Foundations of Geometry Exercises Section 17 Lemma 1 (Exercise 17.2). Let φ :Π ! Π be a map of a Hilbert plane into itself. Assume that AB ' φ(A)φ(B) for any two points A; B. Then φ is a rigid motion. Proof. it is clear that, if A 6= B, then φ(A) 6= φ(B), hence φ is injective. Let us prove that it is also surjective. • If A0;B0 2 Im(φ) then A0B0 2 Im(φ); in fact, a point C0 in A0B0 is uniquely determined by the congruence class of A0C0 and B0C0; let C be the unique point on AB such that AC ' A0C0 and BC ' B0C0. Then φ(C) = C0. • Let A; B; C be non collinear points and let A0;B0;C0 be their images D0 2 Π be any point, and let E0 be a point such that A0 ∗ E0 ∗ B0; then the line D0E0 meets the image of φ in at least two points, by (B4), hence it belongs to the image. We have now to show that φ preserves lines, betweenness and angles. Let A; B; C be three points on a line, and assume that A ∗ B ∗ C. Let Γ and ∆ be circles of radii AB and BC centered at A and C. Then Γ and ∆ are tangent. the point φ(B) will be the unique point of intersection of φ(Γ) and φ(∆) (which are again circles, since φ preserves congruence of segments). Since the two circles are tangent then φ(A); φ(B) and φ(C) are on a line. Moreover, since φ(A)φ(C) ' φ(A)φ(B) + φ(B)φ(C) then φ(A) ∗ φ(B) ∗ φ(C) If \ABC is an angle, then the triangle 4φ(A)φ(B)φ(C) is congruent to the triangle 4ABC by I.8, hence \ABC ' \φ(A)φ(B)φ(C). Exercise (17.3). In a Hilbert plane Π, show: (a) Any rigid motion with at least three noncollinear fixed points must be the identity. (b) Any rigid motion is equal to the product of at most three reflections. 1 Let φ be a rigid motion with three non collinear fixed points A; B; C. Assume that there exists P such that φ(P ) 6= P . Since φ preserves con- gruence of segments, then the three fixed points must belong to the axis of P φ(P ), contradicting the fact that they are not collinear. Let φ be any rigid motion, let A 2 Π be a point, and let σ1 be the reflection sending φ(A) to A (or the identity, if φ(A) = A). Then φ1 = σ1 ◦φ has A as a fixed point. Let B be a point different from A, and let σ2 be the reflection sending B to φ1(B) (or the identity, if φ1(B) = B). Since AB ' Aφ1(B) the axis of σ2 passes by A, which is thus fixed. Therefore φ2 = σ2 ◦ σ1 ◦ φ fixes A and B. Let C be a point not on the line AB. If φ2(C) = C then φ2 is the identity, by part a). Else let σ3 be the reflection sending C to φ2(C). As before, it is easy to check that the axis of the reflection must pass by the fixed points A and B. It follows that φ3 = σ3 ◦ σ2 ◦ σ1 ◦ φ is the identity. Note the we have also shown that Corollary 1. A rigid motion with a point fixed is the product of two re- flections, and a rigid motion with two points fixed is a reflection (or the identity). Lemma 2 (Composition of two reflections). Let σ` and σm be reflections with ` 6= m. Then either ` \ m = O and O is the only fixed point of σ` ◦ σm or ` \ m = ; and σ` ◦ σm has no fixed points. Proof. It is enough to show that a fixed point for σ` ◦ σm has to belong to both lines. Assume that we have a fixed point A 62 ` \ m; it cannot belong to m, otherwise it would be fixed by σm, and so also by σ`, but this is not 0 possible, since A 62 ` \ m. So set A = σm(A). Then m is the perpendicular 0 0 bisector of AA . If A = σ` ◦ σm(A) = σ`(A ) then also ` would be the perpendicular bisector of AA0 and ` = m, a contradiction. Exercise (17.4). In a Hilbert plane Π, define a rotation around a point O to be a rigid motion ρ leaving O fixed and such that for any two points 0 0 0 0 A; B the angles \AOA and \BOB are equal, where ρ(A) = A , ρ(B) = B . Show: (a) For any two points A; A0 with OA ' OA0, there exists a unique rota- tion around O sending A to A0. (b) The set of rotations around a fixed point O, together with the identity, is an abelian subgroup of the group of all rigid motions. (c) Any rotation can be written as the product of two reflections. (d) A rigid motion having exactly one fixed point must be a rotation. 2 (a) Let Γ be the circle with center O and radius OA. Let B be the other point in which the line OA meets Γ, and let α be the angle AOA0 (we are assuming that A; O and A0 are not collinear). Let P be any point different from O; to define θ(P ) we construct the angle α on the ray −−! OP , such that −! • If P 2 OA then α is on the same side of A0 with respect to AB; • If P is on the same side of A0 with respect to AB then α will be on the side opposite to A with respect to OP . −−! • If P 2 OB then α is on the opposite side of A0 with respect to AB; • If P is on the opposite side of A0 with respect to AB then α will be on the same side of A with respect to OP . Then θ(P ) is the unique point on the ray defined by the angle α such that OP ' Oθ(P ). If A; O and A0 are collinear we define θ(P ) to be the intersection of OP with Γ different from P . 0 In both cases it is clear that \AOA ' \P Oθ(P ). To show that θ is a rigid motion it is enough to show that it preserves congruence of segments (by Lemma 1). Given two points P 6= Q one can show that the triangles 4OP Q and 4Oθ(P )θ(Q) are congruent by (C6) by using sums and differences of congruent angles. Notice that, by definition, a rotation different from the identity has only one fixed point. Lemma 3. Let φ :Π ! Π be the composition φ := θ ◦ σ` of the reflection in the line ` and a rotation with center O 2 `. Then φ is a reflection. Proof. If A 6= O is a point of ` then the points of the bisector of AOφ(A) would be fixed by φ. 0 −1 If θ1 and θ2 are two rotations around O sending A to A , then θ2 ◦ θ1 has at least two fixed points. If it were not the identity, then, by the final remark in Exercise 17.3 it would be a reflection σ`, with ` 3 O. In this case θ1 = θ2 ◦ σ` would be a reflection by Lemma 3, a contra- diction. (d) Let φ be a rigid motion with a unique fixed point O. Let A 6= O be a point. Since φ is a rigid motion OA ' Oφ(A). Let θ be the rotation with center O sending A to φ(A). Then θ−1 ◦φ has (at least) two fixed 3 points, O and A, hence it is either the identity or the reflection in the line OA. In the first case we have φ = θ, while in the second, φ would have a line of fixed points by Lemma 3, against the assumptions. (b) Let θ1 and θ2 be two rotations with center O. We need to show that −1 θ = θ2 ◦ θ1 is again a rotation with center O or the identity. Clearly θ(O) = O. If θ had another fixed point A then either θ is the identity or θ is a reflection, by Corollary 1. The second case cannot happen, by Lemma 3. The commutativity of the product follows by sum or difference of congruent angles. (c) Let θ be a rotation around O sending A to A0. Let ` be the line AO 0 and m be the angle bisector of \AOA . Then = σm ◦ σ` fixes O and sends A to A0. By Corollary 1, θ−1 ◦ is the identity or has a line of fixed points. The second case cannot happen, since by Lemma 3, would have a line of fixed points, contradicting Lemma 2. Note that we can decompose θ by choosing any A, so, in an euclidean plane, we can always assume that ` (or m) is parallel to a given line. Lemma 4. Let ABCD be a quadrilateral in which the opposite sides are congruent. Then ABCD is a parallelogram. Proof. Consider a diagonal and use I.8 to prove that the two triangles are congruent. Then use I.27 twice. Exercise (17.5). In a Euclidean plane Π, define a translation to be a rigid motion τ such that for any two points A; B we have AA0 ' BB0, where τ(A) = A0; τ(B) = B0. Show: (a) For any two points A; A0, there exists a unique translation τ such that τ(A) = A0.