Given two direct isometries, say the rotation RP,ϕ and the translation TAB, we know that their product RP,ϕ TAB is a direct isometry, and we may even suspect that it is a rotation RQ,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.
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.
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 TAB 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.
The effect of a translation TUV 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 TAB and TCD be the two translations. The product TCDTAB 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 TCD and TAB are inverses of each other, and so the rotation is actually the identity.
Suppose that TCD TAB = RP,θ. Then
Let P′ = TAB(P). Then TAB = TPP′. Similarly, TCD = TP′P, but then
Thus, the rotation would actually be the identity.
Note that if is not congruent to , then TCD TAB must be a translation T. To pin down this translation geometrically, consider the effect of TCD TAB on a point P. TAB maps P to P′ and TCD maps P′ to P″. Therefore, T = TPP″. 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, TCD TAB maps E to G, and the diagonal is congruent to PP″.
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 TAB RO,θ cannot be either the identity or a translation.
The only possibility left is that the product is a rotation.
In a similar way, we can show that RO,θ TAB cannot be a translation unless RO,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 RQ,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 RO,θ and then back to Q by TAB. 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 RQ,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 TAB RO,θ.
The product RO,θ TAB can be analyzed in a similar way. Here, the center would be a point X such that TAB maps X to Z and RO,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:
W is the center of TAB RQ,30°,
X is the center of RQ,30° TAB,
Y is the center of RQ,−30° TAB,
Z is the center of TAB RQ, −30°.
Example 9.4.4. Given the point Q and rotation RO,θ, where θ is not a multiple of 360°, as in the figure on the left on the following page, find a translation TCD such that
Solution. Again, we exploit the idea that if for some point X we can find its image X′ under TCD, then TCD = TXX′.
We first solve for TCD. Multiply the equation
on the right by the inverse of RO,θ, as follows:
Next use the associative law and the fact that RO,θ RO,−θ = I to obtain TCD:
Now apply TCD to the point O (we chose O because it is fixed under RO,−θ). Then
Letting P be the point RQ,θ(O), we therefore have TCD = TOP, as in the figure on the right at the top of the page.
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 RB,β RA.α when A and B are different, we must first show the following:
Theorem 9.5.1. When A and B are different, RB,β RA,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 RQ,θ to any point we like, as in Example 9.4.4.
Let TCD be the translation such that
Let S be the product RB,β RA,α; that is
Multiply this equation on the left by TCD to get
which implies that
Now multiply this equation on the left by TDC, the inverse of TCD, to get
From the previous section, we know that TDC RA,α+β is a rotation RP,α+β for some point P (unless α + β is a multiple of 360°, in which case TDC RA,α+β = TDC).
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 = RA,β(B), as in the discussion following Example 9.4.4. Applying the inverse transformation TDC to RA,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 RB,β RA,α, as in the following example.
Example 9.5.2. Given RB,β and RA,α, with α and β as shown in the figure below, construe: the center Q of the rotation RB,β RA,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 RB,β RA,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 RA,α to Q′, and Q′ will then be mapped by RB,β back to Q.
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.