The emphasis in these sections in Glencoe and U of Chicago is the glide reflection -- the most often forgotten and most often confusing of the four 2D isometries. There are differences between the U of Chicago and the Glencoe lessons, including most notably the emphasis on coordinates in Glencoe -- just as Glencoe performs the other isometries on the coordinate plane.

I showed my student how if we take two congruent triangles and toss them randomly onto the table, one of the four isometries must map one to the other -- with rotations and glide reflections much more likely than translations and reflections (because the former have more degrees of freedom). My student somewhat understood my point -- but it might have been better if I had cut out the triangles from paper colored differently on each side, so it would have been obvious whether the triangles had the same orientation or not. Or I could have tossed up two letters, say F's, instead. My student wondered why I kept showing him my worksheets with the letter F all over the place -- the reason is that F is an asymmetrical letter. Otherwise we couldn't tell whether the letter had reversed orientation and we'd always be able to map one to the other with only translations and rotations.

But there is something I notice about how glide reflections are defined versus how they are performed on the coordinate plane. We begin with Glencoe's definition of glide reflection:

"A glide reflection is a translation followed by a reflection in a line

*parallel to the translation vector*."

(Emphasis mine.)

This is nearly identical to the U of Chicago definition. The only difference is that the U of Chicago writes "G = T o r_

*m*," where "o" means "following." So the translation is

*following*the reflection -- that is, the reflection is performed

*first*. In Glencoe, however, the translation is first. As it turns out, it doesn't matter much, since in this case the reflection and translation

*commute*-- that is, the composite in either direction is the same glide reflection.

But what I wish to focus on is the phrase "parallel to the translation vector." Both Glencoe and the U of Chicago require that the mirror be parallel to the translation. Yet Glencoe gives examples such as:

Triangle

*XYZ*has vertices

*X*(6, 5),

*Y*(7, -4), and

*Z*(5, -5).

Graph Triangle

*XYZ*and its image after the indicated glide reflection.

Translation: along <1, 2>

Reflection: in

*y*-axis

And we notice immediately that the translation vector <1, 2> is

*not*parallel to the

*y*-axis! So right after defining a glide reflection as the composition of a translation and a reflection with mirror parallel to the translation, the text gives an example where the mirror isn't parallel to the translation. Indeed, we see that the translation and reflection in this case don't commute -- the image of

*X*is (-7, 7) when we perform the translation first but (-5, 7) when we perform the reflection first.

We can see why Glencoe wouldn't actually want the mirror to be parallel to the vector <1, 2>. For reflections in mirrors parallel to the axes are difficult enough to perform -- now imaging trying to reflect in a line that isn't parallel to either axis.

As it turns out, the composite of this translation and reflection is still a glide reflection. But it's possible to determine the

*actual*reflecting line of this glide reflection -- that is, the mirror such that the translation vector is actually parallel to it. We do so by using the Glide Reflection Theorem mentioned in the U of Chicago text:

Glide Reflection Theorem:

Let G = T o r_

*m*be a glide reflection, and let G(

*P*) =

*P'*. Then the midpoint of

*PP'*is on

*m*.

We will observe that this theorem will work for this problem even though we'll be performing the translation first, not the reflection. To use the theorem, we could try using the vertices

*X*,

*Y*,

*Z*given in the actual problem. We perform the translation simply by adding the vectors, while reflecting in the

*y*-axis entails changing the sign of

*x*. So we would obtain the image points

*X*'(-7, 7) and

*Y*(-8, -2) -- only two images are necessary here. The theorem requires us to find the midpoints:

Midpoint of

Midpoint of

Our mirror is the line passing through these points. We could use the Point-Slope Formula to find the equation of this line -- except it won't work, because the slope happens to be undefined. The mirror is the vertical line

*y*= -1/2. Recalling our reflection

*x*--> 2

*m*-

*x*from earlier this week, we see that the reflection of

*X*in this mirror is (-7, 5). So we see that the translation vector must be <0, 2>, which is indeed vertical just like the mirror. (If we had erroneously performed the reflection first, we would have obtained a different mirror --

*y*= +1/2 rather than

*y*= -1/2 -- but the same translation vector.)

