« AB AC « BA BC

’ c¬ ·+d ¬ · = CA

+ +

¬ AB AC · ¬ BA BC ·

After simplification we obtain:

BC CA

d =’

AB + BC + CA

Now we calculate the vector ED:

RAMON GONZALEZ CALVET

232

« 2 AB AC BC CA

BC

ED = D ’ E = ¬ ·

’ +

¬ AB AC · AB + BC + CA

BC

Then the direction of the line ED is given by the vector v:

«2 1 « 1

2 AB BC AC 1

= AB ¬ · + BC ¬ ’ ·

v= ’ + + +

¬ AB · ¬ BC AC ·

AB BC AC AC

When the vector v has the direction AB, the second summand vanishes, ¦AC¦ = ¦BC¦

and the triangle becomes isosceles.

8.7 Let us indicate the sides of the triangle ABC with a, b and c in the following form:

a = BC b = CA c = AB

Suppose without loss of generality that P lies on the side BC and Q on the side AC .

Hence:

P = k B + (1 ’ k ) C Q = l A + (1 ’ l ) C

CP = k CB = ’ k a CQ = l CA = l b

where k and l are real and 0 < k, l < 1. Now the segment PQ is obtained:

PQ = k a + l b

Since the area of the triangle CPQ must be the half of the area of the triangle ABC, it

follows that:

1

(’ k a ) § (l b ) = ’ 1 a § b 1

CP § CQ = CB § CA kl =

’ ’

2 2 2

The substitution into PQ gives:

1

PQ = k a + b

2k

a) If u is the vector of the given direction, PQ is perpendicular when PQ · u = 0 which

results in:

« 1

¬k a + b· · u = 0

2k

whence one obtains:

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 233

b·u b·u a ·u

k= ’ PQ = a ’ +b ’

2a ·u 2a ·u 2b ·u

In this case the solutions only exist when a · u and b · u have different signs. However we

can also choose the point P (or Q) lying on the another side c, what gives an analogous

solution containing the inner product c · u . Note that if a · u and b · u have the same sign,

then c · u have the opposite sign because:

’ c·u=’a·u’b·u

a+b+c=0

what warrants there is always a solution.

b) Let us calculate the square of PQ:

1

PQ 2 = k 2 a 2 + 2 b 2 + a … b

4k

By equating the derivative to zero, one obtains the value of k for which PQ2 is minimum:

b

k=

2a

Then we obtain the segment PQ and its length:

b a

PQ = a +b PQ = a b + a·b

2a 2b

c) Every point may be written as linear combination of the three vertices of the triangle

the sum of the coefficients being equal to the unity:

R = x A + y B + (1 ’ x ’ y )C ” CR = x CA + y CB

where all the coefficients are comprised between 0 and 1:

CR § CB CA § CR

x= y=

0 < x, y, 1 ’ x ’ y < 1

CA § CB CA § CB

Now the point R must lie on the segment PQ, that is, P, Q and R must be aligned. Then

the determinant of their coordinates will vanish:

1’ k

1’ k 0 k

0k

1 1

det (P, Q , R ) = l 0 1’ l = 0 1’ =0

2k 2k

x y 1’ x ’ y y 1’ x ’ y

x

RAMON GONZALEZ CALVET

234

1± 1’8x y 1 1m 1’8x y

2xk2 ’ k + y =0 k= l= =

’ and

4x 2k 4y

There is only solution for a positive discriminant:

1>8xy

The limiting curve is an equilateral hyperbola on the plane x-y. Since the triangle may be

obtained through an affinity, the limiting curve for the triangle is also a hyperbola

although not equilateral.

If the extremes P and Q could move along the prolongations of the sides of the

triangle without limitations, this would be the unique condition. However in this problem

the point P must lie between C and B, and Q must lie between C and A. It means the

additional condition:

1 1± 1’8x y

1 1

< k <1 < l <1 < <1

” ”

2 2 2 4x

From the first solution (root with sign +) we have:

2 x ’1< 1’ 8 x y < 4 x ’1

only existing for x >1/4. For 1/4 < x < 1/2 the left hand side is negative so that the unique

restriction is the right hand side. By squaring it we obtain:

1 1

1 ’ 8 x y < (4 x ’ 1)

2

<x<

0 < 2 x + y ’1

’ for

4 2

For x >1/2 the right hand side is higher than 1 so that the unique restriction is the left hand

side:

1

(2 x ’ 1)2 < 1 ’ 8 x y x + 2 y ’1< 0 x>

’ for

2

For the second solution (root with sign ’) we have:

2 x ’1< ’ 1’ 8 x y < 4 x ’1

only existing for x < 1/2. For 1/4 < x < 1/2 the right hand side is positive so that the

unique restriction is the left hand side:

1 1

(2 x ’ 1)2 > 1 ’ 8 x y ’ x + 2 y ’1> 0 <x<

for

4 2

For x < 1/4 both members are negative and after squaring we have:

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 235

1

(2 x ’ 1)2 > 1 ’ 8 x y > (4 x ’ 1)2 x + 2 y ’1> 0 > 2 x + y ’1 for x <

’

4

All these conditions are plotted in the figure 16.3.

