link.springer.com978-3-642-56127...A few Hints and Solutions to Exercises The reader will find here a mixture of hints, solutions, comments and bib liographical references for …

A few Hints and Solutionsto Exercises

The reader will find here a mixture of hints, solutions, comments and bibliographical references for some of the exercises. These are not completesolutions, even when they are rather detailed. They will exempt the reader

- neither from writing their solutions (some of the hints given here arereally written in telegraphic language)

- nor from taking care of converses. When I write for instance ''the pointP lies on the circle of diameter I J", I do not write ''the locus of pointsP is the circle of diameter I J".

Last remark: the reader will often need to draw a picture to understand thesolution.

Chapter I

Exercises 1.6 and 1.7. Go back to linear algebra.

Exercise 1.15. Use the fixed point theorem (Theorem 2.22).

Exercise 1.23. Assume that

("Ix E E) (3A E K) (f(x) = AX),

in which, a priori, A = Ax depends on x. Check that:

- if dim E = 1, A does not depend on x,- if dim E ~ 2, for any two independent vectors X and y, one has

AX+Y(X + y) = f(x + y) = f(x) + f(y) = AxX + Ayy,

thus Ax = Ay. Deduce that Ax actually does not depend on x.

302

Hence we have

A few Hints and Solutions to Exercises

(3A E K) ('<Ix E E) (f(x) = AX),

this meaning that f is a linear dilatation.

Exercise 1.24. The linear mapping <p maps any vector to a vector that iscollinear. Using the ''trick'' of Exercise 1.23, this is thus a linear dilatation.

Exercise 1.28. Be careful that if (AO,"" An) is a system of barycentriccoordinates of M and if f..L is any nonzero scalar, (f..LAo"", f..LA n ) is also asystem of barycentric coordinates of M.

Exercise 1.29. Consider for instance the barycenter M1 of the system«M,a), (M', -1)); prove that it is invariant and that the mapping that associates M 1 to M is a projection.

Exercise 1.30. In dimensions greater than or equal to 3, you can ''turnaround" a line; the complement is connected. A formal proof: choose anaffine frame whose origin is on the line and whose first vector directs it; weare then in the case of the line Rx {O} in Rx Rn-l; for n-1 ~ 2, Rn-l_{O}is path-connected (this is the place where you ''turn around", clear?), thus

e- D = R x (Rn-

1 - {O})

is path-connected as the product of two path-connected spaces.Similarly, the complement of a complex line in a complex plane is path

connected, as is the complement of 0 in C.

Exercise 1.31. In the case of the centroid, this is a dilatation. In the caseof the orthocenter, you can exhibit three collinear points whose images arenot collinear, for instance remembering that the orthocenter of a right-angledtriangle is the vertex of the right angle.

Exercise 1.33. Ifa subset has two symmetry centers, it is preserved by sometranslation: the composition of two central symmetries is a translation.

Exercise 1.34. Use the composition of central symmetries (jAl 0 ••. 0 (jAn'

that is, a translation or a central symmetry according to the parity of n.

Exercise 1.35. We have

~ --+IfI is defined by BI = GB, the image of I by this dilatation is the midpointJ of AB. The position of B' on I J and the construction are deduced.

A few Hints and Solutions to Exercises 303

Exercise 1.36. You can look for a system of barycentric coordinates for thepoint A" (Figure 20) in the triangle ABG; you can also use the fact that thearea of the triangle AA'G is one third of that of the triangle ABG.

----+ ----+Exercise 1.37. The dilatation of center G' and ratio G'A/G'B transforms

----+ ----+B into A, that of center B' and ratio B'G/ B'A transforms A into G. Youcan use their composition.

The converse statement is a consequence of the direct one.-----t ----+

One has A"G =- A'B, etc., thus A", B" and Gil satisfy the same relationas A', B' and G' do.

The points F, G and I lie on the parallel to A'B' through G, etc., thus I,J and K are three points on the sides of the triangle EFG. Prove that

~.~.~=1IG JE KF

and use Menelaus' theorem.

Exercise 1.38. The simplest thing to do is to use the associativity of thebarycenters. But this is also a consequence of Menelaus' theorem.

Exercise 1.39. You can use Menelaus' theorem six times in the triangleMNP where M = BG' n GA', N = GA' nAB' and P = AB' n BG'.

Exercise 1.40. IfAA', BB' and GG' are concurrent at 0, use Menelaus inthe triangles OAB, OAC, OBC and ABC.

Exercise 1.41. If (3 and 'Yare the points

(3 = BMnPB', 'Y = GMnPG',

the line CC' is the image of PM by a dilatation of center 'Y, hence 'Y, K andI are collinear. Similarly for (3, I and J. Moreover (3P'YM is a parallelogram,thus (3, I and 'Yare collinear.

For the next question, use an affinity about the line BC.

Exercise 1.42. You can use Desargues' theorem to construct another pointof the line, that is on the sheet of paper.

Exercise 1.44. The convex hull of Ao, ... ,AN is the image of

K = {(Xl, ... ,XN) ERN I0 ~ Xi ~ 1, LXi = 1}

304 A few Hints and Solutions to Exercises

(a notorious compact subset of R N ) by the continuous mapping that maps(Xl, ... , XN) to the point M defined by

N--+""'--+AoM = LJ XiAOAi'

i=l

Exercise 1.45. Let n be the dimension of the affine space. Extract n + 1independent points Ao, ... , An of S (use the fact that S is not contained in ahyperplane). Let U be the set of barycenters of these n + 1 points endowedwith strictly positive masses:

U = {M E £ IA;M =Xl~ +... +xnAoA~ avec Xi > 0et LXi < I} .- This is an open subset.- It is not empty since it contains the equibarycenter of the four points.- It is contained in e( S), which consists of the barycenters of the points

of the set S endowed with positive coefficients (Proposition 1-5.6).

We have constructed a nonempty open subset U contained in e(S), the interior of the latter is thus nonempty.

Exercise 1.49. One can use dilatations or even central symmetries andExercise U8.

Exercise 1.50. One gets the bijection by rewriting affine transformationsas compositions of mappings fixing 0 and translations. There is no groupisomorphism as can be seen by the comparison of the centers of the twogroups (and using Exercise 1.49).

Exercise 1.51. Ifnecessary, there is a proof of the "fundamental" theorem,proof that has inspired the statement of this exercise, in [Ber77].

Chapter II

Exercise 11.1. Write that IIAx + YII 2 is positive for all A, that is to say thatthe polynomial

A2 IIxI12 + 2AX . Y + IIYII2

keeps the same sign, or that its discriminant is negative:

(x· y)2 - IIxII2 . IIYII2 ~ 0 or Ix, yl ~ IIxil . IIyll .

This is the Cauchy-Schwarz inequality, equality holds if and only if:

- the polynomial has a real double root A (that is, AX+Y = 0 for some A)- and x . y ~ 0 (due to the square root),

that is, if and only if x and yare collinear and have the same direction.


Exercise 11.3. For all points 0 and G, one has

OA2 = OG2 +GA2 +200· GA OB2 = OG2 +GB2 +2oo.GB

and thus also

The expected equality holds for some point G if only if the latter satisfies,for all point 0,

~ ~ ~

OG· (aGA + (1 - a)GB) = 0,

namely if G is the barycenter of ((A, a), (B, 1 - a)).

Exercise 11.4. It can be shown that cp preserves the barycenters, for instanceusing Exercise 11.3. Denoting the images of the points by " the equality ofExercise II.3 and the preservation of distances imply that one has, for allpoints 0, the equality

aO' A,2 + (1 - a )0'B,2 = 0'G,2 +aG'A,2 + (1 - a )G'B,2.

Ifwe were sure that the point 0' obtained can be any point, this would imply,still using Exercise II.3, that G' is the barycenter of ((A', a), (B', 1- a)), andthis would be enough to assert that cp is affine (Proposition 1-2.8).

It suffices thus to prove that cp is surjective. Fix an affine frame. Tosimplify, let us work in a plane, so that the frame consists of three noncollinearpoints A, Band C. Their images A', B' and C' are not collinear either (dueto the preservation of distances and the triangle inequality). Let N be a pointof the plane. There is a unique point M such that

MA=NA', MB=NB' MC=NC'

and then cp(M) = N.

Exercise 11.5. One can expand Ilf(,xx + fLY) - Aj(x) - fLf(y) 112 or use the

fact that there exist orthonormal bases, or even use Exercise 11.4.

Exercise 11.9. The formula for F(M) is proved exactly as that of Exercise11.3 (that is a special case). The loci are a circle (maybe empty or reducedto a point), a line and, in the last case, a circle if k =F 1.

Exercise 11.13. Use a square exterior to the triangle, one of the sides ofwhich is BC and a dilatation of center A.

--+Exercise 11.14. The vector M M' is constant.


Exercise 11.15. The absolute value ofthe ratio of the dilatation is the ratioR'/ R of the radii. The center is one of the points S of the line 00' of centersthat satisfy SO' / SO = R'/ R. One can then use the last question (or answer)of Exercise II.9. Be careful with the case where the two circles have the sameradius and the case where they have the same center.

Exercise 11.17. If the two lines intersect at A and if the given point doesnot belong to one of the two lines, one can use a circle tangent to the twolines and a dilatation of center A.

Exercise 11.19. Let I be the midpoint of AB. Use the dilatation of ratio---t _

-1/2 and center the centroid G of AMB to prove the equality MH = 201(0 is the center of e). One can also use translations (see [DC51]) or angles(next chapter).

Exercise 11.20. One can begin by determining the dilatations that transform the circumcircle to ABC into e.

Exercise 11.21. Use the ''trick'' of Exercise 1.23.

Exercise 11.23. Let J be the diagonal matrix

Isometries have matrices A that satisfy tAJA = J. The elements of Oq(2)are the matrices of the form

A (c cosht TJ sinht) h R d 2 2

=. h t h were t E an c = "I = 1.sm cTJcos t

Thus the group Oq is not compact when n = 2 (and not more compact ingeneral): it contains a closed noncompact subgroup. The group 0: has twoconnected components ("I = 1) and the group Oq has four.

Chapter III

Exercise 111.2. One can use the properties of the angles at the circumference(Proposition 1.17).

Exercise 111.3. Evaluate the scalar product AB .AG.


Exercise 111.4. The bisectors of the angles at A and B intersect at a point I.This point is equidistant from C A and C B and thus belongs to one of thebisectors of these two lines. It also belongs to the sector defined by the angleat A and to that defined by the angle at B, it is thus inside the triangle andhence on the internal bisector of the angle at C. Analogous reasoning for theexcircle.

Exercise 111.6. This is the vector of the translation you are required tofind!

Exercise 111.8. The affine mappings are the

z t-----+ az + b-Z + c,

the affine transformations those for which lal2 -lbl2 f:: O.

Exercise 111.9. Remember that O+(P) is commutative, that a nontrivialtranslation cannot preserve a bounded subset and that it has infinite orderin the affine group.

Exercise 111.10. Let <PI, .. " <Pn be the elements of G and let M be anypoint of e. The equibarycenter of <PI (M), ... ,<Pn(M) is fixed under all theelements of G.

Exercise 111.12. The ratio of a similarity mapping e to e' is the ratio

k = ~ of their radii. The center A satisfies the inequality ~~ = k. It is

thus:

- on a circle (the circle of diameter 88' where 8 and 8' are the centersof the dilatations mapping e to e', see Exercise ILl5) if R f:: R',

- on the perpendicular bisector of 00' if R = R'.

And of course, you have to study the reverse inclusion...

Exercise 111.15. The composition of two inversions of the same pole is adilatation, not an inversion.

For the conjugacy, this can be done by a computation in complex numbers.There is also a geometric way of reasoning. Assume (this is enough) that thepower p of the inversion I = I(O,p) is positive, and let e be the circle ofinversion. Conjugate I by I' = 1(0., k). Let A and B be a point of theplane and its image by I. Any circle S through A and B is orthogonal to e(Proposition 4.12). Its image S' by I', a circle through the images A' and B'of A and B, is orthogonal to I'(e). Deduce that A' and B' are transformedone into the other by an inversion of circle I'(e). The conjugate I' 0 I 0 I' isan inversion of circle I'(e).


The pole is the center of this circle, that is easily determined: let J = 1(0),we have

~--. ----+---tOA·OB =p=on· OJ

hence the circle nAB passes through J, and therefore the line A'B' (itsimage by I') passes through J' = I'(J), that is, the pole (this being true forall points A and B = I(A)).

The power is determined by a direct computation using these points, it isfound that

See, e.g., [De51].

Exercise 111.16. Let 0 be the pole of the inversion. If A, A' and Bare----+ ----+

collinear, B belongs to the line OAA' hence B' too. Otherwise, OA· OA' isthe power of 0 with respect to the circle AA'B; this is also the power of the

--. ----+inversion and thus we have also 0 B . 0 B', therefore B' also belongs to thiscircle.

Exercise 111.19. You must draw a figure. The point C is fixed, the circleis transformed into the tangent at C, the sides AB and AD are preserved,they intersect the tangent at the images B' and D' of Band D, the lines BCand DC are transformed into the circles of diameters AB' and AD'. .. andall the assertions are easily justified.

Exercise 111.21. It has many failings: it is not defined everywhere, it doesnot preserve collinearity, it is not an involution!

Exercise 111.22. The theorem on the angles at the circumference and thereflection about I J give the equality of angles of lines

(AB, AI) = (AI, AC).

Therefore, the line AI is one of the bisectors of the angle at A of the triangleABC. The points A and I are on both sides of the line BC. The latter thusintersects the segment AI at a point K. Since AI is a chord of e, the pointK is inside the circle e; it is thus on the segment BC. Therefore AI is thebisector that intersects the segment BC, namely the internal bisector.

Exercise 111.23. Assume that the ABC is not right-angled (otherwise, thereis nothing to prove). By symmetry, we have

(DB, DC) = (HC,HB)

and by Chasles' relation,

(HC,HB) = (HC,AB) + (AB,AC) + (AC,HB).


As H C is perpendicular to AB and H B to AC, the equality of angles of lines

(DB, DC) = (AB, AC)

follows, hence A, B, C and D are cyclic.Assume now that the angles of the triangle are acute. The feet of the

altitudes D', E' and F' are then located inside the segments BC, CA and AB.

They are the images of the points D, E and F by the dilatation h ( H, ~ ).

One has

CD=CH=CE,

thus C belongs to the perpendicular bisector of DE. We want to apply theresult of Exercise 111.22. We must thus check that C and F are on both sidesof DE.

The points E' and D' are on both sides of the altitude CF'. Using thedilatation h(H, 2), the points E and D are thus on both sides of CF'. Therefore the line CF intersects the segment ED at a point inside the circle, henceinside the segment FC.

Exercise 111.24. For the existence, one can use the compactness of thesegments AB, BC and CA and the continuity of the function that, with(P, Q, R), associates PQ + QR + RP.

To minimize the perimeter of PQR amounts to minimizing the length ofthe broken line

The four points must be collinear in the order RlRQQl.As the angles at Band C are acute, R l and Ql are on the same side of

BC as A. As the angle at A is acute, RlQl intersects the halflines AB andAC of origin A. Hence Rand Q belong to the segments AB and AC.

To determine P: the triangle ARlQl is isosceles at A and its angle at thevertex is the double of that of ABC. To minimize RlQl amounts to minimizeARl = AP, so that P must be the foot of the altitude.

For the end, use Exercise 111.23.

Exercise 111.25. Use the fact that the altitude through A has length ~ Ra

to get !aRa ~ Area(ABC). To prove the Erdos-Mordell inequality, use thethree inequalities such as

aRa ~ We + eTb.

For the equality case, we must have a2 + b2 = 2ab, etc., that is, a = b = c(the triangle is equilateral) and aRa ~ brb + eTc, etc., that is, P belongs tothe altitudes. Hence P is the center of the triangle.


Conversely, if ABC is equilateral and P is its center, we have Ta = Tb = Tc

(P is the incenter), Ra = Rb = Rc (P is the circumcenter) and Ra = 2T a

(P is the centroid), thus we have indeed the equality.

Exercise 111.26. If AP is a bisector of the angle at A, consider the otherbisector AQ and the orthogonal projections B' and C' of Band Con AQ.It is checked that

PB AB'PC = AC'

(since the lines PA, CC' and BB' are not parallel) and that

AB' AB--=-AC' AC

(since the right-angled triangles ABB' and ACC' are similar).

Exercise 111.28. One can for instance (as this is suggested in [Ber77])prove the last equality criterion by mapping A to A' and B to B' by anisometry and adjust, then prove the two others using the metric relations inthe triangle (Exercises III.2 and III.3).

Exercise 111.30. This is a central symmetry. To determine its center, lookat the images of the contact points of the incircle with the sides of the triangle.

Exercise 111.32. Write the equalities of angles of lines

(AB,AA') = (B'B,B'A')

(CC',CB) = (B'C',B'B)

(CD, CC') = (D'D, D'C')

(AA', AD) = (D'A',D'D)

and (CB,CD) = (AB,AD) if we assume A, B, C and D cyclic. Adding thefive equalities to get (B'A', B'C') = (D' A', D'C'), gives the cyclicity of A',B', C' and D'.

Exercise 111.33. Because of the right angles, the circle of diameter MCcontains the points P and Q and the circle of diameter MB passes throughP and R. We thus have, for all points M

(PM,PQ) = (CM,CQ) = (CM,CA)

and

(PM,PR) = (BM,BR) = (BM,BA).

The points P, Q and R are collinear if and only if the equality (PM, PQ) =

(PM, P R) holds, that is, if and only if the points A, B, C and M are cyclic.


Exercise 111.34. The dilatation h ( M, ~) maps the points P', Q' and R'

on the orthogonal projections P, Q and R of M on the three sides. Thepoints P', Q' and R' are collinear if and only if P, Q and R are, that is(Exercise III.33) if and only if M belongs to the circumcircle of ABC (andthe Steiner line is parallel to the Simson line SM ).

Let D be the symmetric of the orthocenter H with respect to BC. Thisis a point of e (Exercise III.23). Assume that M is distinct from A, B, Cand D. The line M P' intersects the circle at another point pI! (equal to Mif M P' is tangent to e). We have the equality of angles of lines

(HP', MP') = (AP",AD)

from which we deduce that H P' and AP" are parallel.One proves also the equality

(AP", MP') = (SM,MP').

Hence ApI! is parallel to the Simson line and H P' is parallel to ApI!. Wededuce that H P' is the Steiner line of M and thus that the latter passesthrough H.

We still have to check that the Steiner lines of the points A, B, C and Dpass through H. But these are the altitudes for A, Band C and the parallelto the tangent at A through H in the case of D.

Exercise 111.36. Choose A on D1 . The vertex C is deduced from B by arotation of center A and angle ±1r/3; it is thus also on the image D~ of Dzby such a rotation.

Exercise 111.38. Let 0 be the midpoint of BD. One checks that the rotationof center 0 and angle ±1r/2 maps P to S, thus OP = OS and OP -lOS.Similarly, OQ = ORand OQ -lOR, hence the associated linear rotation

---+ ---+maps PR onto SQ.

Therefore the diagonals of PQRS are orthogonal and have the same length.The quadrilateral PQRS is a square if and only if they intersect at theirmidpoints. Let a and (3 be the midpoints of the diagonals PQ and RS, w themidpoint of AC. Notice that the quadrilateral O(3wa is a square. ThereforePQRS is a square if and only if (3 = a, that is, if and only if 0 = w, that is,if and only if ABCD is a parallelogram.

Exercise 111.39. The line OP is the perpendicular bisector of AF but isalso that of E B, hence that of E'B', thus the triangle P B'E' is isosceles ofvertex P. The triangle OAB is equilateral and its median AB' is perpendicular to BO. Therefore P and B' are on the circle of diameter OA and we

312

have


(PB',PO) = (AB',AO)

thus the angle at the vertex of the isosceles triangle PB'E' has measure 7r/3and this triangle is equilateral. Moreover, we have

{B'M= ~OC~ 1E'N= -OD

2{

B'M'=E'N

and thus --+ ~ ~ _ 7r(B'M,E'N) = (OC,OD) == '3 mod 27r.

--+ ~

The linear rotation of angle 7r/3 maps B'M to E'N. The rotation of centerP and angle 7r/3 maps B' to E' and thus M to N, from which it is deducedthat PMN is equilateral.

This exercise can also be solved using complex numbers and the characterization of (direct) equilateral triangles by "a + bj + ej2 = 0".

Exercise 111.40. With the assumption about angles, the point F is insidethe triangle.

Let M be any point. Then M A + M B + M C is the length of the brokenline C'M' +M'M +MC where M' is the image of M by a rotation of centerB and angle ±7r/3. Deduce that, if there exists a solution, it is at F.

Exercise 111.42. For rigid motions: all finite subgroups of the group ofplane rigid motions are abelian (Exercise III.9). For isometries: Qt4 has noindex-2 subgroup.

Exercise 111.45. Consider the triangle (similar to the others) Abc, where bis the orthogonal projection of A on 'D, together with its orthocenter h. Thetriangles ABb and ACe are similar, thus C belongs to the perpendicular toAe at e and similarly H belongs to the perpendicular to Ah at h.

Exercise 111.46. Consider the orthogonal projections m and m' of Son Dand D', the center w of the corresponding circle, the contact points t and t'of the tangents through S. If0 is the midpoint of MM', it is the image ofw by a similarity of center S that maps m to M and m' to M'. Deduce thatSwO and StT are similar and eventually that T and T' belong to wt and wt'.

Exercise 111.48. One can consider the points A and A' opposite to I one and e' respectively and M', the other intersection point of M J and e'.Check that cr(A) = A' and that the triangles IMM' and IAA' are similar.As for the point P, it lies on the circle of diameter I J.


Exercise 111.50. Observe that B' and C' belong to the circle of diameterBC to obtain the relation, which also says that a belongs to the radical axisof the circumcircle of ABC and its Euler circle.

Exercise 111.51. Use Menelaus for the transversals mentioned in the text-- ---+ ---+--and use the cyclicity to write equalities such as AR· BR = CR· DR... then

conclude.

Exercise 111.52. In general, there is an inversion of pole 0 that maps e toe' if and only if there is a dilatation of center 0 that maps e to e'.

Exercise 111.53. The pole is one of the intersection points of the circles ofdiameters AC and BD.

Exercise 111.54. Let 0 and 0' be the centers of the circles, and A one oftheir intersection points. In the triangle 00'A, we have the relation

d2 = R2+ R,2 - 2RR' cos A

and we conclude by preservation of angles.

Exercise 111.55. One may consider the images A', B' and C' of A, BandC by an inversion of pole D. Prove that the inequality is an equality if andonly if the points A', B' and C' are collinear in this order; this is equivalentto saying that A, B, C and D are cyclic "in this order" (that is, to the factthat the quadrilateral ABCD is cyclic and convex).

Exercise 111.56. Consider the image of the line M M' by the inversion---+ ----+

I (P, P A· PM), which is a circle centered on ~ and passing through A and A'.

Exercise 111.58. This is a circle of the pencil spanned by the two circles(the radical axis if k = 1).

Exercise 111.60. Let M be a point of the plane, not on the radical axis~ of~. Choose a point P on~. On the line PM, there is a unique point

----+ --M' such that PM· PM' is the power of P with respect to the circles of thepencil ~. The center of the circle we are looking for is the intersection pointof the perpendicular bisector of M M' and the line of centers of ~.

Exercise 111.61. This is essentially done in the text!

Exercise 111.62. One can use Exercise III.61 and the easy case where e ande' are concentric.


Exercise 111.63. If0 is a point of the plane and H its orthogonal projectionon the line D, the image 1(0, k)(D) is a circle of diameter OHI where HI

-- ---+is the point of OH such that OH . OHI = k. IfD' is another line, consider1(0, k)(D'), and deduce that the locus consists of the two bisectors of D andD' (minus the intersection point of D and D').

The centers of the in- and excircles are thus the poles of the inversionsthat transform the three sides in circles of the same radius. Let 1 be one ofthese four points. The images of the sides are the circles centered at a, f3and 'Y; they intersect at 1 and at the inverses A', B' and C' of the vertices.The inverse of the circumcircle of ABC is the circumcircle of A'B'C'. Onecan then argue like this: the point 1 is the center of the circumcircle of af3'Y,that has the same radius as the three others, the dilatation of center 1 andradius 1/2 maps A', B' and C' on the midpoints of the sides of the triangleaf3'Y. .. Hence the radius of the circumcircle of A'B'C' is twice that of theEuler circle of af3'Y and thus(I) it is equal to that of the circumcircle to af3'Y.

Up to this point, only the poles of the inversions have been used. Fixnow the power. Let 1 be the center of the incircle of ABC, consider theinversion of pole 1 and power k = 0/2 - R2 . The circumcircle is invariantand all the circles images of the sides have the same radius R. Let K be thecontact point of the incircle with the side BC and let K' be its image. one

---+ --has IK' = 2R and IK . IK' = -2Rr from which the expected relation isdeduced. The readers who want can prove analogous relations involving theradii of the excircles (all these relations can also be proved, of course, in asimpler way).

Exercise 111.66. For the end: if the function has an essential singularityat infinity, it cannot be injective. It is not hard to prove, indeed, usingWeierstrass' theorem that the set of points that are reached infinitely manytimes is dense in C (see [Car95]). Otherwise, this is a rational function, andthe injectivity assumption allows us to conclude.

Chapter IV

Exercise IV.I. Its cosine is -1/2.

Exercise IV.4. What is the conjugate of a translation by an affine transformation?

(1) At this point, I assume that the reader has drawn a figure.


Exercise IV.6. Use the corresponding linear result. One checks easily: ifthe composition is a translation, the axes are parallel, if this is a rotation,the axes have a common point and are thus coplanar. .. if the axes are notcoplanar, the composition is a screw displacement.

Exercise IV.7. It is obvious, by definition of the reflections, that -Sp is asymmetry with respect to the line pl... Then, if D and D' are two lines, wehave

SD' 0 SD = S p' 0 Sp,

where P and P' are the planes orthogonal to D and D' and hence this is arotation of axis PnP'. Ifthe composition of the two half-turns is a half-turn,it is thus necessary (and sufficient) that P and P' are orthogonal, namely thatthe lines D and D' are orthogonal; the axis of the composed half-turn is thenthe line perpendicular to D and D'.

That the half-turns are all conjugated is a consequence of the conjugacyprinciple:

/0 SD 0/-1 = Sf(D),

hence, to conjugate S D and SD', it suffices to use a rotation that maps D

to D'.

Exercise IV.IO. There exists a reflection mapping Ul to VI (about theperpendicular bisector plane of U1 and VI) j the vector U2 is mapped to anotherunit vector v2' such that

The perpendicular bisector plane of V2 and V2 thus contains VI; the reflectionabout this plane fixes VI and maps V2 to V2. The composition of the tworeflections is a rotation and has the expected properties.

Exercise IV.12. IfV and ware collinear, this is easy. Otherwise, it is clearthat the result of the computation is a vector of the plane spanned by V andw. The simplest thing to do is to choose a direct orthonormal basis (el, e2, e3)such that

1el = MV' (el, e2) = (v, w)

and to compute in this basis.

Exercise IV.14. If the angles are equal and if A, B, C and D are fourconsecutive vertices, one has AB = CD.

Exercise IV.16. Figure 14 illustrates the property... but, to prove it, onecan also look at Figure 15.


Exercise IV.17. Any isometry that preserves the cube transforms one of thetwo tetrahedra into one of the two tetrahedra. The group of the isometriesof the cube that preserve one of the tetrahedra is an index-2 subgroup (thereare indeed isometries that exchange them). Moreover, any isometry thatpreserves this tetrahedron preserves the cube, hence the group of isometriesof the (abstract) regular tetrahedron is an index-2 subgroup in the isometrygroup of the cube.

Exercise IV.lB. If A' and B' are the images of the two vertices A andB of the tetrahedron by an isometry, we have A'B' = AB. If, moreover,the isometry preserves the tetrahedron, A' and B' must be points of thistetrahedron. But the length AB is the largest possible distance between twopoints of the tetrahedron and it is realized by two points only if these pointsare vertices. Hence A' and B' are vertices of the tetrahedron.

The morphism maps an isometry to the permutation of the vertices that itdetermines. It is injective because an affine mapping is completely determinesby the image of an affine frame. Since the reflection with respect to theperpendicular bisector plane of an edge realizes the transposition of the endsof this edge, the image of our morphism contains all the transpositions. Asthe latter generate the symmetric group, the morphism is surjective.

Exercise IV.20. It is easy to make a list of the rigid motions that preservea cube, but a little less simple to prove that we have forgotten nothing. Theconsideration of the two tetrahedra gives the order of the group "for free" andallows us to stop once we have found twenty-four rigid motions.

The length of a main diagonal is the biggest distance between two pointsof the cube; this is why the isometries preserve the set of main diagonals.This way we have a homomorphism cp from the isometry group of the cubeto the symmetric group 64. It is clear that cp is not injective: the reflectionthat exchanges two parallel faces fixes each main diagonal. To prove that cpnevertheless defines an isomorphism from the rigid motion group to 64 is asomewhat boring exercise. Here we know the order of the rigid motion groupand it suffices to exhibit a rigid motion that induces a transposition of themain diagonals to have the surjectivity of cpo .. and thus also its injectivity.

Exercise IV.21. We know that the composition r is a rotation and that itsaxis passes through A. The point C' ofthe plane BCD defined by C'B = DCis fixed (use the fact that the edges AB and CD are orthogonal). Thus theaxis is AC'. We still have to determine the angle of the rotation. One canconsider the image of Bj it is easy to check that this is the point /3 defined by


---+ ~

Af3 = CB. IfH is the orthogonal projection of Bon AC I, one checks that

BHf3 is a right-angled (at H) triangle and deduces that the angle is ±7r/2.

Exercise IV.22. The triangles GAB, GBC and GCA have the same areabecause the barycentric coordinates of G are (1,1,1) ... but it is also possibleto solve this exercise by transforming ABC into an equilateral triangle by anaffine map of determinant 1 (that preserves the areas)!

Exercise IV.23. The existence of the Gergonne point is also an immediateconsequence of Ceva's theorem.

Exercise IV.24. Let 8 be the south pole. I suggest that the reader drawsa figure in the plane N 8 M. Let pI be the parallel plane to P through 8 andlet J.L be the intersection point of the line N M with Pl. We have

~ .m= N 8 2 = 4R 2

and-- - 1-- ----+ 2Nm·NM= -Nm·NJ.L=2R

2(the dilatation of center N and ratio 1/2 maps pI to P; I am copying theproof of Proposition III-4.8).

Thus cp is indeed the restriction of an inversion, so that it is continuous,it is clearly invertible. .. and its inverse is the restriction to S - {N} of thesame inversion, it is thus also continuous.

For the second part, notice that the triangles OMNand ONM' are similarand one deduces the inequality

---+ --+OM . OMI = ON2 = R2

•

Exercise IV.26. An origin must be chosen to define the longitude! However,the difference of two longitudes is independent of the choice of the origin.With his watch giving Washington time, Gedeon Spilett is able to measurethe difference between the longitude of the island and that of Washington.

The heroes of the "Fur Country" have certainly measured their latitude...but the height of the sun could not inform them on their longitude,

Exercise IV.27. The existence of the point Mo and the fact that d is positivecome from the continuity of the function and the compactness of its source

space. Then the point A is on the side "AM .AM;; < ~" and it suffices to2

check that C is on the other side. Assume thus that there is a point M in Csuch that

- ----+ ~MoM'MoA~2'


If N is a point of the segment MoM at the distance e of Mo, N is in C(use the convexity assumption) and, using the metric relations in the triangleAMoM, it is checked that, for e small enough, one has AN2 < d2 , which iscontrary to the definition of d.

Exercise IV.28. It is clear that 6 is convex, since this is an intersection ofhalf-spaces.

Since 0 is in the interior of C, for e > 0 small enough, the closed ballB(O, e) of center 0 and radius e is contained in C. Then

6 C B (o,~).Indeed, if u E P, one has u . v ~ 1 for all v in P and in particular for all v inB(O, e). Using v = cu II~II, the expected inequality lIull ~ lie is deduced.

The inclusion C C 6 holds by definition. For the inver,:>e inclusion, consider a point A out of C; prove that it does not belong to 6 by separating itfrom C by a plane as Exercise IV.27 allows us to do.

IfP is the convex hull of the finite set {Bl, ... , BN }, write----+~----+

v = OM = L....JAiOBi

with Ai ~ 0 and L Ai = 1, so thatN

p=n{UEElu.oB;~l}i=1

Then P is the intersection of a finite number of closed half-spaces. this is aconvex polyhedron according to the characterization given in Proposition 4.3and to the fact that it is bounded.

Exercise IV.30. Let d be the distance from the center 0 to the faces of P

and let MI,.'" MF be the centers of its faces. The faces of P are the planesof equations

----+ ----t 2OM·OMi=d,

the points of P are the points M such that----+ ----t 2OM·OMi ~d.

As for the dual of pI, this is the polyhedron defined by the inequalities----+ ----tOM· OMi ~ 1.

We just need to use the dilatation of center 0 and ratio d2 •

Exercise IV.3!. This is a parallelogram. In particular, its isometry groupis bigger than that of the quadrilateral we started from and hence of its dual(Exercise IV.29). Thus, they are not images one of the other by a dilatation.


Exercise IV.32. Assume the bathroom has been tiled, or the ball has been

made with hexagonal pieces. Each face has six edges, and each edge belongs

to two faces, thus 6F = 2E. Therefore the Euler relation gives

EF - E + V = "3 - E + V = 2, that is, 3V = 6 + 2E.

The number of edges at each vertex is a number r s that depends on the vertex

and satisfies r s ~ 3, thus

3V ~ LTs = 2E,

which gives the contradiction.

Exercise IV.33. The Euler formula gives twelve pentagons... but saysnothing on the number of hexagons (on a real soccer ball, there are twenty

hexagons). Indeed, if m is the number of hexagons and n the number of

pentagons, we have F = m + n, 2E = 6m + 5n, 3V = 2E and the Euler

formula gives6m + 5n 6m + 5n _ 2

m+n- 2 + 3 -,

that is, n = 12.

Exercise IV.34. Transform the figure by stereographic projection to see a

polyhedron drawn on a sphere and evaluate its numbers of vertices, edgesand faces.

Exercise IV.36. You must draw a figure. The great circles defined by xy,xz intersect the plane x.L at two points u and v such that

y = (cose)x + (sine)uz = (cosb)x + (sinb)v

and u· v = cosa. Evaluate cosa = y. z to obtain the formula. We have thus

\cosa - cosbcoscl .

. b . < 1, that IS, cos(b + e) < cos a < cos(b - e)sm sme

from which the triangle inequality is deduced. See also [Ber77, Chapter 18]for these considerations on the sphere.

Exercise IV.37. Extend 'P to E putting

{

ep(O) = 0

ep(v) = Ilvll 'P (II~II) if v # O.

Notice that ep preserves the scalar product (because x· y = cosd(x,y)), thenuse Exercise II.5.


Exercise IV.38. The two conditions imply the inequality

Icosa - cosbcosel 1.

. b . <SIn SIne

Let a E ]0, 11"[ be such that

cos a - cos bcos ecos a =. b .

SIn SIne

Fix x on S, construct two great circles through x making the angle a, puty and z on these circles at the distances band e of x, and the sphericaltrigonometry formula gives d(x, y) = a. Check the uniqueness.

Exercise IV.39. Draw a small equilateral triangle ABC in U. Let M beits center (intersection point of the perpendicular bisectors); this point isequidistant from A, Band C. Assume that f : U -> R 2 preserves thedistances. Let A', B', C' and M' be the images of A, B, C and M. ThenA'B'C' is equilateral and M' is its center. We prove now that the metricrelations in the two triangles are incompatible:

- in the "flat" triangle, we have A'B' = V3A' M',- but in the spherical triangle, we have AB < V3AM.

To prove this inequality, apply the spherical trigonometry formula (Exercise IV.36). Put a = AB and b = AM; we have

2 . 2 211"cos a = cos b + SIn bcos 3""

=- ~ + ~ cos2 b.2 2

We want to prove that a < V3b. For b small enough, this is a consequenceof the computation

3b2 9b4 3b2 b4

cos(V3b) = 1 - "2 + 24 + o(b5) and cos a = 1 - "2 +"2 + o(b5

),

which gives (exactly) the expected inequality. The case of a very small equilateral triangle is enough (but the inequality also holds true for the largetriangles) .

Exercise IV.40. The hints in the statement should be sufficient, the existence of the equilateral triangle coming from the result of Exercise IV.38.

Exercise IV.4I. This is the dodecahedron constructed on the cube of vertices (±1, ±1, ±1) as in Exercise IV.40 and in Figure 9.

Exercise IV.43. Use the fact that ep is the restriction to :P of an inversion(Exercise IV.24).


Exercise IV.44. See [Cox49] or [Art90] for instance.

321

Exercise IV.45. All the proofs, with other applications of quaternions, canbe found, e.g., in [Per96]' [MT86], [God71] and [Por69].

Chapter V

Exercise V.4. The complement of a line in a projective plane is an affineplane, hence it is connected. The complement of P1(R) in P 1(C) is thecomplement of R in C and therefore, it has two connected components (seeFigure 13).

Exercise V.6. Linear algebra, that's easy!

Exercise V.7. Let f : E -+ E be a linear isomorphism defining the projective transformation 9. A fixed point of 9 is the image of an eigenvectorof f. Complex endomorphisms have eigenvectors (thanks to the fundamentaltheorem of algebra) and the same is true of real endomorphisms of odd dimensional real vector spaces (for a similar reason).

For the counter-example, start from an isomorphism of R 2 without realeigenvalue, a rotation for instance. Here is one: z t-+ -liz (from whichrotation does it come?).

Exercise V.B. The translation in linear algebra is a special case of the(easy to prove and) very classical fact: two endomorphisms of the planeboth having two independent eigenvectors (namely, two diagonalizable endomorphisms) commute if and only if they have the same eigenvectors (namely,are diagonalizable in the same basis).

Exercise V.9. The homographies of P 1 that preserve 00 are the affinetransformations z t-+ az + b (with a =1= 0); those that also preserve 0 are thelinear isomorphisms z t-+ az (always with a =1= 0). The group of homographiespreserving 0 and 00 is thus isomorphic to the multiplicative group K*.

If a and b are two (distinct) points, let 90 be a homography that mapsthem to 0 and 00. Then 9 preserves a and b if and only if 909901 preserves 0and 00. The group of homographies that preserve a and b is thus conjugatedto K* in PGL(2; K), in particular, it is isomorphic to it.


Exercise V.U. Choose a basis (el, e2, e3) of the vector space E defining Pin such a way that m = p(el) and D = P( (e2' e3))' The line m* is the imagein P(E*) of the plane (e:i, e3) (linear forms vanishing on ed.

The subspace of E of equation (ae:i + be3)(U) = 0 intersects the plane(e2, e3) along the line spanned by (b, -a). The incidence is thus the homography associated with the linear isomorphism

(e:i,e3) ~ (e2,e3)(a, b) ....--- (b, -a).

Exercise V.12. The projective hyperplanes come from two (linear) hyperplanes F and F' , and the point m comes from a line D. The hypothesis isthat D is contained neither in F nor in F'. Let 1 : E - E be the projectiononto F ' in the direction D. It defines, by restriction, an isomorphism

IIF:F~F'

and the projective transformation g.

The perspectivity 9 is also the composition of the mappings

H - m* m* - H'andx ~ mx d ~ dnH '

which are, indeed, incidences (by definition for the second one, by duality forthe first one: write H = A* for A E P(E*), x E H is a line of P(E*) throughA and mx its intersection point with m*).

Exercise V.13. It can immediately be checked that I(B) = B, I(M) = A',I(C' ) = N, I(a) = , on the one hand, and that g(B) = B, g(M) = A',g(C') = N on the other. Therefore 9 = 1 and also g(a) =" but g(a) is theintersection point of (3a and B'A, thus a, (3 and, are collinear.

Exercise V.14. One could choose another secant line D' and prove thatthere exists a homography (a perspectivity) from D to D' mapping ai to a~.

Exercise V.15. The incidence m* - D is a homography (Exercise V.U)that maps di to ai.

Exercise V.IB. Use MenelaUs' theorem.

Exercise V.19. One obtains, of course, Thales' theorem. What has beenused is the fact that perspectivities are projective transformations, namelythe fact that the projections are linear mappings... this is the same argumentas in Chapter I (I hope that nobody had a doubt).


Exercise V.20. If BB', ee' and DD' are concurrent at m, we have theequality of cross-ratios (this is the cross-ratio of the four lines mA, mB, meand mD). The converse is a consequence of the direct statement.

Exercise V.21. Consequence of V.16.

Exercise V.22. Let 0 be the center of e. The circles eand e' are orthogonal:::-t ---+ -----+

if and only if 0 A 2 = 0 B 2 = OM· ON, a relation which is equivalent to[A,B,M,N] =-1.

Exercise V.23. The construction is shown in the figure. To check that itworks, one may either consider the affine situation where A is at infinity oruse perspectivities.

Exercise V.24. One can send 0{3 to infinity.

Exercise V.26. There is only one homography that fixes 1 and exchanges°and 00, this is z t--+ liz. One maps first a, b, c, d (in this order) to 00, 0,1,[a,b,c,dj, then (by liz) to 0,00,1, [a,b,c,d]-l. The composition maps (inthis order) b, a, c, d to 00, 0, 1, [a, b, c, dj-l. The latter is thus the cross-ratio[b, a, c, dj. To prove that the other equalities hold, one can argue similarly,with:

- the homography z t--+ [a, b, c, d] -1 z that fixes 00 and °and maps[a, b, c, d] to 1 for the first one,

- the homography z t--+ 1 - z that fixes 00 and exchanges°and 1 for thesecond one.

Exercise V.28. Let p and p' be two distinct points exchanged by g. Let qbe another point and q' be its image. We want to prove that g(q') = q. Wecompute

[p,p',q,q'] = [g(p),g(p'),g(q),g(q')]

= [p', p, q', g(q')], then exchanging the two points in each pair

= [p, p', g(q'), q'], hence the result.

As f is a homography, we have

[aI, a2,a3, aj] = [a~, a~, a;, f(aj)].

Hence the condition [al,a2,a3,aj] = [ai,a2,a~,aj] is equivalent to aj =

f(aj). We use the previous property.Translation in linear algebra: an involutive endomorphism of R 2 that is not

a dilatation and that has a real eigenvalue has two distinct real eigenvalues.


Let >. and J.L be these eigenvalues. Using a basis of eigenvectors and theformulas for the cross-ratio, it is checked that, for any point m,

>.[a, b, m, g(m)] = -.

J.L

In particular, this cross-ratio is constant. The case where 9 is involutive isthe special case where J.L =- >..

Exercise V.29. The image of a circle or a line by a homography is also acircle or a line. The required images are thus determined very easily onceyou know the images of a few points.

Exercise V.30. Writing E = i EB H, the lines of E H are identified with thegraphs of the linear mappings from i to H.

Exercise V.31. The composed mapping g2 0 gIl is a projective transformation; the images are "glued" using a projective frame of H (see Chapter 4in [Ber77]).

Exercise V.32. Let D and D' be two points, a, band c be three linesthrough D, a', b' and d be three lines through D'. The lines joining thepoints an b' and a' n b, and and a' n c, bn c' and b' n c are concurrent.

Exercise V.33. This is a consequence of Pappus' theorem. Notice first that,however short your ruler, it is easy to draw arbitrarily long lines. Call thetwo points you want to join a and 13. Using the ruler, you can draw two longlines through a making a very small angle and two lines through 13 with thesame properties, so that the four lines determine a small quadrilateral. LetC and C' be two opposite vertices of this quadrilateral (see Figure 1). Drawa line D through C that intersects C'a at Band C'j3 at A. Draw a line D'through C' that intersects j3C at A' and aC at B'. Choose them so that thepoints A and B' are close enough and similarly the points A' and B. Thusyou can draw the lines A'Band AB'. They intersect at a point 'Y of the lineaj3 (this is Pappus' theorem). If it is not close enough to a or 13, play thesame game again.


Fig. 1

325

Exercise V.34. Assume that AP, BQ and CR are concurrent at a point Dand compute the cross-ratio of the four points of d:

[P',Q',R',P] = [AP',AQ',AR',AP]

=[AP', AC, AB, AD] (intersection with BC)

=[P',C,B,D'] where D' = APnBC

=[DP',DC,DB,DD']

= [P', R, Q, P] (intersection with d)

= [P,Q,R,P']

and apply the result of Exercise V.28. Incidences and perspectivities can alsobe used in place of intersections; this is strictly equivalent. The converse is aconsequence of the direct statement.

Exercise V.35. The affine Euclidean plane is considered as an affine subspace (z = 1) of a Euclidean vector space of dimension 3. The line at infinitycomes from the plane z = O. A rotation of angle ±1l'/2 in this plane is a linearisomorphism that defines the homography in question.

Exercise V.36. Apply the second theorem of Desargues (Exercise V.34) tothe triangle ABC and the line at infinity. The lines AP, BQ and CR arethe altitudes and the points P', Q' and R' are the points at infinity of thesides of the triangle. They correspond to the directions of the altitudes viathe orthogonality relation, which is an involution, thanks to Exercise V.35.Therefore, the altitudes are concurrent.


Exercise V.37. The Hausdorff property comes from the fact that the set ofunit vectors on any line is compact.

Exercise V.38. Straightforward verification for the product formula. Forthe cubic aspects, draw a cube and call two parallel faces, respectively,A'GB'D, AG'BD', in such a way that AA', BB', GG' and DD' are themain diagonals. For the applications, we just need to be very careful withthe notation.

- For Miquel's theorem, we need to change the names of the points(2)in the statement of Exercise II1.32. The circles e4 and e1 intersect atD and G', e1 and e2 at B and A', e2 and e3 at G and D', e3 and e4

at A and B'.- For Simson's line, project a point d of the plane on the three sides of

a triangle abc at a', b', c' and let, in this order, ABGDA'B'G'D' =

abc'coa'b'cd to check that a', b' and c' are collinear if and only if a, b,c and d are cyclic.

- For the pivot, the notation of Exercise III.35 works (let D be the otherintersection point of the circles AB'G' and A'BG' and put D' = co).

Exercise V.41. Consider Cartesian equations of the lines.

Exercise V.44. One could use the affine result (see Exercise 1.51), or use aprojective but analogous reasoning.

Exercise V.50. To learn more on the Poincare half-plane or for proofs ofthe properties presented here, or to learn even more on hyperbolic geometry,see, e.g., Chapter 19 of [Ber77] and [Cox42].

Chapter VI

Exercise VI.2. This is the dimension of the vector space of symmetricmatrices.

Exercise VI.3. It can be written as the difference of the squares of twolinear forms. Therefore, its rank is ~ 2.

(2lTo every problem its notation.


Exercise VI.5. The bilinear form defined on Rn by

cp(x, y) = (Ax) . y = tyAx

327

is symmetric. There exists thus, thanks to Theorem 8.8, an orthonormal basisthat is orthogonal for cpo Letting P be the matrix of the change of basis, itfollows that t PAP is diagonal. Since P is orthogonal, t P = p-l; we havethus found a change of basis P such that p- 1 AP is diagonal.

Exercise VI.6. Use a basis of eigenvectors for the endomorphism ip-l 0 ;P'.

Exercise VI.9. See Figure 5 of Chapter VIII.

Exercise VI.12. No difference between ellipses and hyperbolas...

Exercise VI.17. No difference between an ellipse and a circle from the pointof view of affine mappings.

Exercise VI.19. One could begin with the case of a circle and then noticethat all the notions involved are affine.

Exercise VI.20. The nicest thing to do is to interpret geometrically theorthogonality condition, considering the circle of center 0 and radius a andthe affinity of ratio b/ a that fixes the major axis of the ellipse and maps the

----+ ~

circle onto the ellipse. The vectors OM and OM' are conjugated diametersif and only if the points M and M' are the images of M 1 and Mf withOM1 ..1 0 Mf. We are left with the case of the circle.

For the second theorem, write the coordinates of M and M ' in terms of ()such that the affix of Ml is eilJ .

Exercise VI.23. A parabola is determined by its focus and its directrix,that is to say by a point and a line not through this point. This figure isunique up to similarity (use Exercise 111.13). The proper conics that are notcircles are determined by focus, directrix and eccentricity, the latter beinginvariant by similarity...

Exercise VI.24. For a parabola, the order of the group is 2. For an ellipseor a hyperbola, this is as for a rectangle. For a circle...

Exercise VI.26. It is then the sign of q that determines the type of theconic.


Exercise VI.27. The answer ised

p= ----:1 + ecosO

where e is the eccentricity and d the distance from F to the directrix.

Exercise VI.29. A segment, a half-line.

Exercise VI.31. An ellipse, a hyperbola (of foci F, F') ... and a line ifF' E e.

Exercise VI.32. The first question can be solved by a computation usingan equation of e. For the second, assume M P is parallel to an asymptote.Let N be a point of M P, n its projection on D and N' its projection on P F.Thanks to the first question and to Thales' theorem, the ratio N N' / N Ftends to 1 when N tends to infinity on MP. But this ratio is the sine of theangle at F in the right-angled triangle NF N'. You still have to check thatthe limit of this angle is the angle at P of M PF and that this is an acuteangle, a contradiction.

Exercise VI.33. Solve the differential equation in 1/f and contemplate thesolution of Exercise VI.27.

Exercise VI.34. The projections of the focus on the tangents to the parabola also lie on the tangent at the vertex (Corollary 2.17), therefore thistangent is the Simson line of the focus (see, if necessary, Exercise III.33), sothat the focus F is on the circumcircle of the triangle determined by the threelines. The directrix is the image of the tangent at the vertex by the dilatationof center F and ratio 2, this is thus the Steiner line of F (see Exercise III.34).

When the focus F tends to a vertex A of the triangle, the Steiner line,which, as all Steiner lines do, passes through the orthocenter, tends to thealtitude of the vertex A and the parabola degenerates into a double line.

Exercise VI.35. Call the lines AFB, AEC, DEF, DCB (have you drawnthe figure?). Using Exercise VI.34, we know that the focus must lie on thecircumcircles of the triangles ABC, CDE, BDF and AEF and that thispoint must have the same Simson line for the four triangles.

The circumcircles of ABC and BDF intersect at B, hence, either theyhave a common point P, or they are tangent. If they were tangent, thelines DF and AC would be parallel, which is contradictory to the generalposition assumption. Let thus P be the second intersection point. It lies onthe circumcircle of ABC, thus its projections PI, P2 and P3 on BC, CA andAB are collinear. It also lies on the circumcircle of BDF, thus its projections


PI, P4 and P3 on B D, DF and FB are collinear. All the expected propertiesare deduced.

Exercise VI.36. See [DeS1], [LH61] and Chapter 17 of [Ber17], wherethese classical results are illustrated and proved.

Exercise VI.37. Use the notation of §1. The conic passes through A andhas the equation

---+ ---+q(AM) + LA(AM) =O.

The line through A directed by u intersects it at the unique simple point Aif and only if q(u) = O. The projective conic has the equation

q(u) + ZLA(U) = O.

It intersects the line at infinity (z = 0) at the points (u,O) suchthat q(u) = o.

Exercise VI.40. Use the classification of real quadrics.

Exercise VI.42. Three points can be chosen on each of the three lines. Nowremember that there is always a quadric through nine points...

Exercise VI.43. The two last pictures of the top line represent conics generating the same pencil. The same is true of the two first pictures of thebottom line.

Exercise VI.44. The intersection points of the two conics are the sameas those of one of the conics with a degenerate conic of the pencil. Thedegenerate conics are found by solving a degree-3 equation (proposition 3.11)and the intersection of these with one of the given conics by solving degree-2equations.

Exercise VI.4S. If the point m is not on the degenerate conic e, its orthogonal is a line that passes through the intersection point of the two lines(this is proved, e.g., in coordinates in which an equation of e is xy = 0). If

the point m lies on the conic...

Exercise VI.46. Let eo be the degenerate conic consisting of the lines xy

and zt, and let, similarly, e1 be that consisting of xz and yt. Consider thepencil they span. The point n is on the polar of m with respect to eo, andon the polar of m with respect to e1 (an application of Exercise VI.45); it isthus orthogonal to m for all the conics of the pencil (all the conics throughx, y, z and t) and in particular to e, the one we are interested in.


Exercise VI.47. Let u be a vector spanning the vectorial line definingthe point A of the projective plane. If ep and ep' are the polar forms oftwo quadratic forms defining two conics spanning the pencil 1', the polars inquestion are the images of the vectorial planes P>..,>..' defined by the equivalence

v E P>..,>..' {::::::::} (Aep + A'ep')(U,v) = O.

Ifep and ep' are nondegenerate, all these planes intersect along the intersectionline u.l Q n U.lQI .

Exercise VI.4S. Degenerate Pascal, this gives Pappus.

Exercise VI.50. Consider an arbitrary line D through e and construct itssecond intersection point f with e using Pascal's theorem. The notationsused here are consistent with those of Figure 18 and of the proof of Pascal'stheorem (Theorem 4.4). The point z = ab n de is well defined, as is the pointu = be n D; the line uz intersects ed at t, and f is found at the intersectionof D and at.

To construct the tangent at f, apply Pascal's theorem to the hexagonfbcdef (with a = f, check that the theorem remains true when replacing afby the tangent at J). Let z' = fb ned, u' = be n fe, t' = u'z' n cd, then thetangent is the line ft'.

Exercise VI.5!. This is a conic through A, B, C and D.

Exercise VI.52. In the projective plane, a pencil of circles is a pencil ofconics through the circular points. One of the degenerate conics of the pencilconsists of two intersecting lines, one of which is the line at infinity and theother the radical axis. The two other degenerate conics are imaginary.

Exercise VI.53. This is a hyperbola, as it intersects the line at infinity.The asymptotes are orthogonal, so that this is a rectangular hyperbola.

Exercise VI.59. The group homomorphism x ~ x2 gives the exact sequence

1 -----? {±I} -----? F; -----? (F;)2-----? 1,

thus there are q ; 1 squares in F~ and q ; 1 in F q. When y varies, the quant-

.1- by2 q + 1 .. . q + 1Ity a2 takes -2-'" among whIch there IS a square, smce 2-2- > q.

The proof of the "normal form" for the quadratic forms proceeds by induction on n. The n = 1 case is clear. The main point is the n = 2 case:the preliminary question gives a vector el such that Q(el) = 1. Let f be anorthogonal vector. IfQ(J) is a square a 2

, put e2 = f ja; the form is x~ + x~.Otherwise, Q(J) = b is not a square, then ajb is a square (F~j(F~)2 has onlytwo elements), a = ba2

, put e2 = af, the form is x~ + ax~. For n > 2, start


with an orthogonal basis C1, ... ,Cn· In the plane (c1, cz) there is an el suchthat Q(eI) = 1, then apply the induction hypothesis to H = ef·

The number a (or better to say, its class in F~/(F~)Z} is the discriminantof the form (Exercise VI.58), thus the two given forms are not equivalent.Moreover, a quadratic form belongs to one or the other type according towhether it has a trivial or nontrivial discriminant.

Exercise VI.61. As M is invertible, the symmetric matrix t M M is positivedefinite; one can take its square root for S (Exercise VI.60) and put 0 =MS- 1.

To prove uniqueness, use the polynomial P. If M = OS = 01S1, thentMM = Sr The matrix SI commutes with S~, thus also with tMM, withP(tM M) and with S. The two symmetric matrices Sand SI are diagonalizable and commute, therefore they are diagonalizable in the same basis andeventually they are equal.

Exercise VI.62. Use the polar decomposition (Exercise VI.61) M = OSand diagonalize SasS = t02DOz.

Exercise VI.64. Two points m and m' of P2(R) are in the same orbit ifthere exists a linear isomorphism 1of R 3 such that 15 = D' and q 0 1= q(D and D' are the two lines in R3 that define m and m'). According to Witt'stheorem (Theorem 8.10), this is equivalent to the existence of an isomorphismf : D -+ D' such that

(q IDI) 0 f = q ID .

Now, on a I-dimensional real space, there are exactly three types of quadratic forms (zero, positive or negative definite). Thus there are three orbitscorresponding to:

- lines (strictly) inside the isotropy cone, where the form is negativedefinite,

- generatrices of the cone, where the form is zero,- lines (strictly) outside the cone, where the form is positive definite.

Exercise VI.66. The proofs of these two theorems can be made very simpleby the use of duality. See, e.g., Chapter 2 of ITab95J.

Exercise VI.69. The only thing you have to be careful about: once youhave chosen two vectors el and it such that cp(el' it) = 1, you have to checkthat the restriction of cp to the orthogonal of the plane they span is stillnondegenerate; then you can use induction to construct the basis.

The dimension of E is even, 2n say; the dimension of F is less than orequal to n.


Chapter VII

Exercise VII.I. Take the origin at A by an affine change of coordinatesthis will not affect the rationality of the solutions. The intersection points of'D t and e are the (x, tx) such that

q(x, tx) + LA(x, tx) = 0

(with the notation of Chapter VI). This is a degree-2 equation in Xj it hasthe root x = 0 (corresponding to the point A). Thus, it has the form

x(a(t)x +b(t)) = 0

where a and b are polynomial of degrees ~ 2 and a(t) =ft O. The coordinatesof the second intersection point are the rational functions

b(t) tb(t)x = - a(t) y = - a(t) .

The case of the unit circle (x2+ y2 = 1) and the point A = (-1, 0) givesthe classical parametrization by ''t = tan B/2" (you are requested to draw afigure!)

1- t2

x=--1 + t2

2tY = 1 + t2·

Exercise VII.2. This is a "nodal cubic", with a node at the origin. A linethrough the origin intersects the cubic at the origin (double root) and atanother point. Ifthe slope of the line is t, one finds

x=t2-1 y=t(e-1).

For the "cuspidal cubic", the parametrization

x = t 2

is found in the very same way.

Exercise VII.3. This is a "figure eight" or rather a "figure infinity". ACartesian equation is

xy = (x2+ y2)2

and (hence) an equation in polar coordinates 2p4 = sin(2B).

Exercise VIlA. This is a lemniscate of Bernoulli. The asymptotes (tangents at 00) are transformed into the tangents at the pole (image of 00), etc.

Exercise VII.5. The two parabolas have equations y = x2 (when t tendsto infinity, x "J t, Y "J t2) and y2 = 8x (when t tends to 0, x "J 1/t2, Y "J 2/t).There is a singular point (obtained for t = 1), which is a cusp point.


Exercise VII.6. Fix two points M and N of 'D and the two correspondingpoints M' and N' of'D'. The assumption is that, for P in 'D, the corresponding point P' satisfies

------+ ----+M'P' MP"===t =~.M'N' MN

Let A be the center of the direct similarity that maps M to M' and N toN' (and thus also P to P'). The triangles AMN and AM'N' are similar,therefore AMM' and ANN' are similar too. IfHM is the orthogonal projection of A to MM', the triangles AHMM and AHNN are similar, thus thereexists a similarity of center A that maps the point M of'D to HM. Let /:1 bethe image of 'D under this similarity. The desired envelope is the parabolawhose focus is A and tangent at the vertex is /:1.

Exercise VII.7. Let Band B' be two points of D and D' respectively suchthat 0 B = a and 0 B' = a and let F be the intersection of the perpendicularbisectors of 0 Band 0 B'. The perpendicular bisector of AA' passes throughF. Let I be the midpoint of AA'. The triangle FAI is isosceles and rightangled; the locus of I is thus the image D" of D by a similarity. The line AA'envelops a parabola of focus F whose tangent at the vertex is the line D".

Exercise VII.S. Using Thales' theorem, it is proved that----+ ----+FB' FB===t = ===tFM FP

and

----+ ----+FB' FP===t = ----+FM Fe

and therefore we also have F p2 = FE. Fe. Hence the point F is fixed. Theenvelope is a parabola with focus at F and whose tangent at the vertex isthe line AP.

Exercise VII.n. The same group as for an equilateral triangle.

Exercise VII.12. Parametrize the "discriminant" curve by the multiple roott, writing

P(X) = (X - t)2(X + 2t),

which gives a and b in terms of t ... and once again the cuspidal cubic. Thesingular point corresponds to the polynomial X 3 , which has a triple root.The two components of the complement correspond to the polynomials thathave:

- a unique real root (the one that contains the points a = 0, b:l 0),- three real roots (the other one). See also Exercise VIII.9.


Exercise VII.14. A line that is parallel to the x-axis, when reflected at thepoint (cosO,sinO) of the unit circle becomes a line of slope tan(20). Differentiate with respect to 0 the equation

sin(20)x - cos(20)y = sin 0

of this line, to find the parametrization

x + iy = ~ (3ei9 _ e3i9 )4

which represents a nephroid.

Exercise VII.16. The radius of the circle e' is R and that of the circle eis r. Fix the origin at the center of the fixed circle e' and parametrize e' byRei9 . Assume that, at the beginning, the point M is at R (0 = 0, this is achoice of origin for 0). The hypothesis ''to roll without slipping" means that

- the circles are tangent;- when the point of tangency is at Rei9 , the point M that would have

run the distance RO along e' has run the same length along e (drawa figure).

A parametrization of the desired curve is easily deduced; this is

(R+r)ei9 ±rei~9,

hence the result.

Exercise VII.17. To find the equations of the cycloid is very similar towhat has been done in Exercise VII. 16. Then, the normal has the equation

. t t Rt' t °xsm 2+ycos 2 - sm 2 =.

The envelope of the normals is parametrized by

x = R(t + sin t) y = R(cost - 1).

This is the image of the cycloid by the translation (7rR, - 2R).

Exercise VII.IS. Let 0 be the center of the circle. With any point M(distinct from A) is associated the second intersection point M' of 'D M withthe circle. If0 is a measure of the angle (MA, MO), 40 is a measure of the

angle (oA,O'M\Let a be a measure of the fixed angle (OA, D). Evaluate the angle~~ 7r Oft-'(M' A, M'M) (you should find a - 0 - 2') then (0 ,0 M), a measure of

which is 20' - 20 -7r. Therefore M' turns twice faster than M in the oppositesense. The envelope is indeed a three-cusped hypocycloid.


Exercise VII.20. The matrix A(s) is orthogonal, thus satisfies, for all s,the relation

t A(s)A(s) = Id.

Differentiate this relation with respect to s to obtain

t (dA ) dAdi(s) A(s) + t A(s) ds (s) = o.

Let B(s) = tA(s)~~(s). The previous relation asserts that B(s) is skew

symmetric.If A(s) is the (orthogonal) matrix that describes the orthonormal basis

(7(S), n(s), b(s)) in the canonical basis, the matrix B(s) expresses the vectors(7'(S), n'(s), b'(s)) in the basis (7(S), n(s), b(s)). It is, indeed, skew symmetric.

Exercise VII.22. The parametrization of the astroid by -3e i8 + e- 3i8 be

comes, after the change of variables () = <p - ~ (and up to a rotation)

3e i<p + e- 3i

<p = 4(cos3 <p + i sin3 <p).

The ellipse is parametrized by (x,y) = (acos<p,bsin<p). The normal atthe point of parameter <p has the equation

ax sin <p - by cos <p - (a 2 - b2 ) sin <p cos <p = o.A parametrization of the evolute is deduced:

a2 _ b2 a2 _ b2

x = a cos3<p y = b sin3

<p.

Exercise VII.23. Differentiate the relation

C(s) = g(s) + p(s)n(s)

with respect to s to obtain the equality

C'(s) = p'(s)n(s)

and to conclude.

Exercise VII.25. The Steiner lines pass through the orthocenter H (Exercise 111.34), their envelope is thus the point H.

The dilatation h ( H, ~) transforms the circumcircle into the Euler circle

e (Exercise 11.20) thus the point J.L lies on e. Moreover

HE 'DM = h(M,2)(SM)

hence h(M, 2)-1(H) = J.L E SM. The point J.L is the midpoint of HM; it is thusprojected on BC at the midpoint of A'MA where A', the end of the altitude


from A, is a (fixed) point of the Euler circle. Moreover, the triangle A'J.LMAis isosceles of vertex J.L; thus the bisector of its angle at J.L is its altitude, henceit is parallel to the altitude from A (a fixed direction).

In conclusion, the Simson line J.LMA of M:

- passes through the point J.L of circle e,- is such that the bisector of the angle (A' J.L, SM) is parallel to the alti-

tude from A.

Just apply now the result of Exercise VILl8.

Chapter VIII

Exercise VIII.2. The curves () = constant are straight lines passing throughthe points of the z-axis, and the curves t = constant are circular helices.

Exercise VIII.3. It must be assumed that x and yare not both zero. Welook for the points (0,0, c) of the z-axis through which the lines containedin the surface pass. In fact, the Plucker conoid consists of two Whitneyumbrellas (see Exercise VIII.lO).

Exercise VIII.4. A hyperboloid of one sheet.

Exercise VIII.7. If the meridian curve is regular, the two vectors are collinear only at the points where it intersects the revolution axis. The parametrizations obtained in Exercise VIII.1 are never regular at these points.However, if the meridians have a tangent vector that is orthogonal to theaxis, the surface is regular and the tangent plane is the plane orthogonal tothe axis (one could use a Cartesian parametrization). Think of the exampleof the sphere.

Exercise VIII.8. The tangent vector at fau is w.The tangent plane at the point of parameter (t, u) is spanned by w(t)

and ')"(t) + uw'(t). It does not depend on u if and only if w'(t) is a linearcombination of ')"(t) and w(t).

Ifthe surface is a cylinder, w is constant hence w' is zero. Ifthis is a cone,w(t) = ')'(t) (taking the origin at the vertex of the cone) and w' = ')". In bothcases, the tangent plane at a point intersects the surface along the generatrixthrough this point. In general, the points of these surfaces are parabolic.


Exercise VIII.9. As in Exercise VII.12, to say that P has a multiple rootis to say that it can be written

P(X) = (X - u)2(X - v)(X +2u +v).

Expressing a, band c in terms of u and v, a parametrization of the "discriminant" surface is deduced. The surface has:

- a line of double points, when the polynomial has two double roots(v = -u),

- two curves of singular points corresponding to the polynomials with atriple root (v = u and v = -3u),

- these three curves intersecting at the point u = v = 0 correspondingto the polynomial X 4 , which has a quadruple root.

Exercise VIII.IO. Why "umbrella"? The points of the surface satisfy theCartesian equation x2 - zy2 = 0, which, conversely, describes the union ofthe surface and the z-axis. The part z < 0 of this axis is considered as thestick of the umbrella.

Exercise VIII.n. All these surfaces are tangent to the plane z = 0 at O.Therefore, no computation is needed. The second derivative is (up to afactor 2) the quadratic part of the equation. Nevertheless, you must draw apicture.

- For z = x 2 , we have a parabolic point, and the surface intersects thetangent plane along a line, staying on the same side of this plane; forz = x2 + y3, still a parabolic point (the second derivative is the same)but the surface has points on both sides of the tangent plane.

- For the other ones, 0 is a planar point. For z = x3 , the surface crossesthe tangent plane, which it intersects along a straight line; for z = x 4 ,

it intersects it along a line, but stays on the same side; for z = x 3

3xy 2, the surface intersects the tangent plane along three concurrentlines, and lies alternatively above and below the plane sectors limitedby these lines (there is a hollow for each of the legs of the monkey andanother one for its tail).

Exercise VIII.13. One shows firstly that, if p is an extremum of h on ~,

then (dh)p(X) = 0 for all X in Tp~. Indeed, use a curve 1 drawn on ~ andsuch that 1(0) = P and 1'(0) = X. Ifp is an extremum of h, 0 is an extremumof h 0 1, thus

(dh)p(X) = (h 0 1)'(0) = O.

Then, (dh)p is zero on Tp~ = Ker(dF)p (Proposition 2.10), hence

(dh)p = >'(dF)p.


On the sphere F-l(O), it is clear that the height function has a maximumat the north pole and a minimum at the south pole. This is confirmed by thestudy above:

(dF)p(X) = 2p· X (dz)p(X) = e3 . X

(this is the third coordinate of X) and (dz)p = A(dF)p if and only if p = ±e3.The condition of independency of (dF)p and (dG)p ensures that the tan

gent planes Ker(dF)p and Ker(dG)p to the two surfaces are distinct. It isthus possible to apply Proposition 2.15: the intersection is a curve r. Thesame argument as above shows that, if pEr is an extremum of h, (dh)p iszero on Tpr = Ker(dF)p n Ker(dG)p. An easy linear algebra lemma allowsus to conclude.

Exercise VIII.14. The tangent plane Tp~ is the kernel of the differentialof

F(x, y, t) = u(t)x + v(t)y + w(t)

at p. This kernel contains the t-axis if and only if of/at is zero. This way,we get again the linear system of §VII-I.

In the case of the cusp, the curve obtained is also the discriminant of thedegree-3 polynomials (in t).

Exercise VIII.15. One can simply say that all the points are planar orparabolic, and therefore that the Gauss curvature is zero. This can also bechecked, noticing that the normal vector is constant along the generatrices.Parametrize r by t f--+ 'Y(t) and the surface by (t, u) f--+ (1 + tu)'Y(t) = f(t, u).One gets

~{ = (1 + u)'Y'(t), ~~ = 'Y(t)

so that one can choose

'Y'(t) /\ 'Y(t)n(t, u) = 1I'Y'(t) /\ 'Y(t) II

which does not depend on u. Hence an/au = 0 and the Gauss curvature iszero.

Exercise VIII.16. The formulas of Appendix (§ 4) easily give the result inthis case:

1(

2 2 2)-2K( )

Xl x2 X3P = -+-+-

ala2a3 a~ a~ a~As can be expected (?), the curvature of the ellipsoids (all the ai'S positive)and that of the hyperboloids of two sheets (two among the ai'S negative) ispositive, while that of the hyperboloids of one sheet, ruled surfaces (one ofthe ai's positive) is negative.


Exercise VIII.17. One has

(

-h'(s) cos 0 )n(s,O) = -h'(s) sinO ,

g'(s)

hence

339

The Gauss curvature is thus

h' (s)h" (s) gil(S)

K(s,O) = g(s)g'(s) =- g(s) .

(1) IfK(s, 0) = 0, g(s) = as + b, up to a change of origin for s, one thushas

.h {g(s) = as {g(s) = belt er orh(s) = ±(J1-a2 )s h(s) = s.

One finds a cone (for lal < 1) or a plane (for lal = 1) in the first case,a cylinder in the second case.

(2) IfK(s,O) = K > 0, g(s) = acos(/Ks + b), one can assume that

{

g(s) = acos(/Ks)

h(s) 18

± V~1---a-2K-sl-'n-2 (-V'K-K-u-)du.

This is a sphere if a 2K = 1, otherwise the meridians are a little bitmore intricate, h being given by an "elliptic integral".

(3) IfK(s,O) < 0, one has g(s) = aexp(J-Ks) + bexp(-J-Ks).

Exercise VIII.18. The curvature of the inner tube of the tire is positive onthe ''tire side" and negative on the "rim side". By computation, parametrizethe circle by arc length

sg(s) = R+rcos-,

rh(s) = rsin ~

rR>r.

Use the formula at the beginning of Exercise VIII.17 and find

8

( ll) __c_o_s..!..r__K S,u =r(R + rcos~)

which is positive for sir in [-~,~] as we expected.


Exercise VIII.19. Fix a point 0 out of E. As the surface is compact, thefunction F(p) = IIpl12 reaches its maximum at some point of E. We showthat, at a point p for which a local maximum of F is reached, we have (for asuitable choice of the normal vector n)

1IIp(X,X) ~-2

Ilpllfor any unit vector X in TpE. In particular, the principal curvatures satisfythis inequality and thus the Gauss curvature is positive.

Notice firstly that TpE = pi., since (dF)p(X) = 2p· X must vanish onTpE. Choose the normal vector in a neighborhood of p in such a way thatn(p) has the opposite direction to p. Let X be a unit vector in TpE. Theplane perpendicular to TpE that contains X is the plane through 0 spannedby p and X. Let "( be a curve drawn on E, parametrized by arc length andsuch that ,,((0) = p, "('(0) = X. We know (Proposition 3.4) that

IIp (X, X) = ("("(O),n(p)).

Consider now the oriented plane Px = (X,p). The point p realizes also amaximum of F on "(. Write

82

"((8) = "((0) + 8r(0) + "2Kx(O)n(O) + 0(8 2)

8 2

= P + 8X + "2Kx(O)n + 0(8 2)

= 8X + (8; Kx(O) -IIPII) n + 0(82

).

Compute

111(8)11 2= IIpl12 + 82(1 - IlpllKx(O)) + 0(8 2).

Ifthis function of 8 has a local maximum at 0, the coefficient of 8 2 is negative,and this gives exactly the expected condition on K x.

Exercise VIII.20. Any surface you construct with a sheet of paper is flat(its Gauss curvature is zero).

Exercise VIII.21. All the points of the sphere are umbilics. For an ellipsoid,using the formulas of the appendix (§ 4), a point is an umbilic if and only ifthe equation in J.L given by the 4 x 4 determinant has a double root. Onefinds, if al > a2 > a3 (namely, up to a change of order, if the ellipsoid is not

a surface of revolution) that the umbilics are the four points of coordinates

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[Wey52] H. WEYL - Symmetry, Princeton University Press, 1952.

Index

In this index, the items are put in alphabetic order of all the words (e. g., "of aprojective space" before "of an affine space"). The numbers are page numbers.There are two kinds of items, something I explain by two examples:

- affine mapping: this is an elementary and basic notion; there are affinemappings everywhere in this text, and the number refers to the pagewhere the notion is defined;

- simultaneous orthogonalization of two quadratic forms: this is a resultthat is useful in many domains of mathematics (a ''transversal'' notion,in technocratic language), and the numbers refer to all the pages wherethis notion is used or mentioned.

On the other hand, the theorems that are only mentioned in the text (ingeneral for cultural reasons) show up in this index if they appear in the textwith a bibliographical reference.

action, 145, 190aerial,263affine

vs projective, 147, 159, 200, 210conic, 185frame, 12group, 19, 35, 40independence, 12, 33isometry, 44, 59line, 10mapping, 14plane, 10projection, 33property, 260quadric, 185space, 7subspace, 10

affinity, 36alternated, 117

bilinear form, 244alternating

group, 106, 130, 133, 140anallagmatic invariant, 108, 241angle

at the circumference, 74geometric, 71of a plane rotation, 55, 78of a rotation, 115of a similarity, 81of two curves, 85of two planes, 121oriented, 1oriented (of lines), 68oriented (of vectors), 66

Antarctic, 135anti-rotation, 114, 116apex, 271Apollonius, 2, 236, 238

348

apparent contour, 299arc length, 261Archimedian, 137area

of a spherical triangle, 121of a time zone, 120signed, 119, 286

astroid, 254, 266asymptote, 191, 260axiom

parallel, 2, 13, 122, 181axis

focal, 194major, 193minor, 193of a proper conic, 193of a rotation, 113, 115of revolution, 269of the Earth, 135radical, 92, 107, 217, 224

ball, see footballBanach,136barycenter, 26

preservation of, 17barycentric coordinates, 36, 119, 178base

locus, 206point, 94

basisdirect, 60orthogonal, 226orthonormal, 52

bathroom, 136Beethoven, 126Belgian story, 238Bernoulli, 262Bezout, 207bifocal property, 195bilinear form, 43, 225

alternated, 244symmetric, 225

biruled surface, 239bisector, 196

external, 72internal, 71perpendicular, 47

bisectors, 70, 199, 219Black, 135Bonnet, 294bowl, 263, see cup of coffeeBrianchon, 243canonical, see naturalcardioid, 254, 263

Index

Cartan, 242decomposition of, 242

Cartesiancoordinates, 12, 30equations, 31parametrization, 274

catenary, 299Cauchy, 44, 58caustic, 249, 258, 263, 264center

of a conic, 188of a quadric, 188, 214of a rotation, 78of a similarity, 80of an excircle, see excenterof curvature, 256of gravity, 27, see centroidof the circumcircle, see circumcenterof the incircle, see incenter

centralconic, 188dilatation, 16quadric, 188

centroid, 27Ceva, 38Chasles, 8, 68, 208, 244chessboard, 174cicumradius, 110circle, 193, 216

great, 120inversion, 84nine point, 62, see circle of Eulerof Euler, 1, 61, 111, 267osculating, 256, 266small, 120

circulargroup, 167, 224helix, 265points, 216

circumcenter, 61circumcircle, 61circumradius, 98classification, 189, 193, 204, 226common perpendicular, 131compactness

of the ellipse, 191, 193of the group O(n), 52of the projective space, 144of the sphere, 140

complementof a conic, 244of a hyperplane, 40of a line, 36, 40, 170

complete quadrilateral, 38, 61completion

projective, 149, 164, 200complex exponential, 54complex numbers

use in plane geometry, 79, 82, 99, 164cone, 243, 271, 289

isotropy, 200of revolution, 238, 270

confocalconics, 243family, 244quadrics, 244

conicaffine, 185, 189central, 188confocal, 243degenerate in a pencil, 207projective, 200proper, 190real, 244section, 238tangential, 243

conjugacy, 100conjugated diameters, 236conjugation, 21, 35

principle of, 21connectedness

of GL(n; R)+, 63of O+(n), 57of the ellipse, 191of the parabola, 193of the projective space, 144

conoid of Pliicker, 296constrained extrema, 298convex, 28, 122

hull, 29, 122coordinates, 12

barycentric, 36, 119, 178Cartesian, 12, 30homogeneous, 157, 179

cosine, 55criteria

for congruence, 103for similarity, 106

cross product, see vector productcross-ratio

and orthogonality, 216and permutations, 164computation, 162of four lines, 172of four collineal points, 161of four points on a conic, 209

Index 349

preservation of the, 161real, 166

cube, 128cubic

cuspidal, 251, 258, 261nodal,261real,244

cup of coffee, 249, see bowlcurvature

radius, 256algebraic, 255center, 256direction of, 290of a cylinder, 289of a cone, 289of a plane, 289of a plane curve, 255of a space curve, 265of a torus of revolution, 299of Gauss, 287of surfaces of revolution, 299of the sphere, 122, 124, 289principal, 290

curve, 258cycloidal, 252parallel, 257parametrized, 247regular, 258

cusp, 299point, 258

cuspidalcubic, 251, 258, 261

cyclicgroup, 130points, 76

cyclicity, 75, 166cycloid, 265cycloidal curve, 252cylinder, 271, 289

of revolution, 270decomposition

of Cartan, 242polar, 242

Desargues, 26,39, 151, 154, 176Descartes, 264diagonalization

of real symmetric matrices, 233simultaneous, 189, 230, 234, 285

diametersconjuated, 236of a circle, 223

diffeomorphism, 85differential of a mapping, 85, 111, 280

350

dihedralgroup, 130, 132, 141

dilatation, 16, 35dimension

of a projective space, 143of an affine space, 8

directbasis, 60similarity, 80, 165

directionof an affine space, 8of curvature, 290

directrix, 194, 219discriminant, 263, 274, 299

of a quadratic form, 241distance

Euclidean, 44hyperbolic, 181intrinsic, 138on the sphere, 138

dodecahedron, 128, 139dot product, see scalar productdouble

line, 185, 192, 207vector product, 131

dualof a conic, 243of a convex, 136of a polyhedron, 125, 136of a quadric, 243

dualityand polarity, 203between convex subsets, 136between polyhedra, 125projective, 153, 172, 242

DUrer, 2, 128, 143, 249Earth, 2, 122, 135, 176Easter Island, 126eccentricity, 194edge, 123ellipse, 190, 211, 266ellipsoid, 214, 269elliptic

integral, 339paraboloid, 214point, 283, 290

envelope, 199,248,299of normals, 257of the Simson lines, 266

epicycloid, 253, 264equation

Cartesian, 31of a quadric, 184

Index

parametric, 30reduced, 193

equibarycenter, 26Erdos, 102

Euclid, 2, 7Euclidean

affine space, 44vector space, 44

Euler, 61, 110, Ill, 123, 284, 293evolute, 257

of a catenary, 299of a cycloid, 265of a parabola, 266of an ellipse, 266

excenter, 98excircle, 98exponential

complex, 54external bisector, 72extrema, 295face, 123Fagnano, 101family

confocal, 244of lines, 249

Fermat, 105

Feuerbach, 110finite

subgroup, 130subgroup of 0+(3), 130, 140

firstcriterion for congruence, 103criterion for similarity, 106fundamental form, 287theorem of Apollonius, 236

flex, 259focal axis, 194focus, 194, 218football, see soccerform

bilinear alternated, 244fundamental, 287fundamental (first), 288fundamental (second), 287homogeneous, 185polar, 225quadratic, 43, 225, 283quadratic (positive definite), 43symmetric bilinear, 43, 225

formulaclass, 140of Euler, 123, 293of Girard, 121, 293

of spherical trigonometry, 138of Taylor, 259, 283to compute the cross-ratio, 162

formulas of Frenet, 265frame

affine, 12projective, 157, 178

Frenet, 265function

holomorpic, 165implicit, 247, 277, 281scalar of Leibniz, 60

fundamentalfirst form, 287second form, 287theorem, 41, 179

Gauss, 229, 285, 291, 293curvature, 287mapping, 286

generatedaffine subspace, 11projective subspace, 147

geodesic, 300geometric surface, 272geopolitical, 135Gergonne, 134Girard, 121, 293glide reflection, 77, 116global property, 248, 293golden number, 139Gram, 62Gram-Schmidt orthonormalization pro-

cess, 62Grassmann, 177Grassmannian, 177great circle, 120group

action, 2, 40, 190, 204affine, 19, 40affine (of the line), 35alternating, 106, 130, 133, 140circular, 167, 224cyclic, 130, 132, 140dihedral, 130, 132, 141linear, 19, 63of a polyhedron, 130of dilatations, 35of isometries, 44of isometries of a figure, 99, 100, 106,

132, 133, 263of Lorentz, 63of projective transformations, 156, 165of similarities, 80

Index

of the cube, 133, 141of the dodecahedron, 139, 141of the tetrahedron, 133, 141operation, see group actionorthogonal, 52projective, 156, 165, 204simple, 137symmetric, 130, 133

Hahn, 136half-plane, 36, 39, 59

of Poincare, 181, 293half-turn, 115harmonic

pencil, 172, 219range, 164

heart, 254, 263helicoid, 296, 300helix

circular, 265hexagon, 137, 182, 209, 243hexagram

mystic,209Hilbert, 7, 238Hobson, 135holomorphic

function, 165mapping, 111

homogeneouscoordinates, 157, 179form, 185

homography, 155, 159, 165, 208horse saddle, see mountain passhull

convex, 29, 122Huygens, 257, 274hyperbola, 190, 211

rectangular, 193hyperbolic

distance, 181geometry, 182paraboloid, 214, 234, 271plane, 182point, 283, 290triangle, 182

hyperboloid, 269of one sheet, 214, 296of revolution, 296of two sheets, 214

hyperplaneat infinity, 149, 160polar, 203projective, 145, 146tangent, 201

351

352

hypocycloid, 253, 264three-cusped, 1, 253, 263, 265, 267

icosahedron, 129image of a quadric, 185incenter, 98incircle, 98inequality

isoperimetric, 248of Cauchy and Schwarz, 44, 58of Ptolemy, 108triangle, 44triangular (on the sphere), 138

inertia law, 228infinity, 91, 148, 150, 159, 164, 221inflexion point, 259inradius, 110integral

elliptic, 339internal bisector, 71intersecting pencil, 94intersection

of a plane and a cone, 238of a quadric and a line, 186of two circles, 217of two conics, 207of two lines, 145of two surfaces, 247, 281

invariantanallagmatic, 241

inversion, 84, 134, 167, 222circle, 84

involution, 34, 84, 175Island

Easter, 126mysterious, 135

isometryaffine, 44, 59group, 44linear, 44, 59

isoperimetric inequality, 248isosceles triangle, 99isotropic vector, 225, 226isotropy cone, 200, 225Jacobi,244Jordan, 248Kepler, 237kidney, 254kinematics, 247latitude, 135law

of Descartes, 264of inertia of Sylvester, 228of Kepler, 237

Index

of Snell, 264Leibniz,60lemniscate of Bernoulli, 262length

are, 261limiting point, 94line

affine, 10, 149, 178at infinity, 148, 151, 152, 210, 216complex projective, 164contained in a quadric, 239double, 185, 192, 207normal to a curve, 256of Simson, 1, 104, 177, 238, 248, 266of Steiner, 104, 238, 267projective, 143, 147, 164real projective, 164

linear group, 19, 63Liouville, 83, 111local property, 248, 293locus

base, 206longitude, 135Lorentz, 63Mobius strip, 180major axis, 193manifold, 293map, 2, 122, 135, 139maple, 297mapping

Gauss, 286affine, 14holomorphic, 11ltangent, 280

massespunctual, 26

matrixof a quadratic form, 225, 233skew-symmetric, 245symmetric, 225, 233

measure, 70Menelaus, 38, 39, 209meridian, 120, 135, 270metamorphosis, 153method of Gauss, 229metric property, 215, 248, 293midpoint, 27minor axis, 193Miquel, 104, 177Mobius, 180, 300Monge, 283monofocal property, 194Mordell, 102

Morley, 267motion

of planets, 237science in, 2

mountain pass, 271, see horse saddlemultiplication of theorems, 154mystic hexagram, 209Napoleon, 105natural,8negative isometry, 50nephroid, 254, 263Newton, 2, 38, 264nine point circle, see circle of Eulernodal cubic, 261nondegenerate, 188, 226

quadratic form, 226norm, 44normal

vector, 254, 284normal to a curve, 256north, 134notation of Monge, 283Nullstellensatz, 185, 238number

golden, 139ocean

polar, 135octahedron, 129, 132Odyssey, 126one point compactification, 140, 148opposite similarity, 80optics, 249, 258, 263orientation, 60

of lines in space, 114, 179oriented angle, 1, 66, 68orthocenter, 61, 176orthogonal

vectors, 226basis, 226group, 52of a pencil, 95of a subspace, 43, 203pencils, 95, 96vectors, 43

orthogonality, 43, 226, 236of two circles, 89, 173of two pencils of circles, 95of two subspaces, 43

orthogonalization, 226simultaneous, 189, 230, 234, 285

orthonormal basis, 52orthonormalization, 62osculating circle, 256, 266

Index

Pappus, 25, 39, 150, 176parabola, 192, 211, 249, 266parabolic

aerial,263point, 284, 298

paraboloid, 269elliptic, 214hyperbolic, 214, 234, 271

parallel, 12, 120, 135, 145, 270axiom, 2, 13, 122, 181curves, 257subspaces, 12

parallelogram, 9, 32rule, 9

parallelsnonexistence of, 122, 145

parameter, 193parametrization

Cartesian, 274of a circle, 261of a conic, 208, 261of a curve, 247of a cuspidal cubic, 261of a hyperbola, 236of a nodal cubic, 261of an algebraic curve, 248of the ellipse, 191

parametrizedcurve, 247surface, 272

Pascal, 103, 108, 209, 240pencil, 93

harmonic, 172, 219intersecting, 94nonintersecting, 94

. of circles, 92, 217, 239of conics, 207, 239, 244of lines, 171, 208of quadrics, 206, 244of real conics, 239tangent, 95

pentagon, 128, 137, 139Perec, 1perpendicular bisector, 47, 59

hyperplane, 59Perrin, 177perspective, 2, 143, 174perspectivity, 171, 176photography, 176pivot, 104, 177planar point, 284, 298plane

affine, 10

353

354

hyperbolic, 182projective, 143, 148real projective, 179rotation, 78tangent, 277tangent to a ruled surface, 297tangent to a surface of revolution, 297

planet, 237Platonic solids, 2, 128Plucker, 177, 215, 296Poincare, 181, 293point

at infinity, 91, 148, 164, 167, 186, 221base, 94circular, 216cusp, 258elliptic, 283, 290hyperbolic, 283, 290limiting, 94of Fermat, 105of Gergonne, 134of inflexion, 259parabolic, 284, 298planar, 284, 298regular, 258, 273singular, 258, 273, 274, 277, 298weighted, 26

polardecomposition, 242form, 225hyperplane, 203ocean, 135of a point with respect to a conic, 203,

240polarity

with respect to a quadric, 203pole

north, 134of a hyperplane with respect to a

quadric, 203of a line with respect to a conic, 203of an inversion, 84

polyhedron, 122dual, 125Platonic, 2, 128regular, 126

porism of Steiner, 109position

general, 280of a curve with respect to its tangent,

259of a parabola with respect to its tan

gent, 236

Index

of a surface with respect to its tangentplane, 282

of an ellipse with respect to its tangent, 191

positive definite quadratic form, 43power

of a point with respect to a circle, 89of an inversion, 84

preservationof alignment, 156of angles, 70, 71, 81, 83, 86, 111, 165of barycenter, 17of circles, 91, 166of collinearity, 16, 41, 179of orientation, 50, 60of orthogonality, 44, 86of the cross-ratio, 161of the norm, 44of the scalar product, 44

principal curvature, 290principle

of conjugation, 21of Huygens, 257, 274

problem of Fagnano, 101product

scalar, 43, 215, 225vector, 117

projectionaffine, 33in projective geometry, see perspectiv

itystereographic, 134, 140, 164

projectivecompletion, 149, 159, 164, 200, 210conic, 200duality, 153, 172, 242frame, 157, 178group, 156, 165, 204hyperplane, 145, 146line, 143, 147, 164plane, 143, 148quadric, 200reader, 185space, 143subspace, 145transformation, 155, 204transformation (of the line), 159, 165

properaffine quadric, 185projective quadric, 200

propertyaffine (of a curve), 248, 260bifocal, 195

global, 248, 293local, 248, 293metric, 215, 248, 293monofocal, 194tangential (of conics), 196, 199, 243,

249pseudosphere, 299Ptolemy, 108punctual masses, 26pyramid, 126, 132Pythagoras, 98quadratic form, 43, 225, 283quadric, 184, 269

affine, 185, 234at infinity, 238central, 188confocal, 244degenerate in a pencil, 207in dimension 3, 234projective, 200proper affine, 185proper projective, 200real, 234, 239

quadrilateralcomplete, 38, 61

quaternions, 141radical axis, 92, 107, 217, 224radius

curvature, 256of the circumcircle, 110, see circumra

diusof the incircle, 110

rain, 2, 264rainbow, 2, 249, 264range

harmonic, 164rank

of a quadratic form, 227, 234of a skew-symmetric matrix, 245

ratioof a dilatation, 16of a similarity, 79

realconic,244cubic, 244quadric, 234, 239

rectangular hyperbola, 193reduced equation, 193reflection, 46, 59, 222

glide, 77, 116regular

point, 258, 273polygon, 126

Index 355

polyhedron, 126surface, 273

relationmetric (in the triangle), 98of Euler, 110

revolution, 269Riemann, 164, 293rigid motion, 50, 58rotation

in space, 113plane, 55, 78

rotatory reflection, 114, 116rule

parallelogram, 9slide, 1

ruled surface, 234, 239, 270, 296, 297ruler, 176saddle, 234, 298

horse, 271, 283monkey, 298

scalarfunction of Leibniz, 60product, 43, 215

Schmidt, 62Schwarz, 44, 58, 180Schwarz lemma, 180screw displacement, 116second

criterion for congruence, 103criterion for similarity, 106fundamental form, 287theorem of Apollonius, 236theorem of Desargues, 176

sectionconic, 238

separation, 135signature of a quadratic form, 228signed area, 119, 286similarity

criteria for, 106direct, 80, 165linear, 79opposite, 80

simple group, 137simplicity of 0+(3), 137Simson, 104, see line of Simsonsine, 55singular point, 258, 273, 274, 277, 298skew-symmetric

bilinear form, 244matrix, 245

slide rule, 1small circle, 120

356

Snell, 264soccer, 137space

affine, 7projective, 143tangent, 201

spannedaffine subspace, 11projective subspace, 147

sphere, 120, 164, 289of Riemann, 164

sphericaltriangle, 120, 138trigonometry, 138

Spilett, 135Steiner, 104, 109, 208, see line of Steinerstereographic projection, 134, 140, 164strip of Mobius, 300structure

of isometries, 46transport of, 9, 67

subspaceaffine, 10generated, 11, 147projective, 145spanned, 11, 147

subspacesparallel, 12

sum of the angles of a triangle, 73, 122,182

summit, 298sun, 135, 248, 264support of a geometric surface, 272surface, 272

geometric, 272of revolution, 269, 296parametrized, 272regular, 273ruled, 234, 239, 270, 296, 297

swallow tail, 274Sylvester, 228symmetric

bilinear form, 43, 225group, 130, 133matrix, 225, 233

symmetry, 34orthogonal, 45

tangenthyperplane, 201mapping, 280plane, 277space, 201to a central conic, 199

Index

to a conic, 188, 243to a curve, 248, 258, 279to a parabola, 198to a quadric, 186vector, 258

Taylor, 259, 283tetrahedron

regular, 128, 132scalene, 36, 133

Thales, 1, 23, 34, 173theorem

fundamental, 41, 179implicit function, 247, 277, 281inverse function, 276of Apollonius, 236of Bezout, 207of Brianchon, 243of Ceva, 38of Chasles, 244of Chasles and Steiner, 208of Desargues, 26, 39, 151, 154of Desargues (second), 176of ErdOs and Mordell, 102of Euler, 285, 289of Feuerbach, 1, 110of Gauss and Bonnet, 294of Hahn and Banach, 136of Jacobi, 244of Jordan, 248of Liouville, 83, 111of Menelaiis, 38, 39, 108, 209of Miquel, 104, 177of Pappus, 25, 39, 150, 176, 240of Pascal, 103, 108, 209, 240of Pythagoras, 98of Thales, 1, 23, 34, 173of the four vertices, 248of the six cross-ratios, 176of Weierstrass, 314of Witt, 213, 231of zeroes, 238remarkable of Gauss, 291

theorema egregiv.m, 291third criterion for congruence, 103time zone, 120topology

of 0+(3), 142, 180of the linear group, 63of the orthogonal group, 52of the projective space, 144of the real projective plane, 179

torsion, 265torus of revolution, 270, 299

tractrix, 299transformation

projective, 155translation, 16transport

by conjugation, 21of structure, 9, 67

transpose, 52triangle

hyperbolic, 182inequality, 44, 58spherical, 120, 138strict inequality, 58

trigonometryplane, 98spherical, 138

type of a conic, 193umbilic, 300umbrella of Whitney, 298underlying vector space, 8

Index 357

using complex numbers, 57, 79, 82, 99,164, 215

vectorisotropic, 225, 226normal, 254, 284tangent, 258

vector spaceEuclidean, 44

vectorialization, 9vertex, 123, 248

of a cone, 271of a parabola, 193

Vinci, 128, 137Vosges, 298wave-front, 258Weierstrass, 314Whitney, 298Witt, 213, 231zeugma, 4, 178

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Nikulin, V. v.; Shafarevich, 1. R.: Geometriesand Groups

Oden, J. J.; Reddy, J. N.: Variational Methodsin Theoretical Mechanics

0ksendal, B.: Stochastic Differential Equations

Poizat, B.: A Course in Model Theory

Polster, B.: A Geometrical Picture Book

POTter, J. R.; Woods, R. G.: Extensions andAbsolutes of Hausdorff Spaces

Radjavi, H.; Rosenthal, P.: Simultaneous Triangularization

Ramsay, A.; Richtmeyer, R. D.: Introductionto Hyperbolic Geometry

Rees, E. G.: Notes on Geometry

Reisel, R. B.: Elementary Theory of MetricSpaces

Rey, w.]. J.: Introduction to Robust andQuasi-Robust Statistical Methods

Ribenboim, P.: Classical Theory ofAlgebraicNumbers

Rickart, C. E.: Natural Function Algebras

Rotman, J.]': Galois Theory

Rubel, L. A.: Entire and Meromorphic Functions

Rybakowski, K. P.: The Homotopy Index andPartial Differential Equations

Sagan, H.: Space-Filling Curves

Samelson, H.: Notes on Lie Algebras

Schiff, J. L.: Normal Families

Sengupta, J. K.: Optimal Decisions underUncertainty

Seroul, R.: Programming for Mathematicians

Seydel, R.: Tools for Computational Finance

Shapiro, J. H.: Composition Operators andClassical Function Theory

Simonnet, M.: Measures and Probabilities

Smith, K. E.; Kahanpaa, L.; Kekalainen, P.;Traves, w,: An Invitation to Algebraic Geometry

Smith, K. T.: Power Series from a Computational Point of View

Smorynski, C.: Logical Number Theory I. AnIntroduction

Stichtenoth, H.: Algebraic Function Fieldsand Codes

Stillwell, J.: Geometry of Surfaces

Stroock, D. w.: An Introduction to the Theory of Large Deviations

Sunder, V. S.: An Invitation to von NeumannAlgebras

Tamme, G.: Introduction to Etale Cohomology

Tondeur, P.: Foliations on Riemannian Manifolds

Verhulst, F.: Nonlinear Differential Equations and Dynamical Systems

Wong, M. w,: Weyl Transforms

Zaanen, A.C.: Continuity, Integration andFourier Theory

Zhang, F.: Matrix Theory

Zong, c.: Sphere Packings

Zong, C.: Strange Phenomena in Convex andDiscrete Geometry

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link.springer.com978-3-642-56127...A few Hints and Solutions to Exercises The reader will find here a mixture of hints, solutions, comments and bib liographical references for …