But there is an even worse example given in Glencoe:

*D*(1, 3) and

*E*(3, -3). Graph

Translation: along <0, -2>

Reflection: in

*x*-axis

Notice that not only are the translation and the mirror not parallel, they are

*perpendicular*. As it turns out, the composite of a translation and a reflection in a mirror perpendicular to the translation isn't even a glide reflection -- it's simply another reflection! To find the mirror, we can use the same theorem as above:

*D'*= (1, -1)

*E'*= (3, 5)

Midpoint of

~~DD~~

Midpoint of

So the reflecting line is

*y*= 1. We notice that the reflection of

*D*(1, 3) in the line

*y*-axis is already the point (1, -1), so the translation vector must be <0, 0>. So we have a "glide reflection" whose translation vector is the zero vector -- that is, we have a reflection without the "glide". Then again, we must admit that Glencoe, in this part of the text, never claims that this is a glide reflection -- indeed, it later gives the composite of a translation and a

*rotation*, which is clearly not a glide reflection as it preserves orientation. But I've seen other texts that claim that the composite of a translation and a reflection whose mirror is perpendicular to the translation vector is a glide reflection.

I could keep going on and on about compositions of reflections. This lesson had me thinking about how to prove that three mirrors suffice for any 2D isometry, and that two mirrors suffice for any 2D isometry that preserves orientation. Clearly the composite of two reflections must be either a translation (if the mirrors are parallel) or a rotation (if they aren't), so it remains only to show that the composite of two translations, two rotations, or one of each is itself a translation or rotation. A trick that I just thought up was to find mirrors

*l*,

*m*, and

*n*such that the first isometry is the composite of reflections in

*l*and

*m*and the second is the composite of reflections in

*m*and

*n*. Then the composite of the two isometries would be the composite of reflections in

*l*and

*n*, therefore two mirrors suffice. It is trivial to find

*l*,

*m*, and

*n*if at least one of the isometries is a rotation, but impossible if they are both translations unless they are in the same direction. (Then again, the composite of two translations is clearly another translation because we can just add the translation vectors.)

Also, the idea that the composite of a translation and a reflection can sometimes be another reflection, as in our second example above, can be important. Notice how we showed that the composite of the translation 2 units down and reflection in the

*x*-axis is reflection in the line

*y*= 1 -- hey, didn't we just have a PARCC question about reflection in the line

*y*= 1 just two days ago? Students might not know how to reflect in the line

*y*= 1, but they should know how to perform the translation 2 units down and the

*x*-axis reflection. As it turns out, rotations about points other than the origin are the composite of some translation and a rotation about the origin of the same rotation angle. In either case, we are assuming that the student can resolve the harder transformation into two easier transformations -- which they won't unless we show them. Notice that so far, actual glide reflections haven't appeared on the PARCC, and I wouldn't be surprised if they don't.

Oops -- that's right, the PARCC. I won't take time to discuss composition of isometries any more because we have to get to the next PARCC question -- a long one, with four parts. Question 23 of the PARCC practice test is on trigonometry and angles of depression:

An unmanned aerial vehicle (UAV) is equipped with cameras used to monitor forest fires. The figure represents a moment in time at which a UAV, at point

*B*, flying at an altitude of 1,000 meters (m) is directly above point

*B*on the forest floor. Point

*A*represents the location of a small fire on the forest floor.

At the moment in time represented by the figure, the angle of depression from the UAV to the fire has a measure of 30 degrees.

Part A

At the moment in time represented by the figure, what is the distance, in meters, from the UAV to the fire?

Part B

What is the distance, to the nearest meter, from the fire to point

*D*?

Part C

Points

*C*and

*E*represent the linear range of view of the camera when it is pointed directly down at point

*D*.

The field of view of the camera is 20 degrees and is represented in the figure by Angle

*CBE*. If the camera takes a picture directly over point

*D*, what is the approximate width of the forest floor that will be captured in the picture?