Figure 16.3

It is easily proved that the limiting lines touch the hyperbola at the points (1/4, 1/2)

and (1/2, 1/4). For both triangles painted with light grey there is a unique solution. In the

region painted with dark grey there are two solutions, that is, there are two segments

passing through the point R and dividing the triangle in two parts with equal area. By

means of an affinity the plot in the figure 16.3 is transformed into any triangle (figure

16.4). The affinity is a linear transformation of the coordinates, which converts the

limiting lines into the medians of the triangle intersecting at the centroid and the

hyperbola into another non equilateral hyperbola:

Figure 16.4

Observe that the medians divide the triangle ABC in six triangles. If a point R lies outside

the shadowed triangles, we can always support the extremes of the segment PQ passing

RAMON GONZALEZ CALVET

236

through R in another pair of sides. This means that the division of the triangle is always

possible provided of the suitable choice of the sides. On the other hand, if we consider all

the possibilities, the points neighbouring the centroid admit three segments.

9. Circles

( A ’ M )2 + ( B ’ M )2 + ( C ’ M )2 = k

9.1

3 M2 ’ 2 M · ( A + B + C ) + A2 + B2 + C2 = k

Introducing the centroid G = ( A + B + C ) / 3:

k 2 ( A 2 + B 2 + C 2 + A · B + B · C + C · A)

M ’ 2 M ·G + G = ’

2 2

3 9

k (B ’ A) + (C ’ B ) + ( A ’ C )

2 2 2

(M ’ G ) 2

=’

3 9

k AB 2 + BC 2 + CA 2

GM = ’ = r2

2

3 9

M runs on a circumference with radius r centred at the centroid of the triangle when k is

higher than the arithmetic mean of the squares of the three sides:

AB 2 + BC 2 + CA 2

k≥

3

9.2 Let A and B be the fixed points and P any point on the searched geometric locus. If the

ratio of distances from P to A and B is constant it holds that:

PA2 = k2 PB2

¦PA¦ = k ¦PB¦ ” with k being a real positive number

A2 ’ 2 A · P + P2 = k2 ( B2 ’ 2 B · P + P2 )

( 1 ’ k2 ) P2 ’ 2 ( A ’ k2 B ) · P = k2 B2 ’ A2

2 P · ( A ’ k 2 B) k 2 B 2 ’ A2

P’ =

2

1’ k2 1’ k2

A ’ k 2B

We indicate by O the expression: O = 2

2 . Adding O to both members we obtain:

1’ k

k 2 B 2 ’ A 2 (A ’ k 2 B )

2

OP = +

2

(1 ’ k 2 )2

1’ k2

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 237

k 2 ( B 2 ’ 2 A · B + A2 ) k 2 AB 2

OP = =

2

(1 ’ k ) (1 ’ k )

22 22

Therefore, the points P form a circumference centred at O, which is a point of the line AB,

with radius:

k AB

OP =

1’ k2

9.3 Let us draw the circumference

circumscribed to the triangle ABC

(figure 16.5) and let D be any point

on this circumference. Let P, Q and R

be the orthogonal projections of P on

the sides AB, BC and CA

respectively. Note that the triangles

ADP and CDR are similar because of

DAB = π ’ BCD = DCR . The

similarity of both triangles is written

as:

Figure 16.5

’1 ’1

DC DA = DR DP

The triangles ADQ and BDR are also similar because of the equality of the inscribed

angles CAD = CBD:

DB DA ’1 = DR DQ ’1

Since the angles RDC = PDA and RDB = QDA, the bisector of the angle BDQ is

also bisector of the angle CDP and RDA, being its direction vector:

DA DR DB DQ DC DP

v= + = + = +

DA DR DB DQ DC DP

Let P', Q' and R' be the reflected points of P, Q and R with respect to the bisector:

DR' = v ’1 DR v

Multiplying by DA we have:

DR' DA = v ’1 DR v DA = ¦DR¦ ¦DA¦

which is an expected result since the reflection of R with respect the bisector of the angle

RDA yields always a point R' aligned with D and A. Analogously for B and C we find that

the points D, B and Q' are aligned and also the points D, C and P':

DQ' DB = v ’1 DQ v DR DQ ’1 DA = v ’1 DR v DA = ¦DR¦ ¦DA¦

RAMON GONZALEZ CALVET

238

DP' DC = v ’1 DP v DR DP ’1 DA = v ’1 DR v DA = ¦DR¦ ¦DA¦

Now we see that the points R', Q' and P' are the transformed of A, B and C under

an inversion with centre D:

DR' DA = DQ' DB = DP' DC

Since D lies on the circumference passing through A, B and C, the inversion transforms

them into aligned points and the reflection preserves this alignment so that P, Q and R are

aligned.

9.4 If m = a + b and n = b +c then:

m 2 n 2 = (a + b ) (b + c ) = (a 2 + a b + b a + b 2 )(b 2 + b c + c b + c 2 )

2 2

= a 2b2 + a 2b c + a 2c b + a 2c 2 + a b3 + a b2c + a b c b + a b c 2 +

+ b a b 2 + b a b c + b a c b + b a c 2 + b 4 + b3c + b 2 c b + b 2 c 2

Taking into account that d = ’(a + b + c) then:

d 2 = (a + b + c ) = a 2 + b 2 + c 2 + a b + b a + b c + c b + c a + a c

2

so we may rewrite the former equality:

m2n 2 = a 2c 2 + b2d 2 + a 2b c + a 2c b + a b c b + a b c 2 + b a b c + b a c 2

We apply the permutative property to some terms:

m2n 2 = a 2c 2 + b2d 2 + a 2b c + a b c a + a b c b + a b c 2 + b a b c + c a b c

in order to arrive at a fully symmetric expression:

m2n2 = a 2c2 + b2d 2 ’ d a b c ’ a b c d

Now we express the product of each pair of vectors using the exponential function of their

angle:

(exp((2π ’ γ ’ ± ) e12 ) + exp((2π ’ β ’ δ ) e12 ))

m2n2 = a 2c2 + b2d 2 ’ a b c d

Adding the arguments of the exponential, simplifying and taking into account that ± + β +

γ + δ = π.

d cos(± + γ )

m2n2 = a 2c2 + b2d 2 ’ 2 a b c

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 239

9.5 Let A, B and C be the midpoints of the sides of any triangle PQR :

Q+R R+P P+Q

A= B= C=

2 2 2

The centre of the circumferences RAB, PBC and QCA will be denoted as D, E and F

respectively. The circumference PBC is obtained from the circumscribed circumference

PQR by means of a homothety with centre P. Then if O is the circumcentre of the triangle

PQR, the centre of the circumference PBC is located at half distance from R:

P+O Q +O

E= F=

2 2

PQ OR PQ OR PQ

EA = + FB = ’ + EF =

2 2 2 2 2

Let Z be the intersection of EA with FB. Then:

Z = a A + ( 1 ’ a ) E = b B + ( 1 ’ b) F

Arranging terms we find:

a EA ’ b FB = EF

The linear decomposition gives:

EF § FB EA § EF

a= ’b=

EA § FB EA § FB

EF § FB FA § FB A E P+Q + R+O

Z= A+ E= + =

EA § FB EA § FB 22 4

which is invariant under cyclic permutation of the vertices P, Q and R. Then the lines EA,

FB and DC intersect in a unique point Z. On the other

Figure 9.8

hand, the point Z lies on the Euler™s line (figure 9.8):

3G + O

Z=

4

9.6 We must prove that the inversion changes the single ratio of three very close points by

its conjugate value, because the orientation of the angles is changed. Let us consider an

inversion with centre O and radius r and let the points A', B', C' be the transformed of A,

B, C under this inversion. Then:

OA' = r2 OA’1 OB' = r2 OB ’1 OC' = r2 OC ’1

A'B' = OB' ’ OA' = r2 (OB ’1 ’ OA ’1) = r2 OA ’1 (OA ’ OB) OB ’1

RAMON GONZALEZ CALVET

240

= ’ r2 OA ’1 AB OB ’1

A'C' = ’ r2 OA ’1 AC OC ’1

In the same way:

Let us calculate the single ratio of the transformed points:

( A', B', C' ) = A'B' A'C' ’1 = OA ’1 AB OB ’1 OC AC ’1 OA =

= OB ’1 AB AC ’1 OC = ( OA + AB ) ’1 AB AC ’1 ( OA + AC )

When B and C come near A, AB and AC tend to zero and the single ratio for the inverse

points becomes the conjugate value of that for the initial points:

( A' , B' , C' ) = OA ’1 AB AC ’1 OA = AC ’1 AB = ( A, B, C ) *

lim

B, C ’ A

10. Cross ratios and related transformations

10.1 Let ABCD be a quadrilateral inscribed in a circle. We must prove the following

equality:

¦BC¦ ¦AD¦ + ¦AB¦ ¦CD¦=¦BD¦ ¦AC¦

which is equivalent to:

BC AD AB CD

+ =1

BD AC BD AC

Now we identify these quotients with cross ratios of four points A, B, C and D on a circle:

(BACD)+(BCAD)=1

equality that always holds because if ( A B C D ) = r then:

1 r ’1

+ =1

r r

When the points do not lie on a circle, the cross ratio is not the quotient of moduli

but a complex number:

BC AD AB CD

exp[e12 (CBD ’ CAD )] + exp[e12 ( ABD ’ ACD )] = 1

BD AC BD AC

According to the triangular inequality, the modulus of the sum of two complex

numbers is lower than or equal to the sum of both moduli:

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 241

BC AD AB CD

+ ≥1 ’ AD + AB CD ≥ BD

BC AC

BD AC BD AC

10.2 If A, B, C and D form a harmonic range their cross ratio is 2:

AC AD ’1 BD BC ’1 = 2

In order to isolate D we must write BD as addition of BA and AD:

AC AD ’1 ( BA + AD ) BC ’1 = 2

AC AD ’1 BA BC ’1 = 2 ’ AC BC ’1

AD ’1 = AC ’1 ( 2 ’ AC BC ’1 ) BC BA ’1 = ( 2 AC ’1 BC ’1 ) BA ’1

AD = AB ( 1 ’ 2 AC ’1 BC ) ’1

10.3 The homography preserves the cross ratio:

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

which may be rewritten as:

A'C' A'D' ’1 ( B'A' + A'D' ) ( B'A' + A'C' ) ’1 = AC AD ’1 ( BA + AD ) ( BA + AC ) ’1

When C and D approach to A, the vectors AC, AD, A'C' and A'D' tend to zero. So the

single ratio of three very close points remains constant:

A'C' A'D' ’1 = AC AD ’1

lim lim

C, D ’ A C, D ’ A

and therefore the angle between tangent vectors of curves. So the homography is a

directly conformal transformation.

