Explaining Isometries: Two Tasks

[jljarrett18]

I am currently working on a task involving isometries. I am confused as to how to explain the next two tasks to my adviser.

B. Any combination of isometries can be accomplished with one of the four fundamental isometries (i.e., translation, reflection, rotation, or glide reflection). Provide a logical argument that demonstrates that, given any pair of reflections, there exists a rotation or translation that produces the same result.

C. Demonstrate, using a counter example, that the product of a pair of isometries A and B is not always commutative (i.e., AB does not always equal BA).

 

[Evgeny.Makarov]

Provide a logical argument that demonstrates that, given any pair of reflections, there exists a rotation or translation that produces the same result.

Suppose the lines that define reflections intersect in point $O$. Take some point $A$, reflect it with respect to the first line to get $A'$, then with respect to the second line to get $A''$. What can be said about $\triangle AOA'$ and $\triangle A'OA''$? What is the relation between the lengths of $OA$, $OA'$ and $OA''$? What is the relation between $\angle AOA''$ and the angle between the two lines?

Do a similar construction when the two lines are parallel. What is your guess about the composition of the two reflections?

Demonstrate, using a counter example, that the product of a pair of isometries A and B is not always commutative (i.e., AB does not always equal BA).

Pick two different isometries, try their action on some point and report the results.

For the future, refer to rules 6 and 11 http://mathhelpboards.com/rules/ (click on the "Expand" button).

 

[jljarrett18]

It can be said that they are the same and all the angles are equal?

 

[Evgeny.Makarov]

It can be said that they are the same and all the angles are equal?

Please be more precise. I named several objects: two triangles, three segment lengths and two angles. What are "they" that are the same? And which angles are equal?

 

[Evgeny.Makarov]

Here is a diagram.

View attachment 2765

The triangles $OAA'$ and $OA'A''$ are not equal, but since $A'$ is the image of $A$ under reflection with respect to $l_1$, $OB_1$ is a median and an altitude in $\triangle OAA'$. Therefore, this triangle is…

 

[jljarrett18]

Triangle B2, A',O and triangle B2,A", O are congruent as well as there angles.
Triangle A',B1, O and Triangle B1, A, O are congruent as well as there angles.

 

[Evgeny.Makarov]

Yes, so $OA=OA'=OA''$, which means that $A''$ is obtained from $A$ by rotation around $O$. What can be said about $\angle AOA''$ and its relationship with $\angle B_1OB_2$? Note that the latter angle is determined only by the two lines and does not depend on $A$.

 

[jljarrett18]

Could we say that B1 and B2 are congruent?

 

[Evgeny.Makarov]

Any two points are congruent.

 

[jljarrett18]

So if any two points are congruent, what does that tell me about how any translations or rotations can equal reflections?

 

[Evgeny.Makarov]

So if any two points are congruent, what does that tell me about how any translations or rotations can equal reflections?

It tells you nothing. Any two points are trivially congruent, i.e., one can be transformed into another by an isometry. The fact that two given points are congruent carries no information, just like the fact that some number $x$ is equal to itself—this goes without saying. In contrast, not every two triangles are congruent, so knowing that some triangles are is a potentially useful information for solving some problem. Also, not every isometry transforms one given point into another, so knowing that one point is transformed into another by a given transformation is a potentially useful information. But I don't see any isometry we discussed that transforms $B_1$ into $B_2$.

I suggest you answer this question.

What can be said about $\angle AOA''$ and its relationship with $\angle B_1OB_2$? Note that the latter angle is determined only by the two lines and does not depend on $A$.

 

[jljarrett18]

Would the answer to my question be as a shape is reflected, a rotation can be made to rotate the shape and then a translation can be used to pick up the shape and place it where the reflection is?

 

[Evgeny.Makarov]

Would the answer to my question be as a shape is reflected, a rotation can be made to rotate the shape and then a translation can be used to pick up the shape and place it where the reflection is?

If I understand correctly, we are proving the following statement (I'll denote it by (*)):

The composition of two reflections with respect to intersecting lines is a rotation.

To say what I think about your quote, I'll compare (*) with the problem of finding solutions of a quadratic equation $ax^2+bx+c=0$, just a metaphor. A correct proof of (*) could be compared to the quadratic formula:
\[
x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.
\]
In contrast, the quote above sounds like the following:

"As we solve the equation, we can compute a square root and then we can divide the result in such a way as to get the solutions of the equation".

Sorry, but this is not nearly precise enough.

Let's start from the beginning. You are given two reflections, which means that you are given two lines. There are two cases: either the lines intersect or they are parallel. If they intersect, the fact is that the composition of reflections with respect to those lines is a rotation. If they are parallel, the composition of reflections is a translation. We are first considering the case when the lines intersect.

So you are given lines $l_1$ and $l_2$ that intersect at point $O$ (see figure in https://driven2services.com/staging/mh/index.php?posts/52547/). We need to find a rotation and prove that reflections w.r.t. $l_1$ and $l_2$ made in turn are equivalent to that rotation. To find a rotation you need to find a center and an angle. Then to prove that two transformations are equivalent (the composition of reflections on the one hand, and the rotation on the other), you need to prove that they transform every point in the same way.

To simplify life, I'll say that the center of the new rotation is $O$. I still suggest that you find the angle. It can be found by inspecting an arbitrary point $A$, which is mapped to $A'$ by the first reflection and and then to $A''$ by the second one. We established in posts #6 and #7 that $OA=AO''$, so the transformation of $A$ into $A''$ does look like a rotation around $O$. But maybe every point is rotated by a different angle. That's why I suggested expressing $\angle AOA''$ in terms of $\angle B_1OB_2$, which does not depend on $A$ but is determined only by $l_1$ and $l_2$. If you show that every point is rotated by the same angle around $O$ when it undergoes the two reflections, then you will prove that the composition of these reflections is indeed a rotation.

 

[jljarrett18]

This is what I have come up with so far:This is what I have come up with so far:

Given any pair of reflections there exists a rotation or translation that produces the same results. I started this task with reflections. I started with an example I created in GSP5.

In this example I created a random triangle and created lines to reflect the triangle across. As you can see number 1 shows my original triangle. Next to get the shape in figure 2 I reflected the triangle across the line. Next figure 3 is the reflection of figure 2 across the second line I created. If you look closely, figure 1 and 3 are translations of each other. So in this example I reflected my original triangle twice and in the end my beginning triangle and end triangle are translations of each other.
Next I moved onto rotations and reflections. Rotations and reflections are related because if you reflect a shape over intersecting lines the shape will have rotated. This is what I created in GSP5.

I first created a triangle and which is triangle 1 in the picture. I then reflected it across the first line, which is not only a reflection but can also be called a rotation of 90 degrees. Next, I rotated my second triangle across the second intersecting line. This triangle is a 180 degree rotation of my first triangle. So from this example we can see that one reflection is equal to a 90 degree rotation and two reflections is equal to a 180 degree rotation or two rotations.

My mentor replied with these comments:
You need to discuss more about the translation scenario, in how two reflections in general will produce a translation. In other words, what relationship is there between the two reflection lines that results in a translation. In the rotation scenario, it is not true when you say this: “which is not only a reflection but can also be called a rotation of 90 degrees.” And again it is not true to say “we can see that one reflection is equal to a 90 degree rotation…or two rotations”. If you labeled your vertices of the triangles, you would see why these statements are not true.

 

[Evgeny.Makarov]

Given any pair of reflections there exists a rotation or translation that produces the same results. I started this task with reflections.

As opposed to what?

My mentor replied with these comments:
You need to discuss more about the translation scenario, in how two reflections in general will produce a translation. In other words, what relationship is there between the two reflection lines that results in a translation. In the rotation scenario, it is not true when you say this: “which is not only a reflection but can also be called a rotation of 90 degrees.” And again it is not true to say “we can see that one reflection is equal to a 90 degree rotation…or two rotations”. If you labeled your vertices of the triangles, you would see why these statements are not true.

Your mentor is correct. Imagine that a plane has two sides: red and blue, with red on top. If you perform a rotation or a translation, the red side is still on top. However, if you perform a reflection, blue side gets on top. If you perform two reflections, red side is again on top, which means that the composition of two reflections can be obtained without a reflection.

 

[Deveno]

Here is my (decidedly naive) take on the situation. I have put it in a spoiler, because I feel you should NOT look at it unless you are totally stumped.

[sp]Suppose we have two non-parallel lines $l_1$ and $l_2$. Call the point they intersect at, $O$.

Call the angle between the two lines (going counter-clockwise), $\theta$. We're going to need some more names, later, I'll introduce the cast of characters as we go along.

Now suppose we have a point, $P$. For the moment, we'll assume $OP$ is at some angle $\phi$ clockwise from $l_1$.

Let's track what happens to $P$, as we reflect first about $l_1$, and then about $l_2$.

Call the point $P$ gets reflected to, $P'$. At this point, the line $OP'$ is at an angle of $2\phi$ with the line $OP$.

Now the line $OP'$ is at some angle $\psi$ (clockwise) from $l_2$. Let's assume (for visualization purposes) that $\phi < \theta$, so that $OP'$ lies between $l_1$ and $l_2$ (you can consider the "other cases" later, on your own).

When we reflect about $l_2$, the point $P'$ gets sent to a new point, let's call it $Q$. As you can readily verify (do this!), the line $OQ$ makes an angle of $\psi$ (counter-clockwise) with $l_2$.

Now when $OP'$ got reflected, the angle between $OP'$ and $OQ$ is clearly $2\psi$. So what angle is $OQ$ from $OP$?

The total angle is: $2\phi + 2\psi$. But $\phi + \psi = \theta$, so the total angle is $2\theta$.

In other words, the angle between $OP$ and $OQ$ doesn't depend on $\phi$, only $\theta$.

Now, one reflection "makes the plane mirrored" (changes the orientation of things). The second reflection "cancels that", so our orientation is preserved, leaving only a change in angle (a rotation).

To get an idea of the "other cases", let's suppose $P$ is "between" $l_1$ and $l_2$. So now $OP$ is at an angle $\phi$ counter-clockwise from $l_1$. So when we reflect, it goes "backwards" -by an angle of $-2\phi$.

Now $OP'$ makes an angle of $\theta + \phi$ with $l_2$. Since we were calling this angle $\psi$, we have:

$\psi = \theta + \phi$, or: $\theta = \psi - \phi$.

When we reflect about $l_2$, the line $OQ$ makes an angle of $2\psi$ with $OP'$. What angle is this line to $OP$?

Well, our original line $OP$ was already $\phi$ towards $l_2$ out of $\theta$. So the original angle between $OP$ and $l_2$ was $\theta - \phi$.

Now, we are $\psi$ "past" $l_2$, so $OQ$ is at an angle of:

$\psi + \theta - \phi = \theta + (\psi - \phi) = \theta + \theta = 2\theta$

with $OP$. Amazing!

Since reflections are isometries, and isometries are, by definition, "distance-preserving", it should be clear that the lengths of the line segments $OP$ and $OQ$ are equal.

It should be suggestive, then, that our two reflections are equivalent to a rotation about the point $O$, with angle equal to twice the angle between $l_1$, and $l_2$.

This is not what I would consider a "proof", but merely a suggestion as to what one is trying TO prove.

Note this only considers the case where $l_1$ and $l_2$ intersect. They might not intersect, or (even more interestingly!) they might be THE SAME line. What do you suppose happens in these two cases?[/sp]

Being more algebraically-minded than a geometer, I would be inclined to write an isometry in THIS form:

$T(x,y) = A\begin{bmatrix}x\\y \end{bmatrix} + \begin{bmatrix}a\\b \end{bmatrix}$

Where $A$ is a 2x2 matrix with $\det(A) = \pm 1$,

because when I calculate something, I like to have some numbers to work with. Of course, to make this work, you need a "formula" for the line you are reflecting about, so you can determine the matrix $A$ and the "translation vector" $(a,b)^T$.

If that formula was, for example:

$l_1: y = mx + b$

I would first translate by $(0,-b)$ to create a "new line" $l_1'$ that goes through the origin, rotate that line so it lies on the $x$-axis, reflect about the $x$-axis, rotate back, and then translate by $(0,b)$.

While this may seem a bit computationally complex, it has the advantage of being able to say PRECISELY what the result of $T(x,y)$ is, for any point $(x,y)$.

 

[ovandagr]

I am currently working on a task involving isometries. I am confused as to how to explain the next two tasks to my adviser.

B. Any combination of isometries can be accomplished with one of the four fundamental isometries (i.e., translation, reflection, rotation, or glide reflection). Provide a logical argument that demonstrates that, given any pair of reflections, there exists a rotation or translation that produces the same result.

C. Demonstrate, using a counter example, that the product of a pair of isometries A and B is not always commutative (i.e., AB does not always equal BA).

If you construct a triangle and then translate and then reflect it, would the outcome (the third image) be the same as if you first reflected, then translated the triangle. Would translation (A)reflection(B) produce the same outcome as reflection(A)translation(B)?