + 1) 4z +3

1

n

1âˆ’

4 (z + 1)

Now we see that at z = âˆ’3/4 the function has a pole. Therefore the radius of convergence

of this series (centred at z = âˆ’1) is ï£¦âˆ’3/4 âˆ’( âˆ’1)ï£¦=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

âˆž

z3 z5

âˆ’ ... = âˆ‘ (âˆ’ 1)

n

sin z = z âˆ’ +

(2 n + 1)!

3! 5 ! n =0

so the Lauren series for the given function is:

z 2 n âˆ’1

sin z âˆž 1 z z3

= âˆ‘ (âˆ’ 1)

n

=âˆ’+ âˆ’ ...

(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<ï£¦zï£¦< âˆž .

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 â†’ df = da e1 + db e 2 dz = dx + dy e12 â†’ dz = dx e1 + dy e 2

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

df1 df2 = dz1 f'(z) dz2 fâ€™(z) = dz1 dz2 [fâ€™(z)]* f'(z) =ï£¦f'(z)ï£¦2 dz1 dz2

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 213

Sinceï£¦f'(z)ï£¦2 is real, both geometric products have the same argument, Î±(df1, df2) = Î±(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 âˆ’1 v d

If v'' is obtained from v' by a rotation through an angle Î±, and z is a unitary complex

number with argument Î± /2 then:

Î± Î±

z = cos + e12 sin

v'' = z âˆ’1 v' z

2 2

Joining both equations:

v'' = z âˆ’1 d âˆ’1 v d z = c âˆ’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 âˆ’1 w' v = v âˆ’1 u âˆ’1 w u v

Then we can write:

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

â‡’ Bâˆ’A=Câˆ’D â‡’ D=Câˆ’B+A

AB = DC

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

ï£¦AB âˆ§ BCï£¦ = ï£¦( 2 e1 âˆ’7 e2 ) âˆ§ ( âˆ’2 e1 âˆ’2 e2 )ï£¦ = ï£¦âˆ’18 e12ï£¦ = 18

The area is:

5.2 The Eulerâ€™s theorem:

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

=CAâˆ’AC+ABâˆ’BA+BCâˆ’CB=2(Câˆ§A+Aâˆ§B+Bâˆ§C)=

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

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

AB âˆ§ AC = 4 e 2 âˆ§ 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 Â· e2 = 4 + 16 + 16 cos Ï€/3 = 28

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

28

d( A, B ) = ï£º ABï£º = d( B, C ) =ï£º BCï£º = 4 d( C, A) = ï£º CAï£º = 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 Â· BC + 2 AB Â· CD + 2 BC Â· CD

= AB2 + BC2 + CD2 + 2 AB Â· BC âˆ’ 2ï£¦ABï£¦ï£¦CDï£¦+ 2 BC Â· 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 âˆ’ AB2 âˆ’ BC2 âˆ’ CD2 + 2 ï£¦ABï£¦ï£¦CDï£¦ = 2 AB Â· BC + 2 BC Â· CD

AD2 âˆ’ AB2 âˆ’ BC2 âˆ’ CD2 + 2 ï£¦ABï£¦ï£¦CDï£¦ = 2 BC Â· ( AB + CD )

Without loss of generality we will suppose that ï£¦ABï£¦>ï£¦CDï£¦:

ï£¦AB + CDï£¦ = ï£¦ABï£¦âˆ’ï£¦CDï£¦

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

have:

AD2 âˆ’AB2 âˆ’BC2 âˆ’CD2 + 2 ï£¦ABï£¦ï£¦CDï£¦ = âˆ’2 ï£¦BCï£¦ ( ï£¦ABï£¦âˆ’ï£¦CDï£¦ ) cos ABC

whence we obtain the angle ABC:

âˆ’ AD 2 + AB 2 + BC 2 + CD 2 âˆ’ 2 AB CD

cos ABC =

( )

AB âˆ’ CD

2 BC

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

2 ï£¦BCï£¦ ( ï£¦ABï£¦âˆ’ï£¦CDï£¦ ) > âˆ’AD2 + AB2 + BC2 + CD2 âˆ’2 ï£¦ABï£¦ï£¦CDï£¦ >

> âˆ’2 ï£¦BCï£¦ (ï£¦ABï£¦âˆ’ï£¦CDï£¦)

