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In a previous problem I asked about the notation used here.. I'm still not sure how to show it even though I now get what it's asking.

The following is an exercise from The Four Pillars of Geometry.

3.6.5. Show that the reflections in lines $L$, $M$, and $N$ (in that order) have the same outcome as reflections in lines $L'$, $M'$, and $N$, where $M'$ is perpendicular to $N$

I've had to ask around to clairify parts of the problem; it should be noted that the lines $L'$ and $M'$ are essentially arbitrary with no relation to $L$ and $M$ except that they appear in the same order when you do the reflections. Also, no picture is given as the lines are general though I've drawn a few pictures to help reason it out.

My main issue might be that I'm not sure what tools to use, so please be as specific as possible in your answer.

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What does this have to do with a glide? –  Gerry Myerson Jul 24 '11 at 5:21
I just added a link to the related question. (I was usure what the problem was stating.) There are 2 other problems, the last asks you to prove that any 3 reflections results in a glide. –  ttt Jul 24 '11 at 5:33
$L'$ and $M'$ are not arbitrary. GIven $L, M, N$ they are to be determined so that the condition $M'$ perpendicular to N is satisfied and so that the reflections in the new lines define the same isometry as the reflections in the old ones. –  Mark Bennet Jul 24 '11 at 5:52
I don't think that this is true in general. The simplest case for me to imagine is the following. Assume that all 6 lines intersect at point $O$. Then the composition of the two first reflections is a rotation about $O$ by an angle that's twice the angle between the two lines. Therefore the orthogonality relation $M'\perp N$ can be basically ignored, because we are free to choose $L'$. Then the third reflection makes the total composite mapping into an arbitrary reflection about a line thru $O$. So you have left out something essential. -1 for asking the same question again? –  Jyrki Lahtonen Jul 24 '11 at 5:59
@Jyrki: It’s not really the same question. The first time was to find out exactly what the question was; now that he understands that, he’d like some help in answering it. –  Brian M. Scott Jul 24 '11 at 6:36

1 Answer 1

up vote 1 down vote accepted

I understand the question as in Mark's comment: $L$, $M$ and $N$ are given lines, and we want to show that there are lines $L'$ and $M'$ with $M'\perp N$ such that reflections in $L'$, $M'$ and $N$ (in that order) have the same effect as reflections in $L$, $M$ and $N$ (in that order).

First, since the reflection in $N$ is last in both cases, we can cancel it, so the task is to show that there are lines $L'$ and $M'$ such that reflections in $L'$ and $M'$ have the same effect as reflections in $L$ and $M$, subject to $M'\perp N$.

This is false. If $L$ and $M$ are parallel, reflections in $L$ and $M$ amount to a translation perpendicular to them by twice their distance, and this can only be reproduced by $L'$ and $M'$ if they, too, are parallel to $L$ and $M$; hence we cannot choose $M'$ perpendicular to $N$ unless $M$ already happens to be.

It's true, however, if $L$ and $M$ aren't parallel. In this case, it follows almost immediately from the hint given in the book right above the exercise:

Reflections in any two lines meeting at the same angle $\theta/2$ at the same point $P$ give the same outcome.

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And if Tony looks carefully at the notation for the next exercise in the book (3.6.6), he'll see that he has to apply the same observation to a different pair of reflections to prove the final theorem. It's a pity the special case doesn't get picked out. The primes in the book aren't standard notation, but I think they may be supposed to give a hint that the mirror lines are being transformed in some way. –  Mark Bennet Jul 24 '11 at 8:20
I'm learning to trackback more often, as here honestly I knew about the hint, and the next exercise, and what all three are trying to say and I still had questions! –  ttt Jul 25 '11 at 5:50
Thanks for pointing out the exception; does this reasoning work for the case where $L$ and $M$ intersect? Choose $M'$ to be the perpendicular to $N$ through the point reflected in $L$ then $M$. Choose $L'$ so that $L'$ and $M'$ intersect so that the initial, final, and intersection define a circle - that is, initial and final points are only $\theta$ apart. Have $L'$ bisect this angle. Done. –  ttt Jul 25 '11 at 5:58
<-- I'm curious if the previous comment worked but it doesn't use the hint. Better to say initial and final point are at angle $\theta$ with the intersection of $L$ and $M$. Place $M'$ through the intersection. Place $L'$ at angle $\theta/2$. Done with hint! –  ttt Jul 25 '11 at 6:04
I don't understand what you're referring to by "the point reflected in $L$ then $M$" and by "the initial, final, and intersection", but it's less complicated than that -- any two lines meeting where $L$ and $M$ meet, at the same angle, give the same rotation, so just turn $L$ and $M$ together such that $M$ becomes perpendicular to $N$. –  joriki Jul 25 '11 at 6:05

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