A few Hints and Solutions to Exercises The reader will find here a mixture of hints, solutions, comments and bib­ liographical references for some of the exercises. These are not complete solutions, even when they are rather detailed. They will exempt the reader - neither from writing their solutions (some of the hints given here are really written in telegraphic language) - nor from taking care of converses. When I write for instance ''the point P lies on the circle of diameter I J", I do not write ''the locus of points P is the circle of diameter I J". Last remark: the reader will often need to draw a picture to understand the solution. 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 A few Hints and Solutions to Exercises Hence we have (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 is collinear. 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 barycentric coordinates of M and if f..L is any nonzero scalar, (f..LAo"", f..LA n ) is also a system of barycentric coordinates of M. Exercise 1.29. Consider for instance the barycenter M 1 of the system «M,a), (M', -1)); prove that it is invariant and that the mapping that as­ sociates M 1 to M is a projection. Exercise 1.30. In dimensions greater than or equal to 3, you can ''turn around" a line; the complement is connected. A formal proof: choose an affine frame whose origin is on the line and whose first vector directs it; we are 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 n 1 e- D = R x (R - - {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 case of the orthocenter, you can exhibit three collinear points whose images are not collinear, for instance remembering that the orthocenter of a right-angled triangle is the vertex of the right angle. Exercise 1.33. Ifa subset has two symmetry centers, it is preserved by some translation: 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 midpoint J 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 the point A" (Figure 20) in the triangle ABG; you can also use the fact that the area 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. You can 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 relation as 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 ~.~.~=1 IG JE KF and use Menelaus' theorem. Exercise 1.38. The simplest thing to do is to use the associativity of the barycenters. But this is also a consequence of Menelaus' theorem. Exercise 1.39. You can use Menelaus' theorem six times in the triangle MNP 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 in the 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 and I 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 point of 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 N (a notorious compact subset of R ) 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 + 1 independent points Ao,... , An of S (use the fact that S is not contained in a hyperplane). Let U be the set of barycenters of these n + 1 points endowed with 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 in­ terior of the latter is thus nonempty. Exercise 1.49. One can use dilatations or even central symmetries and Exercise U8. Exercise 1.50. One gets the bijection by rewriting affine transformations as compositions of mappings fixing 0 and translations. There is no group isomorphism as can be seen by the comparison of the centers of the two groups (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 2 Exercise 11.1. Write that IIAx + YII is positive for all A, that is to say that the polynomial 2 2 2 A IIxI1 + 2AX . Y + IIYII keeps the same sign, or that its discriminant is negative: 2 2 (x· y)2 - IIxII . IIYII ~ 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. A few Hints and Solutions to Exercises 305 Exercise 11.3. For all points 0 and G, one has 2 2 2 2 OA2 = OG2 +GA +200· GA OB = OG +GB +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 instance using Exercise 11.3. Denoting the images of the points by " the equality of Exercise II.3 and the preservation of distances imply that one has, for all points 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)), and this 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. To simplify, let us work in a plane, so that the frame consists ofthree noncollinear points A, Band C. Their images A', B' and C' are not collinear either (due to the preservation of distances and the triangle inequality). Let N be a point of the plane. There is a unique point M such that MA=NA', MB=NB' MC=NC' and then cp( M) = N. 2 Exercise 11.5. One can expand Ilf(,xx + fLY) - Aj(x) - fLf(y) 11 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 Exercise 11.3 (that is a special case). The loci are a circle (maybe empty or reduced to a point), a line and, in the last case, a circle if k =F 1.