R=(1âˆ’pâˆ’q)O+pP+qQ â‡’ RP = ( 1 âˆ’ p âˆ’ q ) OP + q QP

5.5

â‡’ RP âˆ§ PQ = ( 1 âˆ’ p âˆ’ q ) OP âˆ§ PQ

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

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

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

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

(âˆ’ e1 + 3 e 2 ) âˆ§ (3 e1 + e 2 )

AC âˆ§ 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 â‹… AC AB âˆ§ AC

cos Î± = e12 sinÎ± =

=0 = e12

AB AC AB AC

Therefore, Î± = Ï€/2. The angle between two lines is always comprised from âˆ’Ï€/2 to Ï€/2

RAMON GONZALEZ CALVET

216

because a rotation of 2Ï€ around the intersection point does not alter the lines. When the

angle exceeds these boundaries, you may add or subtract Ï€.

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 âˆ’ xD âˆ’ yD ) O + xD P + yD Q

E = (1 âˆ’ xE âˆ’ yE ) O + xE P + yE Q

F = (1 âˆ’ xF âˆ’ yF ) O + xF P + yF Q

1 âˆ’ xD âˆ’ yD xD yD

1 âˆ’ xE âˆ’ yE yE = 0

xE

1 âˆ’ xF âˆ’ 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 âˆ’ x âˆ’ y ) O + x P + y Q

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

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

= ( x ( p1 âˆ’ o1 ) + y (q1 âˆ’ o1 ) + o1 , x ( p2 âˆ’ 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 âˆ’ o1 ) + y ( q1 âˆ’ o1 ) + o1

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

In matrix form:

ï£« x' ï£¶ ï£« p1 âˆ’ o1 q1 âˆ’ o1 ï£¶ ï£« x ï£¶ ï£« o1 ï£¶

ï£¬ ï£·=ï£¬ ï£·ï£¬ ï£· + ï£¬ ï£·

ï£¬ y' ï£· ï£¬ p âˆ’ o q2 âˆ’ o2 ï£· ï£¬ y ï£· ï£¬ o2 ï£·

ï£ ï£¸ï£2 ï£¸ï£ ï£¸ ï£ ï£¸

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 âˆ’ x A âˆ’ y A ) O + x A P + y AQ

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

C = ( 1 âˆ’ xC âˆ’ y C ) O + xC P + yC Q

In matrix form:

ï£« A ï£¶ ï£«1 âˆ’ x A âˆ’ y A y A ï£¶ ï£«O ï£¶

xA

ï£¬ï£·ï£¬ ï£·ï£¬ ï£·

ï£¬ B ï£· = ï£¬1 âˆ’ x B âˆ’ y B xB yB ï£· ï£¬ P ï£·

ï£¬ C ï£· ï£¬1 âˆ’ x âˆ’ y yC ï£· ï£¬ Q ï£·

xC

ï£ï£¸ï£ ï£¸ï£ ï£¸

C C

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

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

In matrix form:

ï£«O ï£¶ ï£« Aï£¶

ï£¬ï£· ï£¬ï£·

D = (1 âˆ’ xD âˆ’ yD y D )ï£¬ P ï£· = ( 1 âˆ’ b âˆ’ c b c)ï£¬ B ï£·

xD

ï£¬Q ï£· ï£¬C ï£·

ï£ï£¸ ï£ï£¸

ï£«1 âˆ’ x A âˆ’ y A yA ï£¶

xA

ï£¬ ï£·

(1 âˆ’ xD âˆ’ yD y D ) = ( 1 âˆ’ b âˆ’ c b c ) ï£¬1 âˆ’ x B âˆ’ y B

xD xB yB ï£·

ï£¬1 âˆ’ x âˆ’ y yC ï£·

xC

ï£ ï£¸

C C

which leads to the following system of equations:

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

ï£²

ï£³ y D = ( 1 âˆ’ b âˆ’ 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 âˆ’ b âˆ’ 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 â‡’ G = D âˆ’ 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' âˆ’ E' + F' â‡’ D'E' = G'F'

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

DE DF âˆ’1 = r â‡’ â‡’ E=(1âˆ’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:

âˆ’1

E' = ( 1 âˆ’ r ) D' + r F' â‡’ 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 âˆ’ xD âˆ’ yD ) O + xD P + yD Q

E = (1 âˆ’ xE âˆ’ yE ) O + xE P + yE Q

F = (1 âˆ’ xF âˆ’ yF ) O + xF P + yF Q

1 âˆ’ xD âˆ’ yD xD yD

1 âˆ’ xE âˆ’ yE yE = 0

xE

1 âˆ’ xF âˆ’ 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 âˆ’x âˆ’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 âˆ’ x âˆ’ y ) O + x P + y Q

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

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

= [ x ( p1 âˆ’ o1 ) + y (q1 âˆ’ o1 ) + o1 , x ( p2 âˆ’ o2 ) + y ( q2 âˆ’ 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 âˆ’ o1 ) + y ( q1 âˆ’ o1 ) + o1

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

In matrix form:

ï£® x ' ï£¹ ï£® p1 âˆ’ o1 q1 âˆ’ o1 ï£¹ ï£® x ï£¹ ï£® o1 ï£¹

ï£¯ y 'ï£º = ï£¯ p âˆ’ o ï£¯ y ï£º + ï£¯o ï£º

q2 âˆ’ o2 ï£º

ï£°ï£» ï£°2 ï£° ï£» ï£° 2ï£»

ï£»

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 âˆ’ x A âˆ’ y A ) O + x A P + y AQ

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

C = ( 1 âˆ’ xC âˆ’ y C ) O + xC P + yC Q

In matrix form:

ï£® Aï£¹ ï£®1 âˆ’ x A âˆ’ y A y A ï£¹ ï£®O ï£¹

xA

ï£¯ B ï£º = ï£¯1 âˆ’ x âˆ’ y y B ï£º ï£¯Pï£º

xB

ï£¯ï£º ï£¯ ï£ºï£¯ ï£º

B B

ï£¯C ï£º ï£¯1 âˆ’ x C âˆ’ y C y C ï£º ï£¯Q ï£º

xC

ï£°ï£» ï£° ï£»ï£° ï£»

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

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

In matrix form:

ï£®O ï£¹ ï£® Aï£¹

y D ] ï£¯ P ï£º = [1 âˆ’ b âˆ’ c b c ] ï£¯ B ï£º

D = [1 âˆ’ x D âˆ’ y D xD

ï£¯ï£º ï£¯ï£º

ï£¯Q ï£º ï£¯C ï£º

ï£°ï£» ï£°ï£»

ï£®1 âˆ’ x A âˆ’ y A yAï£¹

xA

y D ] = [1 âˆ’ b âˆ’ c b c ] ï£¯1 âˆ’ x B âˆ’ y B yB ï£º

[1 âˆ’ x D âˆ’ y D xD xB

ï£¯ ï£º

ï£¯1 âˆ’ x C âˆ’ y C yC ï£º

xC

ï£° ï£»

RAMON GONZALEZ CALVET

220

which leads to the following system of equations:

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

ï£²

ï£³ y D = (1 âˆ’ b âˆ’ 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 âˆ’ b âˆ’ 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 â‡’ G = D âˆ’ 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' âˆ’ E' + F' â‡’ D'E' = G'F'

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

DE DF âˆ’1 = r â‡’ â‡’ E=(1âˆ’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:

âˆ’1

E' = ( 1 âˆ’ r ) D' + r F' â‡’ 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 âˆ’ x A âˆ’ 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 = ( âˆ’1 + 2 x A + y A ) v P + ( âˆ’ 1 + x A + 2 y A ) v Q

And analogously:

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

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

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

The outer product is:

v A âˆ§ v C = ( x C âˆ’ x A âˆ’ ( y C âˆ’ y A ) + 3 ( x A y C âˆ’ x C y A )) v P âˆ§ 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 âˆ§ vC v B âˆ§ v D

( ABCD ) = =

v A âˆ§ v D v B âˆ§ vC

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

=

( x D âˆ’ x A âˆ’ ( y D âˆ’ y A ) + 3 ( x A y D âˆ’ x D y A )) ( x C âˆ’ x B âˆ’ ( y C âˆ’ y B ) + 3 (x B y C âˆ’ 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:

ï£® 1ï£¹ ï£® 1ï£¹

1 1

v A âˆ§ v C = 3 ï£¯ x A âˆ’ , y A âˆ’ ï£º âˆ§ ï£¯ x C âˆ’ , y C âˆ’ ï£º = 3 LA âˆ§ LC

3 3ï£» ï£° 3 3ï£»

ï£°

where LA = A âˆ’ 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 âˆ§ LC LB âˆ§ LD

LA âˆ§ LD LB âˆ§ 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 â€“ y â€“ 1 â‰¡ a' ( â€“ x â€“ y + 1) + b' x + c' y â‡’ a' = âˆ’1 c' = âˆ’2

b' = 0

1 2

a= b=0 c=

â‡’ â‡’ [0, 2/3]

3 3

For the second line:

x â€“ y + 3 â‰¡ a' ( â€“ x â€“ y + 1) + b' x + c' y â‡’ a' = 3 b' = 4 c' = 2

3 4 2

a= b= c=

â‡’ â‡’ [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 â€“ 2 = 0 and y â€“ 1 = 0, whose dual

coordinates are:

â‡’

xâ€“2=0 [1/5, 2/5]

â‡’

yâ€“1=0 [1/2, 0]

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

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

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

b âˆ’ 1/ 2 c

= 4 b + 3c = 2

â‡”

âˆ’4

3

The point (â€“3, â€“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

â‡’

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

â‡’

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] â€“ [4/10, 3/10] = [â€“3/20, 4/20]

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

b âˆ’ 1/ 4 c âˆ’ 2 / 4

8b + 6c = 5

= â‡”

âˆ’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 Â· b

a+b+c=0 â‡’ c=âˆ’aâˆ’b â‡’

c2 = a2 + b2 + 2 ï£¦aï£¦ ï£¦bï£¦ cos(Ï€ âˆ’ Î³ ) = a2 + b2 âˆ’ 2 ï£¦aï£¦ï£¦bï£¦ cos Î³ (law of

cosines)

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

2s=aâˆ§b=bâˆ§c=câˆ§a

ï£¦aï£¦ï£¦bï£¦ sin(Ï€ âˆ’ Î³ ) e12 = ï£¦bï£¦ï£¦cï£¦ sin(Ï€ âˆ’ Î±) e12 = ï£¦cï£¦ï£¦aï£¦ sin(Ï€ âˆ’ Î² ) e12

sinÎ³ sinÎ± sinÎ²

= = (law of sines)

c a b

From here one quickly obtains:

a+b aâˆ’b

sinÎ± + sinÎ² sinÎ± âˆ’ sinÎ²

= =

sinÎ± sinÎ²

a b

By dividing both equations and introducing the identities for the addition and subtraction

of sines we arrive at:

RAMON GONZALEZ CALVET

224

Î±+Î² Î± âˆ’Î²

2 sin

cos

a + b sinÎ± + sinÎ² 2 2

= =

Î±+Î² Î± âˆ’Î²

a âˆ’ b sinÎ± âˆ’ sinÎ²

2 cos sin

2 2

Î±+Î²

tg

a+b 2

= (law of tangents)

Î±âˆ’Î²

aâˆ’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:

Î² âˆ’ 2Î³ + Î±

Î±âˆ’Î² Î² âˆ’Î±

cos(Î± âˆ’ Î² ) + cos(Î² âˆ’ Î³ ) + cos(Î³ âˆ’ Î± ) = 2 cos 2 âˆ’ 1 + cos cos

2 2 2

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

Î±âˆ’Î² Î±âˆ’Î² Î± âˆ’ 2Î³ + Î²

ï£« ï£¶

= 2 cos + cos ï£· âˆ’1

ï£¬ cos

2 2 2

ï£ ï£¸

Î±âˆ’Î² Î± âˆ’Î³ Î³ âˆ’Î²

= 2 cos âˆ’1

cos cos

2 2 2

Î±âˆ’Î² Î² âˆ’Î³ Î³ âˆ’Î±

= 2 cos âˆ’1

cos cos

2 2 2

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

6.3 Using the De Moivreâ€™s identity:

cos 4Î± + e12 sin 4Î± â‰¡ (cos Î± + e12 sin Î± )

4

After developing the right hand side we find:

sin 4Î± â‰¡ 4 cos3Î± sinÎ± âˆ’ 4 cosÎ± sin3Î± cos 4Î± â‰¡ cos4Î± âˆ’ 6 cos2Î± sin2Î± + sin4Î±

4 tg Î± âˆ’ 4 tg 3 Î±

And dividing both identities: tg 4Î± =

1 âˆ’ 6 tg 2 Î± + tg 4 Î±

6.4 Let Î±, Î² and Î³ be the angles of the triangle PAB with vertices A, B and P respectively.

The angle Î³ embracing the arc AB is constant for any point P on the arc AB. By the law of

sines we have:

PA PB PC

= =

sinÎ± sinÎ² sinÎ³

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 225

The sum of both chords is:

sinÎ± + sinÎ²

PA + PB = PC

sinÎ³

Converting the sum of sines into a product we find:

Î±+Î² Î±âˆ’Î² Ï€ âˆ’Î³ Î±âˆ’Î²

2 sin cos 2 sin cos

2 2 = PC 2 2

PA + PB = PC

sinÎ³ sinÎ³

Since Î³ is constant, the maximum is attained for cos(Î± /2 âˆ’ Î² /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:

Î±+Î² Î±âˆ’Î² Î±âˆ’Î²

2 sin cos cos

a+b sinÎ± + sinÎ² 2 2= 2

= =

Î³ Î³ Î³

sinÎ³

c

2 sin cos sin

2 2 2

Following the same way, we also have:

Î±+Î² Î±âˆ’Î² Î±+Î²

2 cos sin sin

aâˆ’b sinÎ± âˆ’ sinÎ² 2 2= 2

= =

Î³ Î³ Î³

sinÎ³

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 Î³ + c sinÎ² and so forth.

6.7 From the double angle identities we have:

Î± Î± Î± Î±

cos Î± â‰¡ cos 2 âˆ’ sin 2 â‰¡ 1 âˆ’ 2 sin 2 â‰¡ 2 cos 2 âˆ’ 1

2 2 2 2

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

Î± 1 âˆ’ cos Î± Î± 1 + cos Î±

â‰¡Â± â‰¡Â±

sin cos

2 2 2 2

Making the quotient:

RAMON GONZALEZ CALVET

226

1 âˆ’ cos Î± 1 âˆ’ cos Î± sinÎ±

tg Î± â‰¡ Â± â‰¡ â‰¡

1 + cos Î± sinÎ± 1 + cosÎ±

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 = âˆ’2 e1 + e2 RP = âˆ’e2

PQ = 2 e1

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

Q'R' = 3 e1 + e2

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

2 and the angle between the directions of homologous sides is

The size ratio is ï£¦rï£¦ =

5Ï€/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 âˆ’ 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 âˆ’1 = ( c b âˆ’1 )* = b âˆ’1 c b2 = c x

â‡’

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

â‡’

Summing both results the Pythagorean theorem is obtained:

a2 + b2 = c ( c âˆ’ 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 Ï€. The opposite similarity implies:

MA AB âˆ’1 = BC âˆ’1 MB MA = BC âˆ’1 MB AB

â‡’

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 227

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

â‡’

Multiplying both expressions we obtain:

MA MC âˆ’1 = BC âˆ’1 MB AB2 MB âˆ’1 BC âˆ’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 âˆ’1 = QC PC âˆ’1 â‡’ QC = BA PA âˆ’1 PC â‡’ ï£¦QCï£¦ = ï£¦BAï£¦ ï£¦PAï£¦âˆ’1 ï£¦PCï£¦

BQ BA âˆ’1 = PA âˆ’1 PB â‡’ BQ = PA âˆ’1 PB BA â‡’ âˆ’1

ï£¦BQï£¦ = ï£¦PAï£¦ ï£¦PBï£¦ ï£¦BAï£¦

Summing both expressions: ï£¦BCï£¦ = ï£¦BQï£¦ + ï£¦QCï£¦ = ï£¦BAï£¦ ( ï£¦PBï£¦ + ï£¦PCï£¦ ) ï£¦PAï£¦ âˆ’1

The three sides of the triangle are equal; therefore: ï£¦PAï£¦ = ï£¦PBï£¦ + ï£¦PCï£¦

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 âˆ’1 A'B' = k

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

OA' = OA k

OA' âˆ’ OA = OA ( 1 âˆ’ k )

AA' ( 1 âˆ’ k ) âˆ’1 = OA

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

O = A âˆ’ AA' ( 1 âˆ’ k ) âˆ’1 = A âˆ’ AA' ( 1 âˆ’ AB âˆ’1 A'B' ) âˆ’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 âˆ’1 PQ = PR' PQ' âˆ’1 â‡’ 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 2Ï€/3:

2Ï€ 2Ï€

t = 12Ï€ / 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 âˆ’ t )

CR = =

3 3

The same argument applies to the other equilateral triangles:

BC ( 1 âˆ’ t ) AB ( 1 âˆ’ t )

BQ = AP =

3 3

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

AB ( 1 âˆ’ t ) BC ( 1 âˆ’ t ) CA ( 1 âˆ’ t )

P = A+ Q=B+ R=C +

3 3 3

Let us calculate the vector PQ:

( C âˆ’ B âˆ’ B + A)( 1 âˆ’ t )

PQ = Q âˆ’ P = B âˆ’ A +

3

A+ B +C

G=

Introducing the centroid:

3

PQ = AB + BG ( 1 âˆ’ t ) = AG â€“ BG t

Analogously:

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 229

QR = BG â€“ CG t RP = CG â€“ AG z

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

PQ t = ( AG â€“ BG t ) t = AG t â€“ 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 âˆ’ CG t

= 3 GG + BG â€“ CG t = BG â€“ 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 âˆ’ 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 + Cï£¹

ï£®

= 3 P 2 âˆ’ 2 P Â· (A + B + C ) +

3 PG = 3 ï£¯ P âˆ’

2

ï£º

3 3

ï£° ï£»

A2 + B 2 + C 2 + 2 A Â· B + 2 B Â· C + 2 C Â· A

= 3 P âˆ’ 2 P Â· (A + B + C ) +

2

3

2 (âˆ’ A âˆ’ B 2 âˆ’ C 2 + A Â· B + B Â· C + C Â· A)

[ ]

2

= ( A âˆ’ P ) + (B âˆ’ P ) + (C âˆ’ P ) +

2 2 2

3

(B âˆ’ A) + (C âˆ’ B ) + ( A âˆ’ C )

2 2 2

= PA 2 + PB 2 + PC 2 âˆ’

3

AB + BC + CA 2

2 2

= PA + PB + PC âˆ’

2 2 2

3

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

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

GB âˆ§ BC

a=

= ( a AB + c CB ) âˆ§ BC = a AB âˆ§ BC â‡’

AB âˆ§ 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 âˆ’ ( a A + b B + c C ) ]2 = P2 âˆ’ 2 P Â· ( a A + b B + c C ) + ( a A + b B + c C )2

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

+2abAÂ·B+2bcBÂ·C+2caCÂ·A

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

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

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

+2abAÂ·B+2bcBÂ·C+2caCÂ·A

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

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

The Leibnizâ€™s theorem is a particular case of the Apolloniusâ€™ 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 Ï€/3, BQ is BC turned Ï€/3, etc, we have:

z = cos Ï€/3 + e12 sin Ï€/3

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

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

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

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

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

+ AC Â· [ ( âˆ’AC âˆ’ 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 Â· QS = ( âˆ’AC + BD ) Â· ( âˆ’AC âˆ’ BD ) + ( âˆ’AC BD âˆ’ BD AC âˆ’ AC2 + BD2 )

2

AC 2 âˆ’ BD 2

+ AC Â· BD =

2

Therefore ï£¦ACï£¦ = ï£¦BDï£¦ â‡” PR Â· 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

ï£«B+C AB 2 AC 2 AB Â· AC

ï£¶ ï£« AB AC ï£¶

m =ï£¬ âˆ’ Aï£· = ï£¬ + ï£·= + +

2

A

ï£2 ï£2 2ï£¸ 4 4 2

ï£¸

AB 2 AC 2 2 AB Â· AC âˆ’ AB 2 âˆ’ AC 2 AB 2 AC 2 BC 2

= + + = + âˆ’

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:

ï£« BA BC ï£¶ ï£« AB AC ï£¶

E =B+bï£¬ ï£·=C +aï£¬ ï£·

+ +

ï£¬ BA ï£· ï£¬ AB AC ï£·

BC ï£¸

ï£ ï£ ï£¸

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

ï£« AB AC ï£¶ ï£« BA BC ï£¶

âˆ’ aï£¬ ï£·+bï£¬ ï£· = BC

+ +

ï£¬ AB ï£· ï£¬ BA BC ï£·

AC ï£¸

ï£ ï£ ï£¸

The linear decomposition yields after simplification:

BC CA

a=âˆ’

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:

ï£« AB AC ï£¶ ï£« BA BC ï£¶

D = A+cï£¬ ï£·=C +d ï£¬ ï£·

+ +

ï£¬ AB AC ï£· ï£¬ BA BC ï£·

ï£ ï£¸ ï£ ï£¸

ñòð. 10 |