Here is a collection of exercises based on those in Tondeur, Chapter 4. Since we have summarized the methods in the lessons, and corrected some errors in the text, the reference to an exercise, section, theorem or example in the text, included in brackets, are advisory. The exercise is given a letter to avoid confusion with the antecedent item in the text. In the exercise, these may have been rephrased, and instructions on how they are to be solved are given. All solutions should be based on the notes, in particular, you need to refer to propositions stated and proved in the lessons. Do not refer to theorems, propositions, lemmas or sections by their number in Tondeur. You may refer to summarizing results in the lessons by their nicknames, such as "Osometry Factorization Theorem"(OFT), "Barytheorem", "Tripod", "Recalibration", "Conjugacy" theorems etc. Also, when properties are in a list you may use the numbering there, though a few well chosen, descriptive words are preferable.
Also note that many of the exercises in Tondeur’s text have been subsumed into the lessons. So there's no point to assigning them as again. Other exercises have been added to simplify your solution of whole groups of exercises.
Use the Barytheorem for isometries and other facts about barycentric coordinates to prove that two isometries $ \alpha $ and $ \omega $ that have identical values on 3 non-collinear points, $ \triangle ABC $, i.e. $ A^\alpha = A^\omega, B^\alpha = B^\omega, C^\alpha = C^\omega $, then $ \alpha = \omega $. Recall that two transformation are the same if they have the same values on every point in the plane.
Prove that a reflection $ \sigma_\ell $ is an involution.
Hint: You may give a one-sentence picture-proof. But to make a precise proof you'll need to set up an argument that does not depend on a particular picture. This is an easy problem, so if your proof is gettin long involved, you don't understand it yet.
Prove that for an isometry $ \alpha $ the identity $ \alpha \sigma_\ell = \sigma_\ell \alpha $ for all lines $ \ell $ implies that $ \alpha = \iota $.
Hint: Use the conjugacy theorem for reflections, which says that $ \alpha \sigma_\ell \alpha^{-1} = \sigma_{(\ell^\alpha)} $. This is proved in the notes, and you do not reprove it here. Next argue that $ \sigma_h = \sigma_k $ implies that $h=k$. This says that the same reflection can't have different mirrors. Finally, if $\alpha(\ell) = \ell$ for all lines, why must $\alpha(X) = X$ for all points as well? Once you have proved this last statement, you're done. $
Draw a figure and give a geometrical argument that shows that a central reflection $ \sigma_Q $ factors into the product of two reflections, $ \sigma_Q = \sigma_a \sigma_b $, where $ a,b $ are any two mirrors intersecting at $ Q $ and perpendicular to each other.
Hint: In this edition of the course we skipped Tondeur's Chapter 2 on dilatations, where central reflections, $\sigma_Q(X) = Q - (X-Q) = 2Q - X $ are introducted. To proceed from here one without this preliminary work on central reflections, you need only this extended hint. The geometrical concept is that given the lens $Q$, a central reflection in $Q$ takes every point $X$ to a point "on the other side of $Q$ as far away again". This is a dumb description until you study the definition above. It reads "from $Q$ reverse the displacement vectore $X-Q$ from $Q$ to $X$, and apply it to $Q$. That algebraically, this happes to have the simple expression $2Q-X$ is interesting, but not particularly useful. This exercise says that reflections in two perpendicular mirrors is a central reflection in the their crossing point. And conversely. But better, you can choose one of the mirrors through $Q$ at will, as long as the second one is perpendicular to the first at $Q$. This is an example of the recalibration theorem. Finally, when you study rotations, such a double reflection in perpendicular mirrors must be a rotation by 180 degrees about $Q$. And in the plane, a central reflection always is such a rotation. (In space, it can't be a rotation because the central reflection of your right hand in point, comes out as your left hand. Spread your fingers, touch thumbs, and rotate one hand 180 degrees.]
Let $ \alpha $ be any isometry and $ \sigma_Q $ a central reflection. Use Exercise D and the conjugacy theorem for reflections to show that its conjugate is given by $ \alpha \sigma_Q \alpha^{-1} = \sigma_{Q^\alpha} $.
Hint: By super-exercise D, $ \alpha \sigma_Q \alpha^{-1} =\alpha \sigma_a \sigma_b \alpha^{-1} =\alpha \sigma_a \alpha^{-1} \alpha \sigma_b \alpha^{-1} = \sigma_{a^\alpha}\sigma_{b^\alpha} $. To clinch the argument you'll need use some properties of isometries regarding angles, as well as something like "the images of two lines cross as the image of their crossing". Sounds like doublespeak, but a formula makes it clear.
Prove that for an isometry $ \alpha $ the identity $ \alpha \sigma_C = \sigma_C \alpha $ for all points $ C $ implies that $ \alpha = \iota $ .
Use the theory for composing reflections developed in the rotations lesson to prove that the product of two reflections in perpendicular lines is the central reflection in their intersection point.
Let $M$ be a point on a line $m$, i.e. $ (Mm)$. Use the conjugacy theorem for reflections to show that $ \sigma_M $ and $ \sigma_m $ commute and find the single reflection equal to their product.
Hint: Use super Exercise D to factor $\sigma_M$ using mirror $m$ and its perpendicular $h$ at $M$. Now it boils down to some algebra. But if you don't use the conjugacy theorem in your proof, the proof is probably incomplete (hence wrong, read the Advice on proofs with gaps.)
Use the conjugacy theorems and other properties of reflections in the notes to prove that $ m = perbis(P,Q)$, the perpendicular bisector of a segment, if and only if (note the iff)
$ sigma_m sigma_Q sigma_m sigma_P = \iota $ |
Use the methods of the lessons to prove that $ k $ is a line parallel to the line $ (AB) $ if and only if (note the iff)
$ sigma_B sigma_k sigma_B sigma_A sigma_k sigma_A = \iota $ |
Use the properties of translations to show that that for two parallel lines $ m $ and $ n $ we have $\sigma_n\sigma_m=\sigma_m\sigma_n$ if and only if $m=n$.
Let $\tau_D $ be a translation and $P$ a point. Prove that $ \tau_D\sigma_P $ is a central reflection $ \sigma_A $ with center $ A= P+\frac{1}{2}D $. Similarly $ \sigma_P\tau_D = \sigma_B $ is a central reflection with center $ B= P-\frac{1}{2}D $. Hint: Use a mirror $ (mP) $ perpendicular to the vector $ D $ to recalibrate each isometry.
Let $ \tau_D $ be a translation and $ \beta $ an osometry (linear isometry). Use the conjugation theorem for translations to show that in this case we do have $ \beta\tau_D\beta^{-1}=\tau_{D^\beta} .$
Given $ \tau_D $ and a point $ Q $. Find two further points $ A, B $ for which $ \tau_D = \sigma_Q \sigma_B = \sigma_A \sigma_Q $ Hint: Use the recalibration theorem for reflection $ \sigma_Q $ to line up its mirrors with $ D $.
Consider triangle $\triangle ABC$ and an arbitrary fourth point $ G $. Prove that $ G $ is the centroid of $\triangle ABC$ if and only if
$ \sigma_G\sigma_C\sigma_G\sigma_B\sigma_G\sigma_A=\iota$. |
Hint: Rewrite the composition of the 6 central reflections as a composition of 3 translations and calculate the effect of this isometry on every point $ X $.
Extra Credit: How does this problem generalize, for example, to a quadrilateral? To an arbitrary closed polygon in the plane?
Consider a triangle $ \triangle ABC $ (oriented counter-clockwise) with (positively oriented) interior angles $\alpha,\beta,\gamma $ at $A,B,C$. Prove that $\rho_{A,2\alpha}\rho_{B,2\beta}\rho_{C,2\gamma}=\iota$.
Hint: Label the lines $ (AB)=c, (BC)=a, (CA)=b $ and factor the rotations into pairs of reflections.
Extra Credit: How does this probem generalize, for example, to a quadrilateral? To an arbitrary closed polygon in the plane?
Prove that the perpendicular bisectors $ a = perbis(BC), b= perbis(CA), c= perbis(AB) $ of the sides of $ \triangle ABC $ are concurrent, using the classification of isometries.
Hint: First, find a fixed point for the composition of the 3 reflections in the perpendicular bisectors. Second, argue that therefore this composition cannot be a glide reflection. Thirdly, observe that the three mirrors cannot be mutually parallel. Fourthly, argue that therefore they must be concurrent.
Consider a triangle $\triangle ABC$ ($A,B,C$ oriented counter-clockwise) with (positively oriented) interior angles $\alpha,\beta,\gamma$ at $A,B,C$. Prove that $\rho_{C,2\gamma}\rho_{B,2\beta}\rho_{A,2\alpha}$ is a translation different from the identity.
Hint: Compare the order of the composition to that in Exercise R. Show that this composition is the square of a glide reflection. Then use the canonical form of a glide reflection to calculate its square.