Part D

The UAV is traveling at 13 meters per second in the direction toward the fire. Suppose the altitude of the UAV is now 800 meters. The new position is represented at point

*F*in the figure.

From its new position at point

*F*, how many minutes, to the nearest tenth of a minute, will it take the UAV to be directly over the fire?

Obviously, we are going to use material from Chapter 14 of the U of Chicago text. For Part A, we are asked to find

*AB*. Since the Angle

*A*of elevation is also 30 degrees, we can use sine:

sin

*A*=

*BD*/

*AB*

sin 30 = 1,000 /

*AB*

*AB*= 1,000 / sin 30

*AB*= 2,000 m

For Part B, we are asked to find

*AD*. Instead of sine, we can use tangent:

tan

*A*=

*BD*/

*AD*

tan 30 = 1,000 /

*AD*

*AD*= 1,000 / tan 30

*AD*= 1,732.1 rounded to 1,732 m.

For Part

*C*, it is implied that Triangle

*CBE*is isosceles, since we expect the range of view of a camera to be symmetrical. This means that we can divide the triangle into two congruent right triangles, namely

*CBD*and

*EBD*. So we cut the angle in half to 10 degrees, use tangent to find

*CD*, and then redouble the answer to find

*CE*:

tan

*CBD*=

*CD*/

*BD*

tan 10 =

*CD*/ 1,000

*CD*= 1,000 tan 10

*CD*= 176.3

*CE*= 352.7 rounded to 353 m

This is choice (B). Sure enough, one of the wrong choices is very close to

*CD*without doubling. I see that another wrong choice occurs if the student forgets to halve 20 degrees, and the remaining wrong choice involves making both errors.

For Part D, we notice that the plane has descended somewhat, so that

*FD*is now 800 m. So we now find the new value of

*AD*, since this equals the distance that the plane must fly:

tan

*A*=

*FD*/

*AD*

tan 30 = 800 /

*AD*

*AD*= 800 tan 30

*AD*= 1,385.6

At 13 minutes per second, it will take about 106.6 seconds for the UAV to reach the fire. But the question asks for minutes. This is about 1.78 minutes, which rounds off to choice (C).

Is this an authentic problem? In some ways, it could be. Since the UAV is

*unmanned*, presumably it is remote-controlled, and so whoever is at the controls can make these calculations. But there are so many places where students can make mistakes. Part D is especially egregious because we see that point

*B*is 1 km above the ground while point

*F*is only 800 m above the ground. Yet they can't both have an angle of depression of 30 degrees -- not

*and*both be directly above the same point

*D*. It would have been more accurate to change the letter

*D*to something like

*G*.

Notice that Part A can be done without a calculator, since sin 30 = 1/2. In fact, the only part that actually requires trig is Part C, since the angle isn't 30, 45, or 60 degrees -- and since it uses the tangent, this problem can be given after Section 14-3 on tangents without Section 14-4 at all.

I might improve this problem by placing Part A in the

*non-calculator*section, to see whether students can figure out the 30-60-90 triangle. Parts B and C could remain to make up a good two-part calculator question. Part D, as currently written, shouldn't see the light of day.

**PARCC Practice EOY Exam Question 23**

**U of Chicago Correspondence: Section 14-3, The Tangent Ratio**

**Key Theorem: definition of tangent**

**In right triangle**

*ABC*with right angle*C*, the tangent of Angle*A*, written tan*A*, is**leg opposite Angle**

*A*/ leg adjacent to Angle*A*

**Common Core Standard:**

CCSS.MATH.CONTENT.HSG.SRT.C.8

Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems.

**Commentary: Again, Question 14 of Section 14-4 is the only question where the phrase "Angle of Depression" actually occurs. Section 15-3 of the text mentions picture angles of a camera, just like Part C in this PARCC question, but there are no trig questions in this section.**

actively fighting policy", the real achieve early discovery, early solve. Forest Fire Detection Cameras

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