Billiards in Near Rectangle - UCLA Department of …yy26/works/billiards_near_rectangle.pdf · r near the rectangle r has a periodic billiard path. Billiards in Near Rectangle August

Billiards in Near Rectangle

Yang Yilong Chang Hai Bin

August 6, 2012

Billiards in Near Rectangle August 6, 2012 1 / 23

Introduction

Billiard ball starts at a point(E), witha given initial direction.

Whenever it hits a side, thetrajectory will satisfy:Angle of incidence= angle of reflection

Ignore cases when: Billiard ball hitsvertex (e.g. J)


Introduction

Billiard ball starts at a point(E), witha given initial direction.

Whenever it hits a side, thetrajectory will satisfy:Angle of incidence= angle of reflection

Ignore cases when: Billiard ball hitsvertex (e.g. J)


Periodic Billiard Path:if the ball comes back to the initial position with initial velocity direction.


Word:After labelling the sides of polygon using numbers/letters, record down thesequence of sides which the periodic path bounces off.

0123 021023



0123

021023



0123 021023


Unfolding:Whenever the ball hits a side, reflect the polygon, and keep the “billiardpath” straight. The path on the new polygon will “corresponds” to thetrajectory in original polygon.






An even word W (word with even number of characters) is an orbit typefor a polygon P iff:

there exists a periodic billiard path that hits the sides of P according tothe order given by W .

Equivalent to saying that:

1 the first and last polygon in the unfolding are related by a translation,AND

2 there exists a path that “stays within” the unfolding, not hitting thevertex.

We will illustrate this using pictures in the next slide.
















Let’s look at unfolding of this periodic path:






Since the line EN forms the same angle with the line AC and A′′C ′′, soAC is parallel to A′′C ′′, hence the last polygon is a translation of the firstpolygon. And the line EN lies “within” the unfolding.


Space of Quadrilaterals

Q, the space of all quadrilaterals (upto similarity/rescaling) ischaracterized by 4 parameters.

3 parameters are not enough to characterize quadrilaterals.

e.g. is not similar to


Space of Quadrilaterals

Q, the space of all quadrilaterals (upto similarity/rescaling) ischaracterized by 4 parameters.

3 parameters are not enough to characterize quadrilaterals.

e.g. is not similar to


So, space of all quadrilaterals Q can be considered as a “subset” of R4.(although some elements might be represented by more than one elementsin R4)

e.g. (65◦, 50◦, 60◦, 47.79◦) and (30◦, 40◦, 32.21◦, 35◦) represent the samequadrilateral below.


So, space of all quadrilaterals Q can be considered as a “subset” of R4.(although some elements might be represented by more than one elementsin R4)

e.g. (65◦, 50◦, 60◦, 47.79◦) and (30◦, 40◦, 32.21◦, 35◦) represent the samequadrilateral below.


Near Square

The square is characterized by thecoordinate

(π4 ,

π4 ,

π4 ,

π4

).

Definition

A quadrilateral is ε-near square iff along one of the diagonal, |ai − π4 | < ε

for i = 1, . . . , 4


Near Square

The square is characterized by thecoordinate

(π4 ,

π4 ,

π4 ,

π4

).

Definition

A quadrilateral is ε-near square iff along one of the diagonal, |ai − π4 | < ε

for i = 1, . . . , 4


Lemma

In the diagram below, if ε < π12 , and |αi − π

4 | < ε for i = 1, . . . , 4, then|βj − π

4 | < 3ε for j = 1, . . . , 4

Consequence: Our definition ofnear-square is not arbitrary/overlyaffected by the choice of diagonal.

As long as one set of angles (e.g. αi )stays near π

4 , the other set of angles(e.g. βj) will not stray too far fromπ4 .


Lemma

In the diagram below, if ε < π12 , and |αi − π

4 | < ε for i = 1, . . . , 4, then|βj − π

4 | < 3ε for j = 1, . . . , 4

Consequence: Our definition ofnear-square is not arbitrary/overlyaffected by the choice of diagonal.

As long as one set of angles (e.g. αi )stays near π

4 , the other set of angles(e.g. βj) will not stray too far fromπ4 .


1st Main Result

Theorem

If q is a quadrilateral that is π107 -near square, then it has a periodic billiard

path.

Theorem

For every rectangle r , there exists an εr > 0, such that every quadrilateralthat is εr near the rectangle r has a periodic billiard path.


1st Main Result

Theorem

If q is a quadrilateral that is π107 -near square, then it has a periodic billiard

path.

Theorem

For every rectangle r , there exists an εr > 0, such that every quadrilateralthat is εr near the rectangle r has a periodic billiard path.


α-β plane

Quick recap on how the proof works:For every quadrilateral, if it is not arectangle, then choose any angleα < π

2 .

Choose the smaller of the twoadjacent angle, call it β.

Every quadrilateral is represented bysome point on the α-β left half planeand origin. (α < π

2 or α = β = π2 )


α-β plane


2 .



2 or α = β = π2 )


α-β plane


2 .



2 or α = β = π2 )


α-β plane

Remark: Being close to(α, β) =

(π2 ,

π2

)does not give you

near square-ness. (e.g. the point(π2 ,

π2

)represents rectangles.)

So, we want to prove: Every pointon this α-β plane, provided thequadrilateral they represent are π

107near square, have periodic billiardpath.


α-β plane

Remark: Being close to(α, β) =

(π2 ,

π2

)does not give you

near square-ness. (e.g. the point(π2 ,

π2

)represents rectangles.)

So, we want to prove: Every pointon this α-β plane, provided thequadrilateral they represent are π

107near square, have periodic billiardpath.


The idea: to chop up the α-β planeinto different cases. For each case(assuming the near-squareness), tryto find a periodic path that satisfyeach case.

For example, in the bolded line indiagram to the right, it represents aright angle adjacent to an acuteangle.

We can relate this case to billiardpath in right triangle, as shown inthe picture below.










Another example: in the shadedregion in diagram to the right, isrepresented by α + 2β > 3

2π.

Which is to say: the angle oppositeα is acute.


Another example: in the shadedregion in diagram to the right, isrepresented by α + 2β > 3

2π.

Which is to say: the angle oppositeα is acute.


We found the following billiard path that satisfy the all quadrilateral in theshaded region, provided the quadrilateral is π

30 near square. We call it the“A” orbit.


The eight regions that we consideredare illustrated in the right: (4regions, 3 lines, and origin).


Why so near square?

e.g. For the right angle-acute anglecase, we just need the quadrilateralto be π

12 near square.

In particular, we need the angle φ tonot exceed π

2


Why so near square?

e.g. For the right angle-acute anglecase, we just need the quadrilateralto be π

12 near square.

In particular, we need the angle φ tonot exceed π

2


Why so near square?

After some calculation, we can show that as long as the quadrilateral is π30

near square, then we can “fit” the billiard path into the unfolding.


Why so near square?

How near square do we need for eachregion (for a billiard path to exists)?Below is a list of the “maximaltolarance”.

Green: π30

Red: > π107

Yellow: π107

Blue: > π56

Gray: π56

Black: π12

Cyan: π12

Origin: > π12


Why so near square?

The following is the unfolding of the periodic path we used to cover theyellow region.

It is hard to “fit” a billiard path within the unfolding. Slight pertubation ofthe quadrilateral could potentially change the unfolding drastically, andhence unable to “fit” a periodic path within the unfolding.


Why so near square?

The following is the unfolding of the periodic path we used to cover theyellow region.It is hard to “fit” a billiard path within the unfolding. Slight pertubation ofthe quadrilateral could potentially change the unfolding drastically, andhence unable to “fit” a periodic path within the unfolding.


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Billiards in Near Rectangle - UCLA Department of …yy26/works/billiards_near_rectangle.pdf · r near the rectangle r has a periodic billiard path. Billiards in Near Rectangle August