10.4 If F is a point on the homology axis we must prove that:

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

As we have seen, a point A' homologous of A is obtained through the equation:

OA' = OA [ 1 ’ r v § FA ( v § FO ) ’1 ] ’1

where O is the centre of homology, F a point on the axis, v the direction vector of the axis

and r the homology ratio. Then the vector FA' is:

FA' = FO + OA' = FO + OA [ 1 ’ r v § FA ( v § FO ) ’1 ] ’1

RAMON GONZALEZ CALVET

242

= [ FO ( 1 ’ r v § FA ( v § FO ) ’1 ) + OA ] [ 1 ’ r v § FA ( v § FO ) ’1 ] ’1

= [ FA ’ r FO v § FA ( v § FO ) ’1 ] [ 1 ’ r v § FA ( v § FO ) ’1 ] ’1

Analogously:

FB' = [ FB ’ r FO v § FB ( v § FO ) ’1 ] [ 1 ’ r v § FB ( v § FO ) ’1 ] ’1

FC' = [ FC ’ r FO v § FC ( v § FO ) ’1 ] [ 1 ’ r v § FC ( v § FO ) ’1 ] ’1

FD' = [ FD ’ r FO v § FD ( v § FO ) ’1 ] [ 1 ’ r v § FD ( v § FO ) ’1 ] ’1

The product of two outer products is a real number so that we must only pay attention to

the order of the first loose vectors FA and FO, etc:

FA' § FC' = [ FA § FC ’ r ( FA § FO v § FC + FO § FC v § FA ) ( v § FO ) ’1 ]

[ 1 ’ r v § FA ( v § FO ) ’1 ] ’1 [ 1 ’ r v § FC ( v § FO ) ’1 ] ’1

Using the identity of the exercise 1.4, we have:

FA' § FC' = [ FA § FC ’ r FA § FC v § FO ( v § FO ) ’1 ]

[ 1 ’ r v § FA ( v § FO ) ’1 ] ’1 [ 1 ’ r v § FC ( v § FO ) ’1 ] ’1

= FA § FC ( 1 ’ r ) [ 1 ’ r v § FA ( v § FO ) ’1 ] ’1 [ 1 ’ r v § FC ( v § FO ) ’1 ] ’1

In the projective cross ratio all the factors except the first outer product are simplified:

FA' § FC' FB' § FD' FA § FC FB § FD

=

FA' § FD' FB' § FC' FA § FD FB § FC

10.5 a) Let us prove that the special conformal transformation is additive:

OP' = ( OP ’1 + v ) ’1 ’ OP' ’1 = OP ’1 + v

OP'' = ( OP' ’1 + w ) ’1 = ( OP ’1 + v + w ) ’1

b) Let us extract the factor OP from OP':

OP' ’1 = ( 1 + v OP ) OP ’1 OP' = OP ( 1 + v OP ) ’1

’

Then we have:

OA' = OA ( 1 + v OA ) ’1 OB' = OB ( 1 + v OB ) ’1

OC' = OC ( 1 + v OC ) ’1 OD' = OD ( 1 + v OD ) ’1

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 243

A'C' = OC' ’ OA' = OC ( 1 + v OC ) ’1 ’ OA ( 1 + v OA ) ’1 =

= [ OC ( 1 + v OA ) ’ OA ( 1 + v OC ) ] ( 1 + v OC ) ’1 ( 1 + v OA ) ’1

= AC ( 1 + v OC ) ’1 ( 1 + v OA ) ’1

Analogously:

B'D' = BD ( 1 + v OB ) ’1 ( 1 + v OD ) ’1

A'D' = AD ( 1 + v OA ) ’1 ( 1 + v OD ) ’1

B'C' = BC ( 1 + v OB ) ’1 ( 1 + v OC ) ’1

From where it follows that the complex cross ratio is preserved and it is a special case of

homography:

A'C' A'D' ’1 B'D' B'C' ’1 = AC ( 1 + v OC ) ’1 ( 1 + v OD ) AD ’1

BD ( 1 + v OD ) ’1 ( 1 + v OC ) ’1 BC ’1 = AC AD ’1 BD BC ’1

10.6 If the points A, B and C are invariant under a certain homography, then for any other

point D' the following equality is fulfilled:

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

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

The simplification of factors yields:

AD ’1 BD = AD' ’1 BD' ’ AD ’1 ( BA + AD ) = AD' ’1 ( BA + AD' ) ’

’1 ’1

AD ’1 = AD' ’1 ’

’ AD BA = AD' BA ’ D = D'

10.7. a) From the definition of antigraphy we have:

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

If C and D approach A, then BC, BD ’ BA:

A' C ' A' D' ’1 = AD ’1 AC

lim lim

C' , D' ’ A C, D ’ A

So the antigraphy is an opposite conformal transformation.

b) The composition of two antigraphies is always a homography because twice the

conjugation of a complex number is the identity, so that the complex cross ratio is

preserved.

c) The inversion conjugates the cross ratio. Consequently an odd number of inversions is

RAMON GONZALEZ CALVET

244

always an antigraphy.

d) If an antigraphy has three invariant points, then all the points lying on the

circumference passing through these points are invariant. Let us prove this statement: if A,

B and C are the invariant points and D belongs to the circumference passing through these

points, then the cross ratio is real:

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

from where D = D' as proved in the exercise 10.6. If D does not belong to this

circumference, D' ≠ D and the cross ratio ( A B C D ) becomes conjugate. A geometric

transformation that preserves a circumference and conjugates the cross ratio can only be

an inversion because there is a unique point D' for each point D fulfilling this condition.

The centre and radius of inversion are those for the circumference ABC.

10.8 Consider the projectivity:

’

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

If A', B' and C' are the images of A, B and C then the coefficients a', b', c' must be

proportional to a, b, c:

ka lb mc

a' = b' = c' =

k a+lb+mc ka+lb+mc ka+lb+mc

where the values are already normalised. Three points are collinear if the determinant of

the barycentric coordinates is zero, what only happens if the initial points are also

collinear:

aD bD cD

k l m aE aE aE

a' D b' D c' D

aF aF aF

a' E = =0

a' E a' E

(k a D + l bD + m c D ) (k a E + l bE + m c E )(k a F + l bF + m c F )

a' F a' F a' F

10.9 In the dual plane the six sides of the hexagon are six points. Since the sides AB, CD

and EF (or their prolongations) pass through P and BC, DE and FA pass through Q, this

means that the dual points are alternatively aligned in two dual lines P and Q. By the

Pappus™ theorem the dual points AD, BE and CF lie on a dual line X, that is, these lines

joining opposite vertices of the hexagon are concurrent in the point X. Moreover, as

proved for the Pappus™ theorem the cross ratios fulfil the equalities in the left hand side:

”

(BE AD FC P) = (PQ AB FE CD) {X, B A F P} = {P, Q A F C}

”

(BE AD FC Q) = (PQ DE CB FA) {X, B A F Q} = {Q, P D C F}

Since the cross ratio of a pencil of lines is equal to the cross ratio of the dual points, the

equalities in the right hand side follow immediately.

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 245

11. Conics

11.1 The starting point is the central equation of a conic:

OA = OQ cos ± + OR sin ± OB = OQ cos β + OR sin β

OC = OQ sin γ + OR sin γ OD = OQ cos δ + OR sin δ

OX = OQ cos χ + OR sin χ

XA = OA ’ OX = OQ ( cos ± ’ cos χ ) + OR ( sin ± ’ sin χ )

XC = OC ’ OA = OQ ( cos γ ’ cos χ ) + OR ( sin γ ’ sin χ )

XA § XC = OQ § OR [ (cos γ ’ cos χ ) (sin ± ’ sin χ ) ’ (cos ± ’ cos χ ) (sin γ ’ sin χ)]

= OQ § OR [ cos γ sin ± ’ cos γ sin χ ’ cos χ sin ± ’ cos ± sin γ

+ cos ± sin χ + cos χ sin γ ]

= OQ § OR [ sin( ± ’ γ ) + sin( γ ’ χ ) + sin( χ ’ ± ) ]

± ’γ γ ’χ χ ’±

= ’ 4 OQ § OR sin sin sin

2 2 2

From where it follows that:

± ’γ β ’δ

sinsin

XA § XC XB § XD 2 2

=

± ’δ β ’γ

XA § XD XB § XC

sin sin

2 2

11.2 The central equation of an ellipse of centre O with semiaxis OQ and OR is:

OP = OQ cos θ + OR sin θ P = O ( 1 ’ cosθ ’ sin θ ) + Q cos θ + R sin θ

’

There is always an affinity transforming an ellipse into a circle:

O' = (0, 0) Q' = (1, 0) R' = (0, 1)

P' = O' ( 1 ’ cos θ ’ sin θ ) + Q' cos θ + R' sin θ = ( cos θ , sin θ )

Let us now consider any point E and a circle. A line passing through E cuts the circle in

the points A and B; another line passing also through E but with different direction cuts

the ellipse in the points C and D. Since the power of a point E with respect to the circle is

constant we have:

RAMON GONZALEZ CALVET

246

EA EB ’1 ¦EB¦2 = EC ED ’1¦ED¦2

EA EB = EC ED

The affinity preserves the single ratio of three aligned points:

EA EB ’1 = E'A' E'B' ’1 EC ED ’1 = E'C' ED' ’1

Then:

E'A' E'B' ’1¦EB¦2 = E'C' E'D' ’1¦ED¦2

2 2

E'B' ED

E'A' E'B'

= 2 2

E'C' E'D' EB E'D'

and this quotient is constant for two given directions because the preservation of the single

ratio implies that the distances in each direction are enlarged by a constant ratio:

EA EB

=

E'A' E'B'

11.3 If a diameter is formed by the midpoints of the chords parallel to the conjugate

diameter, then it is obvious that the tangent to the point where the length of the chord

vanishes is also parallel to the conjugate diameter. This evidence may be proved by

derivation of the central equation:

OP = OQ cos θ + OR sin θ

d OP

= ’ OQ sinθ + OR cos θ

dθ

That is, OP and dOP / dθ (having the direction of the tangent to P) are conjugate radius.

The area a of the parallelogram circumscribed to the ellipse is the outer product of both

conjugate diameters, which is independent of θ :

a = 4 (OQ cosθ + OR sinθ ) § ( ’OQ sinθ + OR cosθ ) = 4 OQ § OR

11.4 The area of a parallelogram formed by two conjugate diameters of the hyperbola is

also independent of ψ:

a = 4 (OQ cosh ψ + OR sinh ψ ) § ( OQ sinh ψ + OR cosh ψ ) = 4 OQ § OR

11.5 Let us prove the statement for a circle. The intersections R and R' of a circumference

with a line passing through P are given by (fig. 9.3):

PR = PF cos ± ’ r 2 ’ PF 2 sin 2±

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 247

PR' = PF cos ± + r 2 ’ PF 2 sin 2±

The point M located between R and R' forming harmonic range with P, R and R' fulfils the

condition:

PM RR'

( P R M R' ) = =2

PR' RM

2 PR PR'

PM =

Then:

PR + PR'

The numerator is the power of P (constant for any line passing through P) and after

operations we obtain the polar equation of the geometric locus of the points M:

PT 2

PM =

PF cos ±

which is the polar line. When ± is the angle TPF (figure 9.2) we have¦PM¦ = ¦PT¦, that

is, the polar passes through the points T of tangency to the circumference.

Now, by means of any projectivity (e.g. a homology) the circle is transformed into

any conic. Since it preserves the projective cross ratio, the points M also form an aligned

harmonic range in the conic, so the polar also passes through the touching points of the

tangents drawn from P.

11.6 Let the points be denoted by:

A = (1, 1) B = (2, 3) C = (1, 4 ) D = (0, 2 ) E = (2, 5)

Then: EA § EC = ’3 EB § ED = ’4 EA § ED = ’5 EB § EC = ’2

The cross ratio is:

EA § EC EB § ED 6

{E , A B C D} = =

EA § ED EB § EC 5

Now a generic point of the conic will be X = (x, y):

XA § XC = ’3 x + 3 XB § XD = x ’ 2 y + 4

XA § XD = ’ x ’ y + 2 XB § XC = ’ x ’ y + 5

Applying the Chasles™ theorem we find the equation of the conic:

RAMON GONZALEZ CALVET

248

(’ 3 x + 3) (x ’ 2 y + 4 ) 6

XA § XC XB XD

= =

XA § XD XB § XC (’ x ’ y + 2 ) (’ x ’ y + 5) 5

After simplifying we arrive at the point equation:

7 x2 + 2 y2 ’ 6 x y + x ’ 4 y = 0

which may be written in the form:

’ 2 «1 ’ x ’

«0 y

1/ 2

¬ ·¬ ·

(1 ’ x ’ y y ) ¬1 / 2 ’ 9 / 2· ¬ ·=0

x 8 x

¬’ 2 ’9/2 ’2 ·¬ ·

y

The inverse matrix of the conic is:

’1

’2 « 145 ’ 40 ’ 55

«0 1/ 2

¬ · 1¬ ·

’ 9 / 2· = ¬ ’ 40 16

¬1 / 2 8 4·

90 ¬

¬’ 2 ’9/2 ’2 · 1·

’ 55 4

So the equation of the tangential conic is:

« 145 ’ 40 ’ 55 ®1 ’ u ’ v

¬ ·

[1 ’ u ’ v u v ] ¬ ’ 40 16 4 · u = 0

¬ ’ 55 1 · v

4

° »

241 u 2 + 256 v 2 + 488 u v ’ 370 u ’ 400 v + 145 = 0

11.7 If we take A, B and C as the base of the points, the coordinates in this base of every

point X will be expressed as:

X = (x A , x B , x C ) x A + x B + xC = 1

If P is the intersection point of the lines AE and BF, then A, P and E are aligned and also

B, P and F, conditions which we may express in coordinates:

1 0 0 0 1 0

(eC f A , e B f C , eC f C )

( p A , p B , pC ) =

pC = 0 pC = 0 ’

pA pB pA pB

f A + e B f C + eC f C

eC

eA eB eC fA fB fC

If Q is the intersection point of the lines AD and CF, then A, Q and D are aligned and also

C, Q and F, conditions which we may express in coordinates:

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 249

1 0 0 0 0 1

(d B f A, d B f B , dC f B )

(q A , q B , qC ) =

qC = 0 qC = 0 ’

qA qB qA qB

f A + d B f B + dC f B

dB

dA dB dC fA fB fC

If R is the intersection point of the lines BD and CE, then B, R and D are aligned and also

C, R and E, conditions which we may express in coordinates:

0 1 0 0 0 1

(d A e A , d A e B , d C e A )

(rA , rB , rC ) =

rC = 0 rC = 0

rA rB rA rB ’

d A e A + d A eB + d C e A

dA dB dC eA eB eC

The points P, Q and R will be aligned if and only if the determinant of their coordinates

vanishes. Setting aside the denominators, we have the first step:

fA eB 1 1 1

1

fC eC fC eC dC

pA pB pC eC f A eB f C eC f C

dC

f 1 1 1

qC ∝ d B f A ∝A ∝

qA qB dB fB dC f B 1

fB dB fB eB dB

rA rB rC d Ae A d AeB dC eA dC 1 1 1

eB

1

fA eA dA

eA dA

In the second step each row has been divided by a product of two coordinates. In the third

step each column has been divided by a coordinate. And finally, after transposition and

exchange of the first and third columns (turning the elements 90º in the matrix) and

multiplying each column by a product of three coordinates we obtain:

1 1 1

fA fB fC

pA pB pC d AeA d B eB d C eC

1 1 1

∝ ∝ dA fA dC fC = 0

qA qB qC dB fB

eA eB eC

rA rB rC eA f A eB f B eC f C

1 1 1

dA dB dC

The last determinant is zero if and only if the point F lies on the conic passing through A,

B, C, D and E because then it fulfils the conic equation:

d AeA d B eB d C eC

d C xC = 0

d AxA d B xB

eA x A eB x B eC x C

11.8 The Brianchon™s theorem is the dual of the Pascal™s theorem. In the dual plane, the

point conic is represented by the tangential conic, the lines tangent to the point conic are

represented by points lying on the tangential conic and the diagonals of the circumscribed

RAMON GONZALEZ CALVET

250

hexagon are represented by the points of intersection P, Q and R of the former exercise.

So the algebraic deduction is the same but in the dual plane.

12. Matrix representation and hyperbolic numbers

12.1 Let us calculate the modulus and argument:

1 4

arg(5 + 4 e1 ) = arg tgh = log 3

5 + 4 e1 = 5 2 ’ 4 2 = 3

2 5

Then:

(( ) ( ))

5 + 4 e1 = 3 log 3 = 3 cosh log 3 + e1 sinh log 3 = 2 + e1

2

The four square roots are:

2 + e1 1 + 2 e1 ’ 2 ’ e1 ’ 1 ’ 2e1

’ 3 ± 9 ’ 4 … 2 … (’ 17 + 3 e1 )

2 z 2 + 3 z ’ 17 + 3 e1 = 0 z=

’

12.2

2…2

We must calculate the square root of the discriminant:

145 ’ 24 e1 = 143 log (11 / 13) = 12 ’ e1

2

So the four solutions are:

9 ’ e1 ’ 15 + e1 ’ 1 ’ 6 e1

z1 = z2 = z3 = z 4 = ’1 + 3 e1

4 4 2

12.3 Let us apply the new formula for the second degree equation:

6 ± 36 ’ 4 … 1 … 5 6 ± 4

z2 ’ 6 z + 5 = 0 z= = and

2 …1 2

6 ± e1 36 ’ 4 … 1 … 5 6 ± 4 e1

z= =

2 …1 2

1 + e1 1 ’ e1

12.4 sin( x + y e1 ) = sin( x + y ) + sin( x ’ y )

2 2

= sin x cos y + e1 cos x sin y

12.5 From the analogous of Moivre™s identity we have:

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 251

(coshψ + e1 sinh ψ )4 ≡ cosh 4ψ + e1 sinh 4ψ

cosh 4ψ ≡ cosh 4 ψ + 6 cosh 2 ψ sinh 2ψ + sinh 4ψ

sinh 4ψ ≡ 4 cosh 3 ψ sinhψ + 4 coshψ sinh 3ψ

12.6 The analytical continuation of the real logarithm is:

1 + e1 1 ’ e1

log( x + y e1 ) = log( x + y ) + log( x ’ y )

2 2

x+ y

log(x 2 ’ y 2 ) + 1 log

e

1

=

x’y

2 2

It may be rewritten in the form:

y

= log x 2 ’ y 2 + e1 arg tgh

x

12.7 For the straight path z = t e1 we have:

1

e1 1

®t 3 2e

« z dz = e1 « t dt = e1 = 1

2 2

° 3 » ’1 3

’ e1 ’1

For the circular path z = cos t + e1 sin t we have:

e1 π /2

(cos t + e1 sin t )2 (’ sin t + e1 cos t ) dt

« z dz = «

2

’ e1 ’π / 2

π /2

® 2e

«

2 2

= cos t + cos 3 t + e1 ¬ sint ’ sin 3 t · =1

3 3 3

» ’π / 2

°

Using the indefinite integral, we find also the same result:

e1

e1

® z3 2e

« z dz = = 1

2

° 3 » ’ e1 3

’ e1

12.8 The proof is analogous to the exercise 3.12: turn the hyperbolic numbers df, dz into

hyperbolic vectors by multiplying them at the left by e2.

13. The hyperbolic or pseudo-Euclidean plane

RAMON GONZALEZ CALVET

252

13.1 If the vertices of the triangle are A = (2, 2), B = (1, 0) and C = (5, 3) then:

AB = (’1, ’2) CA = (’3, ’1)

BC = (4, 3)

AB = 3 e12 BC = 7 CA = 8

From the cosine theorem we have:

9

cosh γ = γ … 0.6264

AB 2 = BC 2 + CA 2 ’ 2 BC CA cosh γ ’ ’

2 14

taking into account that γ = ACB is a positive angle. From the sine theorem we have:

BC CA AB 3 e12

7 8

= = = =

’

sinh ± sinh β sinh γ sinh ± sinh β 5

2 14

whence the angles ± and β follow:

5 e12 π

sinh± = ’ ± … ’0.2027 ’ e12

’

2

26

5 e12 π

sinhβ = ’ β … ’0.4236 ’ e12

’

2

21

The plot helps to choose the right sign of the angles. Anyway a wrong choice would be

revealed by the cosine theorem:

BC 2 = CA 2 + AB 2 ’ 2 CA AB cosh ±

CA 2 = AB 2 + BC 2 ’ 2 AB BC cosh β

13.2 The rotation of the vector is obtained with the multiplication by a hyperbolic

number:

«5 3

v' = v z± = (2 e 2 + e 21 ) (cosh log 2 + e1 sinh log 2 ) = (2 e 2 + e 21 ) ¬ + e1 ·

4 4

13 11

= e 2 + e 21

4 4

The reflection is obtained by multiplying on the right and on the left by the direction

vector of the reflection and its inverse respectively:

13 e 2 ’ 11 e 21

(2 e 2 + e 21 ) (3 e 2 ’ e 21 ) = 1 (3 e 2 ’ e21 )(7 ’ 5 e1 ) =

v' = d ’1 v d = (3 e 2 ’ e 21 )

’1

8 4

The inversion with radius r = 3 is:

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 253

v' = r 2 v ’1 = 9 (2 e 2 + e 21 ) = 3 (2 e 2 + e 21 ) = 6 e 2 + 3 e 21

’1

13.3 The two-point equation of the line y = 2 x + 1 is:

x +1 y

=

1 2

which gives us the direction vector:

v = e1 + 2 e 2

The direction vector is the normal vector of the perpendicular line. The general equation

of a line in the hyperbolic plane is:

n x (x ’ x P ) ’ n y ( y ’ y P ) = 0

n · PR = 0 ’

Note the minus sign since this line lies on the hyperbolic plane. The substitution of the

components of the direction vector of the first line and the coordinates of the point R =

(3, 1) through which the line passes results in the equation:

1 … ( x ’ 3) ’ 2 … ( y ’ 1) = 0 x ’ 2 y ’1= 0

’

The fact that both lines and the first quadrant bisector intersect in a unique point is

circumstantial. Only it is required that the bisectors of both lines be parallel to the

quadrant bisectors.

2 2

13.4 The line y =3 intersects the hyperbola x “ y = 16 in the points R = (’5, 3) and R' =

(5, 3). Then the power of P = (’7, 3) is:

PR PR' = 2 e 2 · 12 e 2 = 24

The line y = “ 3 x “ 18 cuts the hyperbola in the points T = (“5, “3) and T' = (“17/2,

15/2). Then the power of P is:

« 3e 9e

PT PT' = (2 e 2 ’ 6 e 21 ) · ¬ ’ 2 + 21 · = ’3 + 27 = 24

2 2

The line y = “3 x / 7 cuts the hyperbola in the points:

« 7 10 3 10 « 7 10 3 10

S = ¬’ · S' = ¬ ·

,’

, and

¬ 5· ¬5 5·

5

The power is:

« « 7 10

« « 7 10 « 3 10 « 3 10

’ 3 · e 21 ·

PS PS' = ¬ ¬ ’ ’ 3 · e 21 · · ¬ ¬

+ 7 · e2 + ¬ + 7 · e2 + ¬ ’

· · ¬¬ 5 · ¬ ··

¬¬ · ¬5

5 5

RAMON GONZALEZ CALVET

254

147 27

= ’ = 24

5 5

Also the substitution of the coordinates gives 24:

x P ’ y P ’ 16 = 49 ’ 9 ’ 16 = 24

2 2

13.5 From the law of sines we find:

a+b

sinh± sinh± + sinhβ

a

= =

’

sinhβ sinh± ’ sinhβ

a’b

b

Introducing the identity for the addition and subtraction of sines we arrive at the law of

tangents:

±+β

tgh

a+b 2

=

±’β

a’b

tgh

2

13.6 The first triangle has the vertices A = (0, 0), B = (5, 0), C = (5, 3) and the sides:

CA = ’5 e2 ’ 3 e21

AB = 5 e2 BC = 3 e21

The second triangle has the vertices A' = (0, 0), B' = (25, ’15), C' = (16, 0) and the

sides:

A'B' = 25 e2 ’ 15 e21 B'C' = ’9 e2 + 15 e21 C'A' = ’16 e2

Both triangles are directly similar since:

AB’1 A'B' = BC’1 B'C' = CA’1 C'A' = 5 ’ 3 e1 = r

which is the similarity ratio r. The size ratio is the modulus of the similarity ratio:

r = 52 ’ 32 = 4

that is, the second triangle is four times larger than the first. The angle of rotation

between both figures is the argument of the similarity ratio:

« 3

ψ = arg(5 ’ 3 e1 ) = arg tgh¬ ’ · = ’ log 2 = ’0.6931...

5

Now plot the vertices of each triangle in the hyperbolic plane and break your Euclidean

illusions about figures with the same shape.

14. Spherical geometry in the Euclidean space

TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 255

14.1 The shadow of a gnomon follows a hyperbola on a plane (a quadrant) because the

Sun describes approximately a parallel around the North pole during a day. The altitude

of this parallel from the celestial equator is the declination, which changes slowly from

one day to another. The March 21 and September 23 are the equinoxes when the Sun

follows the equator. Since it is a great circle, its central projection is a straight line

named the equinoctial line.

14.2 From the stereographic coordinates we find the Cartesian coordinates of the points:

12 e 2 5 e 3

2 2 1 12 5

A = ’ e1 + e 2 + e3 B=’ + C= e1 + e 3

3 3 3 13 13 13 13

Their outer products are equal to the cosines of the sides:

25

cos a = B · C = a = 1.4223

169

19

cos b = C · A = ’ b = 2.0797

39

19

cos c = A · B = ’ c = 2.0797

39

Now we calculate the bivectors of the planes containing the sides: