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 ∈ Im(φ) then A0B0 ∈ 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 ∈ 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 congruence 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 ∈ 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 reflections, 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 6∈ ` ∩ 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 6∈ ` ∩ 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 rotation 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.

(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 ∈ 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 ∈ 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 4OPQ 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 ∈ `. 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 contradiction.

(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 ψ = σm ◦ σ` fixes O and sends A to A0. By Corollary 1, θ−1 ◦ ψ is the identity or has a line of fixed points. The second case cannot happen, since by Lemma 3, ψ 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.

(b) If τ is a translation, then for any two points A, B we have AB||A0B0 and AA0||BB0.

(d) Any translation is a product of two reﬂections.

(e) Is the group T of translations a normal subgroup of the group G of all rigid motions? Prove yes or no.

(a)-(b) We can assume that A 6= A0, otherwise we can take τ to be the identity. Deﬁne τ as follows: for any point B ∈ Π \ AA0 let ` be the unique line parallel to AA0 and passing by B. We deﬁne B0 = τ(B) to be the

4 unique point on `, on the same side of A0 with respect to the line AB such that BB0 ' AA0. Note that, by Proposition I.33 we have we have AB ' A0B0 and AB||A0B0. If else B ∈ AA0 we set B0 = τ(B) to be the unique point on AA0 such that BB0 ' AA0 and B0 is on the same side of A0 with respect to B if B ∗ A ∗ A0 or B = A, on the opposite side of A with respect to B in the other cases. In this case AB = AB0 = AA0 = BB0, so the lines are all parallel. Moreover, AB ' A0B0 by sum or diﬀerence of congruent segments.

In all case AB ' A0B0; by Lemma 1, τ is then a rigid motion.

Let τ 0 be another translation sending A to A0 and let B be a (general) point not on the line AA0. Since AA0 ' Bτ 0(B) and AB ' A0τ 0(B) the quadrilateral AA0Bτ 0(B) is a parallelogram by Lemma 4, hence τ 0(B) = τ(B). By Exercise (17.3) (a) the composition τ −1 ◦ τ 0 is the identity.

(c) Let τ1 and τ2 be two translations. Let A be any point; we claim that τ1 ◦ τ2(A) = τ2 ◦ τ1(A). If A, τ1(A) and τ2(A) are collinear this follows from addition or subtraction of congruent segments. If this is not the case just consider the parallelogram with vertices A, τ1(A), τ2(A),X and prove that X = τ1 ◦ τ2(A) = τ2 ◦ τ1(A).

Finally we claim that ψ = τ2 ◦ τ1 is a translation. Let A, B be two points. Then AB ' ψ(A)ψ(B) and AB||ψ(A)ψ(B), hence, by I.33 we have also Aψ(A) ' Bψ(B).

(d) Let A be any point, and A0 = τ(A). Let ` be the line perpendicular to AA0 passing by A, and let n be the perpendicular bisector of AA0. We claim that τ = σn ◦ σ`. To prove the claim it is enough to show that τ and σn ◦ σ` agree on three non collinear points. One of them is A. The other two can be taken on n. Note that we can decompose θ by choosing any A, so, in an euclidean plane, we can always assume that ` (or m) passes by a given point.

(e) The subgroup of translations is a normal subgroup. By Exercise 17.3 (b) it is enough to show that ψ := σ` ◦ τ ◦ σ` is a translation, if τ is 00 a translation. Pick A ∈ `, and set A = σ`(τ(A)). We claim that ψ is the unique translation sending A to A00. To prove the claim it is enough to prove that ψ acts as that translation on two points B,C such that A, B, C are not collinear. We can take B = A0 and notice that, calling B0 the intersection of the parallel to AA0 passing by A00 and the parallel to AA00 passing by A0 then B0 ∈ `. We can take as C any point on ` diﬀerent from A.

5 Exercise (17.11). In a Euclidean plane, show that the product of two rotations around diﬀerent points is equal to either a rotation around a third point or a translation. Hint: Show that it has at most one ﬁxed point.

0 0 −1 Let φ = θB ◦ θA, set A = θB(A), B = θA (B) and let `, m be the angle 0 0 bisectors of ∠ABA , ∠BAB . As we have seen in Exercise 17.4 (c), we can write θA = σAB ◦ σm, θB = σ` ◦ σAB Therefore φ = θB ◦ θA = σ` ◦ σAB ◦ σAB ◦ σm = σ` ◦ σm; By Lemma 2 if ` ∩ m 6= 0 this transformation has only one ﬁxed point (the intersection) and thus, by 17.4 (d) it is a rotation. If the two lines are parallel, then φ is a translation (it is the unique translation sending a point A on m to σ`(A), see Exercise 17.5. (d)). Lemma 5. The product of a rotation and a translation is a rotation.

Proof. We can write θ = σ`1 ◦ σ`2 and τ = σm1 ◦ σm2 . By Exercise 17.4 (d) we can choose `1 and `2 so that `2 is parallel to m1. By Exercise 17.5

(d) we can then choose m1 = `2. Therefore θ = σ`1 ◦ m2 (which are not parallel, since `1 and `2 were not parallel). So θ is a rotation by Lemma 2 and Exercise 17.4 (d).

Exercise (17.12). In a Euclidean plane, show that the product of an odd number of reflections cannot be equal to the identity. Hint: Use Exercise 17.11 to reduce products of four reflections to products of two reflections, and proceed by induction.

Let ψ be the product of 2k + 1 reflections. We prove the statement by induction on k. If k = 1 and σ1 ◦ σ2 ◦ σ3 = 1 then σ1 ◦ σ2 = σ3 is a reflection, but we have already observed (Lemma 2) that the product of two (different) reflections has at most one fixed point. So assume that the statement holds for the product of 2k − 1 reflections. The product of four consecutive reflections is equal to the product of two rigid motions, which are either rotations of translations. This product is then equal to a rotation or a translation, which in turn can be written as the product of two reflections. So the original rigid motion can be written as the product of 2k − 1 reflections and induction applies.