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Fun with Isometries

Introduction

Definition: An $isometry$ $\phi$ is a mapping of the Euclidean $\mathbb{R}^2$ plane to itself, such that the distance is preserved.

That is for any two points $A$ and $B$, $d(A,B)=d(\phi(A),\phi(B))$ where $\phi(A)$ and $\phi(B)$ are the images of $A$ and $B$.

Examples of isometries

 $1$. Identity mapping $Id$.

 $2$. Reflection $\sigma_l$ in line $l$.

$3$. Rotation $\rho_{O,\theta} $ about a point $O$ and an angle $\alpha$.

$4$. Translation $\tau_{\overrightarrow{AB}}$ by a vector $AB$ of non-zero length.

 $5$. Glide reflection $\gamma_{l,s}$ with glide axis $l$ and glide length $a$.


Theorems

We now present some theorems on isometries.

Theorem 1[Triangle effect]: An isometry is uniquely determined by the effect of three non-collinear points.


Proof: Given images $\phi(A)$, $\phi(B)$, $\phi(C)$ of three non-collinear points $A$, $B$, $C$ under an isometry $\phi$, we prove that we can always find a unique image of any other point $X$ in the plane.


We use the fact that distance is preserved under an isometry.

Draw a circle $\omega_A$ centred at $\phi(A)$ and radius $d(A,X)$.

As, $d(A,X)=d(\phi(A),\phi(X))$, we have that  $\phi(X)$ lies on $\omega_A$.

Similarly, draw circle $\omega_B$ centred at $\phi(B)$ and radius $d(B,X)$. $\phi(X)$ also lies on this circle.


$\therefore$ $\omega_A$ and $\omega_B$ intersect at $\phi(X)$. If $\omega_A$ and $\omega_B$ have only one point of intersection, then that point is $\phi(X)$ and we are done.

If not, then there would be two intersections of $\omega_A$ and $\omega_B$, and one of them is  $\phi(X)$.

Then, draw circle $\omega_C$, centred at $\phi(C)$ and radius $d(C,X)$. $\phi(X)$ lies on $\omega_C$ as well.

Note $\phi(C)$ can't be equidistant from the two intersections of $\omega_A$ and $\omega_B$, otherwise $\omega_A$, $\omega_B$, $\omega_C$ would be coaxial, and the centres $\phi(A), \phi(B), \phi(C)$ would be collinear, implying $A,B,C$ collinear.

Contradiction!

Hence, choose the unique point which lies on $\omega_1$, $\omega_2$, $\omega_3$ and we are done! 


The following theorem is about expressing any isometry as the composition of reflections.


Theorem 2[Composition as three reflections]: Any isometry $\phi$ is the composition of at most $3$ reflections.

Proof: Choose three non-collinear points $A,B,C$. By theorem 3.1, we only need to find a composition of atmost three reflections which maps $A,B,C$ to $\phi(A),\phi(B),\phi(C)$.


If $\phi(A)=A$, $\phi(B)=B$, $\phi(C)=C$, we are done as $\phi$ is identity mapping.Hence, wlog $\phi(A)\neq A$.

First consider the reflection $R_1$ which maps $A$ to $\phi(A)$. That is reflection over perpendicular bisector of $A\phi(A)$.

Hence, $\phi(A)=R_1(A)$.



If $\phi(B)=R_1(B)$ and $\phi(C)=R_1(C)$, we are done as $\phi = R_1$. Hence, wlog $\phi(B) \neq R_1(B)$.

Second, consider the reflection $R_2$ which maps $R_1(B)$ to $\phi(B)$. That is reflection over perpendicular bisector of $R_1(B)\phi(B)$.

Note that $\phi(A)=R_1(A)$ lies on this perpendicular bisector, as 

$d(\phi(A),\phi(B))=d(A,B)=d(R_1(A),R_1(B))=d(\phi(A),R_1(B))$.


Hence, $\phi(A)=R_1(A) = R_2\circ R_1 (A)$; and also $f(B)=R_2\circ R_1 (B)$.



If $R_2\circ R_1 (C)=C$, we are done as $\phi=R_2\circ R_1$.Hence, let $R_2\circ R_1 (C)\neq \phi(C)$.

Last, consider the reflection $R_3$ which maps $R_2 \circ R_1(C)$ to $\phi(C)$. That is reflection over perpendicular bisector of $R_2 c\circ R_1(C)\phi(C)$.

Note that $f(A)=R_2\circ R_1 (A)$ and $f(B)=R_2\circ R_1 (B)$ lie on this perpendicular bisector of $R_2 \circ R_1(C)\phi(C)$ as : 

$d(\phi(A),\phi(C))=d(A,C)=d(R_2\circ R_1(A),R_2\circ R_1(C))=d(\phi(A),R_2\circ R_1(C))$ and

$d(\phi(B),\phi(C))=d(B,C)=d(R_2\circ R_1 (B), R_2 \circ R_1(C) ) = d(\phi(B),R_2\circ R_1(C))$.


Hence, $\phi(A)=R_2\circ R_1 (A) = R_3 \circ R_2 \circ R_1 (A)$ and $\phi(B)=R_2\circ R_1 (B) = R_3 \circ R_2 \circ R_1 (B)$; and also $\phi(C)=R_3\circ R_2 \circ R_1 (C)$.


Hence, $\phi=R_3 \circ R_2 \circ R_1$.


Hence, we are done! 



Now, Let's see how we can express a rotation as the combination of two reflections.

Theorem 3[Rotation as composition of reflections]: Suppose any two distinct lines $m$ and $l$ intersect at $O$ such that $\angle(m,l)=\theta$, then $$\sigma_l\circ \sigma_m=\rho_{O,2\theta}$$


Proof: Consider any point $P$ whose reflections are easy to track.

The reflections give $$\angle(PO,m)=\angle(\rho_m(P)O,m) \text{ and } \angle(\rho_m(P)O,l)=\angle(\rho_l \circ R_\rho(P)O,m)$$

Hence, $$\angle(PO,R_l \circ R_m(P)O)=2\cdot \angle(m,l)=2\theta$$

Hence, it is easy to observe that the combined isometry $\rho_l \circ \rho_m$ is a rotation about $O$ and angle $2\theta$.



Remark:
A crucial observation is that the choice of the lines does not matter. The only condition needed is that $\angle(m,l)=\theta$.


Here comes another cool theorem, dealing with the combination of two rotations!

Theorem 4[Composition of rotations]: For distinct points $O_1,O_2$ and $2(\theta_1+\theta_2)\neq 360$ $$\boxed{\rho_{O_1,2\theta_1}\circ \rho_{O_2,2\theta_2} = \rho_{O_3,2(\theta_1+\theta_2)}}$$ for point $O_3$ such that $\angle(O_3O_1,O_2O_1)=\theta_1$ and $\angle(O_1O_2,O_3O_2)=\theta_2$.

Proof: We use theorem 3 to prove this.

Let $l$ be the line $O_1O_2$.

Draw line $m$ passing through $O_1$ such that $\angle(m,l)=\theta_1$.

Similarly, draw line $n$ passing through $O_2$ such that $\angle(l,n)=\theta_2$.

Let $m\cap n = O_3$.


Using the Theorem 3 and the remark,

$$\rho_{O_1,2\theta_1}=R_m\circ R_l \text{ and }\rho_{O_2,2\theta_2} =R_l \circ R_n $$ 

$$ \therefore \rho_{O_1,2\theta_1}\circ \rho_{O_2,2\theta_2} = R_m\circ R_l \circ R_l \circ R_n =R_m \circ Id \circ R_n =R_m \circ R_n $$

But $m \cap n = O_3$ and $\angle(m,n)=\theta_1+\theta_2$.

Hence, $$R_m \circ R_n = \rho_{O_3,2(\theta_1+\theta_2)}$$

$$\therefore \rho_{O_1,2\theta_1}\circ \rho_{O_2,2\theta_2} =\rho_{O_3,2(\theta_1+\theta_2)} $$

Hence proved!


The following theorem states that there are only $5$ types of isometries.


Theorem 5[Classification of isometries]:

Any isometry belongs to one of the following $5$ groups.\\

$[0]$. Identity

$[1]$. Reflection

$[2a]$. Rotation

$[2b]$. Translation

$[3].$ Glide reflection


Proof: From Theorem 2, any isometry $\phi$ is a composition of at most 3 reflections.

  •    If $\phi$ is a composition of $0$ reflections, it is easy to observe that $\phi$ is just the identity map.
  •    If $\phi$ is a composition of $1$ reflection in line $l$, again it's obvious that $\phi=\sigma_l$ i.e. it is  itself a reflection in $l$.
  • If $\phi$ is a composition of $2$ reflections in lines $l$ and $m$ , there are two cases.

    $[a]$. If $l$ and $m$ intersect at $O$ and $\angle(m,l)=\theta$, then $\sigma_l \circ \sigma_m= \rho_{O,2\theta}$ from theorem 3, and hence is a rotation.

    $[b]$.If $l$ and $m$ are parallel, let we get that $\sigma_l \circ \sigma_m$ is a translation in the direction perpendicular to the lines; with twice the distance between the lines.

    More precisely, if $A\in m$ and $B\in l$ such that $AB \perp m,l$ then $\sigma_l \circ \sigma_m=\tau_{2\overrightarrow{AB}}$.

    (The proof of $[b]$ is similar to proof of $[a]$)\\

  • If $\phi$ is a composition of $3$ reflections in lines $l,m,n$.

    Note that if $l \parallel m \parallel n$, then we can translate the first two $l$ and $m$, to make $m$ coincide with $n$; without  changing of their composition. Then, $\sigma_{l}\circ \sigma_{m} \circ \sigma_{n} = \sigma_{l'}\circ \sigma_{n}\circ \sigma_{n} = \sigma_{l'}$. Hence the combined effect is a reflection in line $l'$. Hence, we exclude this case.Hence we let all $l,m,n$ be not parallel.



    Then, $m$ is not parallel to at least one of $l,n$,wlog $n$. 

    Then, $\sigma_m \circ \sigma_n =\rho_{m \cap n, 2\angle(m,l)}$ by theorem 3.

    Hence, can now rotate both lines $m,n$ simultaneously  about $m \cap n$ as along as $\angle(m,n)$ is unchanged.

    By an appropriate rotation, we make $m' \perp l$. 



    Then on similar logic, we rotate both simultaneously $l,m'$ about $l\cap m'$ while preserving $\angle (l,m')$ such that $m'' \parallel n'$.



      Then, $$\sigma_l \circ \sigma_m \circ \sigma_n = \sigma_l' \circ \sigma_m'' \circ \sigma n'=\sigma_l' \circ \tau_{\overrightarrow{AB}}  \text{ for some vector} \overrightarrow{AB} \text{ in direction of }l' $$(Here we used theorem 3 and  $[2b]$).



    But, this is just a glide reflection by the defination! Hence, the composition of 2 reflections in non a glide reflection.


    We have exhausted all cases and hence, we are done!


 Finding the centre of rotation

If an isometry $\phi$ is a rotation, then to find its centre :

Consider any two points $A$ and $B$ and their maps $\phi(A)$ and $\phi(B)$.

Then, $O$ is the intersection of perpendicular bisectors of $A\phi(A)$ and $B\phi(B)$.


Problems

We now do some problems using isometries.

The following is a very interesting problem!

Problem 1: A notorious thief amassed a lot of wealth through stealing, and buried the loot in an island. He was on his death bed and he called his son and gave him the directions to find the buried loot. He told him, "Go to the island Thieves' Paradise and look for the pond. From the pond go towards the big banyan tree measuring the distance. Once you reach the tree, turn right and walk the same distance. Mark this point. Again from the pond, walk towards the big neem tree measuring the distance. At the neem tree turn left and walk the same distance. Mark this point. Dig $100$ feet at the midpoint of the marked points and you will get the money." The son goes to the island looking for the pond $−$ but the pond was no longer there even though the banyan tree and neem trees were there. Can he find the money$?$



The son can only see this :



Solution: Let the Banyan tree be $B$ and neem tree be $N$.

Let the variable pond be $P$, and the two marked points after rotations of $P$ about $B$ and $N$ be $B'$ and $N'$.

Let the treasure, which is at midpoint of $B'N'$ be $T$.

We prove that $T$ is dependent only on $B,N$.



Consider the following isometry :

$$\phi=\rho_{N,90}\circ \rho_{B,90}$$


Now construct point $X$ as follows:

Let $X$ be such that $XB=XN$ and $\angle NXB =90$.

Hence, by theorem 3.2, $$\phi=\rho_{X,180}$$

Hence $$\phi \text{ is central symmetry over } X$$



Under this isometry, $\phi(B')=\rho_{N,90}\circ \rho_{B,90}(B')=\rho_{N,90} (P)=N'$.

Hence, $$\phi(B')=N' \implies B'N' \text{ has midpoint }X$$

But $B'N'$ has midpoint $T$, the treasure.

Hence, $$X=T$$


Hence, the treasure is at $X$!

But, we know $X$ as it depends only on $B,N$.

We can find $X$ by the above-mentioned construction and that's where the treasure is!



Problem 2: Given chords $AB$ and $CD$ of a circle, intersecting outside the circle. Find a point $X$ on the circle, such that chords $AX$ and $BX$ cutoff on segment $CD$, a segment $EF$ having a given length $a$.



Solution: Assume that the problem has been solved. That is, consider the required point $X$ and corresponding points $E,F$. We now get conditions on $X,E,F$.\\

Consider the following isometry:

$$\phi=\text{Translation by }a\text{ in direction }CD$$

Under the isometry $\phi$ we get , $E\rightarrow F$, $A\rightarrow A'$ and hence note that $AE \parallel A'F$. 

Also note that $A-E-X$ and $B-F-X$.

Hence, $$\angle(A'FB)=\angle(AXB)=\angle(ACB)$$

But, $A',B$,$\angle(ACB)$ are known to us.

We have now got a condition on $F$ that $\angle A'FB= \angle ACB$


Hence in order to construct $F$: considering chord $A'B$, construct a circular arc that subtends angle $\angle(ACB)$.

The intersection of this circular arc and $CD$ is the prescribed point $F$.

Hence, translating $F$ by length $a$ in direction $DC$ gives $E$.

Hence, we are done!


Problem 3: A billiard ball bounces off a side of a billiard table in such a manner that the two limes along which it moves before and after hitting the sides are equally inclined to the side. Suppose a billiard table were bordered by $n$ lines $l_1,l_2,\dots,l_n$; let $A$ and $B$ be two given points on the billiard table. In what direction should one hit a ball placed at A so that it will bounce consecutively off the lines $l_1,l_2,\dots,l_n$ and then pass through the point $B$



Solution: 




Reflect $B$ over $l_n$ to get $B_n$. Reflect $B_{i+1}$ over $l_{i}$ to get $B_{i}$ for $i=n-1,n-2,\dots,1$.

Let $l_1\cap AB_1=X_1$ and $l_{i}\cap X_{i-1}B_{i} = X_{i}$ for $i=2,3,\dots,n$.

Then, $$A-X_1-X_2-\dots-X_n-B \text{ is the required path.}$$

It can be easily checked that this indeed satisfies the conditions, as shown in the figure.

Problem 4: Let $ABC$ be a given triangle. Draw a line $l$, that meets sides $AB$ and $AC$ in points $P$ and $Q$ such that $BP=PQ=QC$.

Proof: Let $O$ be the intersection of the perpendicular bisector of $BC$ with $arc BAC$.

Then, $\angle PBO=\angle ABO =\angle ACO =\angle QCO$. Also, $PB=QC$ and $OB=OC$.

Hence, $$\triangle PBO \cong \triangle QCO $$

Considering the orientation, it's easy to see that the map $\triangle OPB$ to $\triangle OQC$ is a rotation about $O$; and not a reflection of the perpendicular bisector of $BC$.

But $\angle BOC=\angle BAC=\alpha$, which is known to us.

Hence, $$\rho_{O,\alpha}(B)=C \text{ and }\rho_{O,\alpha}(P)=Q$$

Hence, $$OP=OQ \text{ and } \angle POQ=\alpha$$

Hence, we know the ratio $$k=\frac{PQ}{OP}=2sin\left(\frac{\alpha}{2}\right)$$

But by given condition, $PQ=PB$.

Hence, $$k=\frac{OP}{PB}$$

Note that we know $k,O,B$ and we consider locus of $P$.

  •      If $k=1$ (Hence, $A=60$) the locus of $P$ is a line, which is the perpendicular bisector of $OB$.Hence, $P$ lies on perpendicular bisector of $OB$.

But $P$  also lies on $AB$.


Hence, we can construct $P$ as intersection of $AB$ and perpendicular bisector of $OB$!

And, then by rotating $P$ about $O$ by $60$ gives $Q$.

Working backwards, it is easy to see that this  is the required construction. 

  •  If $k \neq 1$ the locus of $P$ is a circle. (This fact is well known- Apollonian circles).


Hence, we can find $P$ as intersection of $AB$ and the circle that is the locus of points the ratio of whose distances to $O$ and $B$ is equal to $k$.

And then, intersection of $AC$ with  circle centred at $P$ and radius $PB$ gives $Q$.

Working backwards, we see that this is the required construction!. 

(Note that this works for when $AB=AC$ too) 

Hence, we are done!




References




Post by Malay Mahajan.

About the guest blogger: Hello, I am Malay Mahajan, currently in 11th grade, from Mumbai. I love to do Olympiad Math, and play chess, follow sports. I am an INMO Awardee-2022, and am a Sophie-Fellow.



Post set up and modified by Sunaina Pati.

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