§ 1. The Fagnano trajectory

Billiard trajectories

I.F.Fagnano posed in 1775 the following pro- blem : "To inscribe in a given acute-angle triangle

the triangle of a minimum perimeter".

§ 1. The Fagnano trajectory

1.1. Preliminaries

Among the oriented angles in the plane we have 1. the angles between rays or vectors

arrival departure

˛

˛

In radians, the measure ˛ of that angle belongs to R=2Z, which means that ˛ is an equivalence class modulo2. For instance in an

equilateral triangleABC (oriented positively) B C A

the

1

measure of the oriented angle of rays (or vectors). ! AB; !

AC /is

˛ Dn

: : : ; 11 3 ; 5

3 ; 3;7

3 ;13 3 ; : : :o If you divide by2you get an element ofR=Z

1 2˛ Dn

: : : ; 11 6 ; 5

6 ; 6;7

6 ;13 6 ; : : :o 2. the angles between lines

arrival departure

ˇ ˇ

ˇ

ˇ

In radians, the measureˇof that angle belongs toR=Z, which means that˛ is an equivalence class modulo 2. For instance in a rectangle

ABCD (oriented positively) A B

C D

the measure of the oriented angle of lines between the two diagonals. !

AB; ! AC /is ˇDn

: : : ; 11 6 ; 5

6 ; 6;7

6 ;13 6 ; : : :o Think of two parallel lines, their angle is0modulo or

ˇ D fkIk2 Zg .

Theorem 1. Let € be a circle with center O and let A and B be two distinct points belonging to€. A pointM belongs to€ if and only if

.MA; MB/D 1 2. !

OA; ! OB/

A B O

M

€

If you measure the angles on the picture you get˛ D ^AOB

1

^{D 143:35°}

andˇ D^AMB

1

^D108:32°. How does it fit withˇD ¹₂˛?

Définition.Points in an Euclidean plane are said to becocyclic(orconcy- clic) if they lie to a common circle or a common line.

Theorem 2. Four pointsA, B, C andD are cocyclic if and only if the following equality between oriented angles of lines holds :

.CA; CB/D.DA; DB/ (*)

Proof.Suppose the points are on a line, then.CA; CB/D0and.DA; DB/D0.

Thus./holds. Suppose now the points are on a common circle. LetObe the center of that circle. Following the theorem 1 above, we have.CA; CB/ D

1 2. !

OA; !

OB/and similarly.DA; DB/D ¹₂. ! OA; !

OB/, thus./holds.

Conversely, suppose the four points verify ./. Let € be the circle or line ABC. If € is a line, then .CA; CB/ D 0, and from ./ we have .DA; DB/D0which implies thatDbelongs to the lineABand the theorem is proved. If€ is a circle, the converse of theorem 1 shows that D belongs to€.

1.2. The orthic triangle of a triangle

Definition.LetABC be a triangle. LetP be the point ofBC such that AP is the height of that triangle issued fromA. Similarly, letQbe the point ofCAandRthe point ofABsuch thatBQandCRare the two other heights.

The trianglePQR is called the orthic triangle of the triangleABC.

B

A

C R

Q

P H

The three heights meet at a pointH calledorthocenter. Notice that since .PH; P C /Da right angleD.QH; QC /, the pointsH,C,P andQare on a common circle. Similarly, since.RH; RB/ D.PH; PB/, there is a circle going through the four pointsH,B,RandP.

B

A

C R

Q

P H

Theorem.Let PQR be the orthic triangle of a triangleABC, then we have the following equalities of measures of angles

1

CPQD^BPR

1

^AQR

1

^DCQP

1

^ARQ

1

^DBRP

1

B

A

C R

Q

P H

˛⁰ ˛

Proof. Let us call ˛ D .P C; PQ/ and ˛⁰ D .PR; PB/. Since the pointsC,Q, H andP are on common circle, we have˛ D .P C; PQ/ D .H C; HQ/. The lineH C is the same as the lineHRand the lineHQis the same as the lineHB, thus˛ D .HR; HB/. Since the four pointsH,R, B andP are cocyclic.HR; HB/D.PR; PB/thus˛ D˛⁰.

1.3. Solving Fagnanos problem

Definition. Let ABC be an acuteangled triangle. A triangle PQR is inscribed in the triangleABC ifP is a point on the segmentBC,Qa point onCAandRa point onAB.

The perimeterp of a trianglePQR is the sum of the length of the sides of the trianglep D jQRj C jRPj C jPQj.

Theorem.LetABC be an acuteangled triangle. Among all the inscribed inscribed inABC, the triangle with smallest perimeter is the orthic triangle.

Proof.We do the proof in two steps. In the first step we supposeP fixed and we look forQandRsuch that the perimeterp of the inscribed triangle PQRis minimum. In the second step we chooseP alongBC mnimisingp.

First step.LetP1 be the point image ofP in the reflection along the line AB.

B A

P P1

Q

Notice that the symmetry gives us the equality of length AP1 D AP and the equality of angles ^BAP

1

^{1 D PAB}

1

^{and thus PAP}

1

^{1 D 2PAB}

1

^{. Notice}

also that sinceQis a point on the axis of reflection, we have the equality of lengthP1QDPQ.

LetP2be the point image ofP in the reflection along the lineAC.

B

A

P C P1

P2

R

Q

We get similarlyAP2DAP,^PAP

1

^{2 D2PAC}

1

^{andRP2 DPR.}

Consider the triangleAP1P2. The angle^P

2

^1AP2is twice the angle^BAC

1

at the vertex Aof the initial triangle. That angle is constant. The triangle is isoceles sinceAP1 DAP DAP2.

The length of the broken lineP1QRP2is equal to the perimeterpof the inscribed trianglePQRsinceP1QDPQandRP2 DRP and thus

P1QCQRCRP2 DPQCQRCRP Dp

Then for the fixed P the perimeter p will be minimal if the broken line P1QRP2 has minimal length. But we know that the shortest path from P1

toP2is the straight line. Thus to findQandRwe just have to draw the line P1P2 and mark the intersections with the sidesAB andAC.

B

A

P C P1

P2

Q R

Second step.Now we have to choose the pointP on the segmentBC in such a way thatp DP1P2is minimum. For eachP we get a triangleAP1P2. These triangles are all isosceles with same angle at the vertex A, namely 2^BAC

1

. The basisP1P2will be minimum if the length of the equal sides is minimum. But the length of the sides is the length of the segmentAP, and this segment has minimum length when P is the orthogonal projection of Aon the side BC, that is whenP is the foot of the height issued from the vertexA. But then by symmetry reasonsQandRhave also to be the feet of the respective heights and thusPQRhas to be the orthic triangle.QED 1.4. The Fagnanos trajectory

LetABC be an acuteangled triangle and let us take it as the border of a billiard. LetPQRbe the orthic triangle ofABC.

Let us start a trajectory atP in the direction ofQ, when the ball arrives at

1

Qit bounces following the reflection law and since the angles^CQP

1

^and

RQAare equal it goes in the direction ofRand inRit bounces back into the direction ofP coming back to the departure point after having run through3 segments. After that the trajectory do again and again the same travel : The trajectory is said to be3-periodic. Since it follows the border of the triangle solving the Fagnanos problem, it is called the Fagnano trajectory.

B

A

C R

Q

P

˛

˛

ˇ ˇ

1.5. A family of 6-periodic trajectories

Let us first consider parallel trajectories bouncing on a common line

d d

`Ddsin˛

˛ ˛ ˛ ˛

We see that the distance between the two trajectories before and after the shocks with the border are the same : it is an invariant in polygonal billiards.

Let us use the same billiard and start a trajectory from a pointM on the segmentP C strictly betweenP andC, not too far fromP. We put` DPM. Let us start in the direction parallel toPQ. The billiard ball will bounce on the side AC in a pointM1 betweenQandC such that MM1is parallel to PQ. Following the reflection law it will leaveM1 in a direction parallel to QRand meet the sideAB in a pointM2 betweenRandB. Then it goes on toM3onBC betweenP andB such thatM2M3is parallel toRP.

The distanced between the parallel linesMM1 andPQ isd D`sin˛.

This distance will be the same between M1M2 andQR and it will also be the distance betweenM2M3andRP. ThusPM3 D _sin^d_˛ D`.

The trajectory continues after M3 to M 4 on CA, M5 on AB and M6

on BC. By the same reasoning as above we get PM6 D PM3, and thus PM6 D PM. SinceM andM6are on the same side ofP onBC, we have M6 D M. We also have^PM

3

^{6M5 D BMM}

2

¹; thus the trajectory will start again towardsM1. Finally we have proved that the trajectory is6-periodic.

§ 2. Circular billiards

LetDbe limited by a circle. Let us start from a pointAon the circle in a direction such that the angle of the first segment with the half-tangent in the positive direction is ˛ measured in radians. We have two possibilities : the ratio ^˛ is or is not a rational number. We say that˛is or is not-rational.

2.1. ˛is-rational

Let p and q be two integers which are relatively prime and such that

˛ D ^p_q. The numbersp andq have no common divider other than1 and thus the fraction ^p_q cannot be simplified.

O A M1

˛

When the trajectory reaches the border for the first time in a pointM1, we have^AOM

2

^{1 D}2˛. Let us call this angle. Next pointM2will be the image ofM1 in the rotation with centerO and angle. Afterqbounces the image Mq will be such that the oriented angle of rays.OA; OMq/ D q ^2p_q D 2p. ThusMqis the same point asAand the trajectory isq-periodic (and we also see that the star-shaped polygon has turnedptimes around the center.

2.2. ˛is-irrational

The trajectory cannot be periodic since then it would be-rational. But we have

Theorem.The vertices of a trajectory on a circle with angle˛ which is -irrational isequidistributedon the circle.

Definition. Let I be any interval on the circle and denote by jIj the length of that interval. Let.xn/n2N be a sequence of points on the circle and for eachncallk.n/the number of elements among thenfirst of the sequence belonging toI, that is

k.n/ DCardfj 2NIj < n and xj 2 Ig The sequence.xn/n2Nis equidistributed if

nlim!1

k.n/

n D jIj 2

The theorem is a consequence of following theorem of Kronecker and Weyl.

Let us callU the unit circle.

Theorem.If f W U ! R; x 7! f .x/ is integrable function defined on the circle, then

nlim!1

1 n

n 1

X

jD0

f .xj/D 1 2

Z 2 0

f .x/dx

To see that the former theorem follows from this one just takef to be the characterstic function of the intervalI

f .x/D1I.x/D

1 if x 2I 0 if x …I and let the sequence bexnDx0Cn.

Sketch of the proof of the theorem. One may approximate f by trigo- nometric polynomials that is linear combinations of functions coskx and sinkx for k 2 Z. In fact it is easier to go to the complex functions since coskx and sinkx are linear combinations ofe^{i kx}. So the problem reduces to the proof for allkof

nlim!1

1 n

n 1

X

jD0

e^{i k.x0Cj /} D 1 2

Z 2 0

e^{i kx}dx (*)

In the case wherekD0, the left side of the equality reduces to limn!1 1

n

Pn 1

jD01D ¹_nnD1and the righthandside becomes

1 2

R2

0 dx D1. In that case./is verified.

In the case werek ¤0, we have

n 1

X

jD0

e^{i k.x0Cj /} De^{i kx0}

n 1

X

jD0

n e^{i k}oj

But one knows thatPn 1

jD0a^j D ^{1 a}_{1 a}ⁿ for anya¤1. Thus ˇ

ˇ ˇ

n 1

X

jD0

e^{i k.x0Cj /} ˇ ˇ ˇD

ˇ ˇ ˇ

1 e^{i k n} 1 e^{i k}

ˇ ˇ ˇ6

ˇ ˇ ˇ

2 1 e^{i k}

ˇ ˇ ˇ

and then the lefthand side of./tends to0whenn ! 1. The righthandside of./is also0since

Z 2 0

e^{i kx}dx D 1 i k

he^{i kx}i²

0 D 1

i k.1 1/D0

§ 3. Elliptical billiards