## Wednesday, January 24, 2018

### Lesson 9-5: Reflections in Space (Day 95)

Lesson 9-5 of the U of Chicago text is called "Reflections in Space." In the modern Third Edition of the text, reflections in space appear in Lesson 9-7.

Even though we've never officially covered Chapter 9 before, in past years I sometimes mentioned spatial isometries as a special topic during spring break or another holiday period. But this year we're covering Lesson 9-5 as a regular lesson.

This is what I wrote two years ago about today's lesson:

We already discussed translations and rotations for Common Core Geometry, but those transformations applied to the two-dimensional plane. The physical world has three dimensions, and so it's the 3-D transformations that actually apply to physics.

Lesson 9-5 of the U of Chicago text is on reflections in space. I find this to be an interesting topic, but since it doesn't appear on the PARCC, it would be a waste of time to cover it in class. There will be no worksheet for this lesson, since it's not intended to be taught in class. This is why I waited until spring break to blog about this topic. [2018 update: It's a regular lesson now, and so there will be a worksheet for this lesson after all!]

The text begins by defining what it means for a plane in 3-D, rather than a line in 2-D, to be a perpendicular bisector:

"In general, a plane M is the perpendicular bisector of a segment AB if and only if M is perpendicular to AB and M contains the midpoint of AB."

Now we can definition 3-D reflections almost exactly the same way we define 2-D reflections:

"For a point A which is not on a plane M, the reflection image of A over M is the point B if and only if M is the perpendicular bisector of AB. For a point A on a point M, the reflection image of A over M is A itself."

If you think about it, when you (a 3-D figure!) look at yourself in a mirror, the mirror itself isn't a line, but rather a plane. Mirrors in 2-D are lines, while mirrors in 3-D are planes. So now we can define what it means for two 3-D figures to be congruent:

"Two figures F and G in space are congruent figures if and only if G is the image of F under a reflection or composite of reflections."

Most texts don't actually define what it means for two 3-D figures to be congruent. We know that the traditional textbook definition, that congruent figures have corresponding sides and angles congruent, only applies to polygons. It doesn't even apply to circles, much less 3-D figures. But we can simply use the Common Core definition -- two figures are congruent if and only if there exists an isometry (i.e., a composite of reflections) mapping one to the other -- and it instantly applies to all figures, polygons, circles, and 3-D figures.

In 2-D there are only four isometries -- reflections, translations, rotations, and glide reflections. An interesting question is, how many isometries are there in 3-D?

Well, for starters, translations and rotations exist in 3-D. We can define both of these exactly the same way that we do in 2-D -- a translation is the composite of two reflections in parallel planes, while a rotation is the composite of two reflections in intersecting planes.

Notice that every 2-D rotation has a center -- the point of intersection of the reflecting lines. The same thing happens in 3-D, except that the intersection of two planes isn't a point, but a line. Thus, a 3-D rotation has an entire line as its center -- every point on this line is a fixed point of the rotation. But usually, instead of calling the line the center of the rotation, we call it the axis of the rotation. One 3-D object that famously rotates is the earth, and this rotation has an axis -- the line that passes through the North and South Poles. Confusingly, the mirror of a 2-D reflection is often called an axis -- but in some ways, these two definitions are related. One can perform a 2-D reflection by taking a 2-D figure and rotating it 180 degrees about the axis in 3-D (and recall that A Cube has reflected A Square in exactly this manner).

Glide reflections also exist in 3-D -- although these are often called glide planes in 3-D. A glide reflection is the composite of a reflection and a nontrivial translation parallel to the mirror. Notice that there are infinitely many directions to choose from for our translation in 3-D, but if this were a 2-D glide reflection there are only two possible directions for a translation that's parallel to the mirror.

But are there any other isometries in 3-D? Well, we notice that a glide reflection is the composite of two other known isometries, a reflection and a translation. So the next natural possibility to consider is, what if we find the composite of the other two combinations? What is the composite of a reflection and a rotation, or a translation and a rotation?

In 2-D, the composite of a translation and a nontrivial rotation is another rotation. This is possible to prove, as follows: let T be a translation and R be a nontrivial rotation, and G be the composite of T following R. Both translations and rotations preserve orientation, and so their composite G must preserve orientation as well. In 2-D, three mirrors suffice -- that is, every isometry is the composite of at most three reflections. Since G preserves orientation, it must be the composite of an even number of rotations. Therefore G is the composite of two reflections (or the identity transformation -- the transformation that maps every point to itself), and so G is either a translation or a rotation.

Let's try an indirect proof -- assume that G is a translation. That is:

T o R = G

Since T is a translation, it has a translation vector t, and as we're assuming that G is a translation, it must also have a translation vector g. Now let U be the inverse translation of T -- that is, the translation whose vector is -t, the additive inverse of t. We now compose U on both sides:

U o T o R = U o G

Now since U and T are inverses, U o T must be I, the identity transformation:

I o R = U o G

Since I is the identity transformation, I o R must be R:

R = U o G

Notice that G and U are both translations, whose vectors are g and -t, respectively. Then their composite must be another translation V whose vector is g - t. So now we have:

R = V

that is, a rotation equals a translation. Now no rotation can equal a translation (as the former has a fixed point, while the latter has no fixed point) unless both are the identity -- which contradicts the assumption that R is a nontrivial rotation (i.e., not the identity). Therefore, the composite of a translation and a nontrivial rotation isn't a translation, so it must be a rotation. QED

But this proof is invalid in 3-D. This is because the proof uses a step that only works in 2-D -- namely that three mirrors suffice. We must show how many mirrors suffice in 3-D.

Let's recall why mirrors suffice in 2-D. Let G be any 2-D isometry, and let AB, and C be three noncollinear points whose images under G are A'B', and C'. The first mirror maps A to A', the second mirror fixes A' and maps B to B'. It could be that the image of C under both mirrors is already C', otherwise a third mirror maps it to C'. Notice that the proof of the existence of these mirrors is nontrivial and depends on theorems such as the Perpendicular Bisector Theorem, since reflections are defined using perpendicular bisectors.

As it turns out, four mirrors suffice in 3-D. To prove this, we let G be any 3-D isometry, and let ABC, and D be four noncoplanar points. The first mirror maps A to A', the second mirror fixes A' and maps B to B', the third mirror fixes A' and B' and maps C to C', and the fourth, if necessary, fixes A'B', and C' and maps D to D'.

And so this opens the door for there to be a new transformation in 3-D, one that is the composite of a translation and a rotation as well as the composite of four reflections. We can imagine twisting an object like a screw. A screwdriver rotates the screw about its axis, but then it's being translated into the wall in the same direction as that axis. And because of this, this new transformation is often called a screw motion.

We still have one last combination, the composite of a reflection and a rotation. It is subtle why the composite of a reflection and a rotation in 2-D is usually a glide reflection -- why should the composite of a reflection and a rotation equal the composite of a (different) reflection and a translation? And it's even subtler why the composite of a reflection and a rotation may be a new transformation in 3-D.

But the simplest example of this roto-reflection is the inversion map. In 3-D coordinates, we map the point (xyz) to its opposite point (-x, -y, -z). This map is the composite of three reflections -- the mirrors are the three coordinate planes (xyxz, and yz). As the composite of an odd number of reflections, it must reverse orientation. Yet it can't be a reflection, since it has only a single fixed point (0, 0, 0) and not an entire plane. Similarly, it can't be a glide plane because glide planes don't have any fixed points at all.

Roto-reflections are formed when the axis of the rotation intersects the reflecting plane in a single point -- and of course, this single point is the only fixed point of the roto-reflection.

So now we have six isometries in 3-D -- reflections, translations, rotations, glide planes, screw motions, and roto-reflections. Are there any others? As it turns out, these six are all of them -- and the proof depends on the fact that four mirrors suffice in 3-D.

Returning to the U of Chicago text, we have the definition of a reflection-symmetric figure:

"A space figure is F is a reflection-symmetric figure if and only if there is a plane M such that the reflection of F in M is F."

Similarly, a figure can be rotation-symmetric, as well as roto-reflection-symmetric. In the text, figures such as the right cylinder have reflection, rotation, and roto-reflection symmetry.

But a figure can't have translation symmetry unless it's infinite, as translations lack fixed points. So likewise, figures that have glide reflection or screw symmetry must also be infinite, as these transformations are based on translations.

Of course, there exist dilations in 3-D space as well. There is little discussion of similarity in 3-D, except to compare the surface areas and volumes of 3-D figures.

Now that we are reading about the fourth dimension, the next thing we wonder is, what would a reflection in 4-D look like? We follow the same pattern that we followed for 2-D and 3-D -- we begin with the perpendicular bisector of a segment. In 2-D the perpendicular bisector is a 1-D line, in 3-D it's a 2-D plane, and so in 4-D it's a 3-D hyperplane. This hyperplane M is the mirror of a 4-D reflection, mapping each point A to its image B if and only if M is the perpendicular bisector of AB (and of course mapping each point on the hyperplane to itself).

We know that the 3-D isometries include all of the 2-D isometries (translations, rotations, reflections, and glide reflections) plus two new isometries (screws and roto-reflections). So we expect the 4-D isometries to include all the 3-D isometries plus two new ones.

One of these new isometries is called a double rotation, as it is the composite of two rotations. Notice that the center of a 2-D rotation is a 0-D point, and the center (axis) of a 3-D rotation is a 1-D line, and so the axis of a 4-D rotation is a 2-D plane. The double rotation is the composite of two rotations whose respective axes (planes) intersect in a single point. A point inversion is a double rotation.

The other new 4-D isometry is a roto-glide reflection. It is the composite of a roto-reflection (which requires three dimensions) and a translation in the fourth dimension. Just as three mirrors suffice in 2-D and four mirrors suffice in 3-D, we see that five mirrors suffice in 4-D.

OK, let's return to 2018 and continue our reading at Euclid. Of course, Euclid doesn't write about our isometries, and so we proceed directly to the next proposition:

Proposition 8.
If two straight lines are parallel, and one of them is at right angles to any plane, then the remaining one is also at right angles to the same plane.

According to David Joyce, this is essentially a converse to Proposition 6 -- which, as we found out on Monday, is the analog of the Two Perpendiculars Theorem. Therefore Proposition 8 is the analog of the Perpendicular to Parallels Theorem of Lesson 3-5. Again, the old Perpendicular to Parallels Theorem will appear as a step in today's proof.

Given: AB | | CD, AB perp. plane P (with B and D in plane P)
Prove: CD perp. plane P

Proof:
Statements                              Reasons
1. bla, bla, bla                         1. Given
2. AB, CD, BD coplanar         2. Proposition 7 from yesterday
(call it plane Q)
3. Choose E in plane P s.t.     3. Point-Line-Plane, part a (Ruler/Protractor Postulates)
DE perp. BD, DE = AB
4. AB perp. BD, AB perp. BE 4. Definition of a line perpendicular to plane
5. CD perp. BD                       5. Perpendicular to Parallels Theorem [steps 4,1]
6. BD = BD                             6. Reflexive Property of Congruence
7. Triangle ABD = EDB         7. SAS Congruence Theorem [steps 3,4,6]
8. AD = BE                             8. CPCTC
9. AE = AE                             9. Reflexive Property of Congruence
10. Triangle ABE = EDA        10. SSS Congruence Theorems [steps 3,8,9]
11.Angle ABE = Angle EDA  11. CPCTC
12. DE perp. AD                     12. Definition of perpendicular lines [steps 4,11]
13. DE perp. plane (BD, AD) 13. Proposition 4 from last week [steps 3,12]
14. DE perp. plane Q              14. Point-Line-Plane, part f (Q is plane through A, B, D)
15. DE perp. CD                     15. Definition of a line perpendicular to plane [steps 2,14]
16. CD perp. plane (BD, DE) 16. Proposition 4 from last week [steps 5,15]
17. CD perp. plane P              17. Point-Line-Plane, part f (P is plane through B, D, E)

This proof is a bit complex. But since we just learned about isometries today, we might wonder whether isometries can be used to make the proof easier (just as the U of Chicago text uses symmetry to prove the Isosceles Triangle Theorem, for example).

We notice that in step 7, Triangles ABD and EDB are congruent, and in step 10, triangles ABE and EDA are congruent. The isometry that maps both ABD to EDB and  ABE to EDA must switch A with E and B with D. Thus the desired isometry is in fact a rotation of 180 degrees. Its axis passes through the midpoints of  AE and BD (hence it doesn't lie in either plane P or plane Q).

But I'm not quite sure whether this knowledge actually helps to simplify the Proposition 7 proof. So instead, I'm wondering whether our Proposition 4 proof can be simplified using a rotation. In that proof, there are seven pairs of congruent triangles. Every single pair switches A with B and C with D, and G with H, and fixes E and F. Thus we can use a rotation of 180 degrees about axis EF. As soon as we can establish the existence of the rotation mapping A to B and C to D, we can choose our arbitrary line and map G to H on this line. Then we can skip directly to Step 22 -- angles GEF and HEF are a congruent linear pair, hence EF perp. GH, which completes the proof.

Here is the worksheet: