CHAPTER 9

THE PRODUCT OF DIRECT ISOMETRIES

Given two direct isometries, say the rotation R_P,ϕ and the translation T_AB, we know that their product R_P,ϕ T_AB is a direct isometry, and we may even suspect that it is a rotation R_Q,θ. The question is, what is Q and what is θ? This chapter will describe some of the ways that we can determine the values of the parameters that describe the result, in this case, Q and θ.

9.1 Angles

We first need to clear up some possible ambiguities about directed angles.

In Example 8.3.2, we showed that the product of two reflections in nonparallel lines is a rotation whose center is the intersection point of the two lines and whose angle of rotation is twice the angle from the first line to the second line; that is,

where Q = l ∩ m and where α is twice the directed angle from l to m. If l and m are rays, as in (a) below, there is no ambiguity in the meaning of the phrase “the directed angle from l to m.” We can think of it as being 36° or −324°, since we identify the angles of θ and θ + 360n for all integers n.⁸

When we talk about the directed angle from one line to another, however, there are other possible interpretations. In (b) above, in addition to 36° and −324°, the directed angle from l to m can be legitimately interpreted as 216° or −144°. Does this affect the validity of the example? In fact, the answer is no, it does not matter which of the four angles we use. This is illustrated in the following table:

In each case, 2θ is always 72 plus or minus some integral multiple of 360, so the entries in the 2θ column all represent exactly the same angle.

9.2 Fixed Points

Under the identity, every point of the plane is fixed. Under a reflection, each point in the line of reflection is fixed. Under a rotation, the center of rotation is the only fixed point. Translations and glide reflections have no fixed points.

Conversely, given an isometry T, we know from Theorem 8.2.5 that if T fixes each of three noncollinear points, then T must be the identity. The following theorem tells us what we can conclude if we know that either one or two points are fixed.

Theorem 9.2.1. (Fixed Points)

Proof. In each case, these are the only isometries that have (at least) the specified number of fixed points.

It is possible to distinquish between translations and glide reflections in terms of fixed sets.

Theorem 9.2.2. An isometry that fixes exactly one line but does not fix any points is a glide reflection.

Proof. By the previous theorem, the isometry is either a translation or a glide reflection, since it does not fix any points. A translation T_AB fixes all lines parallel to AB, and since this is not the case, we must conclude that the isometry is a glide reflection.

It should be noted that halfturns and translations each fix infinitely many lines. A halfturn fixes each line that contains the center of the turn, and a translation fixes each line parallel to the direction of translation. A halfturn also fixes the center of rotation, as the point common to all of the lines that it fixes.

9.3 The Product of Two Translations

The effect of a translation T_UV is to map each point P to a point P′ such that is congruent to . This means that

Thus, if we know that an isometry T is a translation, we can write

where P′ = T(P). This handy fact is used in the next theorem.

Theorem 9.3.1. The product of two translations is a translation or the identity.

Proof. Let T_AB and T_CD be the two translations. The product T_CDT_AB must be a direct isometry, so it is either the identity, a translation, or a rotation. We will show that the only way the product can be a rotation is if T_CD and T_AB are inverses of each other, and so the rotation is actually the identity.

Suppose that T_CD T_AB = R_P,θ. Then

Let P′ = T_AB(P). Then T_AB = T_PP′. Similarly, T_CD = T_P′P, but then

Thus, the rotation would actually be the identity.

Note that if is not congruent to , then T_CD T_AB must be a translation T. To pin down this translation geometrically, consider the effect of T_CD T_AB on a point P. T_AB maps P to P′ and T_CD maps P′ to P″. Therefore, T = T_PP″. We can construct a directed line segment that is congruent to PP″ by completing the parallelogram EFGH, where and are congruent to and , respectively. Thus, T_CD T_AB maps E to G, and the diagonal is congruent to PP″.

9.4 The Product of a Translation and a Rotation

We will determine the product of a rotation and a translation in two stages. First, we will show that the product is a rotation, and then we will describe how to find the center of rotation and the angle of rotation.

The product of a translation and rotation is a direct isometry, so it is either the identity, a translation, or a rotation. Supposing that is not of length zero and that the angle of rotation is θ, where θ is not a multiple of 360°, a little bit of algebra shows that the product T_AB R_O,θ cannot be either the identity or a translation.

• It cannot be the identity. The reason is that
  
  and since dist(O, O′) = dist(A, B) ≠ 0, it follows that O ≠ O′. Since O is not fixed, the product cannot be the identity.
• It cannot be a translation. Let S = T_AB R_O,θ. Then multiplying both sides of this equation by T_BA (the inverse of T_AB), we get
  
  If S were a translation, then Theorem 9.3.1 tells us that T_BAS (and hence ) is the identity or a translation, which contradicts our assumptions about and θ.

The only possibility left is that the product is a rotation.

In a similar way, we can show that R_O,θ T_AB cannot be a translation unless R_O,θ is the identity or a translation, which is again a contradiction. Thus, we have shown:

Theorem 9.4.1. The product of a nontrivial translation and a rotation is a rotation, unless the angle of rotation is a multiple of 360°.

Although we know that the result has to be a rotation R_Q,ϕ, the theorem does not tell us how to find Q or ϕ. The next example docs this.

Example 9.4.2. Given . the point O. and the angle θ, where θ ≠ n · 180°, find Q and ϕ such that

Solution. The key here is to look for the fixed point Q. In other words, we are looking for the point Q such that

The point Q is transported to Q′ by R_O,θ and then back to Q by T_AB. The solution can be derived from the figure on the right. The points Q and Q′ are on a circle centered at O, with ∠QOQ′ = θ. The segment QQ′ is a chord of a circle, and so its right bisector passes through O.

The center Q can be constructed as follows (see the figure on the right).

Through O, draw a line parallel to , and let P be the point such that . Construct the right bisector l of , and construct the line m through O so that the directed angle from to m is 90° − θ/2. The point where m intersects l is Q.

This constructs the center Q of the rotation R_Q,ϕ. The angle of rotation ϕ is equal to the directed angle θ, and this can be confirmed by noting that ∠OQP = θ and that O is mapped to P by T_AB R_O,θ.

The product R_O,θ T_AB can be analyzed in a similar way. Here, the center would be a point X such that T_AB maps X to Z and R_O,θ maps Z back to X. In the first figure above, this would be the point Q.

Care must also be taken when the angle of rotation is negative.

Example 9.4.3. The figure below represents the centers of the rotations resulting from the four products

The directed segment is as shown. The points W, X, Y, and Z are the centers of rotation, although not necessarily in that order. Determine the correct center for each rotation.

Solution. The centers are as follows:

Example 9.4.4. Given the point Q and rotation R_O,θ, where θ is not a multiple of 360°, as in the figure on the left on the following page, find a translation T_CD such that

Solution. Again, we exploit the idea that if for some point X we can find its image X′ under T_CD, then T_CD = T_XX′.

We first solve for T_CD. Multiply the equation

on the right by the inverse of R_O,θ, as follows:

Next use the associative law and the fact that R_O,θ R_O,−θ = I to obtain T_CD:

Now apply T_CD to the point O (we chose O because it is fixed under R_O,−θ). Then

Letting P be the point R_Q,θ(O), we therefore have T_CD = T_OP, as in the figure on the right at the top of the page.

9.5 The Product of Two Rotations

The product of two rotations with the same center is a third rotation, also with the same center, through an angle that is the sum of the two angles of rotation; that is,

To analyze R_B,β R_A.α when A and B are different, we must first show the following:

Theorem 9.5.1. When A and B are different, R_B,β R_A,α is a rotation through an angle α + β about some point P.

Proof. We relay on the fact that we can use a translation to “move” the center of a rotation R_Q,θ to any point we like, as in Example 9.4.4.

Let T_CD be the translation such that

Let S be the product R_B,β R_A,α; that is

Multiply this equation on the left by T_CD to get

which implies that

Now multiply this equation on the left by T_DC, the inverse of T_CD, to get

From the previous section, we know that T_DC R_A,α+β is a rotation R_P,α+β for some point P (unless α + β is a multiple of 360°, in which case T_DC R_A,α+β = T_DC).

To find the center P when α + β is not a multiple of 360°, we can trace our progress through the preceding equations. For , we can take C = A and D = R_A,β(B), as in the discussion following Example 9.4.4. Applying the inverse transformation T_DC to R_A,α+β then allows us to geometrically construct the center P. A better option is to look for the fixed point P of the product R_B,β R_A,α, as in the following example.

Example 9.5.2. Given R_B,β and R_A,α, with α and β as shown in the figure below, construe: the center Q of the rotation R_B,β R_A,α.

Solution. The figure below shows how to solve this problem. Construct the line AB and an angle ∠XAX′ of size α such that AB is the bisector of ∠XAX′. Construct the angle ∠YAY′ of size β so that AB is the bisector of ∠YAY′. One of the points of intersection of these two angles will be the desired point Q.

Remark. In general, when trying to find the center for the product R_B,β R_A,α, the angles should be drawn so that the directed angle ∠XAB = α/2 and the directed angle ∠ABY = β/2. The point of intersection of the line XA and the line YB will be the center Q of the product.

For example, this is what the construction looks like when α = 30° and β = −60°. The point Q will be mapped by R_A,α to Q′, and Q′ will then be mapped by R_B,β back to Q.

9.6 Problems

1. If ℓ, m, and n are the perpendicular bisectors of the sides AB, BC, and CA, respectively, of ΔABC, then

is a reflection in which line?

2. If R_n R_m R_ℓ is a reflection, show that the lines ℓ, m, and n are concurrent or have a common perpendicular.

3. Find Cartesian equations for lines m and n such that

4. Show that

is a reflection in a line parallel to ℓ.

5. Let C be a point on the line ℓ, and show that

6. Given nonparallel lines AB and CD, show that there is a rotation T such that

7. Show that if S, T, T S, and T⁻¹ S are rotations, then the centers of S, T S, and T⁻¹ S are collinear.

8. In a given acute triangle, inscribe a triangle PQR having minimum perimeter. This is called Fagnano’s problem.

9. Prove Thomsen’s Relation: for any lines a, b, and c, we have

where I is the identity transformation.

10. Show that Thomsen’s Relation is still true if each reflection R_x is replaced by a halfturn H_X; that is, show that if A, B, and C are three distinct points in the plane, then

where I is the identity transformation.

11. If x′ = ax + by + c and y′ = bx − ay + d with a² + b² = 1 are the equations for an isometry T, show that T is a reflection if and only if

12. Find the Cartesian equation of the line m if the equations for a reflection in the line are

13. If the equations for the rotation R_P,θ are

find the center of rotation P and and the angle of rotation θ.

14. If a and b are lines in the plane, show that the following are equivalent:

(a) a = b or a and b are perpendicular,

(b) R_a R_b = R_b R_a,

(c) R_b(a) = a,

(d) (R_b R_a)² = I,

(e) R_b R_a is either the identity or a halfturn.

15. If the isometry H_P is a halfturn, show that given any two perpendicular lines m and n that intersect at the point P, we have H_P = R_m R_n.

16. Let m be a line with equation 2x + y = 1. Find the equations of the transformation R_m.

17. Given a line b and a point A, show that the following conditions are equivalent:

(a) A b,

(b) H_A R_b = R_b H_A,

(c) R_b (A) = A,

(d) H_A(b) = b,

(e) R_b H_A (or H_A R_b) is an involution,

(f) R_b H_A is a reflection in the line through A perpendicular to b.

18. If A ≠ C, show that the following conditions are equivalent:

(a) B is the midpoint of AC,

(b) H_C H_B = H_B H_A,

(c) H_B(A) = C,

(d) H_B H_A H_B = H_C.

19. Show that nonidentity rotations of the plane with different centers do not commute.

20. Let A and B be distinct points in the plane and let S be an isometry. Show that

where C = S(A) and D = S(B).

21. Given a point A and two lines ℓ and m, construct a square ABCD such that B lies on ℓ and D lies on m.

22. Given four distinct points, find a square such that each of the lines containing a side of the square passes through one of the four given points.

Hint. Given A, B, C, and D, we want to find the lines a, b, c, and d. Take P such that R_P,90(B) = C. Let R_P,90 (D) = E. Then take a to be AE.

23. Consider a triangle ΔABC (oriented counterclockwise) with positive angles α, β, γ at A, B, C. Show that

Are there other similar formulas?

24. Construct over each side of ΔABC an equilateral triangle. The centroids G₁, G₂, and G₃ of these triangles form a new triangle, the so-called Napoleon triangle. Show that ΔG₁G₂G₃ is equilateral. This theorem is attributed to Napoleon Bonaparte.

25. Let perpendiculars erected at arbitrary points on the sides of triangle ΔABC meet in pairs at points P, Q, and R. Show that the triangle PQR is similar to the given triangle.

26. Let A and B be two points lying on one side of a line ℓ. Explain how to construct one of the two circles that are tangent to ℓ and pass through A and B.

27. Given points A, B, and P in the plane, construct the reflection of P in the line AB using a Euclidean compass alone.

28. Let ABCD be a square with the vertices in clockwise order. For each of the following translations, find a counterclockwise rotation that brings the image of the translation back to the same physical space as ABCD:

(a) distance AB, direction AB,

(b) distance AB, direction AB,

(c) distance AC, direction AC,

(d) distance AC, direction AC.

29. Let A₀ B₀ C₀ be an equilateral triangle with the vertices in clockwise order. We first rotate it 60° counterclockwise about A₀ to obtain A₁ B₁ C₁, then about B₁ to obtain A₂ B₂ C₂. and finally about C₂ to obtain A₃ B₃ C₃. We continue to rotate about A₃, B₄, C₅, and so on, until A_n B_n C_n occupies the same physical space as A₀ B₀C₀. What is the minimum positive value of n?

30. Let A₀ B₀ C₀ be a triangle with the vertices in counterclockwise order, where ∠A = 40°, ∠B = 60°, and ∠C = 80°. We first rotate it 40° counterclockwise about A₀ to obtain A₁ B₁ C₁, then 60° counterclockwise about B₁ to obtain A₂ B₂ C₂, and finally 80° counterclockwise about C₂ to obtain A₃ B₃ C₃. We continue to rotate about A₃, B₄, C₅, and so on, until A_n B_n C_n occupies the same physical space as A₀ B₀ C₀. What is the minimum positive value of n?

31. A halfturn about a point O is a 180° rotation about the point O. Prove that the composition of:

(a) two halfturns is a translation or the identity;

(b) a translation and a halfturn is a halfturn.

32. Prove that the composition of:

(a) an even number of halfturns is a translation or the identity;

(b) an odd number of halfturns is a halfturn.

33. Let A₀, B₀, O₁, O₂, …, O_n be given points. For 1 ≤ k ≤ n, let A_k B_k be obtained from A_k−1 B_k−1 by a halfturn about O_k.

(a) Prove that A₀ A_n = B₀ B_n if n is even.

(b) What conclusion may be drawn if n is odd?

34. Let A₀, O₁, O₂, …, O_n be given points. For 1 ≤ k ≤ n, let A_k be obtained from A_k−1 by a halfturn about O_k, and let A_{n + k} be obtained from A_{n + k−1} by a halfturn about O_k

(a) Prove that A_2n will coincide with A₀ if n is odd.

(b) Can the same conclusion be drawn if n is even?

35. Let A₀ = B₀, O₁, O₂, …, O_n be given points, For 1 ≤ k ≤ n, let A_k be obtained from A_{k − 1}, by a halfturn about O_k, and let B_k be obtained from B_{k − 1} by a halfturn about O_{n − k + 1}. For what values of n will A_n coincide with B_n?

⁸This is the same as saying that α and β differ by a multiple of 360. In number theory, this is written mathematically as follows:

The expression is called a congruence and is expressed verbally by saying “α is congruent to β modulo 360.” Although the two notions coincide in this specific case, generally there does not have to be any connection between geometric congruence and number theoretical